Integrand size = 26, antiderivative size = 46 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 (2+x)}-\frac {1}{18} \log (1-x)+\frac {1}{48} \log (2-x)+\frac {1}{6} \log (1+x)-\frac {19}{144} \log (2+x) \]
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Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1600, 2099} \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 (x+2)}-\frac {1}{18} \log (1-x)+\frac {1}{48} \log (2-x)+\frac {1}{6} \log (x+1)-\frac {19}{144} \log (x+2) \]
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Rule 1600
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(2+x)^2 \left (2-x-2 x^2+x^3\right )} \, dx \\ & = \int \left (\frac {1}{48 (-2+x)}-\frac {1}{18 (-1+x)}+\frac {1}{6 (1+x)}-\frac {1}{12 (2+x)^2}-\frac {19}{144 (2+x)}\right ) \, dx \\ & = \frac {1}{12 (2+x)}-\frac {1}{18} \log (1-x)+\frac {1}{48} \log (2-x)+\frac {1}{6} \log (1+x)-\frac {19}{144} \log (2+x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \left (\frac {12}{2+x}+24 \log (-1-x)-8 \log (1-x)+3 \log (2-x)-19 \log (2+x)\right ) \]
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Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {1}{12 x +24}-\frac {19 \ln \left (x +2\right )}{144}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{18}+\frac {\ln \left (x -2\right )}{48}\) | \(33\) |
risch | \(\frac {1}{12 x +24}-\frac {19 \ln \left (x +2\right )}{144}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{18}+\frac {\ln \left (x -2\right )}{48}\) | \(33\) |
norman | \(\frac {-\frac {1}{6} x^{2}-\frac {1}{12} x +\frac {1}{12} x^{3}+\frac {1}{6}}{x^{4}-5 x^{2}+4}+\frac {\ln \left (x -2\right )}{48}-\frac {\ln \left (x -1\right )}{18}+\frac {\ln \left (x +1\right )}{6}-\frac {19 \ln \left (x +2\right )}{144}\) | \(54\) |
parallelrisch | \(\frac {3 \ln \left (x -2\right ) x -8 \ln \left (x -1\right ) x +24 \ln \left (x +1\right ) x -19 \ln \left (x +2\right ) x +12+6 \ln \left (x -2\right )-16 \ln \left (x -1\right )+48 \ln \left (x +1\right )-38 \ln \left (x +2\right )}{144 x +288}\) | \(62\) |
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Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {19 \, {\left (x + 2\right )} \log \left (x + 2\right ) - 24 \, {\left (x + 2\right )} \log \left (x + 1\right ) + 8 \, {\left (x + 2\right )} \log \left (x - 1\right ) - 3 \, {\left (x + 2\right )} \log \left (x - 2\right ) - 12}{144 \, {\left (x + 2\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {\log {\left (x - 2 \right )}}{48} - \frac {\log {\left (x - 1 \right )}}{18} + \frac {\log {\left (x + 1 \right )}}{6} - \frac {19 \log {\left (x + 2 \right )}}{144} + \frac {1}{12 x + 24} \]
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Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 \, {\left (x + 2\right )}} - \frac {19}{144} \, \log \left (x + 2\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{18} \, \log \left (x - 1\right ) + \frac {1}{48} \, \log \left (x - 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 \, {\left (x + 2\right )}} - \frac {19}{144} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{18} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{48} \, \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.70 \[ \int \frac {2-x-2 x^2+x^3}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {\ln \left (x+1\right )}{6}-\frac {\ln \left (x-1\right )}{18}+\frac {\ln \left (x-2\right )}{48}-\frac {19\,\ln \left (x+2\right )}{144}+\frac {1}{12\,\left (x+2\right )} \]
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