Optimal. Leaf size=26 \[ \frac {\log \left (2 \left (3-\frac {4}{5} \left (-1+e^3 (2-x)\right )+x\right )\right )}{\log (x)} \]
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Rubi [F] time = 0.31, 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 {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\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} &=\int \frac {\frac {\left (5+4 e^3\right ) \log (x)}{19+4 e^3 (-2+x)+5 x}-\frac {\log \left (\frac {2}{5} \left (19+4 e^3 (-2+x)+5 x\right )\right )}{x}}{\log ^2(x)} \, dx\\ &=\int \left (\frac {5+4 e^3}{\left (19-8 e^3+\left (5+4 e^3\right ) x\right ) \log (x)}-\frac {\log \left (\frac {2}{5} \left (19-8 e^3\right )+\frac {2}{5} \left (5+4 e^3\right ) x\right )}{x \log ^2(x)}\right ) \, dx\\ &=\left (5+4 e^3\right ) \int \frac {1}{\left (19-8 e^3+\left (5+4 e^3\right ) x\right ) \log (x)} \, dx-\int \frac {\log \left (\frac {2}{5} \left (19-8 e^3\right )+\frac {2}{5} \left (5+4 e^3\right ) x\right )}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 23, normalized size = 0.88 \begin {gather*} \frac {\log \left (\frac {2}{5} \left (19+4 e^3 (-2+x)+5 x\right )\right )}{\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.88, size = 18, normalized size = 0.69 \begin {gather*} \frac {\log \left (\frac {8}{5} \, {\left (x - 2\right )} e^{3} + 2 \, x + \frac {38}{5}\right )}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 26, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (5) - \log \left (8 \, x e^{3} + 10 \, x - 16 \, e^{3} + 38\right )}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 21, normalized size = 0.81
method | result | size |
risch | \(\frac {\ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right )}{\ln \relax (x )}\) | \(21\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 30, normalized size = 1.15 \begin {gather*} -\frac {\log \relax (5) - \log \relax (2) - \log \left (x {\left (4 \, e^{3} + 5\right )} - 8 \, e^{3} + 19\right )}{\log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.59, size = 20, normalized size = 0.77 \begin {gather*} \frac {\ln \left (2\,x+\frac {{\mathrm {e}}^3\,\left (8\,x-16\right )}{5}+\frac {38}{5}\right )}{\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.45, size = 22, normalized size = 0.85 \begin {gather*} \frac {\log {\left (2 x + \left (\frac {8 x}{5} - \frac {16}{5}\right ) e^{3} + \frac {38}{5} \right )}}{\log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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