Optimal. Leaf size=25 \[ \log \left (4+e^4-\left (e^x+\frac {2 x}{x+\log (3 x)}\right )^2\right ) \]
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Rubi [F] time = 39.59, 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 {-8 x+2 e^{2 x} x^3+e^x \left (-4 x+4 x^3\right )+\left (8 x+6 e^{2 x} x^2+e^x \left (-4+4 x+8 x^2\right )\right ) \log (3 x)+\left (6 e^{2 x} x+e^x (4+4 x)\right ) \log ^2(3 x)+2 e^{2 x} \log ^3(3 x)}{-e^4 x^3+4 e^x x^3+e^{2 x} x^3+\left (-8 x^2-3 e^4 x^2+8 e^x x^2+3 e^{2 x} x^2\right ) \log (3 x)+\left (-12 x-3 e^4 x+4 e^x x+3 e^{2 x} x\right ) \log ^2(3 x)+\left (-4-e^4+e^{2 x}\right ) \log ^3(3 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 {2 \left (\left (2+e^x\right ) x \left (-2+e^x x^2\right )+\left (4 x+3 e^{2 x} x^2+2 e^x \left (-1+x+2 x^2\right )\right ) \log (3 x)+e^x \left (2+\left (2+3 e^x\right ) x\right ) \log ^2(3 x)+e^{2 x} \log ^3(3 x)\right )}{(x+\log (3 x)) \left (\left (-e^4+4 e^x+e^{2 x}\right ) x^2+2 \left (-4-e^4+2 e^x+e^{2 x}\right ) x \log (3 x)+\left (-4-e^4+e^{2 x}\right ) \log ^2(3 x)\right )} \, dx\\ &=2 \int \frac {\left (2+e^x\right ) x \left (-2+e^x x^2\right )+\left (4 x+3 e^{2 x} x^2+2 e^x \left (-1+x+2 x^2\right )\right ) \log (3 x)+e^x \left (2+\left (2+3 e^x\right ) x\right ) \log ^2(3 x)+e^{2 x} \log ^3(3 x)}{(x+\log (3 x)) \left (\left (-e^4+4 e^x+e^{2 x}\right ) x^2+2 \left (-4-e^4+2 e^x+e^{2 x}\right ) x \log (3 x)+\left (-4-e^4+e^{2 x}\right ) \log ^2(3 x)\right )} \, dx\\ &=2 \int \left (1+\frac {4 x+2 e^x x-e^4 x^3+2 e^x x^3+2 e^x \log (3 x)-4 x \log (3 x)-2 e^x x \log (3 x)+4 e^x x^2 \log (3 x)-8 \left (1+\frac {3 e^4}{8}\right ) x^2 \log (3 x)-2 e^x \log ^2(3 x)+2 e^x x \log ^2(3 x)-12 \left (1+\frac {e^4}{4}\right ) x \log ^2(3 x)-4 \left (1+\frac {e^4}{4}\right ) \log ^3(3 x)}{(x+\log (3 x)) \left (e^4 x^2-4 e^x x^2-e^{2 x} x^2-4 e^x x \log (3 x)-2 e^{2 x} x \log (3 x)+8 \left (1+\frac {e^4}{4}\right ) x \log (3 x)-e^{2 x} \log ^2(3 x)+4 \left (1+\frac {e^4}{4}\right ) \log ^2(3 x)\right )}\right ) \, dx\\ &=2 x+2 \int \frac {4 x+2 e^x x-e^4 x^3+2 e^x x^3+2 e^x \log (3 x)-4 x \log (3 x)-2 e^x x \log (3 x)+4 e^x x^2 \log (3 x)-8 \left (1+\frac {3 e^4}{8}\right ) x^2 \log (3 x)-2 e^x \log ^2(3 x)+2 e^x x \log ^2(3 x)-12 \left (1+\frac {e^4}{4}\right ) x \log ^2(3 x)-4 \left (1+\frac {e^4}{4}\right ) \log ^3(3 x)}{(x+\log (3 x)) \left (e^4 x^2-4 e^x x^2-e^{2 x} x^2-4 e^x x \log (3 x)-2 e^{2 x} x \log (3 x)+8 \left (1+\frac {e^4}{4}\right ) x \log (3 x)-e^{2 x} \log ^2(3 x)+4 \left (1+\frac {e^4}{4}\right ) \log ^2(3 x)\right )} \, dx\\ &=2 x+2 \int \frac {x \left (-4+e^4 x^2-2 e^x \left (1+x^2\right )\right )+\left (3 e^4 x^2+4 x (1+2 x)+e^x \left (-2+2 x-4 x^2\right )\right ) \log (3 x)+\left (-2 e^x (-1+x)+12 x+3 e^4 x\right ) \log ^2(3 x)+\left (4+e^4\right ) \log ^3(3 x)}{(x+\log (3 x)) \left (\left (-e^4+4 e^x+e^{2 x}\right ) x^2+2 \left (-4-e^4+2 e^x+e^{2 x}\right ) x \log (3 x)+\left (-4-e^4+e^{2 x}\right ) \log ^2(3 x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [B] time = 0.46, size = 113, normalized size = 4.52 \begin {gather*} 2 \left (-\log (x+\log (3 x))+\frac {1}{2} \log \left (-e^4 x^2+4 e^x x^2+e^{2 x} x^2-8 x \log (3 x)-2 e^4 x \log (3 x)+4 e^x x \log (3 x)+2 e^{2 x} x \log (3 x)-4 \log ^2(3 x)-e^4 \log ^2(3 x)+e^{2 x} \log ^2(3 x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.75, size = 101, normalized size = 4.04 \begin {gather*} -2 \, \log \left (x + \log \left (3 \, x\right )\right ) + \log \left (\frac {x^{2} e^{4} - x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} e^{x} + {\left (e^{4} - e^{\left (2 \, x\right )} + 4\right )} \log \left (3 \, x\right )^{2} + 2 \, {\left (x e^{4} - x e^{\left (2 \, x\right )} - 2 \, x e^{x} + 4 \, x\right )} \log \left (3 \, x\right )}{e^{4} - e^{\left (2 \, x\right )} + 4}\right ) + \log \left (-e^{4} + e^{\left (2 \, x\right )} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 3.93, size = 184, normalized size = 7.36 \begin {gather*} \log \left (x^{2} e^{4} - x^{2} e^{\left (2 \, x\right )} - 4 \, x^{2} e^{x} + 2 \, x e^{4} \log \relax (3) - 2 \, x e^{\left (2 \, x\right )} \log \relax (3) - 4 \, x e^{x} \log \relax (3) + e^{4} \log \relax (3)^{2} - e^{\left (2 \, x\right )} \log \relax (3)^{2} + 2 \, x e^{4} \log \relax (x) - 2 \, x e^{\left (2 \, x\right )} \log \relax (x) - 4 \, x e^{x} \log \relax (x) + 2 \, e^{4} \log \relax (3) \log \relax (x) - 2 \, e^{\left (2 \, x\right )} \log \relax (3) \log \relax (x) + e^{4} \log \relax (x)^{2} - e^{\left (2 \, x\right )} \log \relax (x)^{2} + 8 \, x \log \relax (3) + 4 \, \log \relax (3)^{2} + 8 \, x \log \relax (x) + 8 \, \log \relax (3) \log \relax (x) + 4 \, \log \relax (x)^{2}\right ) - 2 \, \log \left (x + \log \relax (3) + \log \relax (x)\right ) + \log \left (e^{4} - e^{\left (2 \, x\right )} + 4\right ) - \log \left (-e^{4} + e^{\left (2 \, x\right )} - 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 92, normalized size = 3.68
method | result | size |
risch | \(\ln \left ({\mathrm e}^{2 x}-{\mathrm e}^{4}-4\right )-2 \ln \left (x +\ln \left (3 x \right )\right )+\ln \left (\ln \left (3 x \right )^{2}+\frac {2 x \left (-{\mathrm e}^{2 x}+{\mathrm e}^{4}-2 \,{\mathrm e}^{x}+4\right ) \ln \left (3 x \right )}{-{\mathrm e}^{2 x}+{\mathrm e}^{4}+4}+\frac {\left (-{\mathrm e}^{2 x}+{\mathrm e}^{4}-4 \,{\mathrm e}^{x}\right ) x^{2}}{-{\mathrm e}^{2 x}+{\mathrm e}^{4}+4}\right )\) | \(92\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.74, size = 139, normalized size = 5.56 \begin {gather*} \log \left (-\frac {x^{2} e^{4} + e^{4} \log \relax (3)^{2} + {\left (e^{4} + 4\right )} \log \relax (x)^{2} + 2 \, {\left (e^{4} \log \relax (3) + 4 \, \log \relax (3)\right )} x - {\left (x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2} + 2 \, {\left (x + \log \relax (3)\right )} \log \relax (x) + \log \relax (x)^{2}\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{2} + x \log \relax (3) + x \log \relax (x)\right )} e^{x} + 4 \, \log \relax (3)^{2} + 2 \, {\left (x {\left (e^{4} + 4\right )} + e^{4} \log \relax (3) + 4 \, \log \relax (3)\right )} \log \relax (x)}{x^{2} + 2 \, x \log \relax (3) + \log \relax (3)^{2} + 2 \, {\left (x + \log \relax (3)\right )} \log \relax (x) + \log \relax (x)^{2}}\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 = -0.04 \begin {gather*} \int \frac {2\,{\ln \left (3\,x\right )}^3\,{\mathrm {e}}^{2\,x}-8\,x+2\,x^3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (4\,x-4\,x^3\right )+{\ln \left (3\,x\right )}^2\,\left (6\,x\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (4\,x+4\right )\right )+\ln \left (3\,x\right )\,\left (8\,x+6\,x^2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (8\,x^2+4\,x-4\right )\right )}{4\,x^3\,{\mathrm {e}}^x-{\ln \left (3\,x\right )}^2\,\left (12\,x-3\,x\,{\mathrm {e}}^{2\,x}+3\,x\,{\mathrm {e}}^4-4\,x\,{\mathrm {e}}^x\right )+\ln \left (3\,x\right )\,\left (8\,x^2\,{\mathrm {e}}^x+3\,x^2\,{\mathrm {e}}^{2\,x}-3\,x^2\,{\mathrm {e}}^4-8\,x^2\right )-{\ln \left (3\,x\right )}^3\,\left ({\mathrm {e}}^4-{\mathrm {e}}^{2\,x}+4\right )+x^3\,{\mathrm {e}}^{2\,x}-x^3\,{\mathrm {e}}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 2.95, size = 85, normalized size = 3.40 \begin {gather*} \log {\left (\frac {4 x e^{x}}{x + \log {\left (3 x \right )}} + e^{2 x} + \frac {- x^{2} e^{4} - 2 x e^{4} \log {\left (3 x \right )} - 8 x \log {\left (3 x \right )} - e^{4} \log {\left (3 x \right )}^{2} - 4 \log {\left (3 x \right )}^{2}}{x^{2} + 2 x \log {\left (3 x \right )} + \log {\left (3 x \right )}^{2}} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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