Integrand size = 185, antiderivative size = 32 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{e^{e^{2 \left (-x+\log \left (\frac {4}{x}\right )\right )^2}}+\frac {8}{-3+x}}}{x} \]
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Leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(32)=64\).
Time = 5.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {1608, 27, 2326} \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {\left (e^{-\frac {8}{3-x}} \left (x^4-5 x^3+3 x^2+9 x\right )-e^{-\frac {8}{3-x}} \left (x^3-5 x^2+3 x+9\right ) \log \left (\frac {4}{x}\right )\right ) \exp \left (\exp \left (4^{-4 x} \left (\frac {1}{x}\right )^{-4 x} e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )}\right )\right )}{(3-x)^2 x^2 \left (x-\log \left (\frac {4}{x}\right )-\frac {\log \left (\frac {4}{x}\right )}{x}+1\right )} \]
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Rule 27
Rule 1608
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+\exp \left (e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )\right ) \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{x^2 \left (9-6 x+x^2\right )} \, dx \\ & = \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+\exp \left (e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )\right ) \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{(-3+x)^2 x^2} \, dx \\ & = \frac {\exp \left (\exp \left (4^{-4 x} e^{2 x^2+2 \log ^2\left (\frac {4}{x}\right )} \left (\frac {1}{x}\right )^{-4 x}\right )\right ) \left (e^{-\frac {8}{3-x}} \left (9 x+3 x^2-5 x^3+x^4\right )-e^{-\frac {8}{3-x}} \left (9+3 x-5 x^2+x^3\right ) \log \left (\frac {4}{x}\right )\right )}{(3-x)^2 x^2 \left (1+x-\log \left (\frac {4}{x}\right )-\frac {\log \left (\frac {4}{x}\right )}{x}\right )} \\ \end{align*}
\[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx \]
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Timed out.
\[\int \frac {\left (\left (\left (-4 x^{3}+20 x^{2}-12 x -36\right ) {\mathrm e}^{\frac {8}{-3+x}} \ln \left (\frac {4}{x}\right )+\left (4 x^{4}-20 x^{3}+12 x^{2}+36 x \right ) {\mathrm e}^{\frac {8}{-3+x}}\right ) {\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}} {\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}+\left (-x^{2}-2 x -9\right ) {\mathrm e}^{\frac {8}{-3+x}}\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 \ln \left (\frac {4}{x}\right )^{2}-4 x \ln \left (\frac {4}{x}\right )+2 x^{2}}}}}{x^{4}-6 x^{3}+9 x^{2}}d x\]
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Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x} \]
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Time = 95.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\frac {8}{x - 3}} e^{e^{e^{2 x^{2} - 4 x \log {\left (\frac {4}{x} \right )} + 2 \log {\left (\frac {4}{x} \right )}^{2}}}}}{x} \]
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Time = 0.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {e^{\left (\frac {8}{x - 3} + e^{\left (e^{\left (2 \, x^{2} - 8 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} + 4 \, x \log \left (x\right ) - 8 \, \log \left (2\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2}\right )}\right )}\right )}}{x} \]
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\[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\int { -\frac {{\left (4 \, {\left ({\left (x^{3} - 5 \, x^{2} + 3 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )} \log \left (\frac {4}{x}\right ) - {\left (x^{4} - 5 \, x^{3} + 3 \, x^{2} + 9 \, x\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2} + e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )} + {\left (x^{2} + 2 \, x + 9\right )} e^{\left (\frac {8}{x - 3}\right )}\right )} e^{\left (e^{\left (e^{\left (2 \, x^{2} - 4 \, x \log \left (\frac {4}{x}\right ) + 2 \, \log \left (\frac {4}{x}\right )^{2}\right )}\right )}\right )}}{x^{4} - 6 \, x^{3} + 9 \, x^{2}} \,d x } \]
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Time = 11.69 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {e^{e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}}} \left (e^{\frac {8}{-3+x}} \left (-9-2 x-x^2\right )+e^{e^{2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )}+2 x^2-4 x \log \left (\frac {4}{x}\right )+2 \log ^2\left (\frac {4}{x}\right )} \left (e^{\frac {8}{-3+x}} \left (36 x+12 x^2-20 x^3+4 x^4\right )+e^{\frac {8}{-3+x}} \left (-36-12 x+20 x^2-4 x^3\right ) \log \left (\frac {4}{x}\right )\right )\right )}{9 x^2-6 x^3+x^4} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{{\left (\frac {1}{256}\right )}^x\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x}\right )}^2}\,{\mathrm {e}}^{8\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{8\,\ln \left (2\right )-4\,x}}}\,{\mathrm {e}}^{\frac {8}{x-3}}}{x} \]
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