Integrand size = 92, antiderivative size = 24 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x+\frac {e^{-2 e^x} \left (-5-x+\log ^2(x)\right )^2}{x} \]
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\[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 e^{-2 e^x+x} \left (5+x-\log ^2(x)\right )^2}{x}+\frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2-20 \log (x)-4 x \log (x)+10 \log ^2(x)+4 \log ^3(x)-\log ^4(x)\right )}{x^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{-2 e^x+x} \left (5+x-\log ^2(x)\right )^2}{x} \, dx\right )+\int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2-20 \log (x)-4 x \log (x)+10 \log ^2(x)+4 \log ^3(x)-\log ^4(x)\right )}{x^2} \, dx \\ & = -\left (2 \int \left (\frac {e^{-2 e^x+x} (5+x)^2}{x}-\frac {2 e^{-2 e^x+x} (5+x) \log ^2(x)}{x}+\frac {e^{-2 e^x+x} \log ^4(x)}{x}\right ) \, dx\right )+\int \left (1+\frac {e^{-2 e^x} \left (-25+x^2-20 \log (x)-4 x \log (x)+10 \log ^2(x)+4 \log ^3(x)-\log ^4(x)\right )}{x^2}\right ) \, dx \\ & = x-2 \int \frac {e^{-2 e^x+x} (5+x)^2}{x} \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int \frac {e^{-2 e^x+x} (5+x) \log ^2(x)}{x} \, dx+\int \frac {e^{-2 e^x} \left (-25+x^2-20 \log (x)-4 x \log (x)+10 \log ^2(x)+4 \log ^3(x)-\log ^4(x)\right )}{x^2} \, dx \\ & = x-2 \int \left (10 e^{-2 e^x+x}+\frac {25 e^{-2 e^x+x}}{x}+e^{-2 e^x+x} x\right ) \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int \left (e^{-2 e^x+x} \log ^2(x)+\frac {5 e^{-2 e^x+x} \log ^2(x)}{x}\right ) \, dx+\int \left (\frac {e^{-2 e^x} \left (-25+x^2\right )}{x^2}-\frac {4 e^{-2 e^x} (5+x) \log (x)}{x^2}+\frac {10 e^{-2 e^x} \log ^2(x)}{x^2}+\frac {4 e^{-2 e^x} \log ^3(x)}{x^2}-\frac {e^{-2 e^x} \log ^4(x)}{x^2}\right ) \, dx \\ & = x-2 \int e^{-2 e^x+x} x \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx-4 \int \frac {e^{-2 e^x} (5+x) \log (x)}{x^2} \, dx+4 \int e^{-2 e^x+x} \log ^2(x) \, dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2} \, dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2} \, dx-20 \int e^{-2 e^x+x} \, dx+20 \int \frac {e^{-2 e^x+x} \log ^2(x)}{x} \, dx-50 \int \frac {e^{-2 e^x+x}}{x} \, dx+\int \frac {e^{-2 e^x} \left (-25+x^2\right )}{x^2} \, dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2} \, dx \\ & = x-2 \int e^{-2 e^x+x} x \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int e^{-2 e^x+x} \log ^2(x) \, dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2} \, dx+4 \int \frac {5 \int \frac {e^{-2 e^x}}{x^2} \, dx+\int \frac {e^{-2 e^x}}{x} \, dx}{x} \, dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2} \, dx+20 \int \frac {e^{-2 e^x+x} \log ^2(x)}{x} \, dx-20 \text {Subst}\left (\int e^{-2 x} \, dx,x,e^x\right )-50 \int \frac {e^{-2 e^x+x}}{x} \, dx-(4 \log (x)) \int \frac {e^{-2 e^x}}{x} \, dx-(20 \log (x)) \int \frac {e^{-2 e^x}}{x^2} \, dx+\int \left (e^{-2 e^x}-\frac {25 e^{-2 e^x}}{x^2}\right ) \, dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2} \, dx \\ & = 10 e^{-2 e^x}+x-2 \int e^{-2 e^x+x} x \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int e^{-2 e^x+x} \log ^2(x) \, dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2} \, dx+4 \int \left (\frac {5 \int \frac {e^{-2 e^x}}{x^2} \, dx}{x}+\frac {\int \frac {e^{-2 e^x}}{x} \, dx}{x}\right ) \, dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2} \, dx+20 \int \frac {e^{-2 e^x+x} \log ^2(x)}{x} \, dx-25 \int \frac {e^{-2 e^x}}{x^2} \, dx-50 \int \frac {e^{-2 e^x+x}}{x} \, dx-(4 \log (x)) \int \frac {e^{-2 e^x}}{x} \, dx-(20 \log (x)) \int \frac {e^{-2 e^x}}{x^2} \, dx+\int e^{-2 e^x} \, dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2} \, dx \\ & = 10 e^{-2 e^x}+x-2 \int e^{-2 e^x+x} x \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int e^{-2 e^x+x} \log ^2(x) \, dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2} \, dx+4 \int \frac {\int \frac {e^{-2 e^x}}{x} \, dx}{x} \, dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2} \, dx+20 \int \frac {e^{-2 e^x+x} \log ^2(x)}{x} \, dx+20 \int \frac {\int \frac {e^{-2 e^x}}{x^2} \, dx}{x} \, dx-25 \int \frac {e^{-2 e^x}}{x^2} \, dx-50 \int \frac {e^{-2 e^x+x}}{x} \, dx-(4 \log (x)) \int \frac {e^{-2 e^x}}{x} \, dx-(20 \log (x)) \int \frac {e^{-2 e^x}}{x^2} \, dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2} \, dx+\text {Subst}\left (\int \frac {e^{-2 x}}{x} \, dx,x,e^x\right ) \\ & = 10 e^{-2 e^x}+x+\text {Ei}\left (-2 e^x\right )-2 \int e^{-2 e^x+x} x \, dx-2 \int \frac {e^{-2 e^x+x} \log ^4(x)}{x} \, dx+4 \int e^{-2 e^x+x} \log ^2(x) \, dx+4 \int \frac {e^{-2 e^x} \log ^3(x)}{x^2} \, dx+4 \int \frac {\int \frac {e^{-2 e^x}}{x} \, dx}{x} \, dx+10 \int \frac {e^{-2 e^x} \log ^2(x)}{x^2} \, dx+20 \int \frac {e^{-2 e^x+x} \log ^2(x)}{x} \, dx+20 \int \frac {\int \frac {e^{-2 e^x}}{x^2} \, dx}{x} \, dx-25 \int \frac {e^{-2 e^x}}{x^2} \, dx-50 \int \frac {e^{-2 e^x+x}}{x} \, dx-(4 \log (x)) \int \frac {e^{-2 e^x}}{x} \, dx-(20 \log (x)) \int \frac {e^{-2 e^x}}{x^2} \, dx-\int \frac {e^{-2 e^x} \log ^4(x)}{x^2} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x+\frac {e^{-2 e^x} \left (5+x-\log ^2(x)\right )^2}{x} \]
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Time = 0.99 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
risch | \(x +\frac {\left (\ln \left (x \right )^{4}-2 x \ln \left (x \right )^{2}+x^{2}-10 \ln \left (x \right )^{2}+10 x +25\right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}}}{x}\) | \(37\) |
parallelrisch | \(-\frac {\left (-25-x^{2} {\mathrm e}^{2 \,{\mathrm e}^{x}}-\ln \left (x \right )^{4}+2 x \ln \left (x \right )^{2}-x^{2}+10 \ln \left (x \right )^{2}-10 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}}}{x}\) | \(50\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\frac {{\left (\log \left (x\right )^{4} + x^{2} e^{\left (2 \, e^{x}\right )} - 2 \, {\left (x + 5\right )} \log \left (x\right )^{2} + x^{2} + 10 \, x + 25\right )} e^{\left (-2 \, e^{x}\right )}}{x} \]
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Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x + \frac {\left (x^{2} - 2 x \log {\left (x \right )}^{2} + 10 x + \log {\left (x \right )}^{4} - 10 \log {\left (x \right )}^{2} + 25\right ) e^{- 2 e^{x}}}{x} \]
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=x + \frac {{\left (\log \left (x\right )^{4} - 2 \, {\left (x + 5\right )} \log \left (x\right )^{2} + x^{2} + 25\right )} e^{\left (-2 \, e^{x}\right )}}{x} + 10 \, e^{\left (-2 \, e^{x}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (22) = 44\).
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.50 \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\frac {{\left (e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{4} - 2 \, x e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{2} + x^{2} e^{\left (x - 2 \, e^{x}\right )} + x^{2} e^{x} - 10 \, e^{\left (x - 2 \, e^{x}\right )} \log \left (x\right )^{2} + 10 \, x e^{\left (x - 2 \, e^{x}\right )} + 25 \, e^{\left (x - 2 \, e^{x}\right )}\right )} e^{\left (-x\right )}}{x} \]
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Timed out. \[ \int \frac {e^{-2 e^x} \left (-25+x^2+e^{2 e^x} x^2+e^x \left (-50 x-20 x^2-2 x^3\right )+(-20-4 x) \log (x)+\left (10+e^x \left (20 x+4 x^2\right )\right ) \log ^2(x)+4 \log ^3(x)+\left (-1-2 e^x x\right ) \log ^4(x)\right )}{x^2} \, dx=\int \frac {{\mathrm {e}}^{-2\,{\mathrm {e}}^x}\,\left (4\,{\ln \left (x\right )}^3-\ln \left (x\right )\,\left (4\,x+20\right )+{\ln \left (x\right )}^2\,\left ({\mathrm {e}}^x\,\left (4\,x^2+20\,x\right )+10\right )-{\ln \left (x\right )}^4\,\left (2\,x\,{\mathrm {e}}^x+1\right )+x^2+x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-{\mathrm {e}}^x\,\left (2\,x^3+20\,x^2+50\,x\right )-25\right )}{x^2} \,d x \]
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