Optimal. Leaf size=31 \[ -8+x+\frac {x}{1-\frac {\log ^2\left (\log \left (3 e^3\right )\right )}{(2 x-x \log (3))^2}} \]
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Rubi [A] time = 0.18, antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 13, number of rules used = 6, integrand size = 164, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 1994, 28, 1814, 21, 8} \begin {gather*} \frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+2 x \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 8
Rule 21
Rule 28
Rule 1814
Rule 1994
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^4 (32-64 \log (3))+48 x^4 \log ^2(3)-16 x^4 \log ^3(3)+2 x^4 \log ^4(3)+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {-16 x^4 \log ^3(3)+2 x^4 \log ^4(3)+x^4 \left (32-64 \log (3)+48 \log ^2(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)\right )+x^4 \left (-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 (16-32 \log (3))+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{-8 x^4 \log ^3(3)+x^4 \log ^4(3)+x^4 \left (16-32 \log (3)+24 \log ^2(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 \left (16-32 \log (3)+24 \log ^2(3)\right )+x^4 \left (-8 \log ^3(3)+\log ^4(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 \left (16-32 \log (3)+24 \log ^2(3)-8 \log ^3(3)+\log ^4(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 (2-\log (3))^4-2 x^2 (2-\log (3))^2 \log ^2(3+\log (3))+\log ^4(3+\log (3))} \, dx\\ &=(2-\log (3))^4 \int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{\left (x^2 (2-\log (3))^4-(2-\log (3))^2 \log ^2(3+\log (3))\right )^2} \, dx\\ &=\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+\frac {(2-\log (3))^2 \int \frac {4 x^2 (2-\log (3))^2 \log ^2(3+\log (3))-4 \log ^4(3+\log (3))}{x^2 (2-\log (3))^4-(2-\log (3))^2 \log ^2(3+\log (3))} \, dx}{2 \log ^2(3+\log (3))}\\ &=\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+2 \int 1 \, dx\\ &=2 x+\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 34, normalized size = 1.10 \begin {gather*} x \left (2+\frac {\log ^2(3+\log (3))}{x^2 (-2+\log (3))^2-\log ^2(3+\log (3))}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.78, size = 65, normalized size = 2.10 \begin {gather*} \frac {2 \, x^{3} \log \relax (3)^{2} - 8 \, x^{3} \log \relax (3) + 8 \, x^{3} - x \log \left (\log \relax (3) + 3\right )^{2}}{x^{2} \log \relax (3)^{2} - 4 \, x^{2} \log \relax (3) + 4 \, x^{2} - \log \left (\log \relax (3) + 3\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.22, size = 99, normalized size = 3.19 \begin {gather*} \frac {x \log \left (\log \left (3 \, e^{3}\right )\right )^{2}}{x^{2} \log \relax (3)^{2} - 4 \, x^{2} \log \relax (3) + 4 \, x^{2} - \log \left (\log \left (3 \, e^{3}\right )\right )^{2}} + \frac {2 \, {\left (x \log \relax (3)^{4} - 8 \, x \log \relax (3)^{3} + 24 \, x \log \relax (3)^{2} - 32 \, x \log \relax (3) + 16 \, x\right )}}{\log \relax (3)^{4} - 8 \, \log \relax (3)^{3} + 24 \, \log \relax (3)^{2} - 32 \, \log \relax (3) + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 46, normalized size = 1.48
method | result | size |
risch | \(2 x +\frac {\ln \left (3+\ln \relax (3)\right )^{2} x}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (3+\ln \relax (3)\right )^{2}+4 x^{2}}\) | \(46\) |
norman | \(\frac {\left (2 \ln \relax (3)^{2}-8 \ln \relax (3)+8\right ) x^{3}-\ln \left (3+\ln \relax (3)\right )^{2} x}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (3+\ln \relax (3)\right )^{2}+4 x^{2}}\) | \(61\) |
gosper | \(\frac {x \left (2 x^{2} \ln \relax (3)^{2}-8 x^{2} \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}+8 x^{2}\right )}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}+4 x^{2}}\) | \(68\) |
default | \(2 x +\frac {\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}}{2 \left (2-\ln \relax (3)\right ) \left (-x \ln \relax (3)+\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )+2 x \right )}+\frac {\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}}{2 \left (2-\ln \relax (3)\right ) \left (-x \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )+2 x \right )}\) | \(77\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 41, normalized size = 1.32 \begin {gather*} \frac {x \log \left (\log \left (3 \, e^{3}\right )\right )^{2}}{{\left (\log \relax (3)^{2} - 4 \, \log \relax (3) + 4\right )} x^{2} - \log \left (\log \left (3 \, e^{3}\right )\right )^{2}} + 2 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.80, size = 65, normalized size = 2.10 \begin {gather*} \frac {x\,\left (2\,x^2\,{\ln \relax (3)}^2-{\ln \left (\ln \relax (3)+3\right )}^2-8\,x^2\,\ln \relax (3)+8\,x^2\right )}{x^2\,{\ln \relax (3)}^2-{\ln \left (\ln \relax (3)+3\right )}^2-4\,x^2\,\ln \relax (3)+4\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.71, size = 36, normalized size = 1.16 \begin {gather*} 2 x + \frac {x \log {\left (\log {\relax (3 )} + 3 \right )}^{2}}{x^{2} \left (- 4 \log {\relax (3 )} + \log {\relax (3 )}^{2} + 4\right ) - \log {\left (\log {\relax (3 )} + 3 \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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