Optimal. Leaf size=23 \[ x+\log \left (e^{-\frac {e^{-2+x}}{-9 x+x^2}}+x\right ) \]
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Rubi [F] time = 9.43, 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 {e^{3-x} \left (81 x^2+63 x^3-17 x^4+x^5\right )+e^{-\frac {e^{-2+x}}{-9 x+x^2}} \left (e \left (-9+11 x-x^2\right )+e^{3-x} \left (81 x^2-18 x^3+x^4\right )\right )}{e^{3-x-\frac {e^{-2+x}}{-9 x+x^2}} \left (81 x^2-18 x^3+x^4\right )+e^{3-x} \left (81 x^3-18 x^4+x^5\right )} \, 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 (-9+x)^2 x^2+e^{2+\frac {e^{-2+x}}{(-9+x) x}} (-9+x)^2 x^2 (1+x)-e^x \left (9-11 x+x^2\right )}{e^2 (9-x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=\frac {\int \frac {e^2 (-9+x)^2 x^2+e^{2+\frac {e^{-2+x}}{(-9+x) x}} (-9+x)^2 x^2 (1+x)-e^x \left (9-11 x+x^2\right )}{(9-x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{e^2}\\ &=\frac {\int \left (\frac {e^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}}+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}{1+e^{\frac {e^{-2+x}}{(-9+x) x}} x}-\frac {e^x \left (9-11 x+x^2\right )}{(-9+x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx}{e^2}\\ &=-\frac {\int \frac {e^x \left (9-11 x+x^2\right )}{(-9+x)^2 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{e^2}+\int \frac {1+e^{\frac {e^{-2+x}}{(-9+x) x}}+e^{\frac {e^{-2+x}}{(-9+x) x}} x}{1+e^{\frac {e^{-2+x}}{(-9+x) x}} x} \, dx\\ &=-\frac {\int \left (-\frac {e^x}{9 (-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}+\frac {e^x}{9 (-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}+\frac {e^x}{9 x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}-\frac {e^x}{9 x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx}{e^2}+\int \frac {1+e^{-\frac {e^{-2+x}}{(-9+x) x}}+x}{e^{-\frac {e^{-2+x}}{(-9+x) x}}+x} \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \left (\frac {1+x}{x}-\frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}\right ) \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \frac {1+x}{x} \, dx-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\int \left (1+\frac {1}{x}\right ) \, dx-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ &=x+\log (x)+\frac {\int \frac {e^x}{(-9+x)^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{(-9+x) \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\frac {\int \frac {e^x}{x^2 \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}+\frac {\int \frac {e^x}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx}{9 e^2}-\int \frac {1}{x \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.32, size = 50, normalized size = 2.17 \begin {gather*} \frac {e^2 x+\frac {e^x}{9 x-x^2}+e^2 \log \left (1+e^{\frac {e^{-2+x}}{(-9+x) x}} x\right )}{e^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.02, size = 21, normalized size = 0.91 \begin {gather*} x + \log \left (x + e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, 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*} \int \frac {{\left (x^{5} - 17 \, x^{4} + 63 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )} - {\left ({\left (x^{2} - 11 \, x + 9\right )} e - {\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x + 3\right )}\right )} e^{\left (-\frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x}\right )}}{{\left (x^{4} - 18 \, x^{3} + 81 \, x^{2}\right )} e^{\left (-x - \frac {e^{\left (x - 2\right )}}{x^{2} - 9 \, x} + 3\right )} + {\left (x^{5} - 18 \, x^{4} + 81 \, x^{3}\right )} e^{\left (-x + 3\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.44, size = 49, normalized size = 2.13
method | result | size |
risch | \(x -\frac {{\mathrm e}^{x -2}}{\left (x -9\right ) x}+\frac {{\mathrm e}^{x -2}}{x^{2}-9 x}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e}^{x -2}}{\left (x -9\right ) x}}+x \right )\) | \(49\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{3-x}-81 x \,{\mathrm e}^{3-x}\right ) {\mathrm e}^{x -3}}{x \left (x -9\right )}+\ln \left ({\mathrm e}^{-\frac {{\mathrm e} \,{\mathrm e}^{x -3}}{x^{2}-9 x}}+x \right )\) | \(64\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 67, normalized size = 2.91 \begin {gather*} \frac {9 \, x^{2} e^{2} - 81 \, x e^{2} - e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}} + \log \relax (x) + \log \left (\frac {x e^{\left (\frac {e^{x}}{9 \, {\left (x e^{2} - 9 \, e^{2}\right )}}\right )} + e^{\left (\frac {e^{\left (x - 2\right )}}{9 \, x}\right )}}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.82, size = 20, normalized size = 0.87 \begin {gather*} x+\ln \left (x+{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x-2}}{x\,\left (x-9\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.74, size = 22, normalized size = 0.96 \begin {gather*} x + \log {\left (x + e^{- \frac {e e^{x - 3}}{x^{2} - 9 x}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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