Optimal. Leaf size=29 \[ 6 x-\frac {x (2 x-\log (\log (2)))}{5 \log \left (\log \left (1+\frac {5}{x}\right )\right )} \]
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Rubi [F] time = 0.65, 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+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (\frac {5+x}{x}\right )}+\log \left (\log \left (\frac {5+x}{x}\right )\right ) \left (-4 x+\log (\log (2))+30 \log \left (\log \left (\frac {5+x}{x}\right )\right )\right )}{5 \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (\frac {5+x}{x}\right )}+\log \left (\log \left (\frac {5+x}{x}\right )\right ) \left (-4 x+\log (\log (2))+30 \log \left (\log \left (\frac {5+x}{x}\right )\right )\right )}{\log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx\\ &=\frac {1}{5} \int \left (30+\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {-4 x+\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx\\ &=6 x+\frac {1}{5} \int \frac {-4 x+\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+\int \frac {-2 x+\log (\log (2))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx\\ &=6 x+\frac {1}{5} \int \left (-\frac {4 x}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx+\int \left (-\frac {2}{\log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {10+\log (\log (2))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx\\ &=6 x-\frac {4}{5} \int \frac {x}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx-2 \int \frac {1}{\log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+\frac {1}{5} \log (\log (2)) \int \frac {1}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+(10+\log (\log (2))) \int \frac {1}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 5.05, size = 26, normalized size = 0.90 \begin {gather*} \frac {1}{5} x \left (30+\frac {-2 x+\log (\log (2))}{\log \left (\log \left (\frac {5+x}{x}\right )\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 37, normalized size = 1.28 \begin {gather*} -\frac {2 \, x^{2} - x \log \left (\log \relax (2)\right ) - 30 \, x \log \left (\log \left (\frac {x + 5}{x}\right )\right )}{5 \, \log \left (\log \left (\frac {x + 5}{x}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.37, size = 64, normalized size = 2.21 \begin {gather*} 6 \, x - \frac {2 \, x^{2} \log \left (\frac {x + 5}{x}\right ) - x \log \left (\frac {x + 5}{x}\right ) \log \left (\log \relax (2)\right )}{5 \, {\left (\log \left (x + 5\right ) \log \left (\log \left (\frac {x + 5}{x}\right )\right ) - \log \relax (x) \log \left (\log \left (\frac {x + 5}{x}\right )\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (30 x +150\right ) \ln \left (\frac {5+x}{x}\right ) \ln \left (\ln \left (\frac {5+x}{x}\right )\right )^{2}+\left (\left (5+x \right ) \ln \left (\frac {5+x}{x}\right ) \ln \left (\ln \relax (2)\right )+\left (-4 x^{2}-20 x \right ) \ln \left (\frac {5+x}{x}\right )\right ) \ln \left (\ln \left (\frac {5+x}{x}\right )\right )+5 \ln \left (\ln \relax (2)\right )-10 x}{\left (25+5 x \right ) \ln \left (\frac {5+x}{x}\right ) \ln \left (\ln \left (\frac {5+x}{x}\right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 39, normalized size = 1.34 \begin {gather*} -\frac {2 \, x^{2} - 30 \, x \log \left (\log \left (x + 5\right ) - \log \relax (x)\right ) - x \log \left (\log \relax (2)\right )}{5 \, \log \left (\log \left (x + 5\right ) - \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.25, size = 32, normalized size = 1.10 \begin {gather*} \frac {x\,\left (\ln \left (\ln \relax (2)\right )-2\,x+30\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )\right )}{5\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 24, normalized size = 0.83 \begin {gather*} 6 x + \frac {- 2 x^{2} + x \log {\left (\log {\relax (2 )} \right )}}{5 \log {\left (\log {\left (\frac {x + 5}{x} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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