Optimal. Leaf size=18 \[ -\frac {5 \log (3)}{32 (-3+\log (x)) \log (1+x)} \]
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Rubi [F] time = 0.90, 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 {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+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 {5 \log (3) (-3 x+x \log (x)+(1+x) \log (1+x))}{32 x (1+x) (3-\log (x))^2 \log ^2(1+x)} \, dx\\ &=\frac {1}{32} (5 \log (3)) \int \frac {-3 x+x \log (x)+(1+x) \log (1+x)}{x (1+x) (3-\log (x))^2 \log ^2(1+x)} \, dx\\ &=\frac {1}{32} (5 \log (3)) \int \left (\frac {1}{(1+x) (-3+\log (x)) \log ^2(1+x)}+\frac {1}{x (-3+\log (x))^2 \log (1+x)}\right ) \, dx\\ &=\frac {1}{32} (5 \log (3)) \int \frac {1}{(1+x) (-3+\log (x)) \log ^2(1+x)} \, dx+\frac {1}{32} (5 \log (3)) \int \frac {1}{x (-3+\log (x))^2 \log (1+x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 20, normalized size = 1.11 \begin {gather*} \frac {5 \log (3)}{32 (3-\log (x)) \log (1+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 16, normalized size = 0.89 \begin {gather*} -\frac {5 \, \log \relax (3)}{32 \, {\left (\log \relax (x) - 3\right )} \log \left (x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 20, normalized size = 1.11 \begin {gather*} -\frac {5 \, \log \relax (3)}{32 \, {\left (\log \left (x + 1\right ) \log \relax (x) - 3 \, \log \left (x + 1\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 17, normalized size = 0.94
method | result | size |
risch | \(-\frac {5 \ln \relax (3)}{32 \left (\ln \relax (x )-3\right ) \ln \left (x +1\right )}\) | \(17\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 16, normalized size = 0.89 \begin {gather*} -\frac {5 \, \log \relax (3)}{32 \, {\left (\log \relax (x) - 3\right )} \log \left (x + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.46, size = 110, normalized size = 6.11 \begin {gather*} \frac {\frac {5\,\ln \relax (3)}{32\,x}-\frac {5\,\ln \relax (3)\,\ln \relax (x)}{64\,x}}{\ln \relax (x)-3}-\frac {\frac {5\,\ln \relax (3)}{32\,\left (\ln \relax (x)-3\right )}+\frac {5\,\ln \left (x+1\right )\,\ln \relax (3)\,\left (x+1\right )}{32\,x\,{\left (\ln \relax (x)-3\right )}^2}}{\ln \left (x+1\right )}+\frac {5\,\ln \relax (3)}{64\,x}-\frac {\frac {5\,\left (\ln \relax (3)-2\,x\,\ln \relax (3)\right )}{64\,x}-\frac {5\,\ln \relax (3)\,\ln \relax (x)}{64\,x}}{{\ln \relax (x)}^2-6\,\ln \relax (x)+9} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.26, size = 17, normalized size = 0.94 \begin {gather*} - \frac {5 \log {\relax (3 )}}{\left (32 \log {\relax (x )} - 96\right ) \log {\left (x + 1 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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