Integrand size = 26, antiderivative size = 21 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=2 \log (1-x)-\log (2-x)+\log (1+x) \]
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Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2099} \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=2 \log (1-x)-\log (2-x)+\log (x+1) \]
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Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2-x}+\frac {2}{-1+x}+\frac {1}{1+x}\right ) \, dx \\ & = 2 \log (1-x)-\log (2-x)+\log (1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=2 \log (1-x)-\log (2-x)+\log (1+x) \]
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Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
default | \(\ln \left (x +1\right )+2 \ln \left (x -1\right )-\ln \left (x -2\right )\) | \(18\) |
norman | \(\ln \left (x +1\right )+2 \ln \left (x -1\right )-\ln \left (x -2\right )\) | \(18\) |
risch | \(\ln \left (x +1\right )+2 \ln \left (x -1\right )-\ln \left (x -2\right )\) | \(18\) |
parallelrisch | \(\ln \left (x +1\right )+2 \ln \left (x -1\right )-\ln \left (x -2\right )\) | \(18\) |
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none
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=\log \left (x + 1\right ) + 2 \, \log \left (x - 1\right ) - \log \left (x - 2\right ) \]
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Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=- \log {\left (x - 2 \right )} + 2 \log {\left (x - 1 \right )} + \log {\left (x + 1 \right )} \]
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none
Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=\log \left (x + 1\right ) + 2 \, \log \left (x - 1\right ) - \log \left (x - 2\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=\log \left ({\left | x + 1 \right |}\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) - \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-1-5 x+2 x^2}{2-x-2 x^2+x^3} \, dx=2\,\ln \left (x-1\right )-2\,\mathrm {atanh}\left (\frac {144}{11\,\left (22\,x-50\right )}+\frac {13}{11}\right ) \]
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