Optimal. Leaf size=19 \[ \frac {e^{-\frac {1}{5} e^{-x} x} x}{-2+x} \]
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Rubi [F] time = 0.92, 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 {e^{-x-\frac {e^{-x} x}{5}} \left (-10 e^x+2 x-3 x^2+x^3\right )}{20-20 x+5 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x-\frac {e^{-x} x}{5}} \left (-10 e^x+2 x-3 x^2+x^3\right )}{5 (-2+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-x-\frac {e^{-x} x}{5}} \left (-10 e^x+2 x-3 x^2+x^3\right )}{(-2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {10 e^{-\frac {1}{5} e^{-x} x}}{(-2+x)^2}+\frac {e^{-x-\frac {e^{-x} x}{5}} (-1+x) x}{-2+x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-x-\frac {e^{-x} x}{5}} (-1+x) x}{-2+x} \, dx-2 \int \frac {e^{-\frac {1}{5} e^{-x} x}}{(-2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (e^{-x-\frac {e^{-x} x}{5}}+\frac {2 e^{-x-\frac {e^{-x} x}{5}}}{-2+x}+e^{-x-\frac {e^{-x} x}{5}} x\right ) \, dx-2 \int \frac {e^{-\frac {1}{5} e^{-x} x}}{(-2+x)^2} \, dx\\ &=\frac {1}{5} \int e^{-x-\frac {e^{-x} x}{5}} \, dx+\frac {1}{5} \int e^{-x-\frac {e^{-x} x}{5}} x \, dx+\frac {2}{5} \int \frac {e^{-x-\frac {e^{-x} x}{5}}}{-2+x} \, dx-2 \int \frac {e^{-\frac {1}{5} e^{-x} x}}{(-2+x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 25, normalized size = 1.32 \begin {gather*} \frac {1}{5} e^{-\frac {1}{5} e^{-x} x} \left (5+\frac {10}{-2+x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 23, normalized size = 1.21 \begin {gather*} \frac {x e^{\left (-\frac {1}{5} \, {\left (5 \, x e^{x} + x\right )} e^{\left (-x\right )} + x\right )}}{x - 2} \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 \frac {{\left (x^{3} - 3 \, x^{2} + 2 \, x - 10 \, e^{x}\right )} e^{\left (-\frac {1}{5} \, x e^{\left (-x\right )} - x\right )}}{5 \, {\left (x^{2} - 4 \, x + 4\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 16, normalized size = 0.84
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{-\frac {x \,{\mathrm e}^{-x}}{5}}}{x -2}\) | \(16\) |
norman | \(\frac {x \,{\mathrm e}^{-\frac {x \,{\mathrm e}^{-x}}{5}}}{x -2}\) | \(18\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{5} \, \int \frac {{\left (x^{3} - 3 \, x^{2} + 2 \, x - 10 \, e^{x}\right )} e^{\left (-\frac {1}{5} \, x e^{\left (-x\right )} - x\right )}}{x^{2} - 4 \, x + 4}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.77, size = 21, normalized size = 1.11 \begin {gather*} \frac {x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x-\frac {x\,{\mathrm {e}}^{-x}}{5}}}{x-2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 12, normalized size = 0.63 \begin {gather*} \frac {x e^{- \frac {x e^{- x}}{5}}}{x - 2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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