Optimal. Leaf size=26 \[ \frac {e^{2-e^x-4 x} x^2}{5 \left (e^x+x\right )^2} \]
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Rubi [F] time = 2.65, 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^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, 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^{2-e^x-4 x} x \left (-e^{2 x} x-4 x^2-e^x \left (-2+6 x+x^2\right )\right )}{5 \left (e^x+x\right )^3} \, dx\\ &=\frac {1}{5} \int \frac {e^{2-e^x-4 x} x \left (-e^{2 x} x-4 x^2-e^x \left (-2+6 x+x^2\right )\right )}{\left (e^x+x\right )^3} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 e^{2-e^x-4 x} (-1+x) x^2}{\left (e^x+x\right )^3}-\frac {e^{2-e^x-4 x} x^2}{e^x+x}+\frac {e^{2-e^x-4 x} x \left (2-6 x+x^2\right )}{\left (e^x+x\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx\right )+\frac {1}{5} \int \frac {e^{2-e^x-4 x} x \left (2-6 x+x^2\right )}{\left (e^x+x\right )^2} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} (-1+x) x^2}{\left (e^x+x\right )^3} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx\right )+\frac {1}{5} \int \left (\frac {2 e^{2-e^x-4 x} x}{\left (e^x+x\right )^2}-\frac {6 e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^2}+\frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^2}\right ) \, dx+\frac {2}{5} \int \left (-\frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^3}+\frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^2} \, dx-\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx-\frac {2}{5} \int \frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^3} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^3} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} x}{\left (e^x+x\right )^2} \, dx-\frac {6}{5} \int \frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.47, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{2-e^x-4 x} x^2}{5 \left (e^x+x\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 30, normalized size = 1.15 \begin {gather*} \frac {x^{2} e^{\left (-4 \, x - e^{x} + 2\right )}}{5 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.43, size = 38, normalized size = 1.46 \begin {gather*} \frac {x^{2} e^{2}}{5 \, {\left (x^{2} e^{\left (4 \, x + e^{x}\right )} + 2 \, x e^{\left (5 \, x + e^{x}\right )} + e^{\left (6 \, x + e^{x}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 22, normalized size = 0.85
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{-4 x +2-{\mathrm e}^{x}}}{5 \left ({\mathrm e}^{x}+x \right )^{2}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.87, size = 34, normalized size = 1.31 \begin {gather*} \frac {x^{2} e^{\left (-e^{x} + 2\right )}}{5 \, {\left (x^{2} e^{\left (4 \, x\right )} + 2 \, x e^{\left (5 \, x\right )} + e^{\left (6 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.28, size = 36, normalized size = 1.38 \begin {gather*} \frac {x^2\,{\mathrm {e}}^2\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{5\,\left ({\mathrm {e}}^{6\,x}+2\,x\,{\mathrm {e}}^{5\,x}+x^2\,{\mathrm {e}}^{4\,x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.28, size = 36, normalized size = 1.38 \begin {gather*} \frac {x^{2} e^{2} e^{- e^{x}}}{5 x^{2} e^{4 x} + 10 x e^{5 x} + 5 e^{6 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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