Integrand size = 208, antiderivative size = 38 \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=\left (2-e^{-e^x+\frac {(2+2 x)^2}{e^{2+e^x x}-\log (5)}}\right ) x \]
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Timed out. \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
\[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=\int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx \]
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Time = 56.82 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32
method | result | size |
risch | \(-x \,{\mathrm e}^{-\frac {-{\mathrm e}^{x +{\mathrm e}^{x} x +2}+{\mathrm e}^{x} \ln \left (5\right )+4 x^{2}+8 x +4}{-{\mathrm e}^{{\mathrm e}^{x} x +2}+\ln \left (5\right )}}+2 x\) | \(50\) |
parallelrisch | \(-{\mathrm e}^{-\frac {-{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x} x +2}+{\mathrm e}^{x} \ln \left (5\right )+4 x^{2}+8 x +4}{-{\mathrm e}^{{\mathrm e}^{x} x +2}+\ln \left (5\right )}} x +2 x\) | \(51\) |
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Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.58 \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=-x e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 1\right )} e^{x} + e^{\left (2 \, x\right )} \log \left (5\right ) - e^{\left (x e^{x} + 2 \, x + 2\right )}}{e^{x} \log \left (5\right ) - e^{\left (x e^{x} + x + 2\right )}}\right )} + 2 \, x \]
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Time = 3.65 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=- x e^{\frac {4 x^{2} + 8 x - e^{x} e^{x e^{x} + 2} + e^{x} \log {\left (5 \right )} + 4}{e^{x e^{x} + 2} - \log {\left (5 \right )}}} + 2 x \]
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Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (32) = 64\).
Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.74 \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=-x e^{\left (\frac {4 \, x^{2}}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )} + \frac {e^{x} \log \left (5\right )}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )} + \frac {8 \, x}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )} - \frac {e^{\left (x e^{x} + x + 2\right )}}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )} + \frac {4}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )}\right )} + 2 \, x \]
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\[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=\int { -\frac {{\left (x e^{x} \log \left (5\right )^{2} + {\left (x e^{x} - 1\right )} e^{\left (2 \, x e^{x} + 4\right )} - 2 \, {\left (4 \, x^{2} - {\left (2 \, x^{4} + 6 \, x^{3} + 6 \, x^{2} - x \log \left (5\right ) + 2 \, x\right )} e^{x} + 4 \, x - \log \left (5\right )\right )} e^{\left (x e^{x} + 2\right )} + 8 \, {\left (x^{2} + x\right )} \log \left (5\right ) - \log \left (5\right )^{2}\right )} e^{\left (\frac {4 \, x^{2} + e^{x} \log \left (5\right ) + 8 \, x - e^{\left (x e^{x} + x + 2\right )} + 4}{e^{\left (x e^{x} + 2\right )} - \log \left (5\right )}\right )} - 4 \, e^{\left (x e^{x} + 2\right )} \log \left (5\right ) + 2 \, \log \left (5\right )^{2} + 2 \, e^{\left (2 \, x e^{x} + 4\right )}}{2 \, e^{\left (x e^{x} + 2\right )} \log \left (5\right ) - \log \left (5\right )^{2} - e^{\left (2 \, x e^{x} + 4\right )}} \,d x } \]
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Timed out. \[ \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx=\int \frac {2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}-4\,{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\ln \left (5\right )+{\mathrm {e}}^{-\frac {8\,x+{\mathrm {e}}^x\,\ln \left (5\right )-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,{\mathrm {e}}^x+4\,x^2+4}{\ln \left (5\right )-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}}}\,\left (\ln \left (5\right )\,\left (8\,x^2+8\,x\right )-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\left (8\,x-2\,\ln \left (5\right )-{\mathrm {e}}^x\,\left (4\,x-2\,x\,\ln \left (5\right )+12\,x^2+12\,x^3+4\,x^4\right )+8\,x^2\right )-{\ln \left (5\right )}^2+{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}\,\left (x\,{\mathrm {e}}^x-1\right )+x\,{\mathrm {e}}^x\,{\ln \left (5\right )}^2\right )+2\,{\ln \left (5\right )}^2}{{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}-2\,{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\ln \left (5\right )+{\ln \left (5\right )}^2} \,d x \]
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