Integrand size = 20, antiderivative size = 97 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=-\frac {x}{6 \left (1+x^3\right )}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{12} \log (1-x)-\frac {1}{36} \log (1+x)+\frac {1}{72} \log \left (1-x+x^2\right )+\frac {1}{24} \log \left (1+x+x^2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {482, 536, 206, 31, 648, 632, 210, 642} \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {x}{6 \left (x^3+1\right )}+\frac {1}{72} \log \left (x^2-x+1\right )+\frac {1}{24} \log \left (x^2+x+1\right )-\frac {1}{12} \log (1-x)-\frac {1}{36} \log (x+1) \]
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Rule 31
Rule 206
Rule 210
Rule 482
Rule 536
Rule 632
Rule 642
Rule 648
Rubi steps \begin{align*} \text {integral}& = -\frac {x}{6 \left (1+x^3\right )}+\frac {1}{6} \int \frac {1+2 x^3}{\left (1-x^3\right ) \left (1+x^3\right )} \, dx \\ & = -\frac {x}{6 \left (1+x^3\right )}-\frac {1}{12} \int \frac {1}{1+x^3} \, dx+\frac {1}{4} \int \frac {1}{1-x^3} \, dx \\ & = -\frac {x}{6 \left (1+x^3\right )}-\frac {1}{36} \int \frac {1}{1+x} \, dx-\frac {1}{36} \int \frac {2-x}{1-x+x^2} \, dx+\frac {1}{12} \int \frac {1}{1-x} \, dx+\frac {1}{12} \int \frac {2+x}{1+x+x^2} \, dx \\ & = -\frac {x}{6 \left (1+x^3\right )}-\frac {1}{12} \log (1-x)-\frac {1}{36} \log (1+x)+\frac {1}{72} \int \frac {-1+2 x}{1-x+x^2} \, dx-\frac {1}{24} \int \frac {1}{1-x+x^2} \, dx+\frac {1}{24} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {x}{6 \left (1+x^3\right )}-\frac {1}{12} \log (1-x)-\frac {1}{36} \log (1+x)+\frac {1}{72} \log \left (1-x+x^2\right )+\frac {1}{24} \log \left (1+x+x^2\right )+\frac {1}{12} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {x}{6 \left (1+x^3\right )}+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{12} \log (1-x)-\frac {1}{36} \log (1+x)+\frac {1}{72} \log \left (1-x+x^2\right )+\frac {1}{24} \log \left (1+x+x^2\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=\frac {1}{72} \left (-\frac {12 x}{1+x^3}-2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+6 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-6 \log (1-x)-2 \log (1+x)+\log \left (1-x+x^2\right )+3 \log \left (1+x+x^2\right )\right ) \]
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Time = 0.82 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74
method | result | size |
risch | \(-\frac {x}{6 \left (x^{3}+1\right )}-\frac {\ln \left (x -1\right )}{12}-\frac {\ln \left (x +1\right )}{36}+\frac {\ln \left (x^{2}+x +1\right )}{24}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x +\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}-x +1\right )}{72}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{36}\) | \(72\) |
default | \(\frac {1}{18 x +18}-\frac {\ln \left (x +1\right )}{36}+\frac {-2 x -2}{36 x^{2}-36 x +36}+\frac {\ln \left (x^{2}-x +1\right )}{72}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{36}+\frac {\ln \left (x^{2}+x +1\right )}{24}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (x -1\right )}{12}\) | \(90\) |
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Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=\frac {6 \, \sqrt {3} {\left (x^{3} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left (x^{3} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 3 \, {\left (x^{3} + 1\right )} \log \left (x^{2} + x + 1\right ) + {\left (x^{3} + 1\right )} \log \left (x^{2} - x + 1\right ) - 2 \, {\left (x^{3} + 1\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{3} + 1\right )} \log \left (x - 1\right ) - 12 \, x}{72 \, {\left (x^{3} + 1\right )}} \]
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Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=- \frac {x}{6 x^{3} + 6} - \frac {\log {\left (x - 1 \right )}}{12} - \frac {\log {\left (x + 1 \right )}}{36} + \frac {\log {\left (x^{2} - x + 1 \right )}}{72} + \frac {\log {\left (x^{2} + x + 1 \right )}}{24} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{36} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \]
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Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x}{6 \, {\left (x^{3} + 1\right )}} + \frac {1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{36} \, \log \left (x + 1\right ) - \frac {1}{12} \, \log \left (x - 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x}{6 \, {\left (x^{3} + 1\right )}} + \frac {1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{36} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{12} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx=-\frac {\ln \left (x-1\right )}{12}-\frac {\ln \left (x+1\right )}{36}-\frac {x}{6\,\left (x^3+1\right )}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right ) \]
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