Integrand size = 36, antiderivative size = 19 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=4+\log \left (e+4 x+x^2-\log \left (25 x^2\right )\right ) \]
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\[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{e+4 x+x^2-\log \left (25 x^2\right )}-\frac {2}{x \left (e+4 x+x^2-\log \left (25 x^2\right )\right )}+\frac {2 x}{e+4 x+x^2-\log \left (25 x^2\right )}\right ) \, dx \\ & = -\left (2 \int \frac {1}{x \left (e+4 x+x^2-\log \left (25 x^2\right )\right )} \, dx\right )+2 \int \frac {x}{e+4 x+x^2-\log \left (25 x^2\right )} \, dx+4 \int \frac {1}{e+4 x+x^2-\log \left (25 x^2\right )} \, dx \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\log \left (e+x (4+x)-\log \left (25 x^2\right )\right ) \]
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Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (4 x +{\mathrm e}+x^{2}-\ln \left (25 x^{2}\right )\right )\) | \(19\) |
parallelrisch | \(\ln \left (4 x +{\mathrm e}+x^{2}-\ln \left (25 x^{2}\right )\right )\) | \(19\) |
risch | \(\ln \left (-x^{2}+\ln \left (25 x^{2}\right )-{\mathrm e}-4 x \right )\) | \(21\) |
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\log \left (-x^{2} - 4 \, x - e + \log \left (25 \, x^{2}\right )\right ) \]
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Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\log {\left (- x^{2} - 4 x + \log {\left (25 x^{2} \right )} - e \right )} \]
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Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\log \left (-\frac {1}{2} \, x^{2} - 2 \, x - \frac {1}{2} \, e + \log \left (5\right ) + \log \left (x\right )\right ) \]
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\log \left (-x^{2} - 4 \, x - e + \log \left (25 \, x^{2}\right )\right ) \]
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Time = 11.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {2-4 x-2 x^2}{-e x-4 x^2-x^3+x \log \left (25 x^2\right )} \, dx=\ln \left (4\,x+\mathrm {e}-\ln \left (25\,x^2\right )+x^2\right ) \]
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