Integrand size = 19, antiderivative size = 28 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 (1+x)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (1+x) \]
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Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {27, 90} \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 (x+1)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (x+1) \]
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Rule 27
Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{(-1+x) (1+x)^2} \, dx \\ & = \int \left (\frac {1}{4 (-1+x)}-\frac {1}{2 (1+x)^2}+\frac {3}{4 (1+x)}\right ) \, dx \\ & = \frac {1}{2 (1+x)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{4} \left (\frac {2}{1+x}+\log (-1+x)+3 \log (1+x)\right ) \]
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Time = 0.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\) | \(21\) |
norman | \(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\) | \(21\) |
risch | \(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\) | \(21\) |
parallelrisch | \(\frac {\ln \left (x -1\right ) x +3 \ln \left (x +1\right ) x +2+\ln \left (x -1\right )+3 \ln \left (x +1\right )}{4+4 x}\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {3 \, {\left (x + 1\right )} \log \left (x + 1\right ) + {\left (x + 1\right )} \log \left (x - 1\right ) + 2}{4 \, {\left (x + 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{4} + \frac {3 \log {\left (x + 1 \right )}}{4} + \frac {1}{2 x + 2} \]
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Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 \, {\left (x + 1\right )}} + \frac {3}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 \, {\left (x + 1\right )}} + \frac {3}{4} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{4}+\frac {3\,\ln \left (x+1\right )}{4}+\frac {1}{2\,\left (x+1\right )} \]
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