Integrand size = 123, antiderivative size = 41 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{-8+2^{1-2 e^{-3+x}} 5^{e^{-3+x}} e^x \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}}} \]
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\[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\int \frac {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{x (3+x)} \, dx \\ & = \int \frac {5^{e^{-3+x}} \exp \left (-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{x (3+x)} \, dx \\ & = \int \left (2\ 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}+\frac {2\ 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (-3-2 x+3 x \log \left (\frac {5}{4 x (3+x)}\right )+x^2 \log \left (\frac {5}{4 x (3+x)}\right )\right )}{x (3+x)}\right ) \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \left (-3-2 x+3 x \log \left (\frac {5}{4 x (3+x)}\right )+x^2 \log \left (\frac {5}{4 x (3+x)}\right )\right )}{x (3+x)} \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \left (\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) (-3-2 x) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x (3+x)}+5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \log \left (\frac {5}{4 x (3+x)}\right )\right ) \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) (-3-2 x) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x (3+x)} \, dx+2 \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \log \left (\frac {5}{4 x (3+x)}\right ) \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \left (\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x}-\frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x}\right ) \, dx-2 \int \frac {(-3-2 x) \int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x (3+x)} \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x} \, dx-2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x} \, dx-2 \int \left (\frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{-3-x}-\frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x}\right ) \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx \\ & = 2 \int 5^{e^{-3+x}} \exp \left (-8+x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx+2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{-3-x} \, dx-2 \int \frac {5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}}}{x} \, dx-2 \int \frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{-3-x} \, dx+2 \int \frac {\int \left (\frac {5}{4}\right )^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}} \, dx}{x} \, dx+\left (2 \log \left (\frac {5}{4 x (3+x)}\right )\right ) \int 5^{e^{-3+x}} \exp \left (-11+2 x+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}\right ) \left (\frac {1}{x (12+4 x)}\right )^{e^{-3+x}} \, dx \\ \end{align*}
\[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.91 (sec) , antiderivative size = 164, normalized size of antiderivative = 4.00
\[{\mathrm e}^{2 \,{\mathrm e}^{x -\ln \left (x \right ) {\mathrm e}^{-3+x}+\ln \left (5\right ) {\mathrm e}^{-3+x}-2 \ln \left (2\right ) {\mathrm e}^{-3+x}-{\mathrm e}^{-3+x} \ln \left (3+x \right )} {\mathrm e}^{-\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}} {\mathrm e}^{\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{2} \operatorname {csgn}\left (\frac {i}{3+x}\right )}{2}} {\mathrm e}^{-\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i}{3+x}\right )}{2}}-8}\]
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{\left ({\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (x + 3\right )} - 11 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} \]
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Time = 22.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{2 e^{x} e^{\frac {e^{x} \log {\left (\frac {5}{4 x^{2} + 12 x} \right )}}{e^{3}}} - 8} \]
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Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{\left (2 \, e^{\left (e^{\left (x - 3\right )} \log \left (5\right ) - 2 \, e^{\left (x - 3\right )} \log \left (2\right ) - e^{\left (x - 3\right )} \log \left (x + 3\right ) - e^{\left (x - 3\right )} \log \left (x\right ) + x\right )} - 8\right )} \]
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\[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\int { \frac {2 \, {\left ({\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right ) - {\left (2 \, x + 3\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 3 \, x\right )} e^{\left (x + 3\right )}\right )} \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{x} - 11\right )}}{x^{2} + 3 \, x} \,d x } \]
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Time = 8.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx={\mathrm {e}}^{-8}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x\,{\left (\frac {5}{4\,x^2+12\,x}\right )}^{{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}} \]
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