Optimal. Leaf size=25 \[ \frac {3 e^x}{-1+x}+\frac {2 \log \left (e^x+4 \log (9)\right )}{x} \]
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Rubi [F] time = 0.51, 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 x} \left (-6 x^2+3 x^3\right )+e^x \left (2 x-4 x^2+2 x^3+\left (-24 x^2+12 x^3\right ) \log (9)\right )+\left (e^x \left (-2+4 x-2 x^2\right )+\left (-8+16 x-8 x^2\right ) \log (9)\right ) \log \left (e^x+4 \log (9)\right )}{e^x \left (x^2-2 x^3+x^4\right )+\left (4 x^2-8 x^3+4 x^4\right ) \log (9)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {e^x x \left (2-2 x \left (2+3 e^x+12 \log (9)\right )+x^2 \left (2+3 e^x+12 \log (9)\right )\right )}{(-1+x)^2 \left (e^x+4 \log (9)\right )}-2 \log \left (e^x+4 \log (9)\right )}{x^2} \, dx\\ &=\int \left (\frac {3 e^x (-2+x)}{(-1+x)^2}-\frac {8 \log (9)}{x \left (e^x+4 \log (9)\right )}+\frac {2 \left (x-\log \left (e^x+4 \log (9)\right )\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {x-\log \left (e^x+4 \log (9)\right )}{x^2} \, dx+3 \int \frac {e^x (-2+x)}{(-1+x)^2} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \int \left (\frac {1}{x}-\frac {\log \left (e^x+4 \log (9)\right )}{x^2}\right ) \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \log (x)-2 \int \frac {\log \left (e^x+4 \log (9)\right )}{x^2} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \log (x)+\frac {2 \log \left (e^x+4 \log (9)\right )}{x}-2 \int \frac {e^x}{x \left (e^x+4 \log (9)\right )} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 25, normalized size = 1.00 \begin {gather*} \frac {3 e^x}{-1+x}+\frac {2 \log \left (e^x+4 \log (9)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 2.03, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{x} + 2 \, {\left (x - 1\right )} \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.76, size = 37, normalized size = 1.48 \begin {gather*} \frac {3 \, x e^{x} + 2 \, x \log \left (e^{x} + 8 \, \log \relax (3)\right ) - 2 \, \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 24, normalized size = 0.96
method | result | size |
risch | \(\frac {2 \ln \left ({\mathrm e}^{x}+8 \ln \relax (3)\right )}{x}+\frac {3 \,{\mathrm e}^{x}}{x -1}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{x} + 2 \, {\left (x - 1\right )} \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.36, size = 23, normalized size = 0.92 \begin {gather*} \frac {3\,{\mathrm {e}}^x}{x-1}+\frac {2\,\ln \left (8\,\ln \relax (3)+{\mathrm {e}}^x\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 20, normalized size = 0.80 \begin {gather*} \frac {3 e^{x}}{x - 1} + \frac {2 \log {\left (e^{x} + 8 \log {\relax (3 )} \right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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