Optimal. Leaf size=24 \[ -2+\frac {e^{-4+x}}{\left (1+\frac {1}{-2+x}\right ) x \log (x)} \]
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Rubi [F] time = 3.46, 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^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{e^3 \left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{\left (x^2-2 x^3+x^4\right ) \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{x^2 \left (1-2 x+x^2\right ) \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2\right )+e^{-1+x} \left (-2+6 x-4 x^2+x^3\right ) \log (x)}{(-1+x)^2 x^2 \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} \left (-2+3 x-x^2-2 \log (x)+6 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right )}{(1-x)^2 x^2 \log ^2(x)} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^{-1+x} (2-x)}{(-1+x) x^2 \log ^2(x)}+\frac {e^{-1+x} \left (-2+6 x-4 x^2+x^3\right )}{(-1+x)^2 x^2 \log (x)}\right ) \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x} (2-x)}{(-1+x) x^2 \log ^2(x)} \, dx}{e^3}+\frac {\int \frac {e^{-1+x} \left (-2+6 x-4 x^2+x^3\right )}{(-1+x)^2 x^2 \log (x)} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^{-1+x}}{(-1+x) \log ^2(x)}-\frac {2 e^{-1+x}}{x^2 \log ^2(x)}-\frac {e^{-1+x}}{x \log ^2(x)}\right ) \, dx}{e^3}+\frac {\int \left (\frac {e^{-1+x}}{(-1+x)^2 \log (x)}-\frac {e^{-1+x}}{(-1+x) \log (x)}-\frac {2 e^{-1+x}}{x^2 \log (x)}+\frac {2 e^{-1+x}}{x \log (x)}\right ) \, dx}{e^3}\\ &=\frac {\int \frac {e^{-1+x}}{(-1+x) \log ^2(x)} \, dx}{e^3}-\frac {\int \frac {e^{-1+x}}{x \log ^2(x)} \, dx}{e^3}+\frac {\int \frac {e^{-1+x}}{(-1+x)^2 \log (x)} \, dx}{e^3}-\frac {\int \frac {e^{-1+x}}{(-1+x) \log (x)} \, dx}{e^3}-\frac {2 \int \frac {e^{-1+x}}{x^2 \log ^2(x)} \, dx}{e^3}-\frac {2 \int \frac {e^{-1+x}}{x^2 \log (x)} \, dx}{e^3}+\frac {2 \int \frac {e^{-1+x}}{x \log (x)} \, dx}{e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.26, size = 21, normalized size = 0.88 \begin {gather*} \frac {e^{-4+x} (-2+x)}{(-1+x) x \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 21, normalized size = 0.88 \begin {gather*} \frac {{\left (x - 2\right )} e^{\left (x - 4\right )}}{{\left (x^{2} - x\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 30, normalized size = 1.25 \begin {gather*} \frac {{\left (x e^{x} - 2 \, e^{x}\right )} e^{\left (-3\right )}}{x^{2} e \log \relax (x) - x e \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 21, normalized size = 0.88
method | result | size |
risch | \(\frac {\left (x -2\right ) {\mathrm e}^{x -4}}{\ln \relax (x ) \left (x -1\right ) x}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\left (x - 2\right )} e^{\left (x - 3\right )}}{{\left (x^{2} e - x e\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.60, size = 20, normalized size = 0.83 \begin {gather*} \frac {{\mathrm {e}}^{x-4}\,\left (x-2\right )}{x\,\ln \relax (x)\,\left (x-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.33, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (x - 2\right ) e^{x - 1}}{x^{2} e^{3} \log {\relax (x )} - x e^{3} \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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