Optimal. Leaf size=32 \[ e^{x-\frac {x^2 (1-2 x-\log (5))}{3 \left (1-e^{20}-x\right )}} \]
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Rubi [F] time = 2.47, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}\right ) \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{3 \left (1-e^{20}\right )^2-6 \left (1-e^{20}\right ) x+3 x^2} \, dx\\ &=\int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{3 \left (-1+e^{20}+x\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (3 \left (1-e^{20}\right )^2-4 x^3-2 \left (1-e^{20}\right ) x (4-\log (5))+x^2 \left (10-6 e^{20}-\log (5)\right )\right )}{\left (-1+e^{20}+x\right )^2} \, dx\\ &=\frac {1}{3} \int \left (-4 \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) x-\frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (-1+e^{20}\right )^2 \left (-1+2 e^{20}-\log (5)\right )}{\left (-1+e^{20}+x\right )^2}+2 \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \left (1+e^{20}-\frac {\log (5)}{2}\right )\right ) \, dx\\ &=-\left (\frac {4}{3} \int \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) x \, dx\right )+\frac {1}{3} \left (2+2 e^{20}-\log (5)\right ) \int \exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right ) \, dx+\frac {1}{3} \left (\left (1-e^{20}\right )^2 \left (1-2 e^{20}+\log (5)\right )\right ) \int \frac {\exp \left (\frac {x \left (-3 \left (1-e^{20}\right )-2 x^2+x (4-\log (5))\right )}{-3+3 e^{20}+3 x}\right )}{\left (-1+e^{20}+x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 48, normalized size = 1.50 \begin {gather*} 5^{-\frac {x^2}{3 \left (-1+e^{20}+x\right )}} e^{-\frac {x \left (3-3 e^{20}-4 x+2 x^2\right )}{3 \left (-1+e^{20}+x\right )}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.13, size = 35, normalized size = 1.09 \begin {gather*} e^{\left (-\frac {2 \, x^{3} + x^{2} \log \relax (5) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.38, size = 35, normalized size = 1.09 \begin {gather*} e^{\left (-\frac {2 \, x^{3} + x^{2} \log \relax (5) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 30, normalized size = 0.94
method | result | size |
gosper | \({\mathrm e}^{-\frac {x \left (x \ln \relax (5)+2 x^{2}-3 \,{\mathrm e}^{20}-4 x +3\right )}{3 \left ({\mathrm e}^{20}+x -1\right )}}\) | \(30\) |
risch | \({\mathrm e}^{\frac {\left (-x \ln \relax (5)-2 x^{2}+3 \,{\mathrm e}^{20}+4 x -3\right ) x}{3 \,{\mathrm e}^{20}+3 x -3}}\) | \(31\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-x^{2} \ln \relax (5)+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}+\left ({\mathrm e}^{20}-1\right ) {\mathrm e}^{\frac {-x^{2} \ln \relax (5)+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}}{{\mathrm e}^{20}+x -1}\) | \(95\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.89, size = 115, normalized size = 3.59 \begin {gather*} 5^{\frac {1}{3} \, e^{20} - \frac {1}{3}} e^{\left (-\frac {2}{3} \, x^{2} + \frac {2}{3} \, x e^{20} - \frac {1}{3} \, x \log \relax (5) + \frac {2}{3} \, x - \frac {e^{40} \log \relax (5)}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{20} \log \relax (5)}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{60}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {5 \, e^{40}}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {4 \, e^{20}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {\log \relax (5)}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {1}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {2}{3} \, e^{40} + e^{20} - \frac {1}{3}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.58, size = 86, normalized size = 2.69 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {3\,x}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{-\frac {2\,x^3}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {4\,x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {3\,x\,{\mathrm {e}}^{20}}{3\,x+3\,{\mathrm {e}}^{20}-3}}}{5^{\frac {x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.77, size = 37, normalized size = 1.16 \begin {gather*} e^{\frac {- 2 x^{3} - x^{2} \log {\relax (5 )} + 4 x^{2} - 3 x + 3 x e^{20}}{3 x - 3 + 3 e^{20}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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