Integrand size = 11, antiderivative size = 44 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {x}{16 \left (1+x^2\right )^2}-\frac {3 x}{32 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3 \arctan (x)}{32} \]
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Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5051, 205, 209} \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {3 \arctan (x)}{32}-\frac {3 x}{32 \left (x^2+1\right )}-\frac {x}{16 \left (x^2+1\right )^2}-\frac {\cot ^{-1}(x)}{4 \left (x^2+1\right )^2} \]
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Rule 205
Rule 209
Rule 5051
Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {1}{4} \int \frac {1}{\left (1+x^2\right )^3} \, dx \\ & = -\frac {x}{16 \left (1+x^2\right )^2}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3}{16} \int \frac {1}{\left (1+x^2\right )^2} \, dx \\ & = -\frac {x}{16 \left (1+x^2\right )^2}-\frac {3 x}{32 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3}{32} \int \frac {1}{1+x^2} \, dx \\ & = -\frac {x}{16 \left (1+x^2\right )^2}-\frac {3 x}{32 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3 \arctan (x)}{32} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {x \left (5+3 x^2\right )+8 \cot ^{-1}(x)+3 \left (1+x^2\right )^2 \arctan (x)}{32 \left (1+x^2\right )^2} \]
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Time = 0.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {x}{16 \left (x^{2}+1\right )^{2}}-\frac {3 x}{32 \left (x^{2}+1\right )}-\frac {\operatorname {arccot}\left (x \right )}{4 \left (x^{2}+1\right )^{2}}-\frac {3 \arctan \left (x \right )}{32}\) | \(37\) |
parallelrisch | \(\frac {3 \,\operatorname {arccot}\left (x \right ) x^{4}-3 x^{3}+6 x^{2} \operatorname {arccot}\left (x \right )-5 x -5 \,\operatorname {arccot}\left (x \right )}{32 \left (x^{2}+1\right )^{2}}\) | \(37\) |
parts | \(-\frac {x}{16 \left (x^{2}+1\right )^{2}}-\frac {3 x}{32 \left (x^{2}+1\right )}-\frac {\operatorname {arccot}\left (x \right )}{4 \left (x^{2}+1\right )^{2}}-\frac {3 \arctan \left (x \right )}{32}\) | \(37\) |
risch | \(-\frac {i \ln \left (i x +1\right )}{8 \left (x^{2}+1\right )^{2}}-\frac {-8 i \ln \left (-i x +1\right )-6 i \ln \left (x -i\right ) x^{2}-3 i \ln \left (x -i\right )-3 i \ln \left (x -i\right ) x^{4}+6 i \ln \left (i+x \right ) x^{2}+3 i \ln \left (i+x \right )+3 i \ln \left (i+x \right ) x^{4}+6 x^{3}+8 \pi +10 x}{64 \left (i+x \right ) \left (x^{2}+1\right ) \left (x -i\right )}\) | \(122\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {3 \, x^{3} - {\left (3 \, x^{4} + 6 \, x^{2} - 5\right )} \operatorname {arccot}\left (x\right ) + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (39) = 78\).
Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.00 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=\frac {3 x^{4} \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} - \frac {3 x^{3}}{32 x^{4} + 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} - \frac {5 x}{32 x^{4} + 64 x^{2} + 32} - \frac {5 \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} \]
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {3 \, x^{3} + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {\operatorname {arccot}\left (x\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} - \frac {3}{32} \, \arctan \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {\frac {3}{x} + \frac {5}{x^{3}}}{32 \, {\left (\frac {1}{x^{2}} + 1\right )}^{2}} - \frac {\arctan \left (\frac {1}{x}\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {3}{32} \, \arctan \left (\frac {1}{x}\right ) \]
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Time = 0.73 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.61 \[ \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^3} \, dx=-\frac {3\,\mathrm {atan}\left (x\right )}{32}-\frac {\frac {5\,x}{32}+\frac {\mathrm {acot}\left (x\right )}{4}+\frac {3\,x^3}{32}}{{\left (x^2+1\right )}^2} \]
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