Optimal. Leaf size=33 \[ \frac {e^{-5+x-3 e^{-x^2} x} \left (4 e^{-x}+x\right )}{(-1+x)^2 x} \]
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Rubi [F] time = 10.43, 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 (e^{-x^2} \left (e^{x^2} (-5+x)-3 x\right )-x-x^2\right ) \left (12 x-12 x^2-24 x^3+24 x^4+e^x \left (3 x^2-3 x^3-6 x^4+6 x^5\right )+e^{x^2} \left (4-12 x+e^x \left (-3 x^2+x^3\right )\right )\right )}{-x^2+3 x^3-3 x^4+x^5} \, 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^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} \left (-12 x+12 x^2+24 x^3-24 x^4-e^x \left (3 x^2-3 x^3-6 x^4+6 x^5\right )-e^{x^2} \left (4-12 x+e^x \left (-3 x^2+x^3\right )\right )\right )}{(1-x)^3 x^2} \, dx\\ &=\int \left (-\frac {12 e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )}}{(-1+x)^3}+\frac {12 e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )}}{(-1+x)^3 x}-\frac {24 e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} x}{(-1+x)^3}+\frac {24 e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} x^2}{(-1+x)^3}+\frac {3 e^{x-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} \left (-1+2 x^2\right )}{(-1+x)^2}+\frac {e^{x^2-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} \left (4-12 x-3 e^x x^2+e^x x^3\right )}{(-1+x)^3 x^2}\right ) \, dx\\ &=3 \int \frac {e^{x-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} \left (-1+2 x^2\right )}{(-1+x)^2} \, dx-12 \int \frac {e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )}}{(-1+x)^3} \, dx+12 \int \frac {e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )}}{(-1+x)^3 x} \, dx-24 \int \frac {e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} x}{(-1+x)^3} \, dx+24 \int \frac {e^{-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} x^2}{(-1+x)^3} \, dx+\int \frac {e^{x^2-e^{-x^2} \left (5 e^{x^2}+3 x+e^{x^2} x^2\right )} \left (4-12 x-3 e^x x^2+e^x x^3\right )}{(-1+x)^3 x^2} \, dx\\ &=3 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2} \left (-1+2 x^2\right )}{(1-x)^2} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3} \, dx+12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3 x} \, dx-24 \int \frac {e^{-5-3 e^{-x^2} x-x^2} x}{(-1+x)^3} \, dx+24 \int \frac {e^{-5-3 e^{-x^2} x-x^2} x^2}{(-1+x)^3} \, dx+\int \frac {e^{-5-3 e^{-x^2} x} \left (-4+12 x+3 e^x x^2-e^x x^3\right )}{(1-x)^3 x^2} \, dx\\ &=3 \int \left (2 e^{-5+x-3 e^{-x^2} x-x^2}+\frac {e^{-5+x-3 e^{-x^2} x-x^2}}{(-1+x)^2}+\frac {4 e^{-5+x-3 e^{-x^2} x-x^2}}{-1+x}\right ) \, dx+12 \int \left (\frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3}-\frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2}+\frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x}-\frac {e^{-5-3 e^{-x^2} x-x^2}}{x}\right ) \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3} \, dx-24 \int \left (\frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3}+\frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2}\right ) \, dx+24 \int \left (\frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^3}+\frac {2 e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2}+\frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x}\right ) \, dx+\int \left (\frac {e^{-5+x-3 e^{-x^2} x} (-3+x)}{(-1+x)^3}-\frac {4 e^{-5-3 e^{-x^2} x} (-1+3 x)}{(-1+x)^3 x^2}\right ) \, dx\\ &=3 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx-4 \int \frac {e^{-5-3 e^{-x^2} x} (-1+3 x)}{(-1+x)^3 x^2} \, dx+6 \int e^{-5+x-3 e^{-x^2} x-x^2} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+12 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{-1+x} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{x} \, dx-24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+48 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+\int \frac {e^{-5+x-3 e^{-x^2} x} (-3+x)}{(-1+x)^3} \, dx\\ &=3 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx-4 \int \left (\frac {2 e^{-5-3 e^{-x^2} x}}{(-1+x)^3}-\frac {e^{-5-3 e^{-x^2} x}}{(-1+x)^2}+\frac {e^{-5-3 e^{-x^2} x}}{x^2}\right ) \, dx+6 \int e^{-5+x-3 e^{-x^2} x-x^2} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+12 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{-1+x} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{x} \, dx-24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+48 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+\int \left (-\frac {2 e^{-5+x-3 e^{-x^2} x}}{(-1+x)^3}+\frac {e^{-5+x-3 e^{-x^2} x}}{(-1+x)^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-5+x-3 e^{-x^2} x}}{(-1+x)^3} \, dx\right )+3 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+4 \int \frac {e^{-5-3 e^{-x^2} x}}{(-1+x)^2} \, dx-4 \int \frac {e^{-5-3 e^{-x^2} x}}{x^2} \, dx+6 \int e^{-5+x-3 e^{-x^2} x-x^2} \, dx-8 \int \frac {e^{-5-3 e^{-x^2} x}}{(-1+x)^3} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+12 \int \frac {e^{-5+x-3 e^{-x^2} x-x^2}}{-1+x} \, dx-12 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{x} \, dx-24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+24 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{-1+x} \, dx+48 \int \frac {e^{-5-3 e^{-x^2} x-x^2}}{(-1+x)^2} \, dx+\int \frac {e^{-5+x-3 e^{-x^2} x}}{(-1+x)^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 30, normalized size = 0.91 \begin {gather*} \frac {e^{-5-3 e^{-x^2} x} \left (4+e^x x\right )}{(-1+x)^2 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 46, normalized size = 1.39 \begin {gather*} \frac {{\left (x e^{x} + 4\right )} e^{\left (x^{2} - {\left ({\left (x^{2} + 5\right )} e^{\left (x^{2}\right )} + 3 \, x\right )} e^{\left (-x^{2}\right )}\right )}}{x^{3} - 2 \, x^{2} + x} \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 (24 \, x^{4} - 24 \, x^{3} - 12 \, x^{2} + {\left ({\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 12 \, x + 4\right )} e^{\left (x^{2}\right )} + 3 \, {\left (2 \, x^{5} - 2 \, x^{4} - x^{3} + x^{2}\right )} e^{x} + 12 \, x\right )} e^{\left (-x^{2} + {\left ({\left (x - 5\right )} e^{\left (x^{2}\right )} - 3 \, x\right )} e^{\left (-x^{2}\right )} - x\right )}}{x^{5} - 3 \, x^{4} + 3 \, x^{3} - x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 59, normalized size = 1.79
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{x} x +4\right ) {\mathrm e}^{{\mathrm e}^{x^{2}} {\mathrm e}^{-x^{2}} x -5 \,{\mathrm e}^{x^{2}} {\mathrm e}^{-x^{2}}-3 \,{\mathrm e}^{-x^{2}} x -x}}{\left (x^{2}-2 x +1\right ) x}\) | \(59\) |
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*} \int \frac {{\left (24 \, x^{4} - 24 \, x^{3} - 12 \, x^{2} + {\left ({\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 12 \, x + 4\right )} e^{\left (x^{2}\right )} + 3 \, {\left (2 \, x^{5} - 2 \, x^{4} - x^{3} + x^{2}\right )} e^{x} + 12 \, x\right )} e^{\left (-x^{2} + {\left ({\left (x - 5\right )} e^{\left (x^{2}\right )} - 3 \, x\right )} e^{\left (-x^{2}\right )} - x\right )}}{x^{5} - 3 \, x^{4} + 3 \, x^{3} - x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{-{\mathrm {e}}^{-x^2}\,\left (3\,x-{\mathrm {e}}^{x^2}\,\left (x-5\right )\right )}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-x^2}\,\left (12\,x+{\mathrm {e}}^x\,\left (6\,x^5-6\,x^4-3\,x^3+3\,x^2\right )-12\,x^2-24\,x^3+24\,x^4-{\mathrm {e}}^{x^2}\,\left (12\,x+{\mathrm {e}}^x\,\left (3\,x^2-x^3\right )-4\right )\right )}{-x^5+3\,x^4-3\,x^3+x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.36, size = 34, normalized size = 1.03 \begin {gather*} \frac {\left (x + 4 e^{- x}\right ) e^{\left (- 3 x + \left (x - 5\right ) e^{x^{2}}\right ) e^{- x^{2}}}}{x^{3} - 2 x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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