Optimal. Leaf size=36 \[ 5-\frac {1}{25} e^{\frac {1}{2} \left (-5+\frac {x}{3}\right )^2} \left (1+\frac {e^x}{e^x+x}\right )^2 \]
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Rubi [F] time = 3.58, 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 {1}{18} \left (225-30 x+x^2\right )} \left (e^{3 x} (60-4 x)+15 x^3-x^4+e^{2 x} \left (36+84 x-8 x^2\right )+e^x \left (18 x+57 x^2-5 x^3\right )\right )}{225 e^{3 x}+675 e^{2 x} x+675 e^x x^2+225 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^{\frac {1}{18} (-15+x)^2} \left (-4 e^{3 x} (-15+x)-(-15+x) x^3-4 e^{2 x} \left (-9-21 x+2 x^2\right )-e^x x \left (-18-57 x+5 x^2\right )\right )}{225 \left (e^x+x\right )^3} \, dx\\ &=\frac {1}{225} \int \frac {e^{\frac {1}{18} (-15+x)^2} \left (-4 e^{3 x} (-15+x)-(-15+x) x^3-4 e^{2 x} \left (-9-21 x+2 x^2\right )-e^x x \left (-18-57 x+5 x^2\right )\right )}{\left (e^x+x\right )^3} \, dx\\ &=\frac {1}{225} \int \left (-4 e^{\frac {1}{18} (-15+x)^2} (-15+x)-\frac {18 e^{\frac {1}{18} (-15+x)^2} (-1+x) x^2}{\left (e^x+x\right )^3}-\frac {e^{\frac {1}{18} (-15+x)^2} x \left (54-69 x+x^2\right )}{\left (e^x+x\right )^2}+\frac {4 e^{\frac {1}{18} (-15+x)^2} \left (9-24 x+x^2\right )}{e^x+x}\right ) \, dx\\ &=-\left (\frac {1}{225} \int \frac {e^{\frac {1}{18} (-15+x)^2} x \left (54-69 x+x^2\right )}{\left (e^x+x\right )^2} \, dx\right )-\frac {4}{225} \int e^{\frac {1}{18} (-15+x)^2} (-15+x) \, dx+\frac {4}{225} \int \frac {e^{\frac {1}{18} (-15+x)^2} \left (9-24 x+x^2\right )}{e^x+x} \, dx-\frac {2}{25} \int \frac {e^{\frac {1}{18} (-15+x)^2} (-1+x) x^2}{\left (e^x+x\right )^3} \, dx\\ &=-\frac {4}{25} e^{\frac {1}{18} (15-x)^2}-\frac {1}{225} \int \left (\frac {54 e^{\frac {1}{18} (-15+x)^2} x}{\left (e^x+x\right )^2}-\frac {69 e^{\frac {1}{18} (-15+x)^2} x^2}{\left (e^x+x\right )^2}+\frac {e^{\frac {1}{18} (-15+x)^2} x^3}{\left (e^x+x\right )^2}\right ) \, dx+\frac {4}{225} \int \left (\frac {9 e^{\frac {1}{18} (-15+x)^2}}{e^x+x}-\frac {24 e^{\frac {1}{18} (-15+x)^2} x}{e^x+x}+\frac {e^{\frac {1}{18} (-15+x)^2} x^2}{e^x+x}\right ) \, dx-\frac {2}{25} \int \left (-\frac {e^{\frac {1}{18} (-15+x)^2} x^2}{\left (e^x+x\right )^3}+\frac {e^{\frac {1}{18} (-15+x)^2} x^3}{\left (e^x+x\right )^3}\right ) \, dx\\ &=-\frac {4}{25} e^{\frac {1}{18} (15-x)^2}-\frac {1}{225} \int \frac {e^{\frac {1}{18} (-15+x)^2} x^3}{\left (e^x+x\right )^2} \, dx+\frac {4}{225} \int \frac {e^{\frac {1}{18} (-15+x)^2} x^2}{e^x+x} \, dx+\frac {2}{25} \int \frac {e^{\frac {1}{18} (-15+x)^2} x^2}{\left (e^x+x\right )^3} \, dx-\frac {2}{25} \int \frac {e^{\frac {1}{18} (-15+x)^2} x^3}{\left (e^x+x\right )^3} \, dx+\frac {4}{25} \int \frac {e^{\frac {1}{18} (-15+x)^2}}{e^x+x} \, dx-\frac {6}{25} \int \frac {e^{\frac {1}{18} (-15+x)^2} x}{\left (e^x+x\right )^2} \, dx+\frac {23}{75} \int \frac {e^{\frac {1}{18} (-15+x)^2} x^2}{\left (e^x+x\right )^2} \, dx-\frac {32}{75} \int \frac {e^{\frac {1}{18} (-15+x)^2} x}{e^x+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.71, size = 31, normalized size = 0.86 \begin {gather*} -\frac {e^{\frac {1}{18} (-15+x)^2} \left (2 e^x+x\right )^2}{25 \left (e^x+x\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 43, normalized size = 1.19 \begin {gather*} -\frac {{\left (x^{2} + 4 \, x e^{x} + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (\frac {1}{18} \, x^{2} - \frac {5}{3} \, x + \frac {25}{2}\right )}}{25 \, {\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 [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 32, normalized size = 0.89
method | result | size |
risch | \(-\frac {\left (x^{2}+4 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{\frac {\left (x -15\right )^{2}}{18}}}{25 \left ({\mathrm e}^{x}+x \right )^{2}}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.60, size = 52, normalized size = 1.44 \begin {gather*} -\frac {{\left (x^{2} e^{\frac {25}{2}} + 4 \, x e^{\left (x + \frac {25}{2}\right )} + 4 \, e^{\left (2 \, x + \frac {25}{2}\right )}\right )} e^{\left (\frac {1}{18} \, x^{2}\right )}}{25 \, {\left (x^{2} e^{\left (\frac {5}{3} \, x\right )} + 2 \, x e^{\left (\frac {8}{3} \, x\right )} + e^{\left (\frac {11}{3} \, x\right )}\right )}} \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}}^{\frac {x^2}{18}-\frac {5\,x}{3}+\frac {25}{2}}\,\left ({\mathrm {e}}^{2\,x}\,\left (-8\,x^2+84\,x+36\right )-{\mathrm {e}}^{3\,x}\,\left (4\,x-60\right )+15\,x^3-x^4+{\mathrm {e}}^x\,\left (-5\,x^3+57\,x^2+18\,x\right )\right )}{225\,{\mathrm {e}}^{3\,x}+675\,x\,{\mathrm {e}}^{2\,x}+675\,x^2\,{\mathrm {e}}^x+225\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.28, size = 51, normalized size = 1.42 \begin {gather*} \frac {\left (- x^{2} - 4 x e^{x} - 4 e^{2 x}\right ) e^{\frac {x^{2}}{18} - \frac {5 x}{3} + \frac {25}{2}}}{25 x^{2} + 50 x e^{x} + 25 e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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