Integrand size = 280, antiderivative size = 30 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=\frac {x}{5 \left (x-\log \left (\log \left (\frac {40}{-e^x+\frac {1}{x}}+4 x\right )\right )\right )} \]
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Time = 1.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6820, 12, 6843, 32} \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=-\frac {1}{5 \left (1-\frac {x}{\log \left (\log \left (\frac {4 x \left (11-e^x x\right )}{1-e^x x}\right )\right )}\right )} \]
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Rule 12
Rule 32
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \int \frac {11+e^{2 x} x^2+2 e^x x (-1+5 x)-\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}{5 \left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \left (x-\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {11+e^{2 x} x^2+2 e^x x (-1+5 x)-\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}{\left (11-12 e^x x+e^{2 x} x^2\right ) \log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right ) \left (x-\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )^2} \, dx \\ & = -\left (\frac {1}{5} \text {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {x}{\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )}\right )\right ) \\ & = -\frac {1}{5 \left (1-\frac {x}{\log \left (\log \left (\frac {4 x \left (11-e^x x\right )}{1-e^x x}\right )\right )}\right )} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=-\frac {x}{5 \left (-x+\log \left (\log \left (\frac {4 x \left (-11+e^x x\right )}{-1+e^x x}\right )\right )\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 210, normalized size of antiderivative = 7.00
\[\frac {x}{5 x -5 \ln \left (2 \ln \left (2\right )+\ln \left (x \right )+\ln \left ({\mathrm e}^{x} x -11\right )-\ln \left ({\mathrm e}^{x} x -1\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -11\right )\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x} x -1}\right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right ) \left (-\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )+\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -11\right )}{{\mathrm e}^{x} x -1}\right )\right )}{2}\right )}\]
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=\frac {x}{5 \, {\left (x - \log \left (\log \left (\frac {4 \, {\left (x^{2} e^{x} - 11 \, x\right )}}{x e^{x} - 1}\right )\right )\right )}} \]
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Time = 1.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=- \frac {x}{- 5 x + 5 \log {\left (\log {\left (\frac {4 x^{2} e^{x} - 44 x}{x e^{x} - 1} \right )} \right )}} \]
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Time = 1.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=\frac {x}{5 \, {\left (x - \log \left (2 \, \log \left (2\right ) - \log \left (x e^{x} - 1\right ) + \log \left (x e^{x} - 11\right ) + \log \left (x\right )\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1340 vs. \(2 (27) = 54\).
Time = 4.02 (sec) , antiderivative size = 1340, normalized size of antiderivative = 44.67 \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {11+e^{2 x} x^2+e^x \left (-2 x+10 x^2\right )+\left (-11+12 e^x x-e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )}{\left (55 x^2-60 e^x x^3+5 e^{2 x} x^4\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )+\left (-110 x+120 e^x x^2-10 e^{2 x} x^3\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log \left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )+\left (55-60 e^x x+5 e^{2 x} x^2\right ) \log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right ) \log ^2\left (\log \left (\frac {-44 x+4 e^x x^2}{-1+e^x x}\right )\right )} \, dx=\int \frac {x^2\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (2\,x-10\,x^2\right )-\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )\,\left (x^2\,{\mathrm {e}}^{2\,x}-12\,x\,{\mathrm {e}}^x+11\right )+11}{\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (5\,x^2\,{\mathrm {e}}^{2\,x}-60\,x\,{\mathrm {e}}^x+55\right )\,{\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )}^2-\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (110\,x-120\,x^2\,{\mathrm {e}}^x+10\,x^3\,{\mathrm {e}}^{2\,x}\right )\,\ln \left (\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\right )+\ln \left (-\frac {44\,x-4\,x^2\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x-1}\right )\,\left (5\,x^4\,{\mathrm {e}}^{2\,x}-60\,x^3\,{\mathrm {e}}^x+55\,x^2\right )} \,d x \]
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