Integrand size = 22, antiderivative size = 39 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=-\frac {1}{6} \log (1-2 x)-\frac {1}{6} \log (1-x)+\frac {1}{6} \log (1+x)+\frac {1}{6} \log (1+2 x) \]
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Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1175, 630, 31} \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=-\frac {1}{6} \log (1-2 x)-\frac {1}{6} \log (1-x)+\frac {1}{6} \log (x+1)+\frac {1}{6} \log (2 x+1) \]
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Rule 31
Rule 630
Rule 1175
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{4} \int \frac {1}{-\frac {1}{2}-\frac {x}{2}+x^2} \, dx\right )-\frac {1}{4} \int \frac {1}{-\frac {1}{2}+\frac {x}{2}+x^2} \, dx \\ & = -\left (\frac {1}{6} \int \frac {1}{-1+x} \, dx\right )-\frac {1}{6} \int \frac {1}{-\frac {1}{2}+x} \, dx+\frac {1}{6} \int \frac {1}{\frac {1}{2}+x} \, dx+\frac {1}{6} \int \frac {1}{1+x} \, dx \\ & = -\frac {1}{6} \log (1-2 x)-\frac {1}{6} \log (1-x)+\frac {1}{6} \log (1+x)+\frac {1}{6} \log (1+2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=-\frac {1}{6} \log \left (1-3 x+2 x^2\right )+\frac {1}{6} \log \left (1+3 x+2 x^2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(-\frac {\ln \left (x -1\right )}{6}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -\frac {1}{2}\right )}{6}+\frac {\ln \left (x +\frac {1}{2}\right )}{6}\) | \(26\) |
risch | \(-\frac {\ln \left (2 x^{2}-3 x +1\right )}{6}+\frac {\ln \left (2 x^{2}+3 x +1\right )}{6}\) | \(28\) |
default | \(\frac {\ln \left (1+2 x \right )}{6}-\frac {\ln \left (2 x -1\right )}{6}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{6}\) | \(30\) |
norman | \(\frac {\ln \left (1+2 x \right )}{6}-\frac {\ln \left (2 x -1\right )}{6}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{6}\) | \(30\) |
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=\frac {1}{6} \, \log \left (2 \, x^{2} + 3 \, x + 1\right ) - \frac {1}{6} \, \log \left (2 \, x^{2} - 3 \, x + 1\right ) \]
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Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=- \frac {\log {\left (x^{2} - \frac {3 x}{2} + \frac {1}{2} \right )}}{6} + \frac {\log {\left (x^{2} + \frac {3 x}{2} + \frac {1}{2} \right )}}{6} \]
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Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=\frac {1}{6} \, \log \left (2 \, x + 1\right ) - \frac {1}{6} \, \log \left (2 \, x - 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{6} \, \log \left (x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=\frac {1}{6} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 13.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.38 \[ \int \frac {1-2 x^2}{1-5 x^2+4 x^4} \, dx=\frac {\mathrm {atanh}\left (\frac {3\,x}{2\,x^2+1}\right )}{3} \]
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