Optimal. Leaf size=22 \[ \frac {4}{15 \log (3) \left (\frac {9}{x^2}+\log \left (\frac {2}{x}\right )\right )} \]
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Rubi [F] time = 0.38, 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 {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\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 \left (72+4 x^2\right )}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx\\ &=\int \frac {4 x \left (18+x^2\right )}{15 \log (3) \left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx\\ &=\frac {4 \int \frac {x \left (18+x^2\right )}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{15 \log (3)}\\ &=\frac {4 \int \left (\frac {18 x}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2}+\frac {x^3}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2}\right ) \, dx}{15 \log (3)}\\ &=\frac {4 \int \frac {x^3}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{15 \log (3)}+\frac {24 \int \frac {x}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{5 \log (3)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 25, normalized size = 1.14 \begin {gather*} \frac {4 x^2}{15 \log (3) \left (9+x^2 \log \left (\frac {2}{x}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 24, normalized size = 1.09 \begin {gather*} \frac {4 \, x^{2}}{15 \, {\left (x^{2} \log \relax (3) \log \left (\frac {2}{x}\right ) + 9 \, \log \relax (3)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 21, normalized size = 0.95 \begin {gather*} \frac {4}{15 \, {\left (\log \relax (3) \log \left (\frac {2}{x}\right ) + \frac {9 \, \log \relax (3)}{x^{2}}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 24, normalized size = 1.09
method | result | size |
norman | \(\frac {4 x^{2}}{15 \ln \relax (3) \left (x^{2} \ln \left (\frac {2}{x}\right )+9\right )}\) | \(24\) |
risch | \(\frac {4 x^{2}}{15 \ln \relax (3) \left (x^{2} \ln \left (\frac {2}{x}\right )+9\right )}\) | \(24\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.65, size = 29, normalized size = 1.32 \begin {gather*} \frac {4 \, x^{2}}{15 \, {\left (x^{2} \log \relax (3) \log \relax (2) - x^{2} \log \relax (3) \log \relax (x) + 9 \, \log \relax (3)\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.34, size = 23, normalized size = 1.05 \begin {gather*} \frac {4\,x^2}{15\,\ln \relax (3)\,\left (x^2\,\ln \left (\frac {2}{x}\right )+9\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.13, size = 22, normalized size = 1.00 \begin {gather*} \frac {4 x^{2}}{15 x^{2} \log {\relax (3 )} \log {\left (\frac {2}{x} \right )} + 135 \log {\relax (3 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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