Integrand size = 22, antiderivative size = 57 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=-\frac {1}{1+x}-\arctan (x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2605, 12, 2099, 266, 649, 209} \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=-\arctan (x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{x^2+1}\right )}{x+1}-\frac {1}{2} \log \left (x^2+1\right )-\frac {1}{x+1} \]
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Rule 12
Rule 209
Rule 266
Rule 649
Rule 2099
Rule 2605
Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+\int \frac {4 x}{-1-x+x^4+x^5} \, dx \\ & = -\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+4 \int \frac {x}{-1-x+x^4+x^5} \, dx \\ & = -\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+4 \int \left (\frac {1}{4 (1+x)^2}+\frac {x}{4 \left (-1+x^2\right )}+\frac {-1-x}{4 \left (1+x^2\right )}\right ) \, dx \\ & = -\frac {1}{1+x}-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}+\int \frac {x}{-1+x^2} \, dx+\int \frac {-1-x}{1+x^2} \, dx \\ & = -\frac {1}{1+x}+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\int \frac {1}{1+x^2} \, dx-\int \frac {x}{1+x^2} \, dx \\ & = -\frac {1}{1+x}-\tan ^{-1}(x)+\frac {1}{2} \log \left (1-x^2\right )-\frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{1+x}-\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=\frac {1}{2} \left ((-1+i) \log (i-x)-(1+i) \log (i+x)+\log \left (1-x^2\right )-\frac {2 \left (1+\log \left (\frac {1-x^2}{1+x^2}\right )\right )}{1+x}\right ) \]
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Time = 2.71 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98
method | result | size |
parts | \(-\frac {\ln \left (\frac {-x^{2}+1}{x^{2}+1}\right )}{x +1}+\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}-\arctan \left (x \right )-\frac {1}{x +1}+\frac {\ln \left (x +1\right )}{2}\) | \(56\) |
parallelrisch | \(\frac {-i \ln \left (x +i\right )-i \ln \left (x +i\right ) x -1+i \ln \left (x -i\right )+i \ln \left (x -i\right ) x +x \ln \left (-\frac {x^{2}-1}{x^{2}+1}\right )+x -\ln \left (-\frac {x^{2}-1}{x^{2}+1}\right )}{2+2 x}\) | \(79\) |
risch | \(-\frac {\ln \left (\frac {-x^{2}+1}{x^{2}+1}\right )}{x +1}+\frac {i \ln \left (x -i\right ) x -i \ln \left (x +i\right ) x +i \ln \left (x -i\right )-i \ln \left (x +i\right )-\ln \left (x -i\right ) x -\ln \left (x +i\right ) x +\ln \left (x^{2}-1\right ) x -\ln \left (x -i\right )-\ln \left (x +i\right )+\ln \left (x^{2}-1\right )-2}{2+2 x}\) | \(112\) |
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Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=-\frac {2 \, {\left (x + 1\right )} \arctan \left (x\right ) + {\left (x + 1\right )} \log \left (x^{2} + 1\right ) - {\left (x + 1\right )} \log \left (x^{2} - 1\right ) + 2 \, \log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right ) + 2}{2 \, {\left (x + 1\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=\frac {\log {\left (x^{2} - 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \operatorname {atan}{\left (x \right )} - \frac {4}{4 x + 4} - \frac {\log {\left (\frac {1 - x^{2}}{x^{2} + 1} \right )}}{x + 1} \]
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Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=-\frac {\log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right )}{x + 1} - \frac {1}{x + 1} - \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \]
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Time = 0.33 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=-\frac {\log \left (-\frac {x^{2} - 1}{x^{2} + 1}\right )}{x + 1} - \frac {1}{x + 1} - \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 1.55 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (\frac {1-x^2}{1+x^2}\right )}{(1+x)^2} \, dx=\frac {\ln \left (x^2-1\right )}{2}-\frac {\ln \left (x^2+1\right )}{2}-\mathrm {atan}\left (x\right )-\frac {1}{x+1}+\frac {\ln \left (x^2+1\right )}{x+1}-\frac {\ln \left (1-x^2\right )}{x+1} \]
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