Integrand size = 21, antiderivative size = 24 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\frac {1}{-1+x}+\arctan (x)+\log (1-x)-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1643, 649, 209, 266} \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\arctan (x)-\frac {1}{2} \log \left (x^2+1\right )+\frac {1}{x-1}+\log (1-x) \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{(-1+x)^2}+\frac {1}{-1+x}+\frac {1-x}{1+x^2}\right ) \, dx \\ & = \frac {1}{-1+x}+\log (1-x)+\int \frac {1-x}{1+x^2} \, dx \\ & = \frac {1}{-1+x}+\log (1-x)+\int \frac {1}{1+x^2} \, dx-\int \frac {x}{1+x^2} \, dx \\ & = \frac {1}{-1+x}+\tan ^{-1}(x)+\log (1-x)-\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\frac {1}{-1+x}+\arctan (x)+\log (-1+x)-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {\ln \left (x^{2}+1\right )}{2}+\arctan \left (x \right )+\ln \left (x -1\right )+\frac {1}{x -1}\) | \(21\) |
risch | \(-\frac {\ln \left (x^{2}+1\right )}{2}+\arctan \left (x \right )+\ln \left (x -1\right )+\frac {1}{x -1}\) | \(21\) |
parallelrisch | \(-\frac {i \ln \left (x -i\right ) x -i \ln \left (x +i\right ) x -i \ln \left (x -i\right )+i \ln \left (x +i\right )-2 \ln \left (x -1\right ) x +\ln \left (x -i\right ) x +\ln \left (x +i\right ) x -2+2 \ln \left (x -1\right )-\ln \left (x -i\right )-\ln \left (x +i\right )}{2 \left (x -1\right )}\) | \(85\) |
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\frac {2 \, {\left (x - 1\right )} \arctan \left (x\right ) - {\left (x - 1\right )} \log \left (x^{2} + 1\right ) + 2 \, {\left (x - 1\right )} \log \left (x - 1\right ) + 2}{2 \, {\left (x - 1\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\log {\left (x - 1 \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} + \operatorname {atan}{\left (x \right )} + \frac {1}{x - 1} \]
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Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\frac {1}{x - 1} + \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x - 1\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.39 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\frac {1}{4} \, \pi - \pi \left \lfloor \frac {\pi + 4 \, \arctan \left (x\right )}{4 \, \pi } + \frac {1}{2} \right \rfloor + \frac {1}{x - 1} + \arctan \left (x\right ) - \frac {1}{2} \, \log \left (\frac {2}{x - 1} + \frac {2}{{\left (x - 1\right )}^{2}} + 1\right ) \]
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Time = 9.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-1-2 x+x^2}{(-1+x)^2 \left (1+x^2\right )} \, dx=\ln \left (x-1\right )+\frac {1}{x-1}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \]
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