Integrand size = 13, antiderivative size = 28 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5038, 4946, 272, 36, 29, 31, 5004} \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=-\frac {1}{2} \arctan (x)^2-\frac {\arctan (x)}{x}-\frac {1}{2} \log \left (x^2+1\right )+\log (x) \]
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Rule 29
Rule 31
Rule 36
Rule 272
Rule 4946
Rule 5004
Rule 5038
Rubi steps \begin{align*} \text {integral}& = \int \frac {\arctan (x)}{x^2} \, dx-\int \frac {\arctan (x)}{1+x^2} \, dx \\ & = -\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\int \frac {1}{x \left (1+x^2\right )} \, dx \\ & = -\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = -\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{x}-\frac {\arctan (x)^2}{2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(25\) |
parts | \(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\) | \(25\) |
parallelrisch | \(\frac {-x \arctan \left (x \right )^{2}+2 x \ln \left (x \right )-x \ln \left (x^{2}+1\right )-2 \arctan \left (x \right )}{2 x}\) | \(32\) |
risch | \(\frac {\ln \left (i x +1\right )^{2}}{8}-\frac {\left (\ln \left (-i x +1\right ) x -2 i\right ) \ln \left (i x +1\right )}{4 x}-\frac {-x \ln \left (-i x +1\right )^{2}-8 x \ln \left (x \right )+4 x \ln \left (x^{2}+1\right )+4 i \ln \left (-i x +1\right )}{8 x}\) | \(79\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=-\frac {x \arctan \left (x\right )^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \left (x\right ) + 2 \, \arctan \left (x\right )}{2 \, x} \]
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Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{2} - \frac {\operatorname {atan}{\left (x \right )}}{x} \]
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=-{\left (\frac {1}{x} + \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \]
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\[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=\int { \frac {\arctan \left (x\right )}{{\left (x^{2} + 1\right )} x^{2}} \,d x } \]
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Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\arctan (x)}{x^2 \left (1+x^2\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2+1\right )}{2}-\frac {\mathrm {atan}\left (x\right )}{x}-\frac {{\mathrm {atan}\left (x\right )}^2}{2} \]
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