Integrand size = 11, antiderivative size = 32 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {x}{4 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {\arctan (x)}{4} \]
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Time = 0.02 (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 = {5051, 205, 209} \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {\arctan (x)}{4}-\frac {x}{4 \left (x^2+1\right )}-\frac {\cot ^{-1}(x)}{2 \left (x^2+1\right )} \]
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Rule 205
Rule 209
Rule 5051
Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^{-1}(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 {\cot ^{-1}(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 {\cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {\arctan (x)}{4} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {x+2 \cot ^{-1}(x)+\arctan (x)+x^2 \arctan (x)}{4+4 x^2} \]
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Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75
method | result | size |
parallelrisch | \(\frac {x^{2} \operatorname {arccot}\left (x \right )-x -\operatorname {arccot}\left (x \right )}{4 x^{2}+4}\) | \(24\) |
default | \(-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\operatorname {arccot}\left (x \right )}{2 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{4}\) | \(27\) |
parts | \(-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\operatorname {arccot}\left (x \right )}{2 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{4}\) | \(27\) |
risch | \(-\frac {i \ln \left (i x +1\right )}{4 \left (x^{2}+1\right )}-\frac {-2 i \ln \left (-i x +1\right )+i \ln \left (i+x \right )+i \ln \left (i+x \right ) x^{2}-i \ln \left (x -i\right )-i \ln \left (x -i\right ) x^{2}+2 \pi +2 x}{8 \left (i+x \right ) \left (x -i\right )}\) | \(88\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \operatorname {arccot}\left (x\right ) - x}{4 \, {\left (x^{2} + 1\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\frac {x^{2} \operatorname {acot}{\left (x \right )}}{4 x^{2} + 4} - \frac {x}{4 x^{2} + 4} - \frac {\operatorname {acot}{\left (x \right )}}{4 x^{2} + 4} \]
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Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\operatorname {arccot}\left (x\right )}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{4} \, \arctan \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {\arctan \left (\frac {1}{x}\right )}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{4 \, x {\left (\frac {1}{x^{2}} + 1\right )}} + \frac {1}{4} \, \arctan \left (\frac {1}{x}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\frac {\mathrm {acot}\left (x\right )}{4}-\frac {\frac {x}{4}+\frac {\mathrm {acot}\left (x\right )}{2}}{x^2+1} \]
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