Optimal. Leaf size=24 \[ \frac {x-x^2+\log (x)-\log (\log (x))}{8+2 x+\log (x)} \]
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Rubi [F] time = 1.79, 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 {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(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 {-2 (4+x)-x (1+2 x) \log ^2(x)+\log (x) \left (7+9 x-15 x^2-2 x^3+(1+2 x) \log (\log (x))\right )}{x \log (x) (8+2 x+\log (x))^2} \, dx\\ &=\int \left (\frac {9}{(8+2 x+\log (x))^2}+\frac {7}{x (8+2 x+\log (x))^2}-\frac {15 x}{(8+2 x+\log (x))^2}-\frac {2 x^2}{(8+2 x+\log (x))^2}-\frac {2 (4+x)}{x \log (x) (8+2 x+\log (x))^2}-\frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2}+\frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx\\ &=-\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \frac {4+x}{x \log (x) (8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2} \, dx+\int \frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=-\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \left (\frac {1}{4 x (4+x) \log (x)}-\frac {1}{2 x (8+2 x+\log (x))^2}-\frac {1}{4 x (4+x) (8+2 x+\log (x))}\right ) \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \left (-\frac {2 \left (4+9 x+2 x^2\right )}{(8+2 x+\log (x))^2}+\frac {1+2 x}{8+2 x+\log (x)}\right ) \, dx+\int \left (\frac {2 \log (\log (x))}{(8+2 x+\log (x))^2}+\frac {\log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \frac {1}{x (4+x) (8+2 x+\log (x))} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {4+9 x+2 x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1+2 x}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \left (\frac {1}{4 x (8+2 x+\log (x))}-\frac {1}{4 (4+x) (8+2 x+\log (x))}\right ) \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \left (\frac {4}{(8+2 x+\log (x))^2}+\frac {9 x}{(8+2 x+\log (x))^2}+\frac {2 x^2}{(8+2 x+\log (x))^2}\right ) \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \left (\frac {1}{8+2 x+\log (x)}+\frac {2 x}{8+2 x+\log (x)}\right ) \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=\frac {1}{8} \int \frac {1}{x (8+2 x+\log (x))} \, dx-\frac {1}{8} \int \frac {1}{(4+x) (8+2 x+\log (x))} \, dx-\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx-2 \int \frac {x}{8+2 x+\log (x)} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+4 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+8 \int \frac {1}{(8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+18 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.44, size = 20, normalized size = 0.83 \begin {gather*} -\frac {8+x+x^2+\log (\log (x))}{8+2 x+\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 20, normalized size = 0.83 \begin {gather*} -\frac {x^{2} + x + \log \left (\log \relax (x)\right ) + 8}{2 \, x + \log \relax (x) + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.35, size = 32, normalized size = 1.33 \begin {gather*} -\frac {x^{2} + x + 8}{2 \, x + \log \relax (x) + 8} - \frac {\log \left (\log \relax (x)\right )}{2 \, x + \log \relax (x) + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 33, normalized size = 1.38
method | result | size |
risch | \(-\frac {\ln \left (\ln \relax (x )\right )}{8+2 x +\ln \relax (x )}-\frac {x^{2}+x +8}{8+2 x +\ln \relax (x )}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 20, normalized size = 0.83 \begin {gather*} -\frac {x^{2} + x + \log \left (\log \relax (x)\right ) + 8}{2 \, x + \log \relax (x) + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {2\,x+{\ln \relax (x)}^2\,\left (2\,x^2+x\right )-\ln \relax (x)\,\left (-2\,x^3-15\,x^2+9\,x+7\right )-\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (2\,x+1\right )+8}{x\,{\ln \relax (x)}^3+\left (4\,x^2+16\,x\right )\,{\ln \relax (x)}^2+\left (4\,x^3+32\,x^2+64\,x\right )\,\ln \relax (x)} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 29, normalized size = 1.21 \begin {gather*} \frac {- x^{2} - x - 8}{2 x + \log {\relax (x )} + 8} - \frac {\log {\left (\log {\relax (x )} \right )}}{2 x + \log {\relax (x )} + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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