Integrand size = 16, antiderivative size = 32 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {3 x^2}{4}+\frac {x^6}{4 \left (1-x^4\right )}-\frac {3 \text {arctanh}\left (x^2\right )}{4} \]
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Time = 0.01 (sec) , antiderivative size = 32, 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, 281, 294, 327, 213} \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=-\frac {3 \text {arctanh}\left (x^2\right )}{4}+\frac {3 x^2}{4}+\frac {x^6}{4 \left (1-x^4\right )} \]
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Rule 28
Rule 213
Rule 281
Rule 294
Rule 327
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^9}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^6}{4 \left (1-x^4\right )}+\frac {3}{4} \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,x^2\right ) \\ & = \frac {3 x^2}{4}+\frac {x^6}{4 \left (1-x^4\right )}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,x^2\right ) \\ & = \frac {3 x^2}{4}+\frac {x^6}{4 \left (1-x^4\right )}-\frac {3}{4} \tanh ^{-1}\left (x^2\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {1}{8} \left (2 x^2 \left (2+\frac {1}{1-x^4}\right )+3 \log \left (1-x^2\right )-3 \log \left (1+x^2\right )\right ) \]
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Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {x^{2}}{2}-\frac {x^{2}}{4 \left (x^{4}-1\right )}-\frac {3 \ln \left (x^{2}+1\right )}{8}+\frac {3 \ln \left (x^{2}-1\right )}{8}\) | \(35\) |
default | \(\frac {x^{2}}{2}-\frac {1}{8 \left (x^{2}+1\right )}-\frac {3 \ln \left (x^{2}+1\right )}{8}-\frac {1}{8 \left (x^{2}-1\right )}+\frac {3 \ln \left (x^{2}-1\right )}{8}\) | \(41\) |
norman | \(\frac {-\frac {3}{4} x^{2}+\frac {1}{2} x^{6}}{x^{4}-1}+\frac {3 \ln \left (x -1\right )}{8}+\frac {3 \ln \left (x +1\right )}{8}-\frac {3 \ln \left (x^{2}+1\right )}{8}\) | \(41\) |
parallelrisch | \(\frac {4 x^{6}+3 \ln \left (x -1\right ) x^{4}+3 \ln \left (x +1\right ) x^{4}-3 \ln \left (x^{2}+1\right ) x^{4}-6 x^{2}-3 \ln \left (x -1\right )-3 \ln \left (x +1\right )+3 \ln \left (x^{2}+1\right )}{8 x^{4}-8}\) | \(70\) |
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Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {4 \, x^{6} - 6 \, x^{2} - 3 \, {\left (x^{4} - 1\right )} \log \left (x^{2} + 1\right ) + 3 \, {\left (x^{4} - 1\right )} \log \left (x^{2} - 1\right )}{8 \, {\left (x^{4} - 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {x^{2}}{2} - \frac {x^{2}}{4 x^{4} - 4} + \frac {3 \log {\left (x^{2} - 1 \right )}}{8} - \frac {3 \log {\left (x^{2} + 1 \right )}}{8} \]
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Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} - \frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {3}{8} \, \log \left (x^{2} + 1\right ) + \frac {3}{8} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} - \frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {3}{8} \, \log \left (x^{2} + 1\right ) + \frac {3}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x^9}{1-2 x^4+x^8} \, dx=\frac {x^2}{2}-\frac {x^2}{4\,\left (x^4-1\right )}-\frac {3\,\mathrm {atanh}\left (x^2\right )}{4} \]
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