Optimal. Leaf size=23 \[ \log \left (\log \left (5 \log \left (-\frac {x^2}{5+x}+(-3+x) x \log (3)\right )\right )\right ) \]
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Rubi [F] time = 7.34, 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-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+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 {x^2 (1-17 \log (3))+75 \log (3)-2 x^3 \log (3)+10 x (1-\log (9))}{x \left (75 \log (3)-x^3 \log (3)+5 x (1+\log (3))-x^2 (-1+7 \log (3))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\int \frac {x^2 (1-17 \log (3))+75 \log (3)-2 x^3 \log (3)+10 x (1-\log (9))}{x \left (x^2 (1-7 \log (3))+75 \log (3)-x^3 \log (3)+5 x (1+\log (3))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\int \left (\frac {1}{(-5-x) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}+\frac {1}{x \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}+\frac {1+12 \log ^2(3)-x \left (4 \log ^2(3)+\log (9)-\log ^2(9)\right )-\log (9) (1+\log (729))}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}\right ) \, dx\\ &=\int \frac {1}{(-5-x) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1}{x \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1+12 \log ^2(3)-x \left (4 \log ^2(3)+\log (9)-\log ^2(9)\right )-\log (9) (1+\log (729))}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\int \left (\frac {x \left (-4 \log ^2(3)-\log (9)+\log ^2(9)\right )}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}+\frac {1+12 \log ^2(3)-\log (9) (1+\log (729))}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}\right ) \, dx+\int \frac {1}{(-5-x) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1}{x \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\left (-4 \log ^2(3)-\log (9)+\log ^2(9)\right ) \int \frac {x}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\left (1+12 \log ^2(3)-\log (9) (1+\log (729))\right ) \int \frac {1}{\left (15 \log (3)-x^2 \log (3)+x (1-\log (9))\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1}{(-5-x) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1}{x \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\left (-4 \log ^2(3)-\log (9)+\log ^2(9)\right ) \int \left (\frac {1+\frac {-1+\log (9)}{\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}}}{\left (1-2 x \log (3)-\log (9)-\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}+\frac {1-\frac {-1+\log (9)}{\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}}}{\left (1-2 x \log (3)-\log (9)+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}\right ) \, dx+\left (1+12 \log ^2(3)-\log (9) (1+\log (729))\right ) \int \left (\frac {2 \log (3)}{\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)} \left (1-\log (9)-x \log (9)+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}+\frac {2 \log (3)}{\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)} \left (-1+\log (9)+x \log (9)+\sqrt {1+60 \log ^2(3)-2 \log (9)+\log ^2(9)}\right ) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )}\right ) \, dx+\int \frac {1}{(-5-x) \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx+\int \frac {1}{x \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right ) \log \left (5 \log \left (\frac {x \left (-15 \log (3)+x^2 \log (3)-x (1-\log (9))\right )}{5+x}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [F] time = 4.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-10 x-x^2+\left (-75+20 x+17 x^2+2 x^3\right ) \log (3)}{\left (-5 x^2-x^3+\left (-75 x-5 x^2+7 x^3+x^4\right ) \log (3)\right ) \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right ) \log \left (5 \log \left (\frac {-x^2+\left (-15 x+2 x^2+x^3\right ) \log (3)}{5+x}\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.62, size = 32, normalized size = 1.39 \begin {gather*} \log \left (\log \left (5 \, \log \left (-\frac {x^{2} - {\left (x^{3} + 2 \, x^{2} - 15 \, x\right )} \log \relax (3)}{x + 5}\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 36, normalized size = 1.57 \begin {gather*} \log \left (\log \left (5 \, \log \left (x^{3} \log \relax (3) + 2 \, x^{2} \log \relax (3) - x^{2} - 15 \, x \log \relax (3)\right ) - 5 \, \log \left (x + 5\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x^{3}+17 x^{2}+20 x -75\right ) \ln \relax (3)-x^{2}-10 x}{\left (\left (x^{4}+7 x^{3}-5 x^{2}-75 x \right ) \ln \relax (3)-x^{3}-5 x^{2}\right ) \ln \left (\frac {\left (x^{3}+2 x^{2}-15 x \right ) \ln \relax (3)-x^{2}}{5+x}\right ) \ln \left (5 \ln \left (\frac {\left (x^{3}+2 x^{2}-15 x \right ) \ln \relax (3)-x^{2}}{5+x}\right )\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 34, normalized size = 1.48 \begin {gather*} \log \left (\log \relax (5) + \log \left (\log \left (x^{2} \log \relax (3) + x {\left (2 \, \log \relax (3) - 1\right )} - 15 \, \log \relax (3)\right ) - \log \left (x + 5\right ) + \log \relax (x)\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.54, size = 32, normalized size = 1.39 \begin {gather*} \ln \left (\ln \left (5\,\ln \left (\frac {\ln \relax (3)\,\left (x^3+2\,x^2-15\,x\right )-x^2}{x+5}\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.46, size = 27, normalized size = 1.17 \begin {gather*} \log {\left (\log {\left (5 \log {\left (\frac {- x^{2} + \left (x^{3} + 2 x^{2} - 15 x\right ) \log {\relax (3 )}}{x + 5} \right )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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