Optimal. Leaf size=29 \[ \frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \]
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Rubi [F] time = 5.47, 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 {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{\left (16+4 e^8\right ) x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx\\ &=\int \frac {(-5+x) (x+\log (3)) \left (2-e^4+x-\log (x+\log (3))\right )+\log (x) \left (-2 x^3-10 \left (-2+e^4\right ) \log (3)+x \left (15+e^4 (-10+\log (3))+13 \log (3)\right )+x^2 \left (14+e^4-\log (9)\right )+(-10+x) (x+\log (3)) \log (x+\log (3))\right )}{4 x^3 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {(-5+x) (x+\log (3)) \left (2-e^4+x-\log (x+\log (3))\right )+\log (x) \left (-2 x^3-10 \left (-2+e^4\right ) \log (3)+x \left (15+e^4 (-10+\log (3))+13 \log (3)\right )+x^2 \left (14+e^4-\log (9)\right )+(-10+x) (x+\log (3)) \log (x+\log (3))\right )}{x^3 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {(5-x) (-1+x+\log (3)) \log (x)}{x^2 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {-5+x+10 \log (x)-x \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {(5-x) (-1+x+\log (3)) \log (x)}{x^2 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx+\frac {1}{4} \int \frac {-5+x+10 \log (x)-x \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx\\ &=\frac {1}{4} \int \left (\frac {5 (-1+\log (3)) \log (x)}{x^2 \log (3) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {(-5-\log (3)) \log (x)}{\log ^2(3) (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {\left (5+\log (3)-\log ^2(3)\right ) \log (x)}{x \log ^2(3) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}\right ) \, dx+\frac {1}{4} \int \left (\frac {1}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}+\frac {10 \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}+\frac {5}{x^3 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )}+\frac {\log (x)}{x^2 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {1}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx+\frac {1}{4} \int \frac {\log (x)}{x^2 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )} \, dx+\frac {5}{4} \int \frac {1}{x^3 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )} \, dx+\frac {5}{2} \int \frac {\log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx-\frac {(5 (1-\log (3))) \int \frac {\log (x)}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log (3)}-\frac {(5+\log (3)) \int \frac {\log (x)}{(x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log ^2(3)}+\frac {\left (5+\log (3)-\log ^2(3)\right ) \int \frac {\log (x)}{x \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log ^2(3)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 29, normalized size = 1.00 \begin {gather*} \frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.87, size = 35, normalized size = 1.21 \begin {gather*} \frac {{\left (x - 5\right )} \log \relax (x)}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \relax (3)\right ) + 2 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.84, size = 39, normalized size = 1.34 \begin {gather*} \frac {x \log \relax (x) - 5 \, \log \relax (x)}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \relax (3)\right ) + 2 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 25, normalized size = 0.86
method | result | size |
risch | \(-\frac {\left (x -5\right ) \ln \relax (x )}{4 x^{2} \left ({\mathrm e}^{4}+\ln \left (\ln \relax (3)+x \right )-x -2\right )}\) | \(25\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 32, normalized size = 1.10 \begin {gather*} \frac {{\left (x - 5\right )} \log \relax (x)}{4 \, {\left (x^{3} - x^{2} {\left (e^{4} - 2\right )} - x^{2} \log \left (x + \log \relax (3)\right )\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 = -0.03 \begin {gather*} \int \frac {\ln \left (x+\ln \relax (3)\right )\,\left (5\,x-\ln \relax (3)\,\left (x-5\right )-x^2+\ln \relax (x)\,\left (\ln \relax (3)\,\left (x-10\right )-10\,x+x^2\right )\right )-10\,x+{\mathrm {e}}^4\,\left (5\,x-x^2\right )-\ln \relax (3)\,\left (3\,x+{\mathrm {e}}^4\,\left (x-5\right )-x^2+10\right )-3\,x^2+x^3+\ln \relax (x)\,\left (15\,x-{\mathrm {e}}^4\,\left (10\,x-x^2\right )+\ln \relax (3)\,\left (13\,x+{\mathrm {e}}^4\,\left (x-10\right )-2\,x^2+20\right )+14\,x^2-2\,x^3\right )}{4\,x^4\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^5+16\,x^4\right )+{\ln \left (x+\ln \relax (3)\right )}^2\,\left (4\,x^4+4\,\ln \relax (3)\,x^3\right )-\ln \left (x+\ln \relax (3)\right )\,\left (\ln \relax (3)\,\left (16\,x^3-8\,x^3\,{\mathrm {e}}^4+8\,x^4\right )-8\,x^4\,{\mathrm {e}}^4+16\,x^4+8\,x^5\right )+16\,x^4+16\,x^5+4\,x^6+\ln \relax (3)\,\left (4\,x^3\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^4+16\,x^3\right )+16\,x^3+16\,x^4+4\,x^5\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.44, size = 39, normalized size = 1.34 \begin {gather*} \frac {- x \log {\relax (x )} + 5 \log {\relax (x )}}{- 4 x^{3} + 4 x^{2} \log {\left (x + \log {\relax (3 )} \right )} - 8 x^{2} + 4 x^{2} e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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