List of integrals of irrational functions

The following is a list of integrals (antiderivative functions) of irrational functions. For a complete list of integral functions, see lists of integrals. Throughout this article the constant of integration is omitted for brevity.

Integrals involving r = √a² + x²

$\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)$

$\int r^3 \;dx = \frac{1}{4}xr^3+\frac{3}{8}a^2xr+\frac{3}{8}a^4\ln\left(x+r\right)$

$\int r^5 \; dx = \frac{1}{6}xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln\left(x+r\right)$

$\int x r\;dx=\frac{r^3}{3}$

$\int x r^3\;dx=\frac{r^5}{5}$

$\int x r^{2n+1}\;dx=\frac{r^{2n+3}}{2n+3}$

$\int x^2 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(x+r\right)$

$\int x^2 r^3\;dx= \frac{xr^5}{6}-\frac{a^2xr^3}{24}-\frac{a^4xr}{16}-\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}$

$\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5}$

$\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^2 r^{2n+3}}{2n+3}$

$\int x^4 r\;dx= \frac{x^3r^3}{6}-\frac{a^2xr^3}{8}+\frac{a^4xr}{16}+\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^4 r^3\;dx= \frac{x^3r^5}{8}-\frac{a^2xr^5}{16}+\frac{a^4xr^3}{64}+\frac{3a^6xr}{128}+\frac{3a^8}{128}\ln\left(x+r\right)$

$\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}$

$\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^2 r^7}{7} + \frac{a^4 r^5}{5}$

$\int x^5 r^{2n+1} \; dx = \frac{r^{2n+7}}{2n+7} - \frac{2a^2r^{2n+5}}{2n+5}+\frac{a^4 r^{2n+3}}{2n+3}$

$\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a\, \operatorname{arsinh}\frac{a}{x}$

$\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^5\;dx}{x} = \frac{r^5}{5}+\frac{a^2r^3}{3}+a^4r-a^5\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^7\;dx}{x} = \frac{r^7}{7}+\frac{a^2r^5}{5}+\frac{a^4r^3}{3}+a^6r-a^7\ln\left|\frac{a+r}{x}\right|$

$\int\frac{dx}{r} = \operatorname{arsinh}\frac{x}{a} = \ln\left( \frac{x+r}{a} \right)$

$\int\frac{dx}{r^3} = \frac{x}{a^2r}$

$\int\frac{x\,dx}{r} = r$

$\int\frac{x\,dx}{r^3} = -\frac{1}{r}$

$\int\frac{x^2\;dx}{r} = \frac{x}{2}r-\frac{a^2}{2}\,\operatorname{arsinh}\frac{x}{a} = \frac{x}{2}r-\frac{a^2}{2}\ln\left( \frac {x+r}{a} \right)$

$\int\frac{dx}{xr} = -\frac{1}{a}\,\operatorname{arsinh}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|$

Integrals involving s = √x² - a²

Assume $(x^2>a^2)$, for $(x^2<a^2)$, see next section:

$\int s\;dx = \frac{1}{2}\left( xs-a^{2}\ln(x+s)\right)$

$\int xs\;dx = \frac{1}{3}s^3$

$\int\frac{s\;dx}{x} = s - a\arccos\left|\frac{a}{x}\right|$

$\int\frac{dx}{s} = \ln\left|\frac{x+s}{a}\right|$

Here $\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\,\operatorname{arcosh}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$, where the positive value of $\operatorname{arcosh}\left|\frac{x}{a}\right|$ is to be taken.

$\int\frac{x\;dx}{s} = s$

$\int\frac{x\;dx}{s^3} = -\frac{1}{s}$

$\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}$

$\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}$

$\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}}$

$\int\frac{x^{2m}\;dx}{s^{2n+1}} = -\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int\frac{x^{2m-2}\;dx}{s^{2n-1}}$

$\int\frac{x^2\;dx}{s} = \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^2\;dx}{s^3} = -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s} = \frac{x^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{s}+\frac{3}{2}a^2\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^5} = -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^{2m}\;dx}{s^{2n+1}} = (-1)^{n-m}\frac{1}{a^{2(n-m)}}\sum_{i=0}^{n-m-1}\frac{1}{2(m+i)+1}{n-m-1 \choose i}\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\qquad\mbox{(}n>m\ge0\mbox{)}$

$\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$

$\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]$

$\int\frac{dx}{s^7} =-\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{dx}{s^9} =\frac{1}{a^8}\left[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}\right]$

$\int\frac{x^2\;dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$

$\int\frac{x^2\;dx}{s^7} = \frac{1}{a^4}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]$

Integrals involving u = √a² - x²

$\int u \;dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int xu\;dx = -\frac{1}{3} u^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int x^2u\;dx = -\frac{x}{4} u^3+\frac{a^2}{8}(xu+a^2\arcsin\frac{x}{a}) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{u\;dx}{x} = u-a\ln\left|\frac{a+u}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{dx}{u} = \arcsin\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{x^2\;dx}{u} = \frac{1}{2}\left(-xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int u\;dx = \frac{1}{2}\left(xu-\sgn x\,\operatorname{arcosh}\left|\frac{x}{a}\right|\right) \qquad\mbox{(for }|x|\ge|a|\mbox{)}$

$\int \frac{x}{u}\;dx = -u \qquad\mbox{(}|x|\leq|a|\mbox{)}$

Integrals involving R = √ax² + bx + c

Assume (ax² + bx + c) cannot be reduced to the following expression (px + q)² for some p and q.

$\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(for }a>0\mbox{)}$

$\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\,\operatorname{arsinh}\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad \mbox{(for }a>0\mbox{, }4ac-b^2>0\mbox{)}$

$\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln|2ax+b| \quad \mbox{(for }a>0\mbox{, }4ac-b^2=0\mbox{)}$

$\int\frac{dx}{R} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad \mbox{(for }a<0\mbox{, }4ac-b^2<0\mbox{, }\left|2ax+b\right|<\sqrt{b^2-4ac}\mbox{)}$

$\int\frac{dx}{R^3} = \frac{4ax+2b}{(4ac-b^2)R}$

$\int\frac{dx}{R^5} = \frac{4ax+2b}{3(4ac-b^2)R}\left(\frac{1}{R^2}+\frac{8a}{4ac-b^2}\right)$

$\int\frac{dx}{R^{2n+1}} = \frac{2}{(2n-1)(4ac-b^2)}\left(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int\frac{dx}{R^{2n-1}}\right)$

$\int\frac{x}{R}\;dx = \frac{R}{a}-\frac{b}{2a}\int\frac{dx}{R}$

$\int\frac{x}{R^3}\;dx = -\frac{2bx+4c}{(4ac-b^2)R}$

$\int\frac{x}{R^{2n+1}}\;dx = -\frac{1}{(2n-1)aR^{2n-1}}-\frac{b}{2a}\int\frac{dx}{R^{2n+1}}$

$\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\ln \left|\frac{2\sqrt{c}R+bx+2c}{x}\right|, ~ c > 0$

$\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\operatorname{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right), ~ c < 0$

$\int\frac{dx}{xR}=\frac{1}{\sqrt{-c}}\operatorname{arcsin}\left(\frac{bx+2c}{|x|\sqrt{b^2-4ac}}\right), ~ c < 0, b^2-4ac>0$

$\int\frac{dx}{xR}=-\frac{2}{bx}\left(\sqrt{ax^2+bx}\right), ~ c = 0$

(Click "show" at right to see a proof or "hide" to hide it.) 

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First use substitution with $u \equiv \sqrt{ax+b}$ so that $R = \sqrt{x} u$, and the integral becomes

$2\sqrt{a} \int \frac{du}{(u^2-b)^{\frac{3}{2}}}$

Then make another substitution with $v \equiv \frac{1}{\sqrt{u^2-b}}$ so that the integral becomes

$-2\sqrt{a} \int \frac{vdv}{\sqrt{1+bv^2}}$

The integral then has the form $\int \frac{x\;dx}{r} = r$ as given above and can be readily evaluated.

One can also perform the substitutions in a single step with the less apparent choice $u \equiv \sqrt{\frac{ax + b}{ax}}$, giving

$\int\frac{dx}{xR}= -\frac{2\sqrt{a}}{b} \int du = -\frac{2\sqrt{a}}{b}u$

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$\int\frac{x^2}{R}\;dx = \frac{2ax-3b}{4a^2}R+\frac{3b^2-4ac}{8a^2}\int\frac{dx}{R}$

$\int \frac{dx}{x^{2} R}=- \frac{ R}{cx}-\frac{b}{2c} \int \frac{dx}{x R}$

$\int R\,dx= \frac{2ax+b}{4a} R+ \frac{4ac-b^{2}}{8a} \int \frac{dx}{ R}$

$\int x R\,dx = \frac{R^3}{3a}-\frac{b(2ax+b)}{8a^{2}} R - \frac{b(4ac-b^{2})}{16a^{2}} \int \frac{dx}{ R}$

$\int x^{2} R\,dx= \frac{6ax-5b}{24a^{2}}R^3+\frac{5b^{2}-4ac}{16a^{2}} \int R\,dx$

$\int \frac{ R}{x}\,dx= R+ \frac{b}{2} \int \frac{dx}{ R}+c \int \frac{dx}{x R}$

$\int \frac{ R}{x^{2}}\,dx=- \frac{ R}{x}+a \int \frac{dx}{R^2}+ \frac{b}{2} \int \frac{dx}{ R}$

$\int \frac{x^{2}\,dx}{R^3}= \frac{(2b^{2}-4ac)x+2bc}{a(4ac-b^{2}) R}+ \frac{1}{a} \int \frac{dx}{ R}$

Integrals involving S = √ax + b

$\int S {dx} = \frac{2 S^{3}}{3 a}$

$\int \frac{dx}{S} = \frac{2S}{a}$

$\int \frac{dx}{x S} = \begin{cases} -\frac{2}{\sqrt{b}} \mathrm{arcoth}\left( \frac{S}{\sqrt{b}}\right) & \mbox{(for }b > 0, \quad a x > 0\mbox{)} \\ -\frac{2}{\sqrt{b}} \mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right) & \mbox{(for }b > 0, \quad a x < 0\mbox{)} \\ \frac{2}{\sqrt{-b}} \arctan\left( \frac{S}{\sqrt{-b}}\right) & \mbox{(for }b < 0\mbox{)} \\ \end{cases}$

$\int\frac{S}{x}\,dx = \begin{cases} 2 \left( S - \sqrt{b}\,\mathrm{arcoth}\left( \frac{S}{\sqrt{b}}\right)\right) & \mbox{(for }b > 0, \quad a x > 0\mbox{)} \\ 2 \left( S - \sqrt{b}\,\mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right)\right) & \mbox{(for }b > 0, \quad a x < 0\mbox{)} \\ 2 \left( S - \sqrt{-b} \arctan\left( \frac{S}{\sqrt{-b}}\right)\right) & \mbox{(for }b < 0\mbox{)} \\ \end{cases}$

$\int \frac{x^{n}}{S} dx = \frac{2}{a (2 n + 1)} \left( x^{n} S - b n \int \frac{x^{n - 1}}{S} dx\right)$

$\int x^{n} S dx = \frac{2}{a (2 n + 3)} \left(x^{n} S^{3} - n b \int x^{n - 1} S dx\right)$

$\int \frac{1}{x^{n} S} dx = -\frac{1}{b (n - 1)} \left( \frac{S}{x^{n - 1}} + \left( n - \frac{3}{2}\right) a \int \frac{dx}{x^{n - 1} S}\right)$

References

• Peirce, Benjamin Osgood (1929) [1899]. "Chap. 3". A Short Table of Integrals (3rd revised ed.). Boston: Ginn and Co. pp. 16–30.
• Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1972, Dover: New York. (See chapter 3.)
• Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo, eds. Table of Integrals, Series, and Products. Translated by Scripta Technica, Inc. (8 ed.). Academic Press, Inc. ISBN 0-12-384933-0. LCCN 2014010276. ISBN 978-0-12-384933-5. (Several previous editions as well.)