List of integrals of rational functions

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• $\backslash int\backslash frac\{f\text{'}(x)\}\{f(x)\}\; \backslash ,\; dx=\; \backslash ln\backslash left|\; f(x)\backslash right|\; +\; C$
• $\backslash int\backslash frac\{1\}\{x^2+a^2\}\; \backslash ,\; dx\; =\; \backslash frac\{1\}\{a\}\backslash arctan\backslash frac\{x\}\{a\}\backslash ,\backslash !\; +\; C$
• $\backslash int\; \backslash frac\{dx\}\{x^\{2^n\}\; +\; 1\}\; =\; \backslash frac\{1\}\{2^\{n-1\}\}\backslash sum\_\{k=1\}^\{2^\{n-1\}\}\; \backslash sin\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\backslash right)\; \backslash arctan\backslash left[\backslash left(x\; -\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash right\; )\; \backslash csc\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash right]\; -\; \backslash frac\{1\}\{2\}\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; \backslash ln\; \backslash left\; |\; x^2\; -\; 2\; x\; \backslash cos\; \backslash left(\backslash frac\{2k\; -1\}\{2^n\}\backslash pi\; \backslash right)\; +\; 1\; \backslash right\; |\; +\; C$

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Many of the following antiderivatives have a term of the form ln |ax + b|. Because this is undefined when x = −b / a, the most general form of the antiderivative replaces the constant of integration with a locally constant function."[http://golem.ph.utexas.edu/category/2012/03/reader_survey_logx_c.html Reader Survey: log|x| + C]", Tom Leinster, The n-category Café, March 19, 2012 However, it is conventional to omit this from the notation. For example,

$$\backslash int\backslash frac\{1\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash begin\{cases\}$$

\dfrac{1}{a}\ln(-(ax + b)) + C^- & ax+b<0 \\

\dfrac{1}{a}\ln(ax + b) + C^+ & ax+b>0

\end{cases}

is usually abbreviated as

$$\backslash int\backslash frac\{1\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C,$$

where C is to be understood as notation for a locally constant function of x. This convention will be adhered to in the following.

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• $\backslash int\; (ax\; +\; b)^n\; \backslash ,\; dx=\; \backslash frac\{(ax\; +\; b)^\{n+1\}\}\{a(n\; +\; 1)\}\; +\; C\; \backslash qquad\backslash text\{(for\; \}\; n\backslash neq\; -1\backslash mbox\{)\}$ (Cavalieri's quadrature formula)
• $\backslash int\; x^\{a-1\}\; (1-x)^\{b-1\}\; \backslash ,\; dx=\; \backslash Beta(x;\backslash ,a,b)\; +\; C\; \backslash qquad\backslash text\{(for\; \}\; \backslash operatorname\{Re\}(a),\; \backslash operatorname\{Re\}(b)>0\backslash mbox\{)\}$ (Incomplete beta function)
• $\backslash int\backslash frac\{x\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{x\}\{a\}\; -\; \backslash frac\{b\}\{a^2\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C$
• $\backslash int\backslash frac\{mx\; +\; n\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{m\}\{a\}\; x\; +\; \backslash frac\{an\; -\; bm\}\{a^2\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C$
• $\backslash int\backslash frac\{x\}\{(ax\; +\; b)^2\}\; \backslash ,\; dx=\; \backslash frac\{b\}\{a^2(ax\; +\; b)\}\; +\; \backslash frac\{1\}\{a^2\}\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; C$
• $\backslash int\backslash frac\{x\}\{(ax\; +\; b)^n\}\; \backslash ,\; dx=\; \backslash frac\{a(1\; -\; n)x\; -\; b\}\{a^2(n\; -\; 1)(n\; -\; 2)(ax\; +\; b)^\{n-1\}\}\; +\; C\; \backslash qquad\backslash text\{(for\; \}\; n\backslash not\backslash in\; \backslash \{1,\; 2\backslash \}\backslash mbox\{)\}$
• $\backslash int\; x(ax\; +\; b)^n\; \backslash ,\; dx=\; \backslash frac\{a(n\; +\; 1)x\; -\; b\}\{a^2(n\; +\; 1)(n\; +\; 2)\}\; (ax\; +\; b)^\{n+1\}\; +\; C\; \backslash qquad\backslash text\{(for\; \}n\; \backslash not\backslash in\; \backslash \{-1,\; -2\backslash \}\backslash mbox\{)\}$
• $\backslash int\backslash frac\{x^2\}\{ax\; +\; b\}\; \backslash ,\; dx=\; \backslash frac\{b^2\backslash ln(\backslash left|ax\; +\; b\backslash right|)\}\{a^3\}+\backslash frac\{ax^2\; -\; 2bx\}\{2a^2\}\; +\; C$
• $\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^2\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(ax\; -\; 2b\backslash ln\backslash left|ax\; +\; b\backslash right|\; -\; \backslash frac\{b^2\}\{ax\; +\; b\}\backslash right)\; +\; C$
• $\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^3\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(\backslash ln\backslash left|ax\; +\; b\backslash right|\; +\; \backslash frac\{2b\}\{ax\; +\; b\}\; -\; \backslash frac\{b^2\}\{2(ax\; +\; b)^2\}\backslash right)\; +\; C$
• $\backslash int\backslash frac\{x^2\}\{(ax\; +\; b)^n\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{a^3\}\backslash left(-\backslash frac\{(ax\; +\; b)^\{3-n\}\}\{(n-3)\}\; +\; \backslash frac\{2b\; (ax\; +\; b)^\{2-n\}\}\{(n-2)\}\; -\; \backslash frac\{b^2\; (ax\; +\; b)^\{1-n\}\}\{(n\; -\; 1)\}\backslash right)\; +\; C\; \backslash qquad\backslash text\{(for\; \}\; n\backslash not\backslash in\; \backslash \{1,\; 2,\; 3\backslash \}\backslash mbox\{)\}$
• $\backslash int\backslash frac\{1\}\{x(ax\; +\; b)\}\; \backslash ,\; dx\; =\; -\backslash frac\{1\}\{b\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\; +\; C$
• $\backslash int\backslash frac\{1\}\{x^2(ax+b)\}\; \backslash ,\; dx\; =\; -\backslash frac\{1\}\{bx\}\; +\; \backslash frac\{a\}\{b^2\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\; +\; C$
• $\backslash int\backslash frac\{1\}\{x^2(ax+b)^2\}\; \backslash ,\; dx\; =\; -a\backslash left(\backslash frac\{1\}\{b^2(ax+b)\}\; +\; \backslash frac\{1\}\{ab^2x\}\; -\; \backslash frac\{2\}\{b^3\}\backslash ln\backslash left|\backslash frac\{ax+b\}\{x\}\backslash right|\backslash right)\; +\; C$

{{endplainlist}}

For $a\backslash neq\; 0:$

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• $\backslash int\backslash frac\{1\}\{ax^2+bx+c\}\; dx\; =$

\begin{cases}

\displaystyle \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C & \text{(for }4ac-b^2>0\mbox{)} \\[12pt]

\displaystyle \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C =

\begin{cases}

\displaystyle -\frac{2}{\sqrt{b^2-4ac}}\,\operatorname{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(for }|2ax+b|<\sqrt{b^2-4ac}\mbox{)} \\[6pt]

\displaystyle -\frac{2}{\sqrt{b^2-4ac}}\,\operatorname{arcoth}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(else)}

\end{cases}

& \text{(for }4ac-b^2<0\mbox{)} \\[12pt]

\displaystyle -\frac{2}{2ax+b} + C & \text{(for }4ac-b^2=0\mbox{)}

\end{cases}

• $\backslash int\backslash frac\{x\}\{ax^2+bx+c\}\; \backslash ,\; dx\; =\; \backslash frac\{1\}\{2a\}\backslash ln\backslash left|ax^2+bx+c\backslash right|-\backslash frac\{b\}\{2a\}\backslash int\backslash frac\{dx\}\{ax^2+bx+c\}\; +\; C$
• $\backslash int\backslash frac\{mx+n\}\{ax^2+bx+c\}\; \backslash ,\; dx\; =\; \backslash begin\{cases\}$

\displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C &\text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{2a\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C =

\begin{cases}

\displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\operatorname{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(for }|2ax+b|<\sqrt{b^2-4ac}\mbox{)} \\[6pt]

\displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\operatorname{arcoth}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(else)}

\end{cases}

& \text{(for }4ac-b^2<0\mbox{)} \\[12pt]

\displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} + C = \frac{m}{a}\ln\left|x+\frac{b}{2a}\right|-\frac{2an-bm}{a(2ax+b)} + C &\text{(for }4ac-b^2=0\mbox{)}\end{cases}

• $\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^n\}\; \backslash ,\; dx=\; \backslash frac\{2ax+b\}\{(n-1)(4ac-b^2)(ax^2+bx+c)^\{n-1\}\}+\backslash frac\{(2n-3)2a\}\{(n-1)(4ac-b^2)\}\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^\{n-1\}\}\; \backslash ,\; dx\; +\; C$
• $\backslash int\backslash frac\{x\}\{(ax^2+bx+c)^n\}\; \backslash ,\; dx=\; -\backslash frac\{bx+2c\}\{(n-1)(4ac-b^2)(ax^2+bx+c)^\{n-1\}\}-\backslash frac\{b(2n-3)\}\{(n-1)(4ac-b^2)\}\backslash int\backslash frac\{1\}\{(ax^2+bx+c)^\{n-1\}\}\; \backslash ,\; dx\; +\; C$
• $\backslash int\backslash frac\{1\}\{x(ax^2+bx+c)\}\; \backslash ,\; dx=\; \backslash frac\{1\}\{2c\}\backslash ln\backslash left|\backslash frac\{x^2\}\{ax^2+bx+c\}\backslash right|-\backslash frac\{b\}\{2c\}\backslash int\backslash frac\{1\}\{ax^2+bx+c\}\; \backslash ,\; dx\; +\; C$

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• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.

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• $$

\int x^m \left(a+b\,x^n\right)^p dx =

\frac{x^{m+1} \left(a+b\,x^n\right)^p}{m+n\,p+1}\,+\,

\frac{a\,n\,p}{m+n\,p+1}\int x^m \left(a+b\,x^n\right)^{p-1}dx

• $$

\int x^m \left(a+b\,x^n\right)^p dx =

-\frac{x^{m+1} \left(a+b\,x^n\right)^{p+1}}{a\,n (p+1)}\,+\,

\frac{m+n (p+1)+1}{a\,n (p+1)}\int x^m \left(a+b\,x^n\right)^{p+1}dx

• $$

\int x^m \left(a+b\,x^n\right)^p dx =

\frac{x^{m+1} \left(a+b\,x^n\right)^p}{m+1}\,-\,

\frac{b\,n\,p}{m+1}\int x^{m+n} \left(a+b\,x^n\right)^{p-1}dx

• $$

\int x^m \left(a+b\,x^n\right)^p dx =

\frac{x^{m-n+1} \left(a+b\,x^n\right)^{p+1}}{b\,n (p+1)}\,-\,

\frac{m-n+1}{b\,n (p+1)}\int x^{m-n} \left(a+b\,x^n\right)^{p+1}dx

• $$

\int x^m \left(a+b\,x^n\right)^p dx =

\frac{x^{m-n+1} \left(a+b\,x^n\right)^{p+1}}{b (m+n\,p+1)}\,-\,

\frac{a (m-n+1)}{b (m+n\,p+1)}\int x^{m-n}\left(a+b\,x^n\right)^pdx

• $$

\int x^m \left(a+b\,x^n\right)^p dx =

\frac{x^{m+1} \left(a+b\,x^n\right)^{p+1}}{a (m+1)}\,-\,

\frac{b (m+n (p+1)+1)}{a (m+1)}\int x^{m+n}\left(a+b\,x^n\right)^pdx

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, n and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $(a+b\backslash ,x)^m\; (c+d\backslash ,x)^n\; (e+f\backslash ,x)^p$ by setting B to 0.

{{startplainlist|indent=1}}

• $\backslash begin\{align\}$

&\int (A+B\,x) (a+b\,x)^m (c+d\,x)^n (e+f\,x)^p dx=

-\frac{(A\,b-a\,B)(a+b\,x)^{m+1} (c+d\,x)^n(e+f\,x)^{p+1}}{b (m+1) (a\,f-b\,e)}\,+\,

\frac{1}{b (m+1) (a\,f-b\,e)}\,\cdot \\

&\qquad \int (b\,c(m+1) (A\,f-B\,e)+(A\,b-a\,B) (n\,d\,e+c\,f(p+1))+d(b(m+1) (A\,f-B\,e)+f(n+p+1) (A\,b-a\,B))x)(a+b\,x)^{m+1} (c+d\,x)^{n-1}(e+f\,x)^p dx

\end{align}

• $\backslash begin\{align\}$

&\int (A+B\,x) (a+b\,x)^m (c+d\,x)^n (e+f\,x)^p dx=

\frac{B(a+b\,x)^m (c+d\,x)^{n+1}(e+f\,x)^{p+1}}{d\,f(m+n+p+2)}\,+\,

\frac{1}{d\,f(m+n+p+2)}\,\cdot \\

&\qquad \int (A\,a\,d\,f(m+n+p+2)-B (b\,c\,e\,m+a(d\,e(n+1)+c\,f(p+1)))+(A\,b\,d\,f(m+n+p+2)+B (a\,d\,f\,m-b(d\,e(m+n+1)+c\,f(m+p+1)))) x)(a+b\,x)^{m-1} (c+d\,x)^n(e+f\,x)^p dx

\end{align}

• $\backslash begin\{align\}$

&\int (A+B\,x) (a+b\,x)^m (c+d\,x)^n (e+f\,x)^p dx=

\frac{(A\,b-a\,B)(a+b\,x)^{m+1} (c+d\,x)^{n+1}(e+f\,x)^{p+1}}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}\,+\,

\frac{1}{(m+1)(a\,d-b\,c)(a\,f-b\,e)}\,\cdot \\

&\qquad \int ((m+1) (A (a\,d\,f-b(c\,f+d\,e))+B\,b\,c\,e)-(A\,b-a\,B) (d\,e(n+1)+c\,f(p+1))-d\,f(m+n+p+3) (A\,b-a\,B)x)(a+b\,x)^{m+1} (c+d\,x)^n(e+f\,x)^p dx

\end{align}

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m, p and q toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^q$ and $x^m\backslash left(a+b\backslash ,x^n\backslash right)^p\backslash left(c+d\backslash ,x^n\backslash right)^q$ by setting m and/or B to 0.

{{startplainlist|indent=1}}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

-\frac{(A\,b-a\,B) x^{m+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^q}{a\,b\,n (p+1)}\,+\,

\frac{1}{a\,b\,n (p+1)}\,\cdot \\

&\qquad \int x^m\left(c (A\,b\,n (p+1)+(A\,b-a\,B) (m+1))+d (A\,b\,n (p+1)+(A\,b-a\,B) (m+n\,q+1)) x^n\right)\left(a+b\,x^n\right)^{p+1}\left(c+d\,x^n\right)^{q-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

\frac{B\,x^{m+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^q}{b (m+n (p+q+1)+1)}\,+\,

\frac{1}{b (m+n (p+q+1)+1)}\,\cdot \\

&\qquad \int x^m\left(c ((A\,b-a\,B) (1+m)+A\,b\,n (1+p+q))+(d(A\,b-a\,B) (1+m)+B\,n\,q(b\,c-a\,d)+A\,b\,d\,n (1+p+q))\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^{q-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

-\frac{(A\,b-a\,B) x^{m+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^{q+1}}{a\,n (b\,c-a\,d) (p+1)}\,+\,

\frac{1}{a\,n(b\,c-a\,d)(p+1)}\,\cdot \\

&\qquad \int x^m\left(c(A\,b-a\,B)(m+1)+A\,n (b\,c-a\,d)(p+1)+d(A\,b-a\,B) (m+n (p+q+2)+1) x^n\right)\left(a+b\,x^n\right)^{p+1}\left(c+d\,x^n\right)^qdx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

\frac{B\,x^{m-n+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^{q+1}}{b\,d (m+n (p+q+1)+1)}\,-\,

\frac{1}{b\,d (m+n (p+q+1)+1)}\,\cdot \\

&\qquad \int x^{m-n}\left(a\,B\,c (m-n+1)+(a\,B\,d (m+n\,q+1)-b (-B\,c (m+n\,p+1)+A\,d (m+n (p+q+1)+1))) x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

\frac{A\,x^{m+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^{q+1}}{a\,c (m+1)}\,+\,

\frac{1}{a\,c (m+1)}\,\cdot \\

&\qquad \int x^{m+n}\left(a\,B\,c (m+1)-A (b\,c+a\,d) (m+n+1)-A\,n (b\,c\,p+a\,d\,q)-A\,b\,d (m+n (p+q+2)+1) x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

\frac{A\,x^{m+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^q}{a (m+1)}\,-\,

\frac{1}{a (m+1)}\,\cdot \\

&\qquad \int x^{m+n}\left(c(A\,b-a\,B)(m+1)+A\,n (b\,c (p+1)+a\,d\,q)+d ((A\,b-a\,B) (m+1)+A\,b\,n (p+q+1)) x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^{q-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m\left(A+B\,x^n\right)\left(a+b\,x^n\right)^p\left(c+d\,x^n\right)^qdx=

\frac{(A\,b-a\,B) x^{m-n+1} \left(a+b\,x^n\right)^{p+1} \left(c+d\,x^n\right)^{q+1}}{b\,n (b\,c-a\,d) (p+1)}\,-\,

\frac{1}{b\,n(b\,c-a\,d)(p+1)}\,\cdot \\

&\qquad \int x^{m-n}\left(c(A\,b-a\,B)(m-n+1)+(d(A\,b-a\,B)(m+n\,q+1)-b\,n(B\,c-A\,d)(p+1)) x^n\right)\left(a+b\,x^n\right)^{p+1}\left(c+d\,x^n\right)^qdx

\end{align}

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p$ when $b^2-4\backslash ,a\backslash ,c=0$ by setting m to 0.

{{startplainlist|indent=1}}

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^p}{e(m+1)}\,-\,

\frac{p (d+e\,x)^{m+2}(b+2 c\,x) \left(a+b\,x+c\,x^2\right)^{p-1}}{e^2(m+1)(m+2 p+1)}\,+\,

\frac{p(2 p-1)(2 c\,d-b\,e)}{e^2(m+1)(m+2 p+1)} \int (d+e\,x)^{m+1}\left(a+b\,x+c\,x^2\right)^{p-1}dx

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^p}{e(m+1)}\,-\,

\frac{p (d+e\,x)^{m+2}(b+2\,c\,x)\left(a+b\,x+c\,x^2\right)^{p-1}}{e^2(m+1)(m+2)}\,+\,

\frac{2\,c\,p\,(2\,p-1)}{e^2(m+1)(m+2)} \int (d+e\,x)^{m+2} \left(a+b\,x+c\,x^2\right)^{p-1}dx

• $$

\int (d+e\,x)^m\left(a+b\,x+c\,x^2\right)^pdx=

-\frac{e(m+2 p+2)(d+e\,x)^m \left(a+b\,x+c\,x^2\right)^{p+1}}{(p+1)(2p+1)(2 c\,d-b\,e)}\,+\,

\frac{(d+e\,x)^{m+1}(b+2 c\,x) \left(a+b\,x+c\,x^2\right)^p}{(2p+1)(2 c\,d-b\,e)}\,+\,

\frac{e^2m(m+2 p+2)}{(p+1)(2p+1)(2 c\,d-b\,e)} \int (d+e\,x)^{m-1} \left(a+b\,x+c\,x^2\right)^{p+1}dx

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

-\frac{e\,m(d+e\,x)^{m-1} \left(a+b\,x+c\,x^2\right)^{p+1}}{2c (p+1) (2p+1)}\,+\,

\frac{(d+e\,x)^m(b+2 c\,x)\left(a+b\,x+c\,x^2\right)^p}{2c (2p+1)}\,+\,

\frac{e^2m(m-1)}{2c (p+1) (2p+1)} \int (d+e\,x)^{m-2} \left(a+b\,x+c\,x^2\right)^{p+1}dx

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^p}{e(m+2p+1)}\,-\,

\frac{p(2 c\,d-b\,e)(d+e\,x)^{m+1}(b+2 c\,x)\left(a+b\,x+c\,x^2\right)^{p-1}}{2c\,e^2(m+2 p)(m+2p+1)}\,+\,

\frac{p (2 p-1)(2 c\,d-b\,e)^2}{2c\,e^2(m+2 p)(m+2p+1)} \int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^{p-1}dx

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

-\frac{2c\,e(m+2p+2)(d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^{p+1}}{(p+1) (2 p+1)(2 c\,d-b\,e)^2}\,+\,

\frac{(d+e\,x)^{m+1}(b+2 c\,x)\left(a+b\,x+c\,x^2\right)^p}{(2 p+1)(2 c\,d-b\,e)}\,+\,

\frac{2c\,e^2(m+2p+2)(m+2 p+3)}{(p+1) (2 p+1)(2 c\,d-b\,e)^2} \int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^{p+1}dx

• $$

\int (d+e\,x)^m \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^m (b+2 c\,x)\left(a+b\,x+c\,x^2\right)^p}{2c (m+2p+1)}\,+\,

\frac{m(2 c\,d-b\,e)}{2c (m+2p+1)} \int (d+e\,x)^{m-1}\left(a+b\,x+c\,x^2\right)^pdx

• $$

\int (d+e\,x)^m\left(a+b\,x+c\,x^2\right)^pdx=

-\frac{(d+e\,x)^{m+1} (b+2 c\,x)\left(a+b\,x+c\,x^2\right)^p}{(m+1)(2 c\,d-b\,e)}\,+\,

\frac{2c (m+2p+2)}{(m+1)(2 c\,d-b\,e)} \int (d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^pdx

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $\backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p$ and $(d+e\backslash ,x)^m\; \backslash left(a+b\backslash ,x+c\backslash ,x^2\backslash right)^p$ by setting m and/or B to 0.

{{startplainlist|indent=1}}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} (A\,e (m+2 p+2)-B\,d (2 p+1)+e\,B (m+1) x) \left(a+b\,x+c\,x^2\right)^p}{e^2(m+1) (m+2 p+2)}\,+\,

\frac{1}{e^2(m+1) (m+2 p+2)}p\,\cdot \\

&\qquad \int (d+e\,x)^{m+1} (B (b\,d+2 a\,e+2 a\,e\,m+2 b\,d\,p)-A\,b\,e (m+2 p+2)+(B (2 c\,d+b\,e+b\,e m+4 c\,d\,p)-2 A\,c\,e (m+2 p+2))x)\left(a+b\,x+c\,x^2\right)^{p-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^m (A\,b-2 a\,B-(b\,B-2 A\,c) x)\left(a+b\,x+c\,x^2\right)^{p+1}}{(p+1)\left(b^2-4 a\,c\right) }\,+\,

\frac{1}{(p+1)\left(b^2-4 a\,c\right) }\,\cdot \\

&\qquad \int (d+e\,x)^{m-1}(B (2 a\,e\,m+b\,d (2 p+3))-A (b\,e\,m+2 c\,d (2 p+3))+e(b\,B-2 A\,c) (m+2 p+3) x)\left(a+b\,x+c\,x^2\right)^{p+1}dx

\end{align}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} (A\,c\,e (m+2 p+2)-B (c\,d+2 c\,d\,p-b\,e\,p)+B\,c\,e(m+2 p+1) x)\left(a+b\,x+c\,x^2\right)^p}{c\,e^2(m+2 p+1) (m+2 p+2)}\,-\,

\frac{p}{c\,e^2(m+2 p+1) (m+2 p+2)}\,\cdot \\

&\qquad \int (d+e\,x)^m (A\,c\,e (b\,d-2 a\,e) (m+2 p+2)+B (a\,e (b\,e-2 c\,d\,m+b\,e\,m)+b\,d (b\,e\,p-c\,d-2 c\,d\,p))+ \\

&\qquad \qquad \left(A\,c\,e (2 c\,d-b\,e) (m+2 p+2)-B \left(-b^2 e^2 (m+p+1)+2 c^2 d^2 (1+2 p)+c\,e (b\,d (m-2 p)+2 a\,e (m+2 p+1))\right)\right) x)\left(a+b\,x+c\,x^2\right)^{p-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

\frac{(d+e\,x)^{m+1} \left(A \left(b\,c\,d-b^2 e+2 a\,c\,e\right)-a\,B (2 c\,d-b\,e)+c (A (2 c\,d-b\,e)-B (b\,d-2 a\,e)) x\right)\left(a+b\,x+c\,x^2\right)^{p+1}}{(p+1)\left(b^2-4 a\,c\right) \left(c\,d^2-b\,d\,e+a\,e^2\right)}\,+ \\

&\qquad \frac{1}{(p+1)\left(b^2-4 a\,c\right) \left(c\,d^2-b\,d\,e+a\,e^2\right)}\,\cdot \\

&\qquad \qquad \int (d+e\,x)^m (A \left(b\,c\,d\,e (2 p-m+2)+b^2 e^2 (m+p+2)-2 c^2 d^2 (3+2 p)-2 a\,c\,e^2 (m+2 p+3)\right)- \\

&\qquad \qquad \qquad B (a\,e (b\,e-2 c\,d m+b\,e\,m)+b\,d (-3 c\,d+b\,e-2 c\,d\,p+b\,e\,p))+c\,e(B (b\,d-2 a\,e)-A (2 c\,d-b\,e)) (m+2 p+4) x)\left(a+b\,x+c\,x^2\right)^{p+1}dx

\end{align}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

\frac{B(d+e\,x)^m\left(a+b\,x+c\,x^2\right)^{p+1}}{c(m+2 p+2)}\,+\,

\frac{1}{c(m+2 p+2)}\,\cdot \\

&\qquad \int (d+e\,x)^{m-1} (m(A\,c\,d-a\,B\,e)-d(b\,B-2 A\,c)(p+1) +((B\,c\,d-b\,B\,e+A\,c\,e) m-e(b\,B-2 A\,c)(p+1))x) \left(a+b\,x+c\,x^2\right)^pdx

\end{align}

• $\backslash begin\{align\}$

&\int (d+e\,x)^m (A+B\,x) \left(a+b\,x+c\,x^2\right)^pdx=

-\frac{(B\,d-A\,e) (d+e\,x)^{m+1} \left(a+b\,x+c\,x^2\right)^{p+1}}{(m+1)\left(c\,d^2-b\,d\,e+a\,e^2\right)}\,+\,

\frac{1}{(m+1)\left(c\,d^2-b\,d\,e+a\,e^2\right)}\,\cdot \\

&\qquad \int (d+e\,x)^{m+1} ((A\,c\,d-A\,b\,e+a\,B\,e) (m+1)+b (B\,d-A\,e) (p+1)+c (B\,d-A\,e) (m+2 p+3) x)\left(a+b\,x+c\,x^2\right)^pdx

\end{align}

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p$ when $b^2-4\backslash ,a\backslash ,c=0$ by setting m to 0.

{{startplainlist|indent=1}}

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{ x^{m+1}\left(a+b\,x^n+c\,x^{2 n}\right)^p}{m+2 n\,p+1}\,+\,

\frac{n\,p\,x^{m+1} \left(2 a+b\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}}{(m+1)(m+2 n\,p+1)}\,-\,

\frac{b\,n^2 p (2 p-1)}{(m+1)(m+2 n\,p+1)} \int x^{m+n} \left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{(m+n(2 p-1)+1) x^{m+1}\left(a+b\,x^n+c\,x^{2 n}\right)^p}{(m+1)(m+n+1)}\,+\,

\frac{n\,p\,x^{m+1} \left(2 a+b\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}}{(m+1)(m+n+1)}\,+\,

\frac{2 c\,p\,n^2(2 p-1)}{(m+1)(m+n+1)} \int x^{m+2n} \left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{(m+n(2 p+1)+1) x^{m-n+1}\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{b\,n^2 (p+1) (2p+1)}\,-\,

\frac{x^{m+1} \left(b+2 c\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{b\,n (2p+1)}\,-\,

\frac{(m-n+1)(m+n(2 p+1)+1)}{b\,n^2 (p+1) (2p+1)} \int x^{m-n} \left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}dx

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

-\frac{(m-3 n-2 n\,p+1) x^{m-2n+1}\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{2 c\,n^2(p+1)(2p+1)}\,-\,

\frac{ x^{m-2n+1} \left(2 a+b\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{2 c\,n(2p+1)}\,+\,

\frac{(m-n+1)(m-2n+1)}{2 c\,n^2(p+1)(2p+1)} \int x^{m-2n} \left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}dx

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{x^{m+1}\left(a+b\,x^n+c\,x^{2 n}\right)^p}{m+2 n\,p+1}\,+\,

\frac{n\,p\,x^{m+1} \left(2 a+b\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}}{(m+2 n\,p+1) (m+n(2 p-1)+1)}\,+\,

\frac{2 a\,n^2 p (2 p-1)}{(m+2 n\,p+1) (m+n(2 p-1)+1)} \int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

• $$

\int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

-\frac{(m+n+2 n\,p+1) x^{m+1}\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{2 a\,n^2 (p+1) (2p+1)}\,-\,

\frac{x^{m+1} \left(2 a+b\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{2 a\,n(2p+1)}\,+\,

\frac{(m+n(2 p+1)+1)(m+2 n (p+1)+1)}{2 a\,n^2 (p+1) (2p+1)} \int x^m \left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}dx

• $$

\int x^m\left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{x^{m-n+1} \left(b+2c\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{2c (m+2n\,p+1)}\,-\,

\frac{b (m-n+1)}{2c (m+2n\,p+1)} \int x^{m-n} \left(a+b\,x^n+c\,x^{2 n}\right)^p dx

• $$

\int x^m\left(a+b\,x^n+c\,x^{2 n}\right)^p dx=

\frac{x^{m+1} \left(b+2c\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{b (m+1)}\,-\,

\frac{2c (m+n(2 p+1)+1)}{b (m+1)} \int x^{m+n} \left(a+b\,x^n+c\,x^{2 n}\right)^p dx

{{endplainlist}}

• The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
• These reduction formulas can be used for integrands having integer and/or fractional exponents.
• Special cases of these reductions formulas can be used for integrands of the form $\backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p$ and $x^m\; \backslash left(a+b\backslash ,x^n+c\backslash ,x^\{2\; n\}\backslash right)^p$ by setting m and/or B to 0.

{{startplainlist|indent=1}}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

\frac{x^{m+1} \left(A (m+n (2 p+1)+1)+B (m+1) x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^p}{(m+1) (m+n (2 p+1)+1)}\,+\,

\frac{n\,p}{(m+1) (m+n (2 p+1)+1)}\,\cdot \\

&\qquad \int x^{m+n} \left(2 a\,B (m+1)-A\,b (m+n (2 p+1)+1)+(b\,B (m+1)-2\,A\,c (m+n (2 p+1)+1)) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

\frac{x^{m-n+1} \left(A\,b-2 a\,B-(b\,B-2 A\,c) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{n(p+1) \left(b^2-4 a\,c\right)}\,+\,

\frac{1}{n(p+1) \left(b^2-4 a\,c\right)}\,\cdot \\

&\qquad \int x^{m-n}\left((m-n+1)(2 a\,B-A\,b)+(m+2n (p+1)+1) (b\,B-2 A\,c) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

\frac{x^{m+1} \left(b\,B\,n\,p+A\,c (m+n (2 p+1)+1)+B\,c (m+2 n\,p+1) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^p}{c (m+2 n\,p+1) (m+n (2 p+1)+1)}\,+\,

\frac{n\,p}{c (m+2 n\,p+1) (m+n (2 p+1)+1)}\,\cdot \\

&\qquad \int x^m \left(2 a\,A\,c (m+n (2 p+1)+1)-a\,b\,B (m+1)+\left(2 a\,B\,c (m+2 n\,p+1)+A\,b\,c (m+n (2 p+1)+1)-b^2 B (m+n\,p+1)\right) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p-1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

-\frac{x^{m+1} \left(A\,b^2-a\,b\,B-2 a\,A\,c+(A\,b-2 a\,B) c\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{a\,n(p+1) \left(b^2-4 a\,c\right)}\,+\,

\frac{1}{a\,n(p+1) \left(b^2-4 a\,c\right)}\,\cdot \\

&\qquad \int x^m \left((m+n (p+1)+1) A\,b^2-a\,b\,B(m+1)-2(m+2n (p+1)+1)a\,A\,c+(m+n (2p+3)+1)(A\,b-2 a\,B) c\,x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}dx

\end{align}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

\frac{B\,x^{m-n+1}\left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{c (m+n (2 p+1)+1)}\,-\,

\frac{1}{c (m+n (2 p+1)+1)}\,\cdot \\

&\qquad \int x^{m-n} \left(a\,B (m-n+1)+(b\,B (m+n\,p+1)-A\,c (m+n (2 p+1)+1)) x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx

\end{align}

• $\backslash begin\{align\}$

&\int x^m \left(A+B\,x^n\right) \left(a+b\,x^n+c\,x^{2 n}\right)^pdx=

\frac{A\,x^{m+1} \left(a+b\,x^n+c\,x^{2 n}\right)^{p+1}}{a(m+1)}\,+\,

\frac{1}{a(m+1)}\,\cdot \\

&\qquad \int x^{m+n} \left(a\,B (m+1)-A\,b (m+n (p+1)+1)-A\,c (m+2 n(p+1)+1) x^n\right)\left(a+b\,x^n+c\,x^{2 n}\right)^pdx

\end{align}

{{endplainlist}}

{{reflist}}

{{Lists of integrals}}

Rational functions