Integrand size = 16, antiderivative size = 28 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (1-x^4\right )}+\log (x)-\frac {1}{4} \log \left (1-x^4\right ) \]
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Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {28, 272, 46} \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=\frac {1}{4 \left (1-x^4\right )}-\frac {1}{4} \log \left (1-x^4\right )+\log (x) \]
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Rule 28
Rule 46
Rule 272
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x \left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{(-1+x)^2 x} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{1-x}+\frac {1}{(-1+x)^2}+\frac {1}{x}\right ) \, dx,x,x^4\right ) \\ & = \frac {1}{4 \left (1-x^4\right )}+\log (x)-\frac {1}{4} \log \left (1-x^4\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{4 \left (-1+x^4\right )}+\log (x)-\frac {1}{4} \log \left (1-x^4\right ) \]
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Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {1}{4 \left (x^{4}-1\right )}+\ln \left (x \right )-\frac {\ln \left (x^{4}-1\right )}{4}\) | \(21\) |
norman | \(-\frac {1}{4 \left (x^{4}-1\right )}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}-\frac {\ln \left (x^{2}+1\right )}{4}+\ln \left (x \right )\) | \(33\) |
default | \(\ln \left (x \right )+\frac {1}{16 x +16}-\frac {\ln \left (x +1\right )}{4}-\frac {\ln \left (x^{2}+1\right )}{4}+\frac {1}{8 x^{2}+8}-\frac {1}{16 \left (x -1\right )}-\frac {\ln \left (x -1\right )}{4}\) | \(47\) |
parallelrisch | \(\frac {4 \ln \left (x \right ) x^{4}-\ln \left (x -1\right ) x^{4}-\ln \left (x +1\right ) x^{4}-\ln \left (x^{2}+1\right ) x^{4}-1-4 \ln \left (x \right )+\ln \left (x -1\right )+\ln \left (x +1\right )+\ln \left (x^{2}+1\right )}{4 x^{4}-4}\) | \(66\) |
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Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=-\frac {{\left (x^{4} - 1\right )} \log \left (x^{4} - 1\right ) - 4 \, {\left (x^{4} - 1\right )} \log \left (x\right ) + 1}{4 \, {\left (x^{4} - 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{4} - 1 \right )}}{4} - \frac {1}{4 x^{4} - 4} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{4} \, \log \left (x^{4} - 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=\frac {x^{4} - 2}{4 \, {\left (x^{4} - 1\right )}} + \frac {1}{4} \, \log \left (x^{4}\right ) - \frac {1}{4} \, \log \left ({\left | x^{4} - 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x \left (1-2 x^4+x^8\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^4-1\right )}{4}-\frac {1}{4\,\left (x^4-1\right )} \]
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