Integrand size = 201, antiderivative size = 30 \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=e^{e^{\left (1+x-\log ^2(2)\right ) \left (\frac {18}{(1-x)^2}+\log \left (\frac {x}{2}\right )\right )}} \]
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Timed out. \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=\text {\$Aborted} \]
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Rubi steps Aborted
\[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=\int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx \]
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Time = 122.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{-\frac {\left (\ln \left (2\right )^{2}-x -1\right ) \left (x^{2} \ln \left (\frac {x}{2}\right )-2 x \ln \left (\frac {x}{2}\right )+\ln \left (\frac {x}{2}\right )+18\right )}{\left (-1+x \right )^{2}}}}\) | \(40\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\left (\left (-x^{2}+2 x -1\right ) \ln \left (2\right )^{2}+x^{3}-x^{2}-x +1\right ) \ln \left (\frac {x}{2}\right )-18 \ln \left (2\right )^{2}+18 x +18}{x^{2}-2 x +1}}}\) | \(58\) |
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Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 5.90 \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=e^{\left (\frac {{\left (x^{2} - 2 \, x + 1\right )} e^{\left (-\frac {18 \, \log \left (2\right )^{2} - {\left (x^{3} - {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} - x^{2} - x + 1\right )} \log \left (\frac {1}{2} \, x\right ) - 18 \, x - 18}{x^{2} - 2 \, x + 1}\right )} - 18 \, \log \left (2\right )^{2} + {\left (x^{3} - {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} - x^{2} - x + 1\right )} \log \left (\frac {1}{2} \, x\right ) + 18 \, x + 18}{x^{2} - 2 \, x + 1} + \frac {18 \, \log \left (2\right )^{2} - {\left (x^{3} - {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} - x^{2} - x + 1\right )} \log \left (\frac {1}{2} \, x\right ) - 18 \, x - 18}{x^{2} - 2 \, x + 1}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 2.62 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=e^{e^{\frac {18 x + \left (x^{3} - x^{2} - x + \left (- x^{2} + 2 x - 1\right ) \log {\left (2 \right )}^{2} + 1\right ) \log {\left (\frac {x}{2} \right )} - 18 \log {\left (2 \right )}^{2} + 18}{x^{2} - 2 x + 1}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (25) = 50\).
Time = 2.43 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.07 \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=e^{\left (\frac {1}{2} \, x e^{\left (\log \left (2\right )^{3} - \log \left (2\right )^{2} \log \left (x\right ) - x \log \left (2\right ) + x \log \left (x\right ) - \frac {18 \, \log \left (2\right )^{2}}{x^{2} - 2 \, x + 1} + \frac {36}{x^{2} - 2 \, x + 1} + \frac {18}{x - 1}\right )}\right )} \]
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\[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx=\int { \frac {{\left (x^{4} - 2 \, x^{3} - {\left (x^{3} - 3 \, x^{2} - 33 \, x - 1\right )} \log \left (2\right )^{2} - 18 \, x^{2} + {\left (x^{4} - 3 \, x^{3} + 3 \, x^{2} - x\right )} \log \left (\frac {1}{2} \, x\right ) - 52 \, x - 1\right )} e^{\left (-\frac {18 \, \log \left (2\right )^{2} - {\left (x^{3} - {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} - x^{2} - x + 1\right )} \log \left (\frac {1}{2} \, x\right ) - 18 \, x - 18}{x^{2} - 2 \, x + 1} + e^{\left (-\frac {18 \, \log \left (2\right )^{2} - {\left (x^{3} - {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} - x^{2} - x + 1\right )} \log \left (\frac {1}{2} \, x\right ) - 18 \, x - 18}{x^{2} - 2 \, x + 1}\right )}\right )}}{x^{4} - 3 \, x^{3} + 3 \, x^{2} - x} \,d x } \]
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Time = 13.99 (sec) , antiderivative size = 159, normalized size of antiderivative = 5.30 \[ \int \frac {e^{e^{\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}}+\frac {18+18 x-18 \log ^2(2)+\left (1-x-x^2+x^3+\left (-1+2 x-x^2\right ) \log ^2(2)\right ) \log \left (\frac {x}{2}\right )}{1-2 x+x^2}} \left (-1-52 x-18 x^2-2 x^3+x^4+\left (1+33 x+3 x^2-x^3\right ) \log ^2(2)+\left (-x+3 x^2-3 x^3+x^4\right ) \log \left (\frac {x}{2}\right )\right )}{-x+3 x^2-3 x^3+x^4} \, dx={\mathrm {e}}^{{\left (\frac {1}{2}\right )}^{x+1}\,x^{\frac {x^2}{x-1}-\frac {x+x^2\,{\ln \left (2\right )}^2-2\,x\,{\ln \left (2\right )}^2+{\ln \left (2\right )}^2}{x^2-2\,x+1}+\frac {1}{x^2-2\,x+1}}\,{\mathrm {e}}^{\frac {18}{x^2-2\,x+1}}\,{\mathrm {e}}^{-\frac {2\,x\,{\ln \left (2\right )}^3}{x^2-2\,x+1}}\,{\mathrm {e}}^{\frac {18\,x}{x^2-2\,x+1}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \left (2\right )}^3}{x^2-2\,x+1}}\,{\mathrm {e}}^{\frac {{\ln \left (2\right )}^3}{x^2-2\,x+1}}\,{\mathrm {e}}^{-\frac {18\,{\ln \left (2\right )}^2}{x^2-2\,x+1}}} \]
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