Integrand size = 11, antiderivative size = 22 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=-\frac {\arctan (x)}{x}+x \arctan (x)+\log (x)-\log \left (1+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {5070, 4946, 272, 36, 29, 31, 4930, 266} \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=x \arctan (x)-\frac {\arctan (x)}{x}-\log \left (x^2+1\right )+\log (x) \]
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 272
Rule 4930
Rule 4946
Rule 5070
Rubi steps \begin{align*} \text {integral}& = \int \arctan (x) \, dx+\int \frac {\arctan (x)}{x^2} \, dx \\ & = -\frac {\arctan (x)}{x}+x \arctan (x)+\int \frac {1}{x \left (1+x^2\right )} \, dx-\int \frac {x}{1+x^2} \, dx \\ & = -\frac {\arctan (x)}{x}+x \arctan (x)-\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right ) \\ & = -\frac {\arctan (x)}{x}+x \arctan (x)-\frac {1}{2} \log \left (1+x^2\right )+\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}+x \arctan (x)+\log (x)-\log \left (1+x^2\right ) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=-\frac {\arctan (x)}{x}+x \arctan (x)+\log (x)-\log \left (1+x^2\right ) \]
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Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{x}+x \arctan \left (x \right )+\ln \left (x \right )-\ln \left (x^{2}+1\right )\) | \(23\) |
parts | \(-\frac {\arctan \left (x \right )}{x}+x \arctan \left (x \right )+\ln \left (x \right )-\ln \left (x^{2}+1\right )\) | \(23\) |
parallelrisch | \(\frac {x^{2} \arctan \left (x \right )+x \ln \left (x \right )-x \ln \left (x^{2}+1\right )-\arctan \left (x \right )}{x}\) | \(29\) |
meijerg | \(\ln \left (x \right )-\frac {\arctan \left (\sqrt {x^{2}}\right )}{\sqrt {x^{2}}}-\ln \left (x^{2}+1\right )+\frac {x^{2} \arctan \left (\sqrt {x^{2}}\right )}{\sqrt {x^{2}}}\) | \(40\) |
risch | \(-\frac {i \left (x^{2}-1\right ) \ln \left (i x +1\right )}{2 x}+\frac {i \left (-2 i \ln \left (x \right ) x +2 i \ln \left (x^{2}+1\right ) x +x^{2} \ln \left (-i x +1\right )-\ln \left (-i x +1\right )\right )}{2 x}\) | \(63\) |
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=\frac {{\left (x^{2} - 1\right )} \arctan \left (x\right ) - x \log \left (x^{2} + 1\right ) + x \log \left (x\right )}{x} \]
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Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=x \operatorname {atan}{\left (x \right )} + \log {\left (x \right )} - \log {\left (x^{2} + 1 \right )} - \frac {\operatorname {atan}{\left (x \right )}}{x} \]
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Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx={\left (x - \frac {1}{x}\right )} \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx={\left (x - \frac {1}{x}\right )} \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^2\right ) \arctan (x)}{x^2} \, dx=\ln \left (x\right )-\ln \left (x^2+1\right )-\frac {\mathrm {atan}\left (x\right )}{x}+x\,\mathrm {atan}\left (x\right ) \]
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