Integrand size = 8, antiderivative size = 39 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=-\frac {1}{4 (1+x)}-\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{4} \log (1+x)-\frac {1}{8} \log \left (1+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4972, 724, 815, 266} \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=-\frac {\arctan (x)}{2 (x+1)^2}-\frac {1}{8} \log \left (x^2+1\right )-\frac {1}{4 (x+1)}+\frac {1}{4} \log (x+1) \]
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Rule 266
Rule 724
Rule 815
Rule 4972
Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{2} \int \frac {1}{(1+x)^2 \left (1+x^2\right )} \, dx \\ & = -\frac {1}{4 (1+x)}-\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{4} \int \frac {1-x}{(1+x) \left (1+x^2\right )} \, dx \\ & = -\frac {1}{4 (1+x)}-\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{4} \int \left (\frac {1}{1+x}-\frac {x}{1+x^2}\right ) \, dx \\ & = -\frac {1}{4 (1+x)}-\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{4} \log (1+x)-\frac {1}{4} \int \frac {x}{1+x^2} \, dx \\ & = -\frac {1}{4 (1+x)}-\frac {\arctan (x)}{2 (1+x)^2}+\frac {1}{4} \log (1+x)-\frac {1}{8} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=\frac {1}{8} \left (-\frac {2}{1+x}-\frac {4 \arctan (x)}{(1+x)^2}+2 \log (1+x)-\log \left (1+x^2\right )\right ) \]
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Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {1}{4 \left (1+x \right )}-\frac {\arctan \left (x \right )}{2 \left (1+x \right )^{2}}+\frac {\ln \left (1+x \right )}{4}-\frac {\ln \left (x^{2}+1\right )}{8}\) | \(32\) |
parts | \(-\frac {1}{4 \left (1+x \right )}-\frac {\arctan \left (x \right )}{2 \left (1+x \right )^{2}}+\frac {\ln \left (1+x \right )}{4}-\frac {\ln \left (x^{2}+1\right )}{8}\) | \(32\) |
parallelrisch | \(\frac {2 \ln \left (1+x \right ) x^{2}-\ln \left (x^{2}+1\right ) x^{2}-2+4 \ln \left (1+x \right ) x -2 x \ln \left (x^{2}+1\right )+2 \ln \left (1+x \right )-\ln \left (x^{2}+1\right )-2 x -4 \arctan \left (x \right )}{8 \left (1+x \right )^{2}}\) | \(67\) |
risch | \(\frac {i \ln \left (i x +1\right )}{4 \left (1+x \right )^{2}}-\frac {i \left (2 i \ln \left (1+x \right ) x^{2}-i \ln \left (x^{2}+1\right ) x^{2}+4 i \ln \left (1+x \right ) x -2 i \ln \left (x^{2}+1\right ) x +2 i \ln \left (1+x \right )-i \ln \left (x^{2}+1\right )-2 i x -2 i+2 \ln \left (-i x +1\right )\right )}{8 \left (1+x \right )^{2}}\) | \(97\) |
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Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=-\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (x^{2} + 1\right ) - 2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right ) + 2 \, x + 4 \, \arctan \left (x\right ) + 2}{8 \, {\left (x^{2} + 2 \, x + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (31) = 62\).
Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 3.92 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=\frac {2 x^{2} \log {\left (x + 1 \right )}}{8 x^{2} + 16 x + 8} - \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{8 x^{2} + 16 x + 8} + \frac {4 x \log {\left (x + 1 \right )}}{8 x^{2} + 16 x + 8} - \frac {2 x \log {\left (x^{2} + 1 \right )}}{8 x^{2} + 16 x + 8} - \frac {2 x}{8 x^{2} + 16 x + 8} + \frac {2 \log {\left (x + 1 \right )}}{8 x^{2} + 16 x + 8} - \frac {\log {\left (x^{2} + 1 \right )}}{8 x^{2} + 16 x + 8} - \frac {4 \operatorname {atan}{\left (x \right )}}{8 x^{2} + 16 x + 8} - \frac {2}{8 x^{2} + 16 x + 8} \]
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Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=-\frac {1}{4 \, {\left (x + 1\right )}} - \frac {\arctan \left (x\right )}{2 \, {\left (x + 1\right )}^{2}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \log \left (x + 1\right ) \]
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Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=-\frac {1}{4 \, {\left (x + 1\right )}} - \frac {\arctan \left (x\right )}{2 \, {\left (x + 1\right )}^{2}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.41 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan (x)}{(1+x)^3} \, dx=\frac {\ln \left (x+1\right )}{4}-\frac {\ln \left (x^2+1\right )}{8}-\frac {\frac {x}{4}+\frac {\mathrm {atan}\left (x\right )}{2}+\frac {1}{4}}{{\left (x+1\right )}^2} \]
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