Integrand size = 16, antiderivative size = 46 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=x+\frac {x^2}{2}-\frac {2}{3} \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )+\frac {1}{3} \log (1-x)-\frac {2}{3} \log \left (2+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1643, 649, 209, 266} \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=-\frac {2}{3} \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )+\frac {x^2}{2}-\frac {2}{3} \log \left (x^2+2\right )+x+\frac {1}{3} \log (1-x) \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {1}{3 (-1+x)}+x-\frac {4 (1+x)}{3 \left (2+x^2\right )}\right ) \, dx \\ & = x+\frac {x^2}{2}+\frac {1}{3} \log (1-x)-\frac {4}{3} \int \frac {1+x}{2+x^2} \, dx \\ & = x+\frac {x^2}{2}+\frac {1}{3} \log (1-x)-\frac {4}{3} \int \frac {1}{2+x^2} \, dx-\frac {4}{3} \int \frac {x}{2+x^2} \, dx \\ & = x+\frac {x^2}{2}-\frac {2}{3} \sqrt {2} \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )+\frac {1}{3} \log (1-x)-\frac {2}{3} \log \left (2+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=\frac {1}{6} \left (-9+6 x+3 x^2-4 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )+2 \log (-1+x)-4 \log \left (2+x^2\right )\right ) \]
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Time = 0.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {x^{2}}{2}+x -\frac {2 \ln \left (x^{2}+2\right )}{3}-\frac {2 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{3}+\frac {\ln \left (x -1\right )}{3}\) | \(34\) |
risch | \(\frac {x^{2}}{2}+x -\frac {2 \ln \left (x^{2}+2\right )}{3}-\frac {2 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{3}+\frac {\ln \left (x -1\right )}{3}\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=\frac {1}{2} \, x^{2} - \frac {2}{3} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + x - \frac {2}{3} \, \log \left (x^{2} + 2\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]
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Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=\frac {x^{2}}{2} + x + \frac {\log {\left (x - 1 \right )}}{3} - \frac {2 \log {\left (x^{2} + 2 \right )}}{3} - \frac {2 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{3} \]
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=\frac {1}{2} \, x^{2} - \frac {2}{3} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + x - \frac {2}{3} \, \log \left (x^{2} + 2\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=\frac {1}{2} \, x^{2} - \frac {2}{3} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + x - \frac {2}{3} \, \log \left (x^{2} + 2\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \frac {x^4}{(-1+x) \left (2+x^2\right )} \, dx=x+\frac {\ln \left (x-1\right )}{3}+\ln \left (x-\sqrt {2}\,1{}\mathrm {i}\right )\,\left (-\frac {2}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}\right )-\ln \left (x+\sqrt {2}\,1{}\mathrm {i}\right )\,\left (\frac {2}{3}+\frac {\sqrt {2}\,1{}\mathrm {i}}{3}\right )+\frac {x^2}{2} \]
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