Optimal. Leaf size=27 \[ \frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(4-x)^2 x \log (x)} \]
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Rubi [F] time = 8.30, 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 {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(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 {-((64-16 x) \log (x))-\left (-64 x+16 x^2\right ) \log (x) \log (2 x)-(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{(4-x)^3 x^2 \log ^2(x) \log (2 x)} \, dx\\ &=\int \left (\frac {16 (-1+x \log (2 x))}{(-4+x)^2 x^2 \log (x) \log (2 x)}-\frac {16 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 x^2 \log ^2(x)}\right ) \, dx\\ &=16 \int \frac {-1+x \log (2 x)}{(-4+x)^2 x^2 \log (x) \log (2 x)} \, dx-16 \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 x^2 \log ^2(x)} \, dx\\ &=16 \int \left (\frac {1}{(-4+x)^2 x \log (x)}-\frac {1}{(-4+x)^2 x^2 \log (x) \log (2 x)}\right ) \, dx-16 \int \left (\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{16 (-4+x)^3 \log ^2(x)}-\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{32 (-4+x)^2 \log ^2(x)}+\frac {3 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{256 (-4+x) \log ^2(x)}-\frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{64 x^2 \log ^2(x)}-\frac {3 (-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{256 x \log ^2(x)}\right ) \, dx\\ &=-\left (\frac {3}{16} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx\right )+\frac {3}{16} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {1}{4} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)} \, dx+\frac {1}{2} \int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-16 \int \frac {1}{(-4+x)^2 x^2 \log (x) \log (2 x)} \, dx-\int \frac {(-4+x-4 \log (x)+3 x \log (x)) \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx\\ &=\frac {3}{16} \int \left (\frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)}+\frac {3 \log \left (\frac {e^x}{\log (2 x)}\right )}{\log (x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log (x)}\right ) \, dx-\frac {3}{16} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)}\right ) \, dx+\frac {1}{4} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)}+\frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log (x)}+\frac {3 \log \left (\frac {e^x}{\log (2 x)}\right )}{x \log (x)}\right ) \, dx+\frac {1}{2} \int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)}\right ) \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-16 \int \left (\frac {1}{16 (-4+x)^2 \log (x) \log (2 x)}-\frac {1}{32 (-4+x) \log (x) \log (2 x)}+\frac {1}{16 x^2 \log (x) \log (2 x)}+\frac {1}{32 x \log (x) \log (2 x)}\right ) \, dx-\int \left (-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)}+\frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)}-\frac {4 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)}+\frac {3 x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)}\right ) \, dx\\ &=\frac {3}{16} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log ^2(x)} \, dx-\frac {3}{16} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx+\frac {1}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {1}{2} \int \frac {1}{(-4+x) \log (x) \log (2 x)} \, dx-\frac {1}{2} \int \frac {1}{x \log (x) \log (2 x)} \, dx+\frac {1}{2} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx+\frac {9}{16} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{\log (x)} \, dx-\frac {9}{16} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)} \, dx+\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log ^2(x)} \, dx-\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)} \, dx+\frac {3}{4} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x) \log (x)} \, dx+\frac {3}{2} \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)} \, dx-2 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log ^2(x)} \, dx-2 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 \log (x)} \, dx-3 \int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)} \, dx+4 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx+4 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log (x)} \, dx+16 \int \frac {1}{(-4+x)^2 x \log (x)} \, dx-\int \frac {1}{(-4+x)^2 \log (x) \log (2 x)} \, dx-\int \frac {1}{x^2 \log (x) \log (2 x)} \, dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)} \, dx-\int \frac {x \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^3 \log ^2(x)} \, dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log (x)} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.55, size = 25, normalized size = 0.93 \begin {gather*} \frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 x \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.90, size = 31, normalized size = 1.15 \begin {gather*} \frac {16 \, \log \left (\frac {e^{x}}{\log \relax (2) + \log \relax (x)}\right )}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.37, size = 50, normalized size = 1.85 \begin {gather*} -\frac {16 \, \log \left (\log \relax (2) + \log \relax (x)\right )}{x^{3} \log \relax (x) - 8 \, x^{2} \log \relax (x) + 16 \, x \log \relax (x)} + \frac {16}{x^{2} \log \relax (x) - 8 \, x \log \relax (x) + 16 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.62, size = 333, normalized size = 12.33
method | result | size |
risch | \(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{x \left (x^{2}-8 x +16\right ) \ln \relax (x )}+\frac {-8 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )+8 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}-8 i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )-8 i \pi \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}+8 i \pi \,\mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}-8 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{3}+8 i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{2}+8 i \pi \mathrm {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \relax (2)+2 i \ln \relax (x )}\right )^{3}+8 i \pi +16 \ln \relax (2)-16 \ln \left (2 i \ln \relax (2)+2 i \ln \relax (x )\right )}{x \left (x -4\right )^{2} \ln \relax (x )}\) | \(333\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 30, normalized size = 1.11 \begin {gather*} \frac {16 \, {\left (x - \log \left (\log \relax (2) + \log \relax (x)\right )\right )}}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.39, size = 68, normalized size = 2.52 \begin {gather*} -\frac {\left (\ln \relax (x)\,\left (\frac {48\,x^3-256\,x^2+256\,x}{x^2\,{\left (x-4\right )}^4}-\frac {48\,x-64}{x\,{\left (x-4\right )}^3}\right )-\frac {16}{x\,{\left (x-4\right )}^2}\right )\,\left (x+\ln \left (\frac {1}{\ln \left (2\,x\right )}\right )\right )}{\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.68, size = 34, normalized size = 1.26 \begin {gather*} \frac {16 \log {\left (\frac {e^{x}}{\log {\relax (x )} + \log {\relax (2 )}} \right )}}{x^{3} \log {\relax (x )} - 8 x^{2} \log {\relax (x )} + 16 x \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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