Integrand size = 22, antiderivative size = 27 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}-\frac {7}{34} \log (1-4 x)+\frac {6}{17} \log \left (1+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1643, 649, 209, 266} \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}+\frac {6}{17} \log \left (x^2+1\right )-\frac {7}{34} \log (1-4 x) \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {14}{17 (-1+4 x)}+\frac {3 (1+4 x)}{17 \left (1+x^2\right )}\right ) \, dx \\ & = -\frac {7}{34} \log (1-4 x)+\frac {3}{17} \int \frac {1+4 x}{1+x^2} \, dx \\ & = -\frac {7}{34} \log (1-4 x)+\frac {3}{17} \int \frac {1}{1+x^2} \, dx+\frac {12}{17} \int \frac {x}{1+x^2} \, dx \\ & = \frac {3}{17} \tan ^{-1}(x)-\frac {7}{34} \log (1-4 x)+\frac {6}{17} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3 \arctan (x)}{17}-\frac {7}{34} \log (-1+4 x)+\frac {6}{17} \log \left (17+2 (-1+4 x)+(-1+4 x)^2\right ) \]
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Time = 0.81 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {6 \ln \left (x^{2}+1\right )}{17}+\frac {3 \arctan \left (x \right )}{17}-\frac {7 \ln \left (-1+4 x \right )}{34}\) | \(22\) |
risch | \(\frac {6 \ln \left (x^{2}+1\right )}{17}+\frac {3 \arctan \left (x \right )}{17}-\frac {7 \ln \left (-1+4 x \right )}{34}\) | \(22\) |
parallelrisch | \(-\frac {7 \ln \left (x -\frac {1}{4}\right )}{34}+\frac {6 \ln \left (x -i\right )}{17}-\frac {3 i \ln \left (x -i\right )}{34}+\frac {6 \ln \left (x +i\right )}{17}+\frac {3 i \ln \left (x +i\right )}{34}\) | \(38\) |
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Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left (4 \, x - 1\right ) \]
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Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=- \frac {7 \log {\left (x - \frac {1}{4} \right )}}{34} + \frac {6 \log {\left (x^{2} + 1 \right )}}{17} + \frac {3 \operatorname {atan}{\left (x \right )}}{17} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left (4 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=\frac {3}{17} \, \arctan \left (x\right ) + \frac {6}{17} \, \log \left (x^{2} + 1\right ) - \frac {7}{34} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-1+2 x^2}{(-1+4 x) \left (1+x^2\right )} \, dx=-\frac {7\,\ln \left (x-\frac {1}{4}\right )}{34}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {6}{17}-\frac {3}{34}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {6}{17}+\frac {3}{34}{}\mathrm {i}\right ) \]
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