Optimal. Leaf size=24 \[ 3 x \log \left (-\log \left (\frac {8 x}{5-x}+\log \left (\frac {x}{e}\right )\right )\right ) \]
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Rubi [A] time = 1.95, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 3, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6741, 6742, 2549} \begin {gather*} 3 x \log \left (-\log \left (-\frac {-9 x-(5-x) \log (x)+5}{5-x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2549
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-75-90 x-3 x^2-\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{(5-x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx\\ &=\int \left (\frac {3 \left (25+30 x+x^2\right )}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+3 \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right )\right ) \, dx\\ &=3 \int \frac {25+30 x+x^2}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+3 \int \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )+3 \int \left (\frac {35}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {200}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}\right ) \, dx-3 \int \frac {25+30 x+x^2}{(-5+x) (5-9 x+(-5+x) \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )-3 \int \left (\frac {35}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {200}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}\right ) \, dx+3 \int \frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+105 \int \frac {1}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+600 \int \frac {1}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 24, normalized size = 1.00 \begin {gather*} 3 x \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 2.82, size = 26, normalized size = 1.08 \begin {gather*} 3 \, x \log \left (-\log \left (\frac {{\left (x - 5\right )} \log \left (x e^{\left (-1\right )}\right ) - 8 \, x}{x - 5}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.03, size = 25, normalized size = 1.04 \begin {gather*} 3 \, x \log \left (-\log \left (x \log \relax (x) - 9 \, x - 5 \, \log \relax (x) + 5\right ) + \log \left (x - 5\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.24, size = 148, normalized size = 6.17
method | result | size |
risch | \(3 x \ln \left (\ln \left (x -5\right )-\ln \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )+\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right )+\mathrm {csgn}\left (\frac {i}{x -5}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right )+\mathrm {csgn}\left (i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )\right )\right )}{2}\right )\) | \(148\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 23, normalized size = 0.96 \begin {gather*} 3 \, x \log \left (-\log \left ({\left (x - 5\right )} \log \relax (x) - 9 \, x + 5\right ) + \log \left (x - 5\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.52, size = 28, normalized size = 1.17 \begin {gather*} 3\,x\,\ln \left (-\ln \left (-\frac {8\,x-\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (x-5\right )}{x-5}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 7.45, size = 51, normalized size = 2.12 \begin {gather*} \left (3 x - \frac {5}{2}\right ) \log {\left (- \log {\left (\frac {- 8 x + \left (x - 5\right ) \log {\left (\frac {x}{e} \right )}}{x - 5} \right )} \right )} + \frac {5 \log {\left (\log {\left (\frac {- 8 x + \left (x - 5\right ) \log {\left (\frac {x}{e} \right )}}{x - 5} \right )} \right )}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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