Integrand size = 17, antiderivative size = 31 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2083, 649, 209, 266} \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)-\frac {1}{10} \log \left (4 x^2+1\right )+\frac {1}{5} \log (1-4 x) \]
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Rule 209
Rule 266
Rule 649
Rule 2083
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{5 (-1+4 x)}+\frac {-1-4 x}{5 \left (1+4 x^2\right )}\right ) \, dx \\ & = \frac {1}{5} \log (1-4 x)+\frac {1}{5} \int \frac {-1-4 x}{1+4 x^2} \, dx \\ & = \frac {1}{5} \log (1-4 x)-\frac {1}{5} \int \frac {1}{1+4 x^2} \, dx-\frac {4}{5} \int \frac {x}{1+4 x^2} \, dx \\ & = -\frac {1}{10} \tan ^{-1}(2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \arctan (2 x)+\frac {1}{5} \log (1-4 x)-\frac {1}{10} \log \left (1+4 x^2\right ) \]
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Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {\ln \left (4 x^{2}+1\right )}{10}-\frac {\arctan \left (2 x \right )}{10}+\frac {\ln \left (-1+4 x \right )}{5}\) | \(26\) |
risch | \(-\frac {\ln \left (4 x^{2}+1\right )}{10}-\frac {\arctan \left (2 x \right )}{10}+\frac {\ln \left (-1+4 x \right )}{5}\) | \(26\) |
parallelrisch | \(-\frac {\ln \left (x -\frac {i}{2}\right )}{10}+\frac {i \ln \left (x -\frac {i}{2}\right )}{20}-\frac {\ln \left (x +\frac {i}{2}\right )}{10}-\frac {i \ln \left (x +\frac {i}{2}\right )}{20}+\frac {\ln \left (x -\frac {1}{4}\right )}{5}\) | \(38\) |
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Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left (4 \, x - 1\right ) \]
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Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=\frac {\log {\left (x - \frac {1}{4} \right )}}{5} - \frac {\log {\left (x^{2} + \frac {1}{4} \right )}}{10} - \frac {\operatorname {atan}{\left (2 x \right )}}{10} \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left (4 \, x - 1\right ) \]
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Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=-\frac {1}{10} \, \arctan \left (2 \, x\right ) - \frac {1}{10} \, \log \left (4 \, x^{2} + 1\right ) + \frac {1}{5} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-1+4 x-4 x^2+16 x^3} \, dx=\frac {\ln \left (x-\frac {1}{4}\right )}{5}+\ln \left (x-\frac {1}{2}{}\mathrm {i}\right )\,\left (-\frac {1}{10}+\frac {1}{20}{}\mathrm {i}\right )+\ln \left (x+\frac {1}{2}{}\mathrm {i}\right )\,\left (-\frac {1}{10}-\frac {1}{20}{}\mathrm {i}\right ) \]
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