Integrand size = 148, antiderivative size = 30 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 \left (3+x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right )\right ) \]
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\[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{\left (e^{2 x}-x\right )^2} \, dx \\ & = \int \left (-\frac {20 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x}-\frac {10 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2}+5 \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right )\right ) \, dx \\ & = 5 \int \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right ) \, dx-10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx \\ & = 25 x+10 x^2+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+5 \int \log (x) \, dx-10 \int \left (-\frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2}+\frac {2 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2}\right ) \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx \\ & = 20 x+10 x^2+5 x \log (x)+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2} \, dx \\ \end{align*}
Time = 0.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right ) \]
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Time = 52.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
risch | \(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{-\frac {2}{x -{\mathrm e}^{2 x}}}}+20 x\) | \(32\) |
parallelrisch | \(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{\frac {2}{{\mathrm e}^{2 x}-x}}}+20 x\) | \(32\) |
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]
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Timed out. \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\text {Timed out} \]
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]
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\[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\int { \frac {5 \, {\left (4 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + 5\right )} e^{\left (4 \, x\right )} - 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x e^{\left (2 \, x\right )} - x\right )} e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )} + e^{\left (4 \, x\right )}\right )} e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}\right )} \log \left (x\right )\right )}}{x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}} \,d x } \]
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Time = 8.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=20\,x+5\,x\,\ln \left (x\right )+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2}{x-{\mathrm {e}}^{2\,x}}}}+10\,x^2 \]
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