Integrand size = 13, antiderivative size = 31 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \arctan \left (\frac {x}{2}\right )+\frac {1}{13} \log (3-x)-\frac {1}{26} \log \left (4+x^2\right ) \]
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Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {720, 31, 649, 209, 266} \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \arctan \left (\frac {x}{2}\right )-\frac {1}{26} \log \left (x^2+4\right )+\frac {1}{13} \log (3-x) \]
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Rule 31
Rule 209
Rule 266
Rule 649
Rule 720
Rubi steps \begin{align*} \text {integral}& = \frac {1}{13} \int \frac {1}{-3+x} \, dx+\frac {1}{13} \int \frac {-3-x}{4+x^2} \, dx \\ & = \frac {1}{13} \log (3-x)-\frac {1}{13} \int \frac {x}{4+x^2} \, dx-\frac {3}{13} \int \frac {1}{4+x^2} \, dx \\ & = -\frac {3}{26} \tan ^{-1}\left (\frac {x}{2}\right )+\frac {1}{13} \log (3-x)-\frac {1}{26} \log \left (4+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \arctan \left (\frac {x}{2}\right )-\frac {1}{26} \log \left (13+6 (-3+x)+(-3+x)^2\right )+\frac {1}{13} \log (-3+x) \]
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Time = 0.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {\ln \left (x^{2}+4\right )}{26}-\frac {3 \arctan \left (\frac {x}{2}\right )}{26}+\frac {\ln \left (-3+x \right )}{13}\) | \(22\) |
risch | \(\frac {\ln \left (-3+x \right )}{13}-\frac {\ln \left (9 x^{2}+36\right )}{26}-\frac {3 \arctan \left (\frac {x}{2}\right )}{26}\) | \(24\) |
parallelrisch | \(\frac {\ln \left (-3+x \right )}{13}-\frac {\ln \left (x -2 i\right )}{26}+\frac {3 i \ln \left (x -2 i\right )}{52}-\frac {\ln \left (x +2 i\right )}{26}-\frac {3 i \ln \left (x +2 i\right )}{52}\) | \(38\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \, \arctan \left (\frac {1}{2} \, x\right ) - \frac {1}{26} \, \log \left (x^{2} + 4\right ) + \frac {1}{13} \, \log \left (x - 3\right ) \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=\frac {\log {\left (x - 3 \right )}}{13} - \frac {\log {\left (x^{2} + 4 \right )}}{26} - \frac {3 \operatorname {atan}{\left (\frac {x}{2} \right )}}{26} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \, \arctan \left (\frac {1}{2} \, x\right ) - \frac {1}{26} \, \log \left (x^{2} + 4\right ) + \frac {1}{13} \, \log \left (x - 3\right ) \]
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Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=-\frac {3}{26} \, \arctan \left (\frac {1}{2} \, x\right ) - \frac {1}{26} \, \log \left (x^{2} + 4\right ) + \frac {1}{13} \, \log \left ({\left | x - 3 \right |}\right ) \]
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Time = 9.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(-3+x) \left (4+x^2\right )} \, dx=\frac {\ln \left (x-3\right )}{13}+\ln \left (x-2{}\mathrm {i}\right )\,\left (-\frac {1}{26}+\frac {3}{52}{}\mathrm {i}\right )+\ln \left (x+2{}\mathrm {i}\right )\,\left (-\frac {1}{26}-\frac {3}{52}{}\mathrm {i}\right ) \]
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