Integrand size = 16, antiderivative size = 27 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=\frac {x}{4 \left (1-x^4\right )}-\frac {\arctan (x)}{8}-\frac {\text {arctanh}(x)}{8} \]
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Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {28, 294, 218, 212, 209} \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=-\frac {\arctan (x)}{8}-\frac {\text {arctanh}(x)}{8}+\frac {x}{4 \left (1-x^4\right )} \]
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Rule 28
Rule 209
Rule 212
Rule 218
Rule 294
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^4}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {x}{4 \left (1-x^4\right )}+\frac {1}{4} \int \frac {1}{-1+x^4} \, dx \\ & = \frac {x}{4 \left (1-x^4\right )}-\frac {1}{8} \int \frac {1}{1-x^2} \, dx-\frac {1}{8} \int \frac {1}{1+x^2} \, dx \\ & = \frac {x}{4 \left (1-x^4\right )}-\frac {1}{8} \tan ^{-1}(x)-\frac {1}{8} \tanh ^{-1}(x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=\frac {1}{16} \left (-\frac {4 x}{-1+x^4}-2 \arctan (x)+\log (1-x)-\log (1+x)\right ) \]
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Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {x}{4 \left (x^{4}-1\right )}-\frac {\arctan \left (x \right )}{8}-\frac {\ln \left (x +1\right )}{16}+\frac {\ln \left (x -1\right )}{16}\) | \(28\) |
default | \(-\frac {1}{16 \left (x +1\right )}-\frac {\ln \left (x +1\right )}{16}+\frac {x}{8 x^{2}+8}-\frac {\arctan \left (x \right )}{8}-\frac {1}{16 \left (x -1\right )}+\frac {\ln \left (x -1\right )}{16}\) | \(42\) |
parallelrisch | \(-\frac {i \ln \left (x +i\right ) x^{4}-i \ln \left (x -i\right ) x^{4}-\ln \left (x -1\right ) x^{4}+\ln \left (x +1\right ) x^{4}-i \ln \left (x +i\right )+i \ln \left (x -i\right )+\ln \left (x -1\right )-\ln \left (x +1\right )+4 x}{16 \left (x^{4}-1\right )}\) | \(79\) |
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=-\frac {2 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) + {\left (x^{4} - 1\right )} \log \left (x + 1\right ) - {\left (x^{4} - 1\right )} \log \left (x - 1\right ) + 4 \, x}{16 \, {\left (x^{4} - 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=- \frac {x}{4 x^{4} - 4} + \frac {\log {\left (x - 1 \right )}}{16} - \frac {\log {\left (x + 1 \right )}}{16} - \frac {\operatorname {atan}{\left (x \right )}}{8} \]
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none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=-\frac {x}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \arctan \left (x\right ) - \frac {1}{16} \, \log \left (x + 1\right ) + \frac {1}{16} \, \log \left (x - 1\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=-\frac {x}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \arctan \left (x\right ) - \frac {1}{16} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {x^4}{1-2 x^4+x^8} \, dx=-\frac {\mathrm {atan}\left (x\right )}{8}-\frac {\mathrm {atanh}\left (x\right )}{8}-\frac {x}{4\,\left (x^4-1\right )} \]
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