Integrand size = 118, antiderivative size = 26 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=5 (4-x) \log \left (x+e^x x \left (2 x+\frac {2}{\log (\log (2))}\right )\right ) \]
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\[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=\int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5 (-4+x) \left (\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )\right )}{x \left (\log (\log (2))+2 e^x (1+x \log (\log (2)))\right )}-5 \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )\right ) \, dx \\ & = -\left (5 \int \frac {(-4+x) \left (\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )\right )}{x \left (\log (\log (2))+2 e^x (1+x \log (\log (2)))\right )} \, dx\right )-5 \int \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right ) \, dx \\ & = -5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \int \frac {\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )}{\log (\log (2))+2 e^x (1+x \log (\log (2)))} \, dx-5 \int \left (-\frac {(-4+x) \log (\log (2)) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}+\frac {(4-x) \left (-1-x^2 \log (\log (2))-x (1+2 \log (\log (2)))\right )}{x (1+x \log (\log (2)))}\right ) \, dx \\ & = -5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )-5 \int \frac {(4-x) \left (-1-x^2 \log (\log (2))-x (1+2 \log (\log (2)))\right )}{x (1+x \log (\log (2)))} \, dx+5 \int \left (-\frac {x \log (\log (2)) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}+\frac {1+x^2 \log (\log (2))+x (1+2 \log (\log (2)))}{1+x \log (\log (2))}\right ) \, dx+(5 \log (\log (2))) \int \frac {(-4+x) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx \\ & = -5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \int \frac {1+x^2 \log (\log (2))+x (1+2 \log (\log (2)))}{1+x \log (\log (2))} \, dx-5 \int \left (-2-\frac {4}{x}+x+\frac {-1-4 \log (\log (2))}{1+x \log (\log (2))}\right ) \, dx-(5 \log (\log (2))) \int \frac {x (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx+(5 \log (\log (2))) \int \left (-\frac {3}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}+\frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}-\frac {1+4 \log (\log (2))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}\right ) \, dx \\ & = 10 x-\frac {5 x^2}{2}+20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \left (4+\frac {1}{\log (\log (2))}\right ) \log (1+x \log (\log (2)))+5 \int \left (2+x+\frac {1}{-1-x \log (\log (2))}\right ) \, dx+(5 \log (\log (2))) \int \frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2))) \int \left (\frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}+\frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}-\frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}\right ) \, dx-(15 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2)) (1+4 \log (\log (2)))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx \\ & = 20 x+20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \left (4+\frac {1}{\log (\log (2))}\right ) \log (1+x \log (\log (2)))-\frac {5 \log (1+x \log (\log (2)))}{\log (\log (2))}-(5 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx+(5 \log (\log (2))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx-(15 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2)) (1+4 \log (\log (2)))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx \\ \end{align*}
Time = 0.96 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+20 \log \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 3.94 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15
method | result | size |
default | \(20 \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x -5 x \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )\) | \(56\) |
norman | \(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\) | \(59\) |
parallelrisch | \(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\) | \(59\) |
risch | \(-5 x \ln \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )-5 x \ln \left (x \right )+\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) \operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}+5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}-\frac {5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{3}}{2}-5 i \pi x -5 x \ln \left (2\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x +20 \ln \left (x^{2} \ln \left (\ln \left (2\right )\right )+x \right )+20 \ln \left ({\mathrm e}^{x}+\frac {\ln \left (\ln \left (2\right )\right )}{2+2 x \ln \left (\ln \left (2\right )\right )}\right )\) | \(242\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, {\left (x - 4\right )} \log \left (\frac {2 \, x e^{x} + {\left (2 \, x^{2} e^{x} + x\right )} \log \left (\log \left (2\right )\right )}{\log \left (\log \left (2\right )\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (22) = 44\).
Time = 0.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.54 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=- 5 x \log {\left (\frac {2 x e^{x} + \left (2 x^{2} e^{x} + x\right ) \log {\left (\log {\left (2 \right )} \right )}}{\log {\left (\log {\left (2 \right )} \right )}} \right )} + 20 \log {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x \right )} + 20 \log {\left (e^{x} + \frac {\log {\left (\log {\left (2 \right )} \right )}}{2 x \log {\left (\log {\left (2 \right )} \right )} + 2} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )\right ) - 5 \, x \log \left (x\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x \log \left (\log \left (2\right )\right ) + 1\right ) + 20 \, \log \left (x\right ) + 20 \, \log \left (\frac {2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )}{2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )}}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, x^{2} e^{x} \log \left (\log \left (2\right )\right ) + 2 \, x e^{x} + x \log \left (\log \left (2\right )\right )\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (2 \, x e^{x} \log \left (\log \left (2\right )\right ) + 2 \, e^{x} + \log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x\right ) \]
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Time = 10.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )+2\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x\,\ln \left (\ln \left (2\right )\right )}{x\,\ln \left (\ln \left (2\right )\right )+1}\right )+20\,\ln \left (\ln \left (\ln \left (2\right )\right )\,x^2+x\right )-5\,x\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )\,\left (x+2\,x^2\,{\mathrm {e}}^x\right )+2\,x\,{\mathrm {e}}^x}{\ln \left (\ln \left (2\right )\right )}\right ) \]
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