Optimal. Leaf size=28 \[ e^{-e^{1+x \left (-5 e^{-x}+x^2\right )}+\frac {3}{\log (3)}} \]
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Rubi [F] time = 1.99, 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 \exp \left (-x+e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )+\frac {3-e^{e^{-x} \left (-5 x+e^x \left (1+x^3\right )\right )} \log (3)}{\log (3)}\right ) \left (5-5 x-3 e^x x^2\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) \left (5-5 x-3 e^x x^2\right ) \, dx\\ &=\int \left (5 \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right )-5 \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x-3 \exp \left (1-e^{1-5 e^{-x} x+x^3}-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x^2\right ) \, dx\\ &=-\left (3 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x^2 \, dx\right )+5 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) \, dx-5 \int \exp \left (1-e^{1-5 e^{-x} x+x^3}-x-5 e^{-x} x+x^3+\frac {3}{\log (3)}\right ) x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.21, size = 26, normalized size = 0.93 \begin {gather*} e^{-e^{1-5 e^{-x} x+x^3}+\frac {3}{\log (3)}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.69, size = 76, normalized size = 2.71 \begin {gather*} e^{\left (-{\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x + \frac {{\left ({\left ({\left (x^{3} - x + 1\right )} \log \relax (3) + 3\right )} e^{x} - 5 \, x \log \relax (3) - e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} + x\right )} \log \relax (3)\right )} e^{\left (-x\right )}}{\log \relax (3)}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -{\left (3 \, x^{2} e^{x} + 5 \, x - 5\right )} e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )} - x - \frac {e^{\left ({\left ({\left (x^{3} + 1\right )} e^{x} - 5 \, x\right )} e^{\left (-x\right )}\right )} \log \relax (3) - 3}{\log \relax (3)}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 31, normalized size = 1.11
method | result | size |
norman | \({\mathrm e}^{\frac {-\ln \relax (3) {\mathrm e}^{\left (\left (x^{3}+1\right ) {\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}+3}{\ln \relax (3)}}\) | \(31\) |
risch | \({\mathrm e}^{-\frac {\ln \relax (3) {\mathrm e}^{\left ({\mathrm e}^{x} x^{3}+{\mathrm e}^{x}-5 x \right ) {\mathrm e}^{-x}}-3}{\ln \relax (3)}}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.93, size = 23, normalized size = 0.82 \begin {gather*} e^{\left (\frac {3}{\log \relax (3)} - e^{\left (x^{3} - 5 \, x e^{\left (-x\right )} + 1\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.40, size = 25, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{\frac {3}{\ln \relax (3)}}\,{\mathrm {e}}^{-{\mathrm {e}}^{x^3}\,\mathrm {e}\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^{-x}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.54, size = 26, normalized size = 0.93 \begin {gather*} e^{\frac {- e^{\left (- 5 x + \left (x^{3} + 1\right ) e^{x}\right ) e^{- x}} \log {\relax (3 )} + 3}{\log {\relax (3 )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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