Integrand size = 19, antiderivative size = 34 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=\frac {3}{2 (1-x)}-\frac {5}{4} \log (1-x)+2 \log (x)-\frac {3}{4} \log (1+x) \]
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Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1626} \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=\frac {3}{2 (1-x)}-\frac {5}{4} \log (1-x)+2 \log (x)-\frac {3}{4} \log (x+1) \]
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Rule 1626
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{2 (-1+x)^2}-\frac {5}{4 (-1+x)}+\frac {2}{x}-\frac {3}{4 (1+x)}\right ) \, dx \\ & = \frac {3}{2 (1-x)}-\frac {5}{4} \log (1-x)+2 \log (x)-\frac {3}{4} \log (1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=-\frac {3}{2 (-1+x)}-\frac {5}{4} \log (1-x)+2 \log (x)-\frac {3}{4} \log (1+x) \]
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Time = 0.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
default | \(2 \ln \left (x \right )-\frac {3 \ln \left (x +1\right )}{4}-\frac {3}{2 \left (x -1\right )}-\frac {5 \ln \left (x -1\right )}{4}\) | \(25\) |
norman | \(2 \ln \left (x \right )-\frac {3 \ln \left (x +1\right )}{4}-\frac {3}{2 \left (x -1\right )}-\frac {5 \ln \left (x -1\right )}{4}\) | \(25\) |
risch | \(2 \ln \left (x \right )-\frac {3 \ln \left (x +1\right )}{4}-\frac {3}{2 \left (x -1\right )}-\frac {5 \ln \left (x -1\right )}{4}\) | \(25\) |
parallelrisch | \(\frac {8 \ln \left (x \right ) x -5 \ln \left (x -1\right ) x -3 \ln \left (x +1\right ) x -6-8 \ln \left (x \right )+5 \ln \left (x -1\right )+3 \ln \left (x +1\right )}{4 x -4}\) | \(45\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=-\frac {3 \, {\left (x - 1\right )} \log \left (x + 1\right ) + 5 \, {\left (x - 1\right )} \log \left (x - 1\right ) - 8 \, {\left (x - 1\right )} \log \left (x\right ) + 6}{4 \, {\left (x - 1\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=2 \log {\left (x \right )} - \frac {5 \log {\left (x - 1 \right )}}{4} - \frac {3 \log {\left (x + 1 \right )}}{4} - \frac {3}{2 x - 2} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=-\frac {3}{2 \, {\left (x - 1\right )}} - \frac {3}{4} \, \log \left (x + 1\right ) - \frac {5}{4} \, \log \left (x - 1\right ) + 2 \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=-\frac {3}{2 \, {\left (x - 1\right )}} + 2 \, \log \left ({\left | -\frac {1}{x - 1} - 1 \right |}\right ) - \frac {3}{4} \, \log \left ({\left | -\frac {2}{x - 1} - 1 \right |}\right ) \]
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Time = 9.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {2+x^2}{(-1+x)^2 x (1+x)} \, dx=2\,\ln \left (x\right )-\frac {3\,\ln \left (x+1\right )}{4}-\frac {5\,\ln \left (x-1\right )}{4}-\frac {3}{2\,\left (x-1\right )} \]
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