Optimal. Leaf size=23 \[ e^{24} \left (5-x-\left (x+\log (3)+\frac {4}{\log (x)}\right )^2\right ) \]
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Rubi [B] time = 0.45, antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 13, number of rules used = 8, integrand size = 136, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6688, 6742, 2302, 30, 2353, 2297, 2298} \begin {gather*} -e^{24} x^2-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-e^{24} x (1+\log (9))-\frac {8 e^{24} \log (3)}{\log (x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2297
Rule 2298
Rule 2302
Rule 2353
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=e^{24} \int \frac {\left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx\\ &=e^{24} \int \frac {32+8 (x+\log (3)) \log (x)-8 x \log ^2(x)-x (1+2 x+\log (9)) \log ^3(x)}{x \log ^3(x)} \, dx\\ &=e^{24} \int \left (-1-2 x-\log (9)+\frac {32}{x \log ^3(x)}+\frac {8 (x+\log (3))}{x \log ^2(x)}-\frac {8}{\log (x)}\right ) \, dx\\ &=-e^{24} x^2-e^{24} x (1+\log (9))+\left (8 e^{24}\right ) \int \frac {x+\log (3)}{x \log ^2(x)} \, dx-\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (32 e^{24}\right ) \int \frac {1}{x \log ^3(x)} \, dx\\ &=-e^{24} x^2-e^{24} x (1+\log (9))-8 e^{24} \text {li}(x)+\left (8 e^{24}\right ) \int \left (\frac {1}{\log ^2(x)}+\frac {\log (3)}{x \log ^2(x)}\right ) \, dx+\left (32 e^{24}\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=-e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-8 e^{24} \text {li}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log ^2(x)} \, dx+\left (8 e^{24} \log (3)\right ) \int \frac {1}{x \log ^2(x)} \, dx\\ &=-e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-8 e^{24} \text {li}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (8 e^{24} \log (3)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)}\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.02, size = 53, normalized size = 2.30 \begin {gather*} -e^{24} x-e^{24} x^2-e^{24} x \log (9)-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 46, normalized size = 2.00 \begin {gather*} -\frac {{\left (2 \, x e^{24} \log \relax (3) + {\left (x^{2} + x\right )} e^{24}\right )} \log \relax (x)^{2} + 8 \, {\left (x e^{24} + e^{24} \log \relax (3)\right )} \log \relax (x) + 16 \, e^{24}}{\log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.31, size = 55, normalized size = 2.39 \begin {gather*} -\frac {x^{2} e^{24} \log \relax (x)^{2} + 2 \, x e^{24} \log \relax (3) \log \relax (x)^{2} + x e^{24} \log \relax (x)^{2} + 8 \, x e^{24} \log \relax (x) + 8 \, e^{24} \log \relax (3) \log \relax (x) + 16 \, e^{24}}{\log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 39, normalized size = 1.70
method | result | size |
default | \({\mathrm e}^{24} \left (-2 x \ln \relax (3)-x^{2}-x -\frac {8 \ln \relax (3)}{\ln \relax (x )}-\frac {8 x}{\ln \relax (x )}-\frac {16}{\ln \relax (x )^{2}}\right )\) | \(39\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.56, size = 35, normalized size = 1.52 \begin {gather*} -\frac {{\left ({\left (x^{2} + x {\left (2 \, \log \relax (3) + 1\right )}\right )} \log \relax (x)^{2} + 8 \, {\left (x + \log \relax (3)\right )} \log \relax (x) + 16\right )} e^{24}}{\log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.66, size = 54, normalized size = 2.35 \begin {gather*} -x^2\,{\mathrm {e}}^{24}-\frac {16\,{\mathrm {e}}^{24}+8\,{\mathrm {e}}^{24}\,\ln \relax (3)\,\ln \relax (x)}{{\ln \relax (x)}^2}-\frac {x\,\left ({\mathrm {e}}^{24}\,\left (2\,\ln \relax (3)+1\right )\,{\ln \relax (x)}^2+8\,{\mathrm {e}}^{24}\,\ln \relax (x)\right )}{{\ln \relax (x)}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.30, size = 51, normalized size = 2.22 \begin {gather*} - x^{2} e^{24} + x \left (- 2 e^{24} \log {\relax (3 )} - e^{24}\right ) + \frac {\left (- 8 x e^{24} - 8 e^{24} \log {\relax (3 )}\right ) \log {\relax (x )} - 16 e^{24}}{\log {\relax (x )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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