Integrand size = 143, antiderivative size = 30 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=-e^x+x+\frac {x}{\log \left (\frac {x}{e^4+2 x+(3+x)^2+\log (x)}\right )} \]
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\[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=\int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-8 \left (1+\frac {e^4}{8}\right )+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ & = \int \left (-e^x+\frac {9 \left (1+\frac {e^4}{9}\right )}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}+\frac {8 x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}+\frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}+\frac {\log (x)}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}+\frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )}+\frac {-8-e^4}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )}+\frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )}+\frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )}\right ) \, dx \\ & = 8 \int \frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\left (-8-e^4\right ) \int \frac {1}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\left (9+e^4\right ) \int \frac {1}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx-\int e^x \, dx+\int \frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {\log (x)}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ & = -e^x+8 \int \frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\left (-8-e^4\right ) \int \frac {1}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\left (9+e^4\right ) \int \frac {1}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \left (1+\frac {-9-e^4-8 x-x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right ) \, dx+\int \frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ & = -e^x+x+8 \int \frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\left (-8-e^4\right ) \int \frac {1}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\left (9+e^4\right ) \int \frac {1}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {-9-e^4-8 x-x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ & = -e^x+x+8 \int \frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\left (-8-e^4\right ) \int \frac {1}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\left (9+e^4\right ) \int \frac {1}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \left (\frac {9 \left (1+\frac {e^4}{9}\right )}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)}+\frac {8 x}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)}+\frac {x^2}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)}\right ) \, dx+\int \frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ & = -e^x+x+8 \int \frac {x}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)} \, dx+8 \int \frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\left (-8-e^4\right ) \int \frac {1}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\left (9+e^4\right ) \int \frac {1}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)} \, dx+\left (9+e^4\right ) \int \frac {1}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {x^2}{-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)} \, dx+\int \frac {x^2}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)} \, dx+\int \frac {\log (x)}{\left (-9 \left (1+\frac {e^4}{9}\right )-8 x-x^2-\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {x^2}{\left (9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx+\int \frac {1}{\log \left (\frac {x}{9 \left (1+\frac {e^4}{9}\right )+8 x+x^2+\log (x)}\right )} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=-e^x+x+\frac {x}{\log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 36.96 (sec) , antiderivative size = 173, normalized size of antiderivative = 5.77
\[x -{\mathrm e}^{x}+\frac {2 i x}{\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9}\right )^{3}-2 i \ln \left (\ln \left (x \right )+{\mathrm e}^{4}+x^{2}+8 x +9\right )+2 i \ln \left (x \right )}\]
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Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=\frac {{\left (x - e^{x}\right )} \log \left (\frac {x}{x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9}\right ) + x}{\log \left (\frac {x}{x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9}\right )} \]
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Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=x + \frac {x}{\log {\left (\frac {x}{x^{2} + 8 x + \log {\left (x \right )} + 9 + e^{4}} \right )}} - e^{x} \]
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Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=\frac {{\left (x - e^{x}\right )} \log \left (x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9\right ) - x \log \left (x\right ) + e^{x} \log \left (x\right ) - x}{\log \left (x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9\right ) - \log \left (x\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).
Time = 0.48 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=\frac {x \log \left (x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9\right ) - e^{x} \log \left (x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9\right ) - x \log \left (x\right ) + e^{x} \log \left (x\right ) - x}{\log \left (x^{2} + 8 \, x + e^{4} + \log \left (x\right ) + 9\right ) - \log \left (x\right )} \]
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Time = 13.80 (sec) , antiderivative size = 167, normalized size of antiderivative = 5.57 \[ \int \frac {-8-e^4+x^2-\log (x)+\left (9+e^4+8 x+x^2+\log (x)\right ) \log \left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )+\left (9+e^4+8 x+x^2+e^x \left (-9-e^4-8 x-x^2\right )+\left (1-e^x\right ) \log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )}{\left (9+e^4+8 x+x^2+\log (x)\right ) \log ^2\left (\frac {x}{9+e^4+8 x+x^2+\log (x)}\right )} \, dx=\frac {\frac {x\,\left (120\,x+{\mathrm {e}}^4+16\,x\,{\mathrm {e}}^4+6\,x^2\,{\mathrm {e}}^4+47\,x^2-2\,x^4+7\right )}{2\,x^2-1}+\frac {x\,\ln \left (x\right )\,\left (6\,x^2+16\,x+1\right )}{2\,x^2-1}}{{\mathrm {e}}^4+\ln \left (x\right )-x^2+8}-{\mathrm {e}}^x-\frac {2\,x+4}{x^2-\frac {1}{2}}-x+\frac {x-\frac {x\,\ln \left (\frac {x}{8\,x+{\mathrm {e}}^4+\ln \left (x\right )+x^2+9}\right )\,\left (8\,x+{\mathrm {e}}^4+\ln \left (x\right )+x^2+9\right )}{{\mathrm {e}}^4+\ln \left (x\right )-x^2+8}}{\ln \left (\frac {x}{8\,x+{\mathrm {e}}^4+\ln \left (x\right )+x^2+9}\right )} \]
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