Optimal. Leaf size=24 \[ \log (x)-\frac {3 \left (5+\frac {x}{x-x^4}\right ) \log (x)}{\log (6)} \]
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Rubi [A] time = 0.38, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 14, number of rules used = 10, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.204, Rules used = {12, 1594, 28, 6742, 266, 44, 261, 43, 2335, 260} \begin {gather*} -\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {18 \log (x)}{\log (6)}+\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 28
Rule 43
Rule 44
Rule 260
Rule 261
Rule 266
Rule 1594
Rule 2335
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x-2 x^4+x^7} \, dx}{\log (6)}\\ &=\frac {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x \left (1-2 x^3+x^6\right )} \, dx}{\log (6)}\\ &=\frac {\int \frac {-18+33 x^3-15 x^6+\left (1-2 x^3+x^6\right ) \log (6)-9 x^3 \log (x)}{x \left (-1+x^3\right )^2} \, dx}{\log (6)}\\ &=\frac {\int \left (-\frac {18}{x \left (-1+x^3\right )^2}+\frac {33 x^2}{\left (-1+x^3\right )^2}-\frac {15 x^5}{\left (-1+x^3\right )^2}+\frac {\log (6)}{x}-\frac {9 x^2 \log (x)}{\left (-1+x^3\right )^2}\right ) \, dx}{\log (6)}\\ &=\log (x)-\frac {9 \int \frac {x^2 \log (x)}{\left (-1+x^3\right )^2} \, dx}{\log (6)}-\frac {15 \int \frac {x^5}{\left (-1+x^3\right )^2} \, dx}{\log (6)}-\frac {18 \int \frac {1}{x \left (-1+x^3\right )^2} \, dx}{\log (6)}+\frac {33 \int \frac {x^2}{\left (-1+x^3\right )^2} \, dx}{\log (6)}\\ &=\frac {11}{\left (1-x^3\right ) \log (6)}+\log (x)-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {3 \int \frac {x^2}{-1+x^3} \, dx}{\log (6)}-\frac {5 \operatorname {Subst}\left (\int \frac {x}{(-1+x)^2} \, dx,x,x^3\right )}{\log (6)}-\frac {6 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^2 x} \, dx,x,x^3\right )}{\log (6)}\\ &=\frac {11}{\left (1-x^3\right ) \log (6)}+\log (x)-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}-\frac {\log \left (1-x^3\right )}{\log (6)}-\frac {5 \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx,x,x^3\right )}{\log (6)}-\frac {6 \operatorname {Subst}\left (\int \left (\frac {1}{1-x}+\frac {1}{(-1+x)^2}+\frac {1}{x}\right ) \, dx,x,x^3\right )}{\log (6)}\\ &=\log (x)-\frac {18 \log (x)}{\log (6)}-\frac {3 x^3 \log (x)}{\left (1-x^3\right ) \log (6)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {\frac {3 \log (x)}{-1+x^3}+(-15+\log (6)) \log (x)}{\log (6)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 31, normalized size = 1.29 \begin {gather*} -\frac {{\left (15 \, x^{3} - {\left (x^{3} - 1\right )} \log \relax (6) - 18\right )} \log \relax (x)}{{\left (x^{3} - 1\right )} \log \relax (6)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {{\left (\log \relax (6) - 15\right )} \log \relax (x) + \frac {3 \, \log \relax (x)}{x^{3} - 1}}{\log \relax (6)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 40, normalized size = 1.67
method | result | size |
norman | \(\frac {\frac {\left (18-\ln \relax (6)\right ) \ln \relax (x )}{\ln \relax (6)}-\frac {\left (-\ln \relax (6)+15\right ) x^{3} \ln \relax (x )}{\ln \relax (6)}}{x^{3}-1}\) | \(40\) |
risch | \(\frac {3 \ln \relax (x )}{\left (\ln \relax (2)+\ln \relax (3)\right ) \left (x^{3}-1\right )}+\frac {\ln \relax (x ) \ln \relax (3)}{\ln \relax (2)+\ln \relax (3)}+\frac {\ln \relax (x ) \ln \relax (2)}{\ln \relax (2)+\ln \relax (3)}-\frac {15 \ln \relax (x )}{\ln \relax (2)+\ln \relax (3)}\) | \(55\) |
default | \(\frac {\ln \relax (6) \ln \relax (x )-\frac {i \sqrt {3}\, \ln \relax (x ) \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right )}{3}+\frac {i \sqrt {3}\, \ln \relax (x ) \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right )}{3}+\frac {\ln \relax (x ) x}{x -1}+\frac {\ln \relax (x ) \left (i \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right ) \sqrt {3}\, x^{2}-i \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right ) \sqrt {3}\, x^{2}+i \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right ) \sqrt {3}\, x -i \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right ) \sqrt {3}\, x +i \sqrt {3}\, \ln \left (\frac {i \sqrt {3}-2 x -1}{-1+i \sqrt {3}}\right )-i \sqrt {3}\, \ln \left (\frac {i \sqrt {3}+2 x +1}{1+i \sqrt {3}}\right )+6 x^{2}+3 x \right )}{3 x^{2}+3 x +3}-18 \ln \relax (x )}{\ln \relax (6)}\) | \(290\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 114, normalized size = 4.75 \begin {gather*} -\frac {{\left (\frac {1}{x^{3} - 1} + \log \left (x^{2} + x + 1\right ) + \log \left (x - 1\right ) - 3 \, \log \relax (x)\right )} \log \relax (6) + {\left (\frac {1}{x^{3} - 1} - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right )\right )} \log \relax (6) - \frac {2 \, \log \relax (6)}{x^{3} - 1} - \frac {9 \, \log \relax (x)}{x^{3} - 1} + 3 \, \log \left (x^{3} - 1\right ) - 3 \, \log \left (x^{3}\right ) - 3 \, \log \left (x^{2} + x + 1\right ) - 3 \, \log \left (x - 1\right ) + 54 \, \log \relax (x)}{3 \, \log \relax (6)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.58, size = 31, normalized size = 1.29 \begin {gather*} -\frac {\ln \relax (x)\,\left (\ln \relax (6)-x^3\,\ln \relax (6)+15\,x^3-18\right )}{\ln \relax (6)\,\left (x^3-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 26, normalized size = 1.08 \begin {gather*} \frac {\left (-15 + \log {\relax (6 )}\right ) \log {\relax (x )}}{\log {\relax (6 )}} + \frac {3 \log {\relax (x )}}{x^{3} \log {\relax (6 )} - \log {\relax (6 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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