Optimal. Leaf size=32 \[ e^{e^{3+\frac {1}{-5+\frac {1}{e^x+x}}}-e^2 \left (1+\frac {x}{2}\right )^2} \]
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Rubi [F] time = 51.66, 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 (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2 \left (1-5 e^x-5 x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\exp \left (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{\left (1-5 e^x-5 x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (1+e^x\right )}{\left (1-5 e^x-5 x\right )^2}-\exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (2+x)\right ) \, dx\\ &=-\left (\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (2+x) \, dx\right )+\int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (1+e^x\right )}{\left (1-5 e^x-5 x\right )^2} \, dx\\ &=-\left (\frac {1}{2} \int \left (2 \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )+\exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x\right ) \, dx\right )+\int \left (-\frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (-6+5 x)}{5 \left (-1+5 e^x+5 x\right )^2}+\frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{5 \left (-1+5 e^x+5 x\right )}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (-6+5 x)}{\left (-1+5 e^x+5 x\right )^2} \, dx\right )+\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{5} \int \left (-\frac {6 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{\left (-1+5 e^x+5 x\right )^2}+\frac {5 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x}{\left (-1+5 e^x+5 x\right )^2}\right ) \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx+\frac {6}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{\left (-1+5 e^x+5 x\right )^2} \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx-\int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x}{\left (-1+5 e^x+5 x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.49, size = 37, normalized size = 1.16 \begin {gather*} e^{\frac {1}{4} e^2 \left (4 e^{\frac {4}{5}+\frac {1}{5-25 e^x-25 x}}-(2+x)^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.75, size = 48, normalized size = 1.50 \begin {gather*} e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {{\left (14 \, x - 3\right )} e^{2} + 14 \, e^{\left (x + 2\right )}}{{\left (5 \, x - 1\right )} e^{2} + 5 \, e^{\left (x + 2\right )}}\right )}\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 -\frac {{\left ({\left (25 \, x^{3} + 40 \, x^{2} - 19 \, x + 2\right )} e^{2} + 25 \, {\left (x + 2\right )} e^{\left (2 \, x + 2\right )} + 10 \, {\left (5 \, x^{2} + 9 \, x - 2\right )} e^{\left (x + 2\right )} - 2 \, {\left (e^{x} + 1\right )} e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )}}{2 \, {\left (25 \, x^{2} + 10 \, {\left (5 \, x - 1\right )} e^{x} - 10 \, x + 25 \, e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 41, normalized size = 1.28
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {14 \,{\mathrm e}^{x}+14 x -3}{5 \,{\mathrm e}^{x}+5 x -1}}-\frac {x^{2} {\mathrm e}^{2}}{4}-{\mathrm e}^{2} x -{\mathrm e}^{2}}\) | \(41\) |
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}{2} \, \int \frac {{\left ({\left (25 \, x^{3} + 40 \, x^{2} - 19 \, x + 2\right )} e^{2} + 25 \, {\left (x + 2\right )} e^{\left (2 \, x + 2\right )} + 10 \, {\left (5 \, x^{2} + 9 \, x - 2\right )} e^{\left (x + 2\right )} - 2 \, {\left (e^{x} + 1\right )} e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )}}{25 \, x^{2} + 10 \, {\left (5 \, x - 1\right )} e^{x} - 10 \, x + 25 \, e^{\left (2 \, x\right )} + 1}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.60, size = 67, normalized size = 2.09 \begin {gather*} {\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^2}{4}}\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {14\,x}{5\,x+5\,{\mathrm {e}}^x-1}}\,{\mathrm {e}}^{\frac {14\,{\mathrm {e}}^x}{5\,x+5\,{\mathrm {e}}^x-1}}\,{\mathrm {e}}^{-\frac {3}{5\,x+5\,{\mathrm {e}}^x-1}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.01, size = 36, normalized size = 1.12 \begin {gather*} e^{\left (- \frac {x^{2}}{4} - x - 1\right ) e^{2} + e^{\frac {14 x + 14 e^{x} - 3}{5 x + 5 e^{x} - 1}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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