Integrand size = 37, antiderivative size = 23 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \left (1+e^2\right )^2}{\log \left (\frac {4-\frac {x}{4}}{x}\right )} \]
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\[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) x \log ^2\left (\frac {16-x}{4 x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx\right ) \\ & = -\left (\left (400 \left (1+e^2\right )^2\right ) \int \left (\frac {1}{16 (-16+x) \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )}-\frac {1}{16 x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )}\right ) \, dx\right ) \\ & = -\left (\left (25 \left (1+e^2\right )^2\right ) \int \frac {1}{(-16+x) \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx\right )+\left (25 \left (1+e^2\right )^2\right ) \int \frac {1}{x \log ^2\left (-\frac {1}{4}+\frac {4}{x}\right )} \, dx \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \left (1+e^2\right )^2}{\log \left (-\frac {1}{4}+\frac {4}{x}\right )} \]
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Time = 0.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {400 \,{\mathrm e}^{4}+800 \,{\mathrm e}^{2}+400}{16 \ln \left (-\frac {1}{4}+\frac {4}{x}\right )}\) | \(25\) |
default | \(-\frac {-400 \,{\mathrm e}^{4}-800 \,{\mathrm e}^{2}-400}{16 \ln \left (-\frac {1}{4}+\frac {4}{x}\right )}\) | \(25\) |
parallelrisch | \(-\frac {-400 \,{\mathrm e}^{4}-800 \,{\mathrm e}^{2}-400}{16 \ln \left (-\frac {x -16}{4 x}\right )}\) | \(26\) |
norman | \(\frac {25 \,{\mathrm e}^{4}+50 \,{\mathrm e}^{2}+25}{\ln \left (\frac {16-x}{4 x}\right )}\) | \(27\) |
risch | \(\frac {25 \,{\mathrm e}^{4}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {50 \,{\mathrm e}^{2}}{\ln \left (\frac {16-x}{4 x}\right )}+\frac {25}{\ln \left (\frac {16-x}{4 x}\right )}\) | \(51\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 + 50 e^{2} + 25 e^{4}}{\log {\left (\frac {4 - \frac {x}{4}}{x} \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=-\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{2 \, \log \left (2\right ) + \log \left (x\right ) - \log \left (-x + 16\right )} \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {25 \, {\left (e^{4} + 2 \, e^{2} + 1\right )}}{\log \left (-\frac {x - 16}{4 \, x}\right )} \]
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Time = 13.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-400-800 e^2-400 e^4}{\left (-16 x+x^2\right ) \log ^2\left (\frac {16-x}{4 x}\right )} \, dx=\frac {50\,{\mathrm {e}}^2+25\,{\mathrm {e}}^4+25}{\ln \left (-\frac {x-16}{4\,x}\right )} \]
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