Optimal. Leaf size=30 \[ 5-\frac {6 \left (x+2 \left (-5+x+\log (5)+\frac {x}{-\frac {3}{x}+\log \left (x^2\right )}\right )\right )}{x} \]
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Rubi [F] time = 0.56, 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 {-540+36 x^2+24 x^3+108 \log (5)+(360 x-72 x \log (5)) \log \left (x^2\right )+\left (-60 x^2+12 x^2 \log (5)\right ) \log ^2\left (x^2\right )}{9 x^2-6 x^3 \log \left (x^2\right )+x^4 \log ^2\left (x^2\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 {12 \left (3 x^2+2 x^3+9 (-5+\log (5))-6 x (-5+\log (5)) \log \left (x^2\right )+x^2 (-5+\log (5)) \log ^2\left (x^2\right )\right )}{x^2 \left (3-x \log \left (x^2\right )\right )^2} \, dx\\ &=12 \int \frac {3 x^2+2 x^3+9 (-5+\log (5))-6 x (-5+\log (5)) \log \left (x^2\right )+x^2 (-5+\log (5)) \log ^2\left (x^2\right )}{x^2 \left (3-x \log \left (x^2\right )\right )^2} \, dx\\ &=12 \int \left (\frac {-5+\log (5)}{x^2}+\frac {3+2 x}{\left (-3+x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {12 (5-\log (5))}{x}+12 \int \frac {3+2 x}{\left (-3+x \log \left (x^2\right )\right )^2} \, dx\\ &=\frac {12 (5-\log (5))}{x}+12 \int \left (\frac {3}{\left (-3+x \log \left (x^2\right )\right )^2}+\frac {2 x}{\left (-3+x \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {12 (5-\log (5))}{x}+24 \int \frac {x}{\left (-3+x \log \left (x^2\right )\right )^2} \, dx+36 \int \frac {1}{\left (-3+x \log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 26, normalized size = 0.87 \begin {gather*} 12 \left (\frac {5-\log (5)}{x}-\frac {x}{-3+x \log \left (x^2\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 38, normalized size = 1.27 \begin {gather*} -\frac {12 \, {\left (x^{2} + {\left (x \log \relax (5) - 5 \, x\right )} \log \left (x^{2}\right ) - 3 \, \log \relax (5) + 15\right )}}{x^{2} \log \left (x^{2}\right ) - 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 23, normalized size = 0.77 \begin {gather*} -\frac {12 \, x}{x \log \left (x^{2}\right ) - 3} - \frac {12 \, {\left (\log \relax (5) - 5\right )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 27, normalized size = 0.90
method | result | size |
risch | \(\frac {60}{x}-\frac {12 \ln \relax (5)}{x}-\frac {12 x}{x \ln \left (x^{2}\right )-3}\) | \(27\) |
default | \(-\frac {12 \ln \relax (5)}{x}+\frac {-180+60 x \ln \left (x^{2}\right )-12 x^{2}}{x \left (x \ln \left (x^{2}\right )-3\right )}\) | \(37\) |
norman | \(\frac {\left (60-12 \ln \relax (5)\right ) x \ln \left (x^{2}\right )-12 x^{2}-180+36 \ln \relax (5)}{x \left (x \ln \left (x^{2}\right )-3\right )}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.70, size = 33, normalized size = 1.10 \begin {gather*} -\frac {12 \, {\left (2 \, x {\left (\log \relax (5) - 5\right )} \log \relax (x) + x^{2} - 3 \, \log \relax (5) + 15\right )}}{2 \, x^{2} \log \relax (x) - 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.07, size = 47, normalized size = 1.57 \begin {gather*} -\frac {12\,x^3-x\,\left (36\,\ln \relax (5)-180\right )+x^2\,\ln \left (x^2\right )\,\left (12\,\ln \relax (5)-60\right )}{x^3\,\ln \left (x^2\right )-3\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.12, size = 20, normalized size = 0.67 \begin {gather*} - \frac {12 x}{x \log {\left (x^{2} \right )} - 3} - \frac {-60 + 12 \log {\relax (5 )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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