Integrand size = 15, antiderivative size = 29 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=-\frac {x}{2 \left (1+x^2\right )}+\frac {\arctan (x)}{2}+\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1818, 649, 209, 266} \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=\frac {\arctan (x)}{2}-\frac {x}{2 \left (x^2+1\right )}+\frac {1}{2} \log \left (x^2+1\right ) \]
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Rule 209
Rule 266
Rule 649
Rule 1818
Rubi steps \begin{align*} \text {integral}& = -\frac {x}{2 \left (1+x^2\right )}-\frac {1}{2} \int \frac {-1-2 x}{1+x^2} \, dx \\ & = -\frac {x}{2 \left (1+x^2\right )}+\frac {1}{2} \int \frac {1}{1+x^2} \, dx+\int \frac {x}{1+x^2} \, dx \\ & = -\frac {x}{2 \left (1+x^2\right )}+\frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} \left (-\frac {x}{1+x^2}+\arctan (x)+\log \left (1+x^2\right )\right ) \]
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Time = 0.49 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {x}{2 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(24\) |
meijerg | \(-\frac {x}{2 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(24\) |
risch | \(-\frac {x}{2 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(24\) |
parallelrisch | \(-\frac {i \ln \left (x -i\right ) x^{2}-i \ln \left (x +i\right ) x^{2}-2 \ln \left (x -i\right ) x^{2}-2 \ln \left (x +i\right ) x^{2}+i \ln \left (x -i\right )-i \ln \left (x +i\right )-2 \ln \left (x -i\right )-2 \ln \left (x +i\right )+2 x}{4 \left (x^{2}+1\right )}\) | \(86\) |
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none
Time = 0.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 1\right )} \arctan \left (x\right ) + {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - x}{2 \, {\left (x^{2} + 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=- \frac {x}{2 x^{2} + 2} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]
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none
Time = 0.39 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=-\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=-\frac {x}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {x \left (1+x+x^2\right )}{\left (1+x^2\right )^2} \, dx=\frac {\ln \left (x^2+1\right )}{2}+\frac {\mathrm {atan}\left (x\right )}{2}-\frac {x}{2\,\left (x^2+1\right )} \]
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