Optimal. Leaf size=26 \[ \left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \]
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Rubi [F] time = 3.76, 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^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{\left (-5 x+x^2\right ) \log \left (e^{-x} x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (-5+6 x-x^2\right ) \log \left (25-10 x+x^2\right )+\left (2 x+e^{e^x+x} \left (5 x-x^2\right ) \log (16)\right ) \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=\int \left (-e^{e^x+x} \log (16) \log \left (\log \left (e^{-x} x\right )\right )+\frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {5 e^{e^x} \log (16)-6 e^{e^x} x \log (16)+e^{e^x} x^2 \log (16)-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \frac {-e^{e^x} \left (5-6 x+x^2\right ) \log (16)+\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )-2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(5-x) x \log \left (e^{-x} x\right )} \, dx\\ &=-\left (\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx\right )+\int \left (\frac {e^{e^x} (-1+x) \log (16)}{x \log \left (e^{-x} x\right )}+\frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=\log (16) \int \frac {e^{e^x} (-1+x)}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {-5 \log \left ((-5+x)^2\right )+6 x \log \left ((-5+x)^2\right )-x^2 \log \left ((-5+x)^2\right )+2 x \log \left (e^{-x} x\right ) \log \left (\log \left (e^{-x} x\right )\right )}{(-5+x) x \log \left (e^{-x} x\right )} \, dx\\ &=\log (16) \int \left (\frac {e^{e^x}}{\log \left (e^{-x} x\right )}-\frac {e^{e^x}}{x \log \left (e^{-x} x\right )}\right ) \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \frac {\frac {\left (5-6 x+x^2\right ) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}-2 \log \left (\log \left (e^{-x} x\right )\right )}{5-x} \, dx\\ &=\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx+\int \left (-\frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}+\frac {2 \log \left (\log \left (e^{-x} x\right )\right )}{-5+x}\right ) \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {(-1+x) \log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \left (\frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )}-\frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=2 \int \frac {\log \left (\log \left (e^{-x} x\right )\right )}{-5+x} \, dx+\log (16) \int \frac {e^{e^x}}{\log \left (e^{-x} x\right )} \, dx-\log (16) \int \frac {e^{e^x}}{x \log \left (e^{-x} x\right )} \, dx-\log (16) \int e^{e^x+x} \log \left (\log \left (e^{-x} x\right )\right ) \, dx-\int \frac {\log \left ((-5+x)^2\right )}{\log \left (e^{-x} x\right )} \, dx+\int \frac {\log \left ((-5+x)^2\right )}{x \log \left (e^{-x} x\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 26, normalized size = 1.00 \begin {gather*} \left (-e^{e^x} \log (16)+\log \left ((-5+x)^2\right )\right ) \log \left (\log \left (e^{-x} x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.84, size = 37, normalized size = 1.42 \begin {gather*} -{\left (4 \, e^{\left (x + e^{x}\right )} \log \relax (2) - e^{x} \log \left (x^{2} - 10 \, x + 25\right )\right )} e^{\left (-x\right )} \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, {\left (x^{2} - 6 \, x + 5\right )} e^{\left (e^{x}\right )} \log \relax (2) - 2 \, {\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \relax (2) - x\right )} \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (x e^{\left (-x\right )}\right )\right ) - {\left (x^{2} - 6 \, x + 5\right )} \log \left (x^{2} - 10 \, x + 25\right )}{{\left (x^{2} - 5 \, x\right )} \log \left (x e^{\left (-x\right )}\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.42, size = 220, normalized size = 8.46
method | result | size |
risch | \(\left (-4 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{x}}+2 \ln \left (x -5\right )\right ) \ln \left (\ln \relax (x )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (x -5\right )^{2}\right ) \left (\mathrm {csgn}\left (i \left (x -5\right )\right )^{2}-2 \,\mathrm {csgn}\left (i \left (x -5\right )^{2}\right ) \mathrm {csgn}\left (i \left (x -5\right )\right )+\mathrm {csgn}\left (i \left (x -5\right )^{2}\right )^{2}\right ) \ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+2 i \ln \relax (x )\right )}{2}\right )}{2}\) | \(220\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.70, size = 23, normalized size = 0.88 \begin {gather*} -2 \, {\left (2 \, e^{\left (e^{x}\right )} \log \relax (2) - \log \left (x - 5\right )\right )} \log \left (-x + \log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.71, size = 32, normalized size = 1.23 \begin {gather*} \ln \left (x^2-10\,x+25\right )\,\ln \left (\ln \relax (x)-x\right )-4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \relax (2)\,\ln \left (\ln \relax (x)-x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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