Optimal. Leaf size=26 \[ 4 e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2 \]
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Rubi [F] time = 1.33, 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 {1}{5} \exp \left (-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )\right ) \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \exp \left (-x-\frac {2}{5} e^{-x} \left (5 e^{2 x}-4 x+5 e^x x\right )\right ) \left (32 x^2-40 e^{2 x} x^2-32 x^3+e^x \left (40 x-40 x^2\right )\right ) \, dx\\ &=\frac {1}{5} \int 8 e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x \left (-5 e^x (-1+x)-5 e^{2 x} x-4 (-1+x) x\right ) \, dx\\ &=\frac {8}{5} \int e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x \left (-5 e^x (-1+x)-5 e^{2 x} x-4 (-1+x) x\right ) \, dx\\ &=\frac {8}{5} \int \left (-5 e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} (-1+x) x-5 e^{-2 e^x-x+\frac {8 e^{-x} x}{5}} x^2-4 e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} (-1+x) x^2\right ) \, dx\\ &=-\left (\frac {32}{5} \int e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} (-1+x) x^2 \, dx\right )-8 \int e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} (-1+x) x \, dx-8 \int e^{-2 e^x-x+\frac {8 e^{-x} x}{5}} x^2 \, dx\\ &=-\left (\frac {32}{5} \int \left (-e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x^2+e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x^3\right ) \, dx\right )-8 \int e^{-2 e^x-x+\frac {8 e^{-x} x}{5}} x^2 \, dx-8 \int \left (-e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x+e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2\right ) \, dx\\ &=\frac {32}{5} \int e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x^2 \, dx-\frac {32}{5} \int e^{-2 e^x-3 x+\frac {8 e^{-x} x}{5}} x^3 \, dx+8 \int e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x \, dx-8 \int e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2 \, dx-8 \int e^{-2 e^x-x+\frac {8 e^{-x} x}{5}} x^2 \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.35, size = 26, normalized size = 1.00 \begin {gather*} 4 e^{-2 e^x-2 x+\frac {8 e^{-x} x}{5}} x^2 \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.94, size = 29, normalized size = 1.12 \begin {gather*} 4 \, x^{2} e^{\left (-\frac {1}{5} \, {\left (15 \, x e^{x} - 8 \, x + 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} + x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 31, normalized size = 1.19 \begin {gather*} 4 \, x^{2} e^{\left (-\frac {1}{5} \, {\left (5 \, x e^{x} - 8 \, x + 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )} - x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 28, normalized size = 1.08
method | result | size |
risch | \(4 x^{2} {\mathrm e}^{-\frac {2 \left (5 \,{\mathrm e}^{2 x}+5 \,{\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{-x}}{5}}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 21, normalized size = 0.81 \begin {gather*} 4 \, x^{2} e^{\left (\frac {8}{5} \, x e^{\left (-x\right )} - 2 \, x - 2 \, e^{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.30, size = 22, normalized size = 0.85 \begin {gather*} 4\,x^2\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {8\,x\,{\mathrm {e}}^{-x}}{5}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.28, size = 27, normalized size = 1.04 \begin {gather*} 4 x^{2} e^{- 2 \left (x e^{x} - \frac {4 x}{5} + e^{2 x}\right ) e^{- x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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