Integrand size = 13, antiderivative size = 34 \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {1}{4 \left (1+x^2\right )}-\frac {x \arctan (x)}{2 \left (1+x^2\right )}+\frac {\arctan (x)^2}{4} \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5054, 5004} \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {x \arctan (x)}{2 \left (x^2+1\right )}+\frac {\arctan (x)^2}{4}-\frac {1}{4 \left (x^2+1\right )} \]
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Rule 5004
Rule 5054
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{4 \left (1+x^2\right )}-\frac {x \arctan (x)}{2 \left (1+x^2\right )}+\frac {1}{2} \int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {1}{4 \left (1+x^2\right )}-\frac {x \arctan (x)}{2 \left (1+x^2\right )}+\frac {\arctan (x)^2}{4} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {-1-2 x \arctan (x)+\left (1+x^2\right ) \arctan (x)^2}{4 \left (1+x^2\right )} \]
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Time = 0.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {1}{4 \left (x^{2}+1\right )}-\frac {x \arctan \left (x \right )}{2 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )^{2}}{4}\) | \(29\) |
parts | \(-\frac {1}{4 \left (x^{2}+1\right )}-\frac {x \arctan \left (x \right )}{2 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )^{2}}{4}\) | \(29\) |
risch | \(-\frac {\ln \left (i x +1\right )^{2}}{16}+\frac {\left (x^{2} \ln \left (-i x +1\right )+\ln \left (-i x +1\right )+2 i x \right ) \ln \left (i x +1\right )}{8 x^{2}+8}-\frac {x^{2} \ln \left (-i x +1\right )^{2}+\ln \left (-i x +1\right )^{2}+4 i x \ln \left (-i x +1\right )+4}{16 \left (x +i\right ) \left (x -i\right )}\) | \(101\) |
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none
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - 2 \, x \arctan \left (x\right ) - 1}{4 \, {\left (x^{2} + 1\right )}} \]
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Exception generated. \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\text {Exception raised: RecursionError} \]
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none
Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {1}{2} \, {\left (\frac {x}{x^{2} + 1} - \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} + 1}{4 \, {\left (x^{2} + 1\right )}} \]
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\[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {x^{2} \arctan \left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]
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Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {x^2 \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {{\mathrm {atan}\left (x\right )}^2}{4}-\frac {\frac {x\,\mathrm {atan}\left (x\right )}{2}+\frac {1}{4}}{x^2+1} \]
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