Integrand size = 11, antiderivative size = 32 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {x}{4 \left (1+x^2\right )}+\frac {\arctan (x)}{4}-\frac {\arctan (x)}{2 \left (1+x^2\right )} \]
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Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5050, 205, 209} \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=-\frac {\arctan (x)}{2 \left (x^2+1\right )}+\frac {\arctan (x)}{4}+\frac {x}{4 \left (x^2+1\right )} \]
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Rule 205
Rule 209
Rule 5050
Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (x)}{2 \left (1+x^2\right )}+\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx \\ & = \frac {x}{4 \left (1+x^2\right )}-\frac {\arctan (x)}{2 \left (1+x^2\right )}+\frac {1}{4} \int \frac {1}{1+x^2} \, dx \\ & = \frac {x}{4 \left (1+x^2\right )}+\frac {\arctan (x)}{4}-\frac {\arctan (x)}{2 \left (1+x^2\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {x+\left (-1+x^2\right ) \arctan (x)}{4 \left (1+x^2\right )} \]
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Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69
method | result | size |
parallelrisch | \(\frac {x^{2} \arctan \left (x \right )+x -\arctan \left (x \right )}{4 x^{2}+4}\) | \(22\) |
default | \(\frac {x}{4 x^{2}+4}+\frac {\arctan \left (x \right )}{4}-\frac {\arctan \left (x \right )}{2 \left (x^{2}+1\right )}\) | \(27\) |
parts | \(\frac {x}{4 x^{2}+4}+\frac {\arctan \left (x \right )}{4}-\frac {\arctan \left (x \right )}{2 \left (x^{2}+1\right )}\) | \(27\) |
risch | \(\frac {i \ln \left (i x +1\right )}{4 x^{2}+4}-\frac {i \left (2 \ln \left (-i x +1\right )-\ln \left (x +i\right ) x^{2}-\ln \left (x +i\right )+x^{2} \ln \left (x -i\right )+\ln \left (x -i\right )+2 i x \right )}{8 \left (x -i\right ) \left (x +i\right )}\) | \(79\) |
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none
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \arctan \left (x\right ) + x}{4 \, {\left (x^{2} + 1\right )}} \]
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Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {x^{2} \operatorname {atan}{\left (x \right )}}{4 x^{2} + 4} + \frac {x}{4 x^{2} + 4} - \frac {\operatorname {atan}{\left (x \right )}}{4 x^{2} + 4} \]
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none
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\arctan \left (x\right )}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{4} \, \arctan \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\arctan \left (x\right )}{2 \, {\left (x^{2} + 1\right )}} + \frac {1}{4} \, \arctan \left (x\right ) \]
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Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {x \arctan (x)}{\left (1+x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (x\right )}{4}+\frac {\frac {x}{4}-\frac {\mathrm {atan}\left (x\right )}{2}}{x^2+1} \]
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