Integrand size = 13, antiderivative size = 22 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1-x^2\right ) \]
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Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1598, 272, 46} \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{2} \log \left (1-x^2\right )+\log (x) \]
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Rule 46
Rule 272
Rule 1598
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^3 \left (1-x^2\right )} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(1-x) x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{1-x}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1-x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1-x^2\right ) \]
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Time = 2.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (x^{2}-1\right )}{2}\) | \(17\) |
default | \(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\) | \(21\) |
norman | \(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\) | \(21\) |
meijerg | \(-\frac {1}{2 x^{2}}+\ln \left (x \right )+\frac {i \pi }{2}-\frac {\ln \left (-x^{2}+1\right )}{2}\) | \(23\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) x^{2}-\ln \left (1+x \right ) x^{2}-\ln \left (-1+x \right ) x^{2}-1}{2 x^{2}}\) | \(33\) |
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Time = 0.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {x^{2} \log \left (x^{2} - 1\right ) - 2 \, x^{2} \log \left (x\right ) + 1}{2 \, x^{2}} \]
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Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{2} - 1 \right )}}{2} - \frac {1}{2 x^{2}} \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 \, x^{2}} - \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) + \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {x^{2} + 1}{2 \, x^{2}} + \frac {1}{2} \, \log \left (x^{2}\right ) - \frac {1}{2} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2-1\right )}{2}-\frac {1}{2\,x^2} \]
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