Integrand size = 58, antiderivative size = 24 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5 e^{-x-4 \left (-2+\frac {e^{e^x}}{3}\right ) x^2} \]
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Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {12, 2306, 6838} \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5 e^{\frac {1}{3} \left (-4 e^{e^x} x^2+24 x^2-3 x\right )} \]
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Rule 12
Rule 2306
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \exp \left (\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )\right ) \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx \\ & = \frac {1}{3} \int \frac {5}{4} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx \\ & = \frac {5}{12} \int e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx \\ & = 5 e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )} \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5 e^{-\frac {1}{3} x \left (3+4 \left (-6+e^{e^x}\right ) x\right )} \]
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Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75
method | result | size |
risch | \(5 \,{\mathrm e}^{-\frac {x \left (4 x \,{\mathrm e}^{{\mathrm e}^{x}}-24 x +3\right )}{3}}\) | \(18\) |
norman | \(4 \,{\mathrm e}^{-\frac {4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{3}+\ln \left (\frac {5}{4}\right )+8 x^{2}-x}\) | \(23\) |
parallelrisch | \(4 \,{\mathrm e}^{-\frac {4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{3}+\ln \left (\frac {5}{4}\right )+8 x^{2}-x}\) | \(23\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=4 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x + \log \left (\frac {5}{4}\right )\right )} \]
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Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5 e^{- \frac {4 x^{2} e^{e^{x}}}{3} + 8 x^{2} - x} \]
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Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x\right )} \]
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=4 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x + \log \left (\frac {5}{4}\right )\right )} \]
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Time = 9.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx=5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{8\,x^2}\,{\mathrm {e}}^{-\frac {4\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{3}} \]
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