Integrand size = 16, antiderivative size = 29 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=\frac {x^3}{4 \left (1-x^4\right )}+\frac {3 \arctan (x)}{8}-\frac {3 \text {arctanh}(x)}{8} \]
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Time = 0.01 (sec) , antiderivative size = 29, 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, 304, 209, 212} \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=\frac {3 \arctan (x)}{8}-\frac {3 \text {arctanh}(x)}{8}+\frac {x^3}{4 \left (1-x^4\right )} \]
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Rule 28
Rule 209
Rule 212
Rule 294
Rule 304
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^6}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {x^3}{4 \left (1-x^4\right )}+\frac {3}{4} \int \frac {x^2}{-1+x^4} \, dx \\ & = \frac {x^3}{4 \left (1-x^4\right )}-\frac {3}{8} \int \frac {1}{1-x^2} \, dx+\frac {3}{8} \int \frac {1}{1+x^2} \, dx \\ & = \frac {x^3}{4 \left (1-x^4\right )}+\frac {3}{8} \tan ^{-1}(x)-\frac {3}{8} \tanh ^{-1}(x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=\frac {1}{16} \left (-\frac {4 x^3}{-1+x^4}+6 \arctan (x)+3 \log (1-x)-3 \log (1+x)\right ) \]
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Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {x^{3}}{4 \left (x^{4}-1\right )}+\frac {3 \arctan \left (x \right )}{8}+\frac {3 \ln \left (x -1\right )}{16}-\frac {3 \ln \left (x +1\right )}{16}\) | \(30\) |
default | \(-\frac {1}{16 \left (x +1\right )}-\frac {3 \ln \left (x +1\right )}{16}-\frac {x}{8 \left (x^{2}+1\right )}+\frac {3 \arctan \left (x \right )}{8}-\frac {1}{16 \left (x -1\right )}+\frac {3 \ln \left (x -1\right )}{16}\) | \(42\) |
parallelrisch | \(\frac {-3 i \ln \left (x -i\right ) x^{4}+3 i \ln \left (x +i\right ) x^{4}+3 \ln \left (x -1\right ) x^{4}-3 \ln \left (x +1\right ) x^{4}-4 x^{3}+3 i \ln \left (x -i\right )-3 i \ln \left (x +i\right )-3 \ln \left (x -1\right )+3 \ln \left (x +1\right )}{16 x^{4}-16}\) | \(84\) |
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=-\frac {4 \, x^{3} - 6 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) + 3 \, {\left (x^{4} - 1\right )} \log \left (x + 1\right ) - 3 \, {\left (x^{4} - 1\right )} \log \left (x - 1\right )}{16 \, {\left (x^{4} - 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=- \frac {x^{3}}{4 x^{4} - 4} + \frac {3 \log {\left (x - 1 \right )}}{16} - \frac {3 \log {\left (x + 1 \right )}}{16} + \frac {3 \operatorname {atan}{\left (x \right )}}{8} \]
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none
Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=-\frac {x^{3}}{4 \, {\left (x^{4} - 1\right )}} + \frac {3}{8} \, \arctan \left (x\right ) - \frac {3}{16} \, \log \left (x + 1\right ) + \frac {3}{16} \, \log \left (x - 1\right ) \]
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none
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=-\frac {x^{3}}{4 \, {\left (x^{4} - 1\right )}} + \frac {3}{8} \, \arctan \left (x\right ) - \frac {3}{16} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {3}{16} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x^6}{1-2 x^4+x^8} \, dx=\frac {3\,\mathrm {atan}\left (x\right )}{8}-\frac {3\,\mathrm {atanh}\left (x\right )}{8}-\frac {x^3}{4\,\left (x^4-1\right )} \]
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