Integrand size = 211, antiderivative size = 35 \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=e^{-\frac {x}{3 \left (-e^x+x \left (-4-\frac {1}{x}+x\right )\right )}} \left (2+\frac {x}{\log (\log (4))}\right ) \]
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\[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=\int \frac {\exp \left (\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}\right ) \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \left (3 e^{2 x}+25 x-23 x^3+3 x^4+3 \left (1+\frac {2}{3} \log (\log (4))\right )+2 x^2 (21+\log (\log (4)))+e^x \left (6+25 x-7 x^2+2 \log (\log (4))-2 x \log (\log (4))\right )\right )}{3 \left (1+e^x+4 x-x^2\right )^2 \log (\log (4))} \, dx \\ & = \frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \left (3 e^{2 x}+25 x-23 x^3+3 x^4+3 \left (1+\frac {2}{3} \log (\log (4))\right )+2 x^2 (21+\log (\log (4)))+e^x \left (6+25 x-7 x^2+2 \log (\log (4))-2 x \log (\log (4))\right )\right )}{\left (1+e^x+4 x-x^2\right )^2} \, dx}{3 \log (\log (4))} \\ & = \frac {\int \left (3 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}}-\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \left (3-6 x+x^2\right ) (x+2 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (-1+x) (x+2 \log (\log (4)))}{-1-e^x-4 x+x^2}\right ) \, dx}{3 \log (\log (4))} \\ & = -\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \left (3-6 x+x^2\right ) (x+2 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (-1+x) (x+2 \log (\log (4)))}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))} \\ & = \frac {\int \left (\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{-1-e^x-4 x+x^2}+\frac {2 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \log (\log (4))}{1+e^x+4 x-x^2}+\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x (-1+2 \log (\log (4)))}{-1-e^x-4 x+x^2}\right ) \, dx}{3 \log (\log (4))}-\frac {\int \left (\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^4}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {2 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^3 (-3+\log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}+\frac {6 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x \log (\log (4))}{\left (-1-e^x-4 x+x^2\right )^2}-\frac {3 e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2 (-1+4 \log (\log (4)))}{\left (-1-e^x-4 x+x^2\right )^2}\right ) \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))} \\ & = \frac {2}{3} \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}}}{1+e^x+4 x-x^2} \, dx-2 \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x}{\left (-1-e^x-4 x+x^2\right )^2} \, dx-\left (-4+\frac {1}{\log (\log (4))}\right ) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{\left (-1-e^x-4 x+x^2\right )^2} \, dx-\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^4}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {\int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^2}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))}+\frac {\int e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} \, dx}{\log (\log (4))}+\frac {(2 (3-\log (\log (4)))) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x^3}{\left (-1-e^x-4 x+x^2\right )^2} \, dx}{3 \log (\log (4))}+\frac {(-1+2 \log (\log (4))) \int \frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} x}{-1-e^x-4 x+x^2} \, dx}{3 \log (\log (4))} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=\frac {e^{\frac {x}{3 \left (1+e^x+4 x-x^2\right )}} (3 x+6 \log (\log (4)))}{3 \log (\log (4))} \]
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Time = 91.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (3 \,{\mathrm e}^{x}-3 x^{2}+12 x +3\right ) \ln \left (\frac {2 \ln \left (2 \ln \left (2\right )\right )+x}{\ln \left (2 \ln \left (2\right )\right )}\right )+x}{3 \,{\mathrm e}^{x}-3 x^{2}+12 x +3}}\) | \(53\) |
risch | \({\mathrm e}^{\frac {-3 \ln \left (\frac {2 \ln \left (2\right )+2 \ln \left (\ln \left (2\right )\right )+x}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )}\right ) x^{2}+3 \ln \left (\frac {2 \ln \left (2\right )+2 \ln \left (\ln \left (2\right )\right )+x}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )}\right ) {\mathrm e}^{x}+12 \ln \left (\frac {2 \ln \left (2\right )+2 \ln \left (\ln \left (2\right )\right )+x}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )}\right ) x +3 \ln \left (\frac {2 \ln \left (2\right )+2 \ln \left (\ln \left (2\right )\right )+x}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )}\right )+x}{3 \,{\mathrm e}^{x}-3 x^{2}+12 x +3}}\) | \(118\) |
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Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.51 \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=e^{\left (\frac {3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )} \log \left (\frac {x + 2 \, \log \left (2 \, \log \left (2\right )\right )}{\log \left (2 \, \log \left (2\right )\right )}\right ) - x}{3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )}}\right )} \]
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Time = 3.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=e^{\frac {x + \left (- 3 x^{2} + 12 x + 3 e^{x} + 3\right ) \log {\left (\frac {x + 2 \log {\left (2 \log {\left (2 \right )} \right )}}{\log {\left (2 \log {\left (2 \right )} \right )}} \right )}}{- 3 x^{2} + 12 x + 3 e^{x} + 3}} \]
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Time = 0.70 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=\frac {{\left (x + 2 \, \log \left (2\right ) + 2 \, \log \left (\log \left (2\right )\right )\right )} e^{\left (-\frac {x}{3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )}}\right )}}{\log \left (2\right ) + \log \left (\log \left (2\right )\right )} \]
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\[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=\int { \frac {{\left (3 \, x^{4} - 23 \, x^{3} + 42 \, x^{2} - {\left (7 \, x^{2} - 25 \, x - 6\right )} e^{x} + 2 \, {\left (x^{2} - {\left (x - 1\right )} e^{x} + 1\right )} \log \left (2 \, \log \left (2\right )\right ) + 25 \, x + 3 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (\frac {3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )} \log \left (\frac {x + 2 \, \log \left (2 \, \log \left (2\right )\right )}{\log \left (2 \, \log \left (2\right )\right )}\right ) - x}{3 \, {\left (x^{2} - 4 \, x - e^{x} - 1\right )}}\right )}}{3 \, {\left (x^{5} - 8 \, x^{4} + 14 \, x^{3} + 8 \, x^{2} + x e^{\left (2 \, x\right )} - 2 \, {\left (x^{3} - 4 \, x^{2} - x\right )} e^{x} + 2 \, {\left (x^{4} - 8 \, x^{3} + 14 \, x^{2} - 2 \, {\left (x^{2} - 4 \, x - 1\right )} e^{x} + 8 \, x + e^{\left (2 \, x\right )} + 1\right )} \log \left (2 \, \log \left (2\right )\right ) + x\right )}} \,d x } \]
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Timed out. \[ \int \frac {e^{\frac {x+\left (3+3 e^x+12 x-3 x^2\right ) \log \left (\frac {x+2 \log (\log (4))}{\log (\log (4))}\right )}{3+3 e^x+12 x-3 x^2}} \left (3+3 e^{2 x}+25 x+42 x^2-23 x^3+3 x^4+e^x \left (6+25 x-7 x^2\right )+\left (2+e^x (2-2 x)+2 x^2\right ) \log (\log (4))\right )}{3 x+3 e^{2 x} x+24 x^2+42 x^3-24 x^4+3 x^5+e^x \left (6 x+24 x^2-6 x^3\right )+\left (6+6 e^{2 x}+48 x+84 x^2-48 x^3+6 x^4+e^x \left (12+48 x-12 x^2\right )\right ) \log (\log (4))} \, dx=\int \frac {{\mathrm {e}}^{\frac {x+\ln \left (\frac {x+2\,\ln \left (2\,\ln \left (2\right )\right )}{\ln \left (2\,\ln \left (2\right )\right )}\right )\,\left (12\,x+3\,{\mathrm {e}}^x-3\,x^2+3\right )}{12\,x+3\,{\mathrm {e}}^x-3\,x^2+3}}\,\left (25\,x+3\,{\mathrm {e}}^{2\,x}+\ln \left (2\,\ln \left (2\right )\right )\,\left (2\,x^2-{\mathrm {e}}^x\,\left (2\,x-2\right )+2\right )+{\mathrm {e}}^x\,\left (-7\,x^2+25\,x+6\right )+42\,x^2-23\,x^3+3\,x^4+3\right )}{3\,x+3\,x\,{\mathrm {e}}^{2\,x}+\ln \left (2\,\ln \left (2\right )\right )\,\left (48\,x+6\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (-12\,x^2+48\,x+12\right )+84\,x^2-48\,x^3+6\,x^4+6\right )+24\,x^2+42\,x^3-24\,x^4+3\,x^5+{\mathrm {e}}^x\,\left (-6\,x^3+24\,x^2+6\,x\right )} \,d x \]
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