Integrand size = 13, antiderivative size = 49 \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1-x^2\right )+\frac {1}{12} \log \left (1+x^2+x^4\right ) \]
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Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {281, 298, 31, 648, 632, 210, 642} \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1-x^2\right )+\frac {1}{12} \log \left (x^4+x^2+1\right ) \]
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Rule 31
Rule 210
Rule 281
Rule 298
Rule 632
Rule 642
Rule 648
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,x^2\right ) \\ & = \frac {1}{6} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{6} \log \left (1-x^2\right )+\frac {1}{12} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{6} \log \left (1-x^2\right )+\frac {1}{12} \log \left (1+x^2+x^4\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1-x^2\right )+\frac {1}{12} \log \left (1+x^2+x^4\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.49 \[ \int \frac {x^3}{1-x^6} \, dx=\frac {1}{12} \left (-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \log (1-x)-2 \log (1+x)+\log \left (1-x+x^2\right )+\log \left (1+x+x^2\right )\right ) \]
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Time = 4.47 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{12}-\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (x^{2}-1\right )}{6}\) | \(43\) |
meijerg | \(-\frac {x^{4} \left (\ln \left (1-\left (x^{6}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{3}}+\left (x^{6}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{2+\left (x^{6}\right )^{\frac {1}{3}}}\right )\right )}{6 \left (x^{6}\right )^{\frac {2}{3}}}\) | \(63\) |
default | \(-\frac {\ln \left (-1+x \right )}{6}+\frac {\ln \left (x^{2}+x +1\right )}{12}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (1+x \right )}{6}+\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{6}\) | \(66\) |
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Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{6} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {x^3}{1-x^6} \, dx=- \frac {\log {\left (x^{2} - 1 \right )}}{6} + \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]
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Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{6} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \log \left (x^{4} + x^{2} + 1\right ) - \frac {1}{6} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{1-x^6} \, dx=-\frac {\ln \left (x^2-1\right )}{6}+\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}+\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}+\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]
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