Integrand size = 194, antiderivative size = 29 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=e^{-2 e^{e^x}} x \left (e^{\frac {4}{-1+x}}-\log (-1+x)\right )^2 \]
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Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(29)=58\).
Time = 0.70 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {27, 2326} \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\frac {e^{-x-2 e^{e^x}} \left (e^{x-\frac {8}{1-x}} \left (x^3-2 x^2+x\right )+e^x \left (x^3-2 x^2+x\right ) \log ^2(x-1)-2 e^{x-\frac {4}{1-x}} \left (x^3-2 x^2+x\right ) \log (x-1)\right )}{(1-x)^2} \]
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Rule 27
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{(-1+x)^2} \, dx \\ & = \frac {e^{-2 e^{e^x}-x} \left (e^{-\frac {8}{1-x}+x} \left (x-2 x^2+x^3\right )-2 e^{-\frac {4}{1-x}+x} \left (x-2 x^2+x^3\right ) \log (-1+x)+e^x \left (x-2 x^2+x^3\right ) \log ^2(-1+x)\right )}{(1-x)^2} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=e^{-2 e^{e^x}} x \left (e^{\frac {4}{-1+x}}-\log (-1+x)\right )^2 \]
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Time = 212.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\left ({\mathrm e}^{\frac {8}{-1+x}}-2 \,{\mathrm e}^{\frac {4}{-1+x}} \ln \left (-1+x \right )+\ln \left (-1+x \right )^{2}\right ) x \,{\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{x}}}\) | \(38\) |
parallelrisch | \(\frac {\left (4 \ln \left (-1+x \right )^{2} x -8 x \,{\mathrm e}^{\frac {4}{-1+x}} \ln \left (-1+x \right )+4 \,{\mathrm e}^{\frac {8}{-1+x}} x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{x}}}}{4}\) | \(47\) |
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Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.59 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx={\left (x e^{\left (\frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1}\right )} \log \left (x - 1\right )^{2} - 2 \, x e^{\left (\frac {x^{2} - x + 8}{x - 1} + \frac {x^{2} - x + 4}{x - 1}\right )} \log \left (x - 1\right ) + x e^{\left (\frac {2 \, {\left (x^{2} - x + 8\right )}}{x - 1}\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1} - 2 \, e^{\left (e^{\left (-\frac {x^{2} - x + 8}{x - 1} + \frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1}\right )}\right )}\right )} \]
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Timed out. \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\text {Timed out} \]
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none
Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=-{\left (2 \, x e^{\left (\frac {4}{x - 1}\right )} \log \left (x - 1\right ) - x \log \left (x - 1\right )^{2} - x e^{\left (\frac {8}{x - 1}\right )}\right )} e^{\left (-2 \, e^{\left (e^{x}\right )}\right )} \]
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\[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\int { \frac {{\left ({\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 1\right )^{2} + {\left (x^{2} - 10 \, x + 1\right )} e^{\left (\frac {8}{x - 1}\right )} - 2 \, {\left (x^{2} - x\right )} e^{\left (\frac {4}{x - 1}\right )} - 2 \, {\left ({\left (x^{3} - 2 \, x^{2} + x\right )} e^{x} \log \left (x - 1\right )^{2} - 2 \, {\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + \frac {4}{x - 1}\right )} \log \left (x - 1\right ) + {\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + \frac {8}{x - 1}\right )}\right )} e^{\left (e^{x}\right )} + 2 \, {\left (x^{2} - {\left (x^{2} - 6 \, x + 1\right )} e^{\left (\frac {4}{x - 1}\right )} - x\right )} \log \left (x - 1\right )\right )} e^{\left (-2 \, e^{\left (e^{x}\right )}\right )}}{x^{2} - 2 \, x + 1} \,d x } \]
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Time = 12.70 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx={\mathrm {e}}^{-2\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (x\,{\ln \left (x-1\right )}^2-2\,x\,{\mathrm {e}}^{\frac {4}{x-1}}\,\ln \left (x-1\right )+x\,{\mathrm {e}}^{\frac {8}{x-1}}\right ) \]
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