Integrand size = 11, antiderivative size = 12 \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \log \left (1-x^2\right ) \]
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Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {266} \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \log \left (1-x^2\right ) \]
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Rule 266
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2} \log \left (1-x^2\right ) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \log \left (1-x^2\right ) \]
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Time = 0.45 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {\ln \left (x^{2}-1\right )}{2}\) | \(9\) |
derivativedivides | \(-\frac {\ln \left (-x^{2}+1\right )}{2}\) | \(11\) |
meijerg | \(-\frac {\ln \left (-x^{2}+1\right )}{2}\) | \(11\) |
default | \(-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\) | \(14\) |
norman | \(-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\) | \(14\) |
parallelrisch | \(-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\) | \(14\) |
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none
Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \, \log \left (x^{2} - 1\right ) \]
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Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{1-x^2} \, dx=- \frac {\log {\left (x^{2} - 1 \right )}}{2} \]
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none
Time = 0.21 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \, \log \left (x^{2} - 1\right ) \]
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none
Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {x}{1-x^2} \, dx=-\frac {1}{2} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x}{1-x^2} \, dx=-\frac {\ln \left (x^2-1\right )}{2} \]
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