Integrand size = 88, antiderivative size = 33 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=5+x+\frac {e^{2 e^{2+\frac {4}{x}-x}+2 x (x+\log (3))}}{\log ^2(\log (5))} \]
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Time = 0.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 14, 2326} \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=\frac {2\ 9^x \log (3) \exp \left (-((1-2 x) x)+x+2 e^{-x+\frac {4}{x}+2}\right )}{\log (9) \log ^2(\log (5))}+x \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2} \, dx}{\log ^2(\log (5))} \\ & = \frac {\int \left (\frac {2\ 9^x e^{2 e^{2+\frac {4}{x}-x}+x (-1+2 x)} \left (-4 e^{2+\frac {4}{x}}-e^{2+\frac {4}{x}} x^2+2 e^x x^3+e^x x^2 \log (3)\right )}{x^2}+\log ^2(\log (5))\right ) \, dx}{\log ^2(\log (5))} \\ & = x+\frac {2 \int \frac {9^x e^{2 e^{2+\frac {4}{x}-x}+x (-1+2 x)} \left (-4 e^{2+\frac {4}{x}}-e^{2+\frac {4}{x}} x^2+2 e^x x^3+e^x x^2 \log (3)\right )}{x^2} \, dx}{\log ^2(\log (5))} \\ & = x+\frac {2\ 9^x e^{2 e^{2+\frac {4}{x}-x}+x-(1-2 x) x} \log (3)}{\log (9) \log ^2(\log (5))} \\ \end{align*}
Time = 0.99 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=x+\frac {2\ 9^x e^{2 e^{2+\frac {4}{x}-x}+x+x (-1+2 x)} \log (3)}{\log (9) \log ^2(\log (5))} \]
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Time = 0.92 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x +\frac {3^{2 x} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {x^{2}-2 x -4}{x}}+2 x^{2}}}{\ln \left (\ln \left (5\right )\right )^{2}}\) | \(37\) |
parallelrisch | \(\frac {x \ln \left (\ln \left (5\right )\right )^{2}+{\mathrm e}^{2 \,{\mathrm e}^{-\frac {x^{2}-2 x -4}{x}}+2 x \ln \left (3\right )+2 x^{2}}}{\ln \left (\ln \left (5\right )\right )^{2}}\) | \(40\) |
norman | \(\frac {\ln \left (\ln \left (5\right )\right ) x^{2}+\frac {x \,{\mathrm e}^{2 \,{\mathrm e}^{\frac {-x^{2}+2 x +4}{x}}+2 x \ln \left (3\right )+2 x^{2}}}{\ln \left (\ln \left (5\right )\right )}}{x \ln \left (\ln \left (5\right )\right )}\) | \(51\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=\frac {x \log \left (\log \left (5\right )\right )^{2} + e^{\left (2 \, x^{2} + 2 \, x \log \left (3\right ) + 2 \, e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )}}{\log \left (\log \left (5\right )\right )^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=x + \frac {e^{2 x^{2} + 2 x \log {\left (3 \right )} + 2 e^{\frac {- x^{2} + 2 x + 4}{x}}}}{\log {\left (\log {\left (5 \right )} \right )}^{2}} \]
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Time = 0.40 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=\frac {x \log \left (\log \left (5\right )\right )^{2} + e^{\left (2 \, x^{2} + 2 \, x \log \left (3\right ) + 2 \, e^{\left (-x + \frac {4}{x} + 2\right )}\right )}}{\log \left (\log \left (5\right )\right )^{2}} \]
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\[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=\int { \frac {x^{2} \log \left (\log \left (5\right )\right )^{2} + 2 \, {\left (2 \, x^{3} + x^{2} \log \left (3\right ) - {\left (x^{2} + 4\right )} e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )} e^{\left (2 \, x^{2} + 2 \, x \log \left (3\right ) + 2 \, e^{\left (-\frac {x^{2} - 2 \, x - 4}{x}\right )}\right )}}{x^{2} \log \left (\log \left (5\right )\right )^{2}} \,d x } \]
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Time = 12.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 e^{\frac {4+2 x-x^2}{x}}+2 x^2+2 x \log (3)} \left (4 x^3+e^{\frac {4+2 x-x^2}{x}} \left (-8-2 x^2\right )+2 x^2 \log (3)\right )+x^2 \log ^2(\log (5))}{x^2 \log ^2(\log (5))} \, dx=x+\frac {3^{2\,x}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{4/x}}\,{\mathrm {e}}^{2\,x^2}}{{\ln \left (\ln \left (5\right )\right )}^2} \]
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