Integrand size = 13, antiderivative size = 24 \[ \int \frac {x^9}{1-x^8} \, dx=-\frac {x^2}{2}+\frac {\arctan \left (x^2\right )}{4}+\frac {\text {arctanh}\left (x^2\right )}{4} \]
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Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {281, 327, 218, 212, 209} \[ \int \frac {x^9}{1-x^8} \, dx=\frac {\arctan \left (x^2\right )}{4}+\frac {\text {arctanh}\left (x^2\right )}{4}-\frac {x^2}{2} \]
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Rule 209
Rule 212
Rule 218
Rule 281
Rule 327
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{1-x^4} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{2}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{2}+\frac {1}{4} \tan ^{-1}\left (x^2\right )+\frac {1}{4} \tanh ^{-1}\left (x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {x^9}{1-x^8} \, dx=-\frac {x^2}{2}-\frac {1}{4} \arctan \left (\frac {1}{x^2}\right )-\frac {1}{8} \log \left (1-x^2\right )+\frac {1}{8} \log \left (1+x^2\right ) \]
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Time = 3.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
risch | \(-\frac {x^{2}}{2}+\frac {\arctan \left (x^{2}\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{8}-\frac {\ln \left (x^{2}-1\right )}{8}\) | \(29\) |
default | \(-\frac {x^{2}}{2}-\frac {\ln \left (-1+x \right )}{8}-\frac {\ln \left (1+x \right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}+\frac {\arctan \left (x^{2}\right )}{4}\) | \(33\) |
parallelrisch | \(-\frac {x^{2}}{2}-\frac {\ln \left (1+x \right )}{8}-\frac {\ln \left (-1+x \right )}{8}+\frac {\ln \left (x -i\right )}{8}+\frac {\ln \left (x +i\right )}{8}-\frac {i \ln \left (x^{2}-i\right )}{8}+\frac {i \ln \left (x^{2}+i\right )}{8}\) | \(53\) |
meijerg | \(\frac {\left (-1\right )^{\frac {3}{4}} \left (4 x^{2} \left (-1\right )^{\frac {1}{4}}+\frac {x^{2} \left (-1\right )^{\frac {1}{4}} \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{8}\right )^{\frac {1}{4}}\right )\right )}{\left (x^{8}\right )^{\frac {1}{4}}}\right )}{8}\) | \(56\) |
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {x^9}{1-x^8} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {x^9}{1-x^8} \, dx=- \frac {x^{2}}{2} - \frac {\log {\left (x^{2} - 1 \right )}}{8} + \frac {\log {\left (x^{2} + 1 \right )}}{8} + \frac {\operatorname {atan}{\left (x^{2} \right )}}{4} \]
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {x^9}{1-x^8} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {x^9}{1-x^8} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 6.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {x^9}{1-x^8} \, dx=\frac {\mathrm {atan}\left (x^2\right )}{4}+\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {x^2}{2} \]
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