Integrand size = 24, antiderivative size = 23 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \arctan \left (\frac {x}{2}\right )+\arctan (x)+\frac {1}{2} \log \left (4+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1687, 1180, 209, 1261, 640, 31} \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \arctan \left (\frac {x}{2}\right )+\arctan (x)+\frac {1}{2} \log \left (x^2+4\right ) \]
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Rule 31
Rule 209
Rule 640
Rule 1180
Rule 1261
Rule 1687
Rubi steps \begin{align*} \text {integral}& = \int \frac {1-2 x^2}{4+5 x^2+x^4} \, dx+\int \frac {x \left (1+x^2\right )}{4+5 x^2+x^4} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1+x}{4+5 x+x^2} \, dx,x,x^2\right )-3 \int \frac {1}{4+x^2} \, dx+\int \frac {1}{1+x^2} \, dx \\ & = -\frac {3}{2} \arctan \left (\frac {x}{2}\right )+\arctan (x)+\frac {1}{2} \text {Subst}\left (\int \frac {1}{4+x} \, dx,x,x^2\right ) \\ & = -\frac {3}{2} \arctan \left (\frac {x}{2}\right )+\arctan (x)+\frac {1}{2} \log \left (4+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \arctan \left (\frac {x}{2}\right )+\arctan (x)+\frac {1}{2} \log \left (4+x^2\right ) \]
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Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {3 \arctan \left (\frac {x}{2}\right )}{2}+\arctan \left (x \right )+\frac {\ln \left (x^{2}+4\right )}{2}\) | \(18\) |
risch | \(-\frac {3 \arctan \left (\frac {x}{2}\right )}{2}+\arctan \left (x \right )+\frac {\ln \left (x^{2}+4\right )}{2}\) | \(18\) |
parallelrisch | \(\frac {\ln \left (x -2 i\right )}{2}+\frac {3 i \ln \left (x -2 i\right )}{4}-\frac {i \ln \left (x -i\right )}{2}+\frac {i \ln \left (x +i\right )}{2}+\frac {\ln \left (x +2 i\right )}{2}-\frac {3 i \ln \left (x +2 i\right )}{4}\) | \(48\) |
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Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=\frac {\log {\left (x^{2} + 4 \right )}}{2} - \frac {3 \operatorname {atan}{\left (\frac {x}{2} \right )}}{2} + \operatorname {atan}{\left (x \right )} \]
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Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) \]
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Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\frac {3}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) \]
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Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {1+x-2 x^2+x^3}{4+5 x^2+x^4} \, dx=-\mathrm {atan}\left (\frac {1305}{4\,\left (144\,x-162\right )}+\frac {9}{8}\right )+\ln \left (x-2{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {3}{4}{}\mathrm {i}\right )+\ln \left (x+2{}\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {3}{4}{}\mathrm {i}\right ) \]
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