Integrand size = 13, antiderivative size = 23 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan (x)-\frac {\arctan (x)^2}{2}-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5036, 4930, 266, 5004} \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=-\frac {1}{2} \arctan (x)^2+x \arctan (x)-\frac {1}{2} \log \left (x^2+1\right ) \]
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Rule 266
Rule 4930
Rule 5004
Rule 5036
Rubi steps \begin{align*} \text {integral}& = \int \arctan (x) \, dx-\int \frac {\arctan (x)}{1+x^2} \, dx \\ & = x \arctan (x)-\frac {\arctan (x)^2}{2}-\int \frac {x}{1+x^2} \, dx \\ & = x \arctan (x)-\frac {\arctan (x)^2}{2}-\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan (x)-\frac {\arctan (x)^2}{2}-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
default | \(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(20\) |
parallelrisch | \(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(20\) |
parts | \(x \arctan \left (x \right )-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(20\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}+\frac {i \left (-x +\frac {i \ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{2}+\frac {\ln \left (-i x +1\right )^{2}}{8}+\frac {i x \ln \left (-i x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(67\) |
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none
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan \left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \operatorname {atan}{\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} \]
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none
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx={\left (x - \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=x \arctan \left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 \arctan (x)}{1+x^2} \, dx=-\frac {{\mathrm {atan}\left (x\right )}^2}{2}+x\,\mathrm {atan}\left (x\right )-\frac {\ln \left (x^2+1\right )}{2} \]
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