Integrand size = 25, antiderivative size = 36 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=6 \arctan (x)-5 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (1+x^2\right )+\log \left (2+x^2\right ) \]
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Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6857, 649, 209, 266} \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=6 \arctan (x)-5 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (x^2+1\right )+\log \left (x^2+2\right ) \]
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Rule 209
Rule 266
Rule 649
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {6-x}{1+x^2}+\frac {2 (-5+x)}{2+x^2}\right ) \, dx \\ & = 2 \int \frac {-5+x}{2+x^2} \, dx+\int \frac {6-x}{1+x^2} \, dx \\ & = 2 \int \frac {x}{2+x^2} \, dx+6 \int \frac {1}{1+x^2} \, dx-10 \int \frac {1}{2+x^2} \, dx-\int \frac {x}{1+x^2} \, dx \\ & = 6 \arctan (x)-5 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (1+x^2\right )+\log \left (2+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=6 \arctan (x)-5 \sqrt {2} \arctan \left (\frac {x}{\sqrt {2}}\right )-\frac {1}{2} \log \left (1+x^2\right )+\log \left (2+x^2\right ) \]
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Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89
method | result | size |
default | \(6 \arctan \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}+\ln \left (x^{2}+2\right )-5 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}\) | \(32\) |
risch | \(6 \arctan \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}+\ln \left (x^{2}+2\right )-5 \arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=-5 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 6 \, \arctan \left (x\right ) + \log \left (x^{2} + 2\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=- \frac {\log {\left (x^{2} + 1 \right )}}{2} + \log {\left (x^{2} + 2 \right )} + 6 \operatorname {atan}{\left (x \right )} - 5 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )} \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=-5 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 6 \, \arctan \left (x\right ) + \log \left (x^{2} + 2\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=-5 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 6 \, \arctan \left (x\right ) + \log \left (x^{2} + 2\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.56 \[ \int \frac {2-4 x^2+x^3}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx=\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{2}-3{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+3{}\mathrm {i}\right )+\ln \left (x-\sqrt {2}\,1{}\mathrm {i}\right )\,\left (1+\frac {\sqrt {2}\,5{}\mathrm {i}}{2}\right )-\ln \left (x+\sqrt {2}\,1{}\mathrm {i}\right )\,\left (-1+\frac {\sqrt {2}\,5{}\mathrm {i}}{2}\right ) \]
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