Optimal. Leaf size=27 \[ \frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \]
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Rubi [F] time = 14.73, 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^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \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 {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{x (-5+5 x) \log (-1+x) \log ^2(x)} \, dx\\ &=\int \left (\frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{5 (-1+x) \log (-1+x) \log ^2(x)}-\frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{5 x \log (-1+x) \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{(-1+x) \log (-1+x) \log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{x \log (-1+x) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (8 \left (4-5 x+x^2\right ) \log (-1+x) \log (\log (-1+x))-x \log (x) \left (8 (-4+x)+\left (-8+12 x-5 x^2+x^3\right ) \log (-1+x) \log (\log (-1+x))\right )\right )}{(1-x) \log (-1+x) \log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-8 \left (4-5 x+x^2\right ) \log (-1+x) \log (\log (-1+x))+x \log (x) \left (8 (-4+x)+\left (-8+12 x-5 x^2+x^3\right ) \log (-1+x) \log (\log (-1+x))\right )\right )}{x \log (-1+x) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {8 e^{5+\frac {x^2}{16}} (-4+x) x}{(-1+x) \log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {8 e^{5+\frac {x^2}{16}} (-4+x)}{\log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} (-1+x) \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} (-1+x) \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} (-4+x)}{\log (-1+x) \log (x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} (-4+x) x}{(-1+x) \log (-1+x) \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}-\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)}-\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)}\right ) \, dx-\frac {8}{5} \int \left (-\frac {4 e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} x}{\log (-1+x) \log (x)}\right ) \, dx+\frac {8}{5} \int \left (-\frac {3 e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)}-\frac {3 e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} x}{\log (-1+x) \log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)} \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)} \, dx+\frac {1}{5} \int \left (-\frac {8 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}+\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{x \log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}-\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 27, normalized size = 1.00 \begin {gather*} \frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.98, size = 22, normalized size = 0.81 \begin {gather*} \frac {8 \, {\left (x - 4\right )} e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 37, normalized size = 1.37 \begin {gather*} \frac {8 \, {\left (x e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right ) - 4 \, e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )\right )}}{5 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 23, normalized size = 0.85
method | result | size |
risch | \(\frac {8 \ln \left (\ln \left (x -1\right )\right ) {\mathrm e}^{\frac {x^{2}}{16}+5} \left (x -4\right )}{5 \ln \relax (x )}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.52, size = 26, normalized size = 0.96 \begin {gather*} \frac {8 \, {\left (x e^{5} - 4 \, e^{5}\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \relax (x)} \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 {\ln \left (\ln \left (x-1\right )\right )\,\left (\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\left (8\,x^2-40\,x+32\right )+\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \relax (x)\,\left (-x^4+5\,x^3-12\,x^2+8\,x\right )\right )+{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \relax (x)\,\left (32\,x-8\,x^2\right )}{\ln \left (x-1\right )\,{\ln \relax (x)}^2\,\left (5\,x-5\,x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.86, size = 31, normalized size = 1.15 \begin {gather*} \frac {\left (8 x \log {\left (\log {\left (x - 1 \right )} \right )} - 32 \log {\left (\log {\left (x - 1 \right )} \right )}\right ) e^{\frac {x^{2}}{16} + 5}}{5 \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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