Integrand size = 104, antiderivative size = 25 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x}{3-\frac {e^4}{x^2}+x-\log \left (\frac {64}{25 x}\right )} \]
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\[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{\left (e^4-x^2 (3+x)+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx \\ & = \int \left (\frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )}-\frac {x^2 \left (2 e^4+x^2+x^3\right )}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}\right ) \, dx \\ & = \int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \frac {x^2 \left (2 e^4+x^2+x^3\right )}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx \\ & = \int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \left (\frac {2 e^4 x^2}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}+\frac {x^4}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}+\frac {x^5}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2}\right ) \, dx \\ & = -\left (\left (2 e^4\right ) \int \frac {x^2}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx\right )+\int \frac {x^2}{-e^4+3 x^2+x^3-x^2 \log \left (\frac {64}{25 x}\right )} \, dx-\int \frac {x^4}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx-\int \frac {x^5}{\left (e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )\right )^2} \, dx \\ \end{align*}
Time = 0.91 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {x^3}{e^4-3 x^2-x^3+x^2 \log \left (\frac {64}{25 x}\right )} \]
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Time = 1.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24
method | result | size |
risch | \(-\frac {x^{3}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(31\) |
derivativedivides | \(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\) | \(32\) |
default | \(-\frac {262144}{\frac {262144 \,{\mathrm e}^{4}}{x^{3}}+\frac {262144 \ln \left (\frac {64}{25 x}\right )}{x}-\frac {786432}{x}-262144}\) | \(32\) |
parallelrisch | \(-\frac {x^{3}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(33\) |
norman | \(\frac {3 x^{2}-x^{2} \ln \left (\frac {64}{25 x}\right )-{\mathrm e}^{4}}{x^{2} \ln \left (\frac {64}{25 x}\right )-x^{3}+{\mathrm e}^{4}-3 x^{2}}\) | \(52\) |
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x^{3}}{x^{3} - x^{2} \log \left (\frac {64}{25 \, x}\right ) + 3 \, x^{2} - e^{4}} \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=- \frac {x^{3}}{- x^{3} + x^{2} \log {\left (\frac {64}{25 x} \right )} - 3 x^{2} + e^{4}} \]
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Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=\frac {x^{3}}{x^{3} + x^{2} {\left (2 \, \log \left (5\right ) - 6 \, \log \left (2\right ) + 3\right )} + x^{2} \log \left (x\right ) - e^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {1}{\frac {\log \left (\frac {64}{25 \, x}\right )}{x} - \frac {3}{x} + \frac {e^{4}}{x^{3}} - 1} \]
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Time = 10.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-3 e^4 x^2+2 x^4-x^4 \log \left (\frac {64}{25 x}\right )}{e^8+9 x^4+6 x^5+x^6+e^4 \left (-6 x^2-2 x^3\right )+\left (2 e^4 x^2-6 x^4-2 x^5\right ) \log \left (\frac {64}{25 x}\right )+x^4 \log ^2\left (\frac {64}{25 x}\right )} \, dx=-\frac {x^3}{{\mathrm {e}}^4-3\,x^2-x^3+x^2\,\ln \left (\frac {64}{25\,x}\right )} \]
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