Integrand size = 47, antiderivative size = 24 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \]
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Timed out. \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\text {\$Aborted} \]
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Rubi steps \begin{align*} \text {integral}& = \int 2^{2+4 x^2} x \left (e^x x\right )^{4 x^2} \left (\frac {\log (3)}{9}\right )^{4 x^2} \left (1+x+2 \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(24)=48\).
Time = 0.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=-\frac {4 e^{x^2 \left (-\log \left (\frac {6561}{16}\right )+4 \log (\log (3))\right )} \left (e^x x\right )^{4 x^2} \left (-\log \left (\frac {9}{2}\right )+\log (\log (3))\right )}{\log \left (\frac {6561}{16}\right )-4 \log (\log (3))} \]
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Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62
method | result | size |
parallelrisch | \({\mathrm e}^{4 x^{2} \ln \left (\frac {2 x \ln \left (3\right ) {\mathrm e}^{x}}{9}\right )}\) | \(15\) |
risch | \(x^{4 x^{2}} \ln \left (3\right )^{4 x^{2}} 16^{x^{2}} \left (\frac {1}{6561}\right )^{x^{2}} \left ({\mathrm e}^{x}\right )^{4 x^{2}} {\mathrm e}^{-2 i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} x \right ) x^{2} \left (-\operatorname {csgn}\left (i {\mathrm e}^{x} x \right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{x} x \right )+\operatorname {csgn}\left (i x \right )\right )}\) | \(81\) |
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Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=\left (\frac {2}{9} \, x e^{x} \log \left (3\right )\right )^{4 \, x^{2}} \]
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Time = 81.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{4 x^{2} \log {\left (\frac {2 x e^{x} \log {\left (3 \right )}}{9} \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (13) = 26\).
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} \log \left (3\right ) + 4 \, x^{2} \log \left (2\right ) + 4 \, x^{2} \log \left (x\right ) + 4 \, x^{2} \log \left (\log \left (3\right )\right )\right )} \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} \log \left (3\right ) + 4 \, x^{2} \log \left (2 \, x \log \left (3\right )\right )\right )} \]
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Time = 14.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (e^x x\right )^{4 x^2} \left (\frac {2 \log (3)}{9}\right )^{4 x^2} \left (4 x+4 x^2+8 x \log \left (\frac {2}{9} e^x x \log (3)\right )\right ) \, dx=x^{4\,x^2}\,{\mathrm {e}}^{4\,x^3}\,{\left (\frac {16\,{\ln \left (3\right )}^4}{6561}\right )}^{x^2} \]
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