Integrand size = 16, antiderivative size = 25 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {x^2}{4 \left (1-x^4\right )}-\frac {\text {arctanh}\left (x^2\right )}{4} \]
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Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {28, 281, 294, 213} \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {x^2}{4 \left (1-x^4\right )}-\frac {\text {arctanh}\left (x^2\right )}{4} \]
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Rule 28
Rule 213
Rule 281
Rule 294
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^5}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{4 \left (1-x^4\right )}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{4 \left (1-x^4\right )}-\frac {1}{4} \tanh ^{-1}\left (x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {1}{8} \left (-\frac {2 x^2}{-1+x^4}+\log \left (1-x^2\right )-\log \left (1+x^2\right )\right ) \]
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Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {x^{2}}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x^{2}-1\right )}{8}-\frac {\ln \left (x^{2}+1\right )}{8}\) | \(30\) |
norman | \(-\frac {x^{2}}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}-\frac {\ln \left (x^{2}+1\right )}{8}\) | \(34\) |
default | \(-\frac {1}{8 \left (x^{2}+1\right )}-\frac {\ln \left (x^{2}+1\right )}{8}-\frac {1}{8 \left (x^{2}-1\right )}+\frac {\ln \left (x^{2}-1\right )}{8}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (x -1\right ) x^{4}+\ln \left (x +1\right ) x^{4}-\ln \left (x^{2}+1\right ) x^{4}-2 x^{2}-\ln \left (x -1\right )-\ln \left (x +1\right )+\ln \left (x^{2}+1\right )}{8 x^{4}-8}\) | \(61\) |
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Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (19) = 38\).
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {2 \, x^{2} + {\left (x^{4} - 1\right )} \log \left (x^{2} + 1\right ) - {\left (x^{4} - 1\right )} \log \left (x^{2} - 1\right )}{8 \, {\left (x^{4} - 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=- \frac {x^{2}}{4 x^{4} - 4} + \frac {\log {\left (x^{2} - 1 \right )}}{8} - \frac {\log {\left (x^{2} + 1 \right )}}{8} \]
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none
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]
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none
Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 8.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {x^2}{4\,\left (x^4-1\right )} \]
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