Integrand size = 70, antiderivative size = 27 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=e^{\frac {4 \left (2 x+\frac {2 (5-x)}{\frac {1}{2}+x}+\log (5)\right )}{x}} \]
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\[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=\int \frac {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2 \left (1+4 x+4 x^2\right )} \, dx \\ & = \int \frac {\exp \left (\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}\right ) \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2 (1+2 x)^2} \, dx \\ & = \int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right ) \left (16 x^2 (2-\log (5))-4 (20+\log (5))-16 x (20+\log (5))\right )}{x^2 (1+2 x)^2} \, dx \\ & = \int \left (\frac {352 \exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{(1+2 x)^2}-\frac {4 \exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right ) (20+\log (5))}{x^2}\right ) \, dx \\ & = 352 \int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{(1+2 x)^2} \, dx-(4 (20+\log (5))) \int \frac {\exp \left (\frac {80+16 x^2-8 x (1-\log (5))+\log (625)}{x (1+2 x)}\right )}{x^2} \, dx \\ \end{align*}
Time = 2.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=\frac {4\ 625^{\frac {1}{x}} e^{\frac {80-8 x+16 x^2}{x+2 x^2}} \log (5)}{\log (625)} \]
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Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
gosper | \({\mathrm e}^{\frac {8 x \ln \left (5\right )+16 x^{2}+4 \ln \left (5\right )-8 x +80}{\left (1+2 x \right ) x}}\) | \(31\) |
risch | \({\mathrm e}^{\frac {8 x \ln \left (5\right )+16 x^{2}+4 \ln \left (5\right )-8 x +80}{\left (1+2 x \right ) x}}\) | \(31\) |
parallelrisch | \({\mathrm e}^{\frac {\left (8 x +4\right ) \ln \left (5\right )+16 x^{2}-8 x +80}{\left (1+2 x \right ) x}}\) | \(31\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {\left (8 x +4\right ) \ln \left (5\right )+16 x^{2}-8 x +80}{2 x^{2}+x}}+2 x^{2} {\mathrm e}^{\frac {\left (8 x +4\right ) \ln \left (5\right )+16 x^{2}-8 x +80}{2 x^{2}+x}}}{\left (1+2 x \right ) x}\) | \(78\) |
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=e^{\left (\frac {4 \, {\left (4 \, x^{2} + {\left (2 \, x + 1\right )} \log \left (5\right ) - 2 \, x + 20\right )}}{2 \, x^{2} + x}\right )} \]
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Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=e^{\frac {16 x^{2} - 8 x + \left (8 x + 4\right ) \log {\left (5 \right )} + 80}{2 x^{2} + x}} \]
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Time = 0.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=e^{\left (\frac {4 \, \log \left (5\right )}{x} - \frac {176}{2 \, x + 1} + \frac {80}{x} + 8\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (24) = 48\).
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx=e^{\left (\frac {16 \, x^{2}}{2 \, x^{2} + x} + \frac {8 \, x \log \left (5\right )}{2 \, x^{2} + x} - \frac {8 \, x}{2 \, x^{2} + x} + \frac {4 \, \log \left (5\right )}{2 \, x^{2} + x} + \frac {80}{2 \, x^{2} + x}\right )} \]
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Time = 17.96 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {80-8 x+16 x^2+(4+8 x) \log (5)}{x+2 x^2}} \left (-80-320 x+32 x^2+\left (-4-16 x-16 x^2\right ) \log (5)\right )}{x^2+4 x^3+4 x^4} \, dx={625}^{1/x}\,{\mathrm {e}}^{\frac {80}{2\,x^2+x}}\,{\mathrm {e}}^{-\frac {8}{2\,x+1}}\,{\mathrm {e}}^{\frac {16\,x}{2\,x+1}} \]
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