Integrand size = 6, antiderivative size = 35 \[ \int x \arctan (x)^2 \, dx=-x \arctan (x)+\frac {\arctan (x)^2}{2}+\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {4946, 5036, 4930, 266, 5004} \[ \int x \arctan (x)^2 \, dx=\frac {1}{2} x^2 \arctan (x)^2+\frac {\arctan (x)^2}{2}-x \arctan (x)+\frac {1}{2} \log \left (x^2+1\right ) \]
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Rule 266
Rule 4930
Rule 4946
Rule 5004
Rule 5036
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \arctan (x)^2-\int \frac {x^2 \arctan (x)}{1+x^2} \, dx \\ & = \frac {1}{2} x^2 \arctan (x)^2-\int \arctan (x) \, dx+\int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -x \arctan (x)+\frac {\arctan (x)^2}{2}+\frac {1}{2} x^2 \arctan (x)^2+\int \frac {x}{1+x^2} \, dx \\ & = -x \arctan (x)+\frac {\arctan (x)^2}{2}+\frac {1}{2} x^2 \arctan (x)^2+\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int x \arctan (x)^2 \, dx=\frac {1}{2} \left (-2 x \arctan (x)+\left (1+x^2\right ) \arctan (x)^2+\log \left (1+x^2\right )\right ) \]
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Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86
method | result | size |
default | \(-x \arctan \left (x \right )+\frac {\arctan \left (x \right )^{2}}{2}+\frac {x^{2} \arctan \left (x \right )^{2}}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(30\) |
parallelrisch | \(-x \arctan \left (x \right )+\frac {\arctan \left (x \right )^{2}}{2}+\frac {x^{2} \arctan \left (x \right )^{2}}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(30\) |
parts | \(-x \arctan \left (x \right )+\frac {\arctan \left (x \right )^{2}}{2}+\frac {x^{2} \arctan \left (x \right )^{2}}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(30\) |
risch | \(-\frac {\left (\frac {x^{2}}{2}+\frac {1}{2}\right ) \ln \left (i x +1\right )^{2}}{4}-\frac {\left (-x^{2} \ln \left (-i x +1\right )-2 i x -\ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{4}-\frac {x^{2} \ln \left (-i x +1\right )^{2}}{8}-\frac {\ln \left (-i x +1\right )^{2}}{8}-\frac {i x \ln \left (-i x +1\right )}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\) | \(99\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int x \arctan (x)^2 \, dx=\frac {1}{2} \, {\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - x \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \arctan (x)^2 \, dx=\frac {x^{2} \operatorname {atan}^{2}{\left (x \right )}}{2} - x \operatorname {atan}{\left (x \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} \]
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Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int x \arctan (x)^2 \, dx=\frac {1}{2} \, x^{2} \arctan \left (x\right )^{2} - {\left (x - \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \arctan (x)^2 \, dx=\frac {1}{2} \, x^{2} \arctan \left (x\right )^{2} - x \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \arctan (x)^2 \, dx=\frac {\ln \left (x^2+1\right )}{2}+\frac {{\mathrm {atan}\left (x\right )}^2}{2}+\frac {x^2\,{\mathrm {atan}\left (x\right )}^2}{2}-x\,\mathrm {atan}\left (x\right ) \]
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