Optimal. Leaf size=23 \[ 9 \left (1-4 e^{-\frac {25 e^{3-x}}{x}}+x\right )^2 \]
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Rubi [F] time = 3.74, 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^{-\frac {50 e^{3-x}}{x}} \left (e^{3-x} (7200+7200 x)+e^{\frac {50 e^{3-x}}{x}} \left (18 x^2+18 x^3\right )+e^{\frac {25 e^{3-x}}{x}} \left (-72 x^2+e^{3-x} \left (-1800-3600 x-1800 x^2\right )\right )\right )}{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 {18 e^{-\frac {50 e^{3-x}}{x}-x} \left (e^{\frac {25 e^{3-x}}{x}+x} x^2-100 e^3 (1+x)\right ) \left (-4+e^{\frac {25 e^{3-x}}{x}} (1+x)\right )}{x^2} \, dx\\ &=18 \int \frac {e^{-\frac {50 e^{3-x}}{x}-x} \left (e^{\frac {25 e^{3-x}}{x}+x} x^2-100 e^3 (1+x)\right ) \left (-4+e^{\frac {25 e^{3-x}}{x}} (1+x)\right )}{x^2} \, dx\\ &=18 \int \left (e^{-\frac {25 e^{3-x}}{x}} \left (-4+e^{\frac {25 e^{3-x}}{x}}+e^{\frac {25 e^{3-x}}{x}} x\right )-\frac {100 e^{3-\frac {50 e^{3-x}}{x}-x} (1+x) \left (-4+e^{\frac {25 e^{3-x}}{x}}+e^{\frac {25 e^{3-x}}{x}} x\right )}{x^2}\right ) \, dx\\ &=18 \int e^{-\frac {25 e^{3-x}}{x}} \left (-4+e^{\frac {25 e^{3-x}}{x}}+e^{\frac {25 e^{3-x}}{x}} x\right ) \, dx-1800 \int \frac {e^{3-\frac {50 e^{3-x}}{x}-x} (1+x) \left (-4+e^{\frac {25 e^{3-x}}{x}}+e^{\frac {25 e^{3-x}}{x}} x\right )}{x^2} \, dx\\ &=18 \int \left (1-4 e^{-\frac {25 e^{3-x}}{x}}+x\right ) \, dx-1800 \int \frac {e^{3-\frac {50 e^{3-x}}{x}-x} (1+x) \left (-4+e^{\frac {25 e^{3-x}}{x}} (1+x)\right )}{x^2} \, dx\\ &=18 x+9 x^2-72 \int e^{-\frac {25 e^{3-x}}{x}} \, dx-1800 \int \left (-\frac {4 e^{3-\frac {50 e^{3-x}}{x}-x} (1+x)}{x^2}+\frac {e^{3-\frac {25 e^{3-x}}{x}-x} (1+x)^2}{x^2}\right ) \, dx\\ &=18 x+9 x^2-72 \int e^{-\frac {25 e^{3-x}}{x}} \, dx-1800 \int \frac {e^{3-\frac {25 e^{3-x}}{x}-x} (1+x)^2}{x^2} \, dx+7200 \int \frac {e^{3-\frac {50 e^{3-x}}{x}-x} (1+x)}{x^2} \, dx\\ &=18 x+9 x^2-72 \int e^{-\frac {25 e^{3-x}}{x}} \, dx-1800 \int \left (e^{3-\frac {25 e^{3-x}}{x}-x}+\frac {e^{3-\frac {25 e^{3-x}}{x}-x}}{x^2}+\frac {2 e^{3-\frac {25 e^{3-x}}{x}-x}}{x}\right ) \, dx+7200 \int \left (\frac {e^{3-\frac {50 e^{3-x}}{x}-x}}{x^2}+\frac {e^{3-\frac {50 e^{3-x}}{x}-x}}{x}\right ) \, dx\\ &=18 x+9 x^2-72 \int e^{-\frac {25 e^{3-x}}{x}} \, dx-1800 \int e^{3-\frac {25 e^{3-x}}{x}-x} \, dx-1800 \int \frac {e^{3-\frac {25 e^{3-x}}{x}-x}}{x^2} \, dx-3600 \int \frac {e^{3-\frac {25 e^{3-x}}{x}-x}}{x} \, dx+7200 \int \frac {e^{3-\frac {50 e^{3-x}}{x}-x}}{x^2} \, dx+7200 \int \frac {e^{3-\frac {50 e^{3-x}}{x}-x}}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 1.00, size = 47, normalized size = 2.04 \begin {gather*} 18 \left (8 e^{-\frac {50 e^{3-x}}{x}}+e^{-\frac {25 e^{3-x}}{x}} (-4-4 x)+x+\frac {x^2}{2}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.35, size = 53, normalized size = 2.30 \begin {gather*} 9 \, {\left ({\left (x^{2} + 2 \, x\right )} e^{\left (\frac {50 \, e^{\left (-x + 3\right )}}{x}\right )} - 8 \, {\left (x + 1\right )} e^{\left (\frac {25 \, e^{\left (-x + 3\right )}}{x}\right )} + 16\right )} e^{\left (-\frac {50 \, e^{\left (-x + 3\right )}}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.74, size = 96, normalized size = 4.17 \begin {gather*} 9 \, {\left (x^{2} e^{\left (-x + 3\right )} + 2 \, x e^{\left (-x + 3\right )} - 8 \, x e^{\left (-\frac {x^{2} - 3 \, x + 25 \, e^{\left (-x + 3\right )}}{x}\right )} + 16 \, e^{\left (-\frac {x^{2} - 3 \, x + 50 \, e^{\left (-x + 3\right )}}{x}\right )} - 8 \, e^{\left (-\frac {x^{2} - 3 \, x + 25 \, e^{\left (-x + 3\right )}}{x}\right )}\right )} e^{\left (x - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 42, normalized size = 1.83
method | result | size |
risch | \(9 x^{2}+18 x +\left (-72 x -72\right ) {\mathrm e}^{-\frac {25 \,{\mathrm e}^{3-x}}{x}}+144 \,{\mathrm e}^{-\frac {50 \,{\mathrm e}^{3-x}}{x}}\) | \(42\) |
norman | \(\frac {\left (144 x -72 x \,{\mathrm e}^{\frac {25 \,{\mathrm e}^{3-x}}{x}}+9 x^{3} {\mathrm e}^{\frac {50 \,{\mathrm e}^{3-x}}{x}}-72 \,{\mathrm e}^{\frac {25 \,{\mathrm e}^{3-x}}{x}} x^{2}+18 \,{\mathrm e}^{\frac {50 \,{\mathrm e}^{3-x}}{x}} x^{2}\right ) {\mathrm e}^{-\frac {50 \,{\mathrm e}^{3-x}}{x}}}{x}\) | \(93\) |
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*} 9 \, x^{2} + 18 \, x + 144 \, e^{\left (-\frac {50 \, e^{\left (-x + 3\right )}}{x}\right )} - 18 \, \int \frac {4 \, {\left (25 \, x^{2} e^{3} + x^{2} e^{x} + 50 \, x e^{3} + 25 \, e^{3}\right )} e^{\left (-x - \frac {25 \, e^{\left (-x + 3\right )}}{x}\right )}}{x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.35, size = 42, normalized size = 1.83 \begin {gather*} 18\,x+144\,{\mathrm {e}}^{-\frac {50\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}-{\mathrm {e}}^{-\frac {25\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}}\,\left (72\,x+72\right )+9\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 5.73, size = 36, normalized size = 1.57 \begin {gather*} 9 x^{2} + 18 x + \left (- 72 x - 72\right ) e^{- \frac {25 e^{3 - x}}{x}} + 144 e^{- \frac {50 e^{3 - x}}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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