Integrand size = 24, antiderivative size = 68 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=-\frac {1}{x}+\arctan (1-x)-\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\frac {1}{2} \log \left (2-2 x+x^2\right ) \]
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Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2605, 12, 6860, 648, 631, 210, 642} \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=\arctan (1-x)-\frac {1}{2} \log \left (x^2-2 x+2\right )-\frac {1}{x}-\frac {\log \left (\frac {1-(1-x)^2}{(x-1)^2+1}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2} \]
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Rule 12
Rule 210
Rule 631
Rule 642
Rule 648
Rule 2605
Rule 6860
Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\int \frac {4 (1-x)}{(2-x) x^2 \left (2-2 x+x^2\right )} \, dx \\ & = -\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+4 \int \frac {1-x}{(2-x) x^2 \left (2-2 x+x^2\right )} \, dx \\ & = -\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+4 \int \left (\frac {1}{8 (-2+x)}+\frac {1}{4 x^2}+\frac {1}{8 x}-\frac {x}{4 \left (2-2 x+x^2\right )}\right ) \, dx \\ & = -\frac {1}{x}-\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\int \frac {x}{2-2 x+x^2} \, dx \\ & = -\frac {1}{x}-\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\frac {1}{2} \int \frac {-2+2 x}{2-2 x+x^2} \, dx-\int \frac {1}{2-2 x+x^2} \, dx \\ & = -\frac {1}{x}-\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\frac {1}{2} \log \left (2-2 x+x^2\right )-\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-x\right ) \\ & = -\frac {1}{x}+\tan ^{-1}(1-x)-\frac {\log \left (\frac {1-(1-x)^2}{1+(-1+x)^2}\right )}{x}+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\frac {1}{2} \log \left (2-2 x+x^2\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=-\frac {1}{x}+\arctan (1-x)+\frac {1}{2} \log (2-x)+\frac {\log (x)}{2}-\frac {\log \left (-\frac {(-2+x) x}{2-2 x+x^2}\right )}{x}-\frac {1}{2} \log \left (2-2 x+x^2\right ) \]
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Time = 1.99 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\ln \left (\frac {x \left (2-x \right )}{x^{2}-2 x +2}\right )}{x}-\frac {1}{x}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (x^{2}-2 x +2\right )}{2}-\arctan \left (-1+x \right )+\frac {\ln \left (-2+x \right )}{2}\) | \(57\) |
parts | \(-\frac {\ln \left (\frac {1-\left (-1+x \right )^{2}}{1+\left (-1+x \right )^{2}}\right )}{x}-\frac {1}{x}+\frac {\ln \left (x \right )}{2}-\frac {\ln \left (x^{2}-2 x +2\right )}{2}-\arctan \left (-1+x \right )+\frac {\ln \left (-2+x \right )}{2}\) | \(59\) |
risch | \(-\frac {\ln \left (\frac {1-\left (-1+x \right )^{2}}{1+\left (-1+x \right )^{2}}\right )}{x}+\frac {i \ln \left (x -1-i\right ) x -i \ln \left (x -1+i\right ) x -\ln \left (x -1-i\right ) x -\ln \left (x -1+i\right ) x +\ln \left (x^{2}-2 x \right ) x -2}{2 x}\) | \(82\) |
parallelrisch | \(-\frac {12+6 i \ln \left (x -1+i\right ) x -6 i \ln \left (x -1-i\right ) x -14 \ln \left (x \right ) x -14 \ln \left (-2+x \right ) x +14 \ln \left (x -1-i\right ) x +14 \ln \left (x -1+i\right ) x +8 \ln \left (-\frac {x \left (-2+x \right )}{x^{2}-2 x +2}\right ) x +3 x +12 \ln \left (-\frac {x \left (-2+x \right )}{x^{2}-2 x +2}\right )}{12 x}\) | \(100\) |
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Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=-\frac {2 \, x \arctan \left (x - 1\right ) + x \log \left (x^{2} - 2 \, x + 2\right ) - x \log \left (x^{2} - 2 \, x\right ) + 2 \, \log \left (-\frac {x^{2} - 2 \, x}{x^{2} - 2 \, x + 2}\right ) + 2}{2 \, x} \]
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Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=\frac {\log {\left (x^{2} - 2 x \right )}}{2} - \frac {\log {\left (x^{2} - 2 x + 2 \right )}}{2} - \operatorname {atan}{\left (x - 1 \right )} - \frac {\log {\left (\frac {1 - \left (x - 1\right )^{2}}{\left (x - 1\right )^{2} + 1} \right )}}{x} - \frac {1}{x} \]
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Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=-\frac {\log \left (-\frac {{\left (x - 1\right )}^{2} - 1}{{\left (x - 1\right )}^{2} + 1}\right )}{x} - \frac {1}{x} - \arctan \left (x - 1\right ) - \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 2\right ) + \frac {1}{2} \, \log \left (x - 2\right ) + \frac {1}{2} \, \log \left (x\right ) \]
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Time = 0.37 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=-\frac {\log \left (-\frac {{\left (x - 1\right )}^{2} - 1}{{\left (x - 1\right )}^{2} + 1}\right )}{x} - \frac {1}{x} - \arctan \left (x - 1\right ) - \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 2\right ) + \frac {1}{2} \, \log \left ({\left | x - 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]
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Time = 1.59 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {\log \left (\frac {1-(-1+x)^2}{1+(-1+x)^2}\right )}{x^2} \, dx=\frac {\ln \left (x\,\left (x-2\right )\right )}{2}-\mathrm {atan}\left (x-1\right )-\frac {\ln \left (x^2-2\,x+2\right )}{2}-\frac {\ln \left (2\,x-x^2\right )}{x}+\frac {\ln \left (x^2-2\,x+2\right )}{x}-\frac {1}{x} \]
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