Integrand size = 15, antiderivative size = 38 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (5+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {12, 815, 649, 209, 266} \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )-\frac {1}{6} \log \left (x^2+5\right )+\frac {1}{3} \log (1-x) \]
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Rule 12
Rule 209
Rule 266
Rule 649
Rule 815
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {x}{(-1+x) \left (5+x^2\right )} \, dx \\ & = 2 \int \left (\frac {1}{6 (-1+x)}+\frac {5-x}{6 \left (5+x^2\right )}\right ) \, dx \\ & = \frac {1}{3} \log (1-x)+\frac {1}{3} \int \frac {5-x}{5+x^2} \, dx \\ & = \frac {1}{3} \log (1-x)-\frac {1}{3} \int \frac {x}{5+x^2} \, dx+\frac {5}{3} \int \frac {1}{5+x^2} \, dx \\ & = \frac {1}{3} \sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (5+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=2 \left (\frac {1}{6} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )+\frac {1}{6} \log (1-x)-\frac {1}{12} \log \left (5+x^2\right )\right ) \]
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Time = 0.77 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {\ln \left (x^{2}+5\right )}{6}+\frac {\arctan \left (\frac {x \sqrt {5}}{5}\right ) \sqrt {5}}{3}+\frac {\ln \left (x -1\right )}{3}\) | \(28\) |
risch | \(-\frac {\ln \left (x^{2}+5\right )}{6}+\frac {\arctan \left (\frac {x \sqrt {5}}{5}\right ) \sqrt {5}}{3}+\frac {\ln \left (x -1\right )}{3}\) | \(28\) |
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + 5 \right )}}{6} + \frac {\sqrt {5} \operatorname {atan}{\left (\frac {\sqrt {5} x}{5} \right )}}{3} \]
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{3}-\ln \left (x-\sqrt {5}\,1{}\mathrm {i}\right )\,\left (\frac {1}{6}+\frac {\sqrt {5}\,1{}\mathrm {i}}{6}\right )+\ln \left (x+\sqrt {5}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {5}\,1{}\mathrm {i}}{6}\right ) \]
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