Optimal. Leaf size=25 \[ \log \left (-x+\frac {25 \log ^2(3)}{e^8 \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )}\right ) \]
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Rubi [F] time = 3.97, 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 {50 \log ^2(3)+e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{-25 x \log ^2(3) \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )+e^8 x^2 \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\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 {-50 \log ^2(3)-e^8 x \log \left (-\frac {x}{2}\right ) \log ^3\left (5 \log \left (-\frac {x}{2}\right )\right )}{x \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx\\ &=\int \left (\frac {1}{x}-\frac {2}{x \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )}+\frac {25 \log ^2(3) \log \left (-\frac {x}{2}\right )+2 e^8 x \log \left (5 \log \left (-\frac {x}{2}\right )\right )}{x \log \left (-\frac {x}{2}\right ) \left (-25 \log ^2(3)+e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )}\right ) \, dx\\ &=\log (x)-2 \int \frac {1}{x \log \left (-\frac {x}{2}\right ) \log \left (5 \log \left (-\frac {x}{2}\right )\right )} \, dx+\int \frac {25 \log ^2(3) \log \left (-\frac {x}{2}\right )+2 e^8 x \log \left (5 \log \left (-\frac {x}{2}\right )\right )}{x \log \left (-\frac {x}{2}\right ) \left (-25 \log ^2(3)+e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx\\ &=\log (x)-2 \operatorname {Subst}\left (\int \frac {1}{x \log (5 x)} \, dx,x,\log \left (-\frac {x}{2}\right )\right )+\int \left (-\frac {2 e^8 \log \left (5 \log \left (-\frac {x}{2}\right )\right )}{\log \left (-\frac {x}{2}\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )}+\frac {25 \log ^2(3)}{x \left (-25 \log ^2(3)+e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )}\right ) \, dx\\ &=\log (x)-2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (5 \log \left (-\frac {x}{2}\right )\right )\right )-\left (2 e^8\right ) \int \frac {\log \left (5 \log \left (-\frac {x}{2}\right )\right )}{\log \left (-\frac {x}{2}\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx+\left (25 \log ^2(3)\right ) \int \frac {1}{x \left (-25 \log ^2(3)+e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx\\ &=\log (x)-2 \log \left (\log \left (5 \log \left (-\frac {x}{2}\right )\right )\right )-\left (2 e^8\right ) \int \frac {\log \left (5 \log \left (-\frac {x}{2}\right )\right )}{\log \left (-\frac {x}{2}\right ) \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx+\left (25 \log ^2(3)\right ) \int \frac {1}{x \left (-25 \log ^2(3)+e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.40, size = 38, normalized size = 1.52 \begin {gather*} -2 \log \left (\log \left (5 \log \left (-\frac {x}{2}\right )\right )\right )+\log \left (25 \log ^2(3)-e^8 x \log ^2\left (5 \log \left (-\frac {x}{2}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 40, normalized size = 1.60 \begin {gather*} \log \left (-\frac {1}{2} \, x\right ) + \log \left (\frac {x e^{8} \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{2} - 25 \, \log \relax (3)^{2}}{x}\right ) - 2 \, \log \left (\log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x e^{8} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{3} + 50 \, \log \relax (3)^{2}}{x^{2} e^{8} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )^{3} - 25 \, x \log \relax (3)^{2} \log \left (-\frac {1}{2} \, x\right ) \log \left (5 \, \log \left (-\frac {1}{2} \, x\right )\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 36, normalized size = 1.44
method | result | size |
risch | \(\ln \relax (x )-2 \ln \left (\ln \left (5 \ln \left (-\frac {x}{2}\right )\right )\right )+\ln \left (\ln \left (5 \ln \left (-\frac {x}{2}\right )\right )^{2}-\frac {25 \ln \relax (3)^{2} {\mathrm e}^{-8}}{x}\right )\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 74, normalized size = 2.96 \begin {gather*} \log \relax (x) + \log \left (\frac {{\left (x e^{8} \log \relax (5)^{2} + 2 \, x e^{8} \log \relax (5) \log \left (-\log \relax (2) + \log \left (-x\right )\right ) + x e^{8} \log \left (-\log \relax (2) + \log \left (-x\right )\right )^{2} - 25 \, \log \relax (3)^{2}\right )} e^{\left (-8\right )}}{x}\right ) - 2 \, \log \left (\log \relax (5) + \log \left (-\log \relax (2) + \log \left (-x\right )\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.63, size = 138, normalized size = 5.52 \begin {gather*} 2\,\ln \left (\frac {4\,x\,{\mathrm {e}}^8-25\,{\ln \left (-\frac {x}{2}\right )}^2\,{\ln \relax (3)}^2}{x}\right )+\ln \left (16\,x\,\ln \left (-\frac {x}{2}\right )\,{\mathrm {e}}^{16}-400\,\ln \left (-\frac {x}{2}\right )\,{\mathrm {e}}^8\,{\ln \relax (3)}^2\right )-2\,\ln \left (5625\,\ln \left (5\,\ln \left (-\frac {x}{2}\right )\right )\,{\ln \left (-\frac {x}{2}\right )}^2\,{\ln \relax (3)}^4-900\,x\,\ln \left (5\,\ln \left (-\frac {x}{2}\right )\right )\,{\mathrm {e}}^8\,{\ln \relax (3)}^2\right )-\ln \left (x-25\,{\mathrm {e}}^{-8}\,{\ln \relax (3)}^2\right )+4\,\ln \relax (x)+\ln \left (\frac {25\,{\ln \relax (3)}^2-x\,{\mathrm {e}}^8\,{\left (\ln \relax (5)+\ln \left (\ln \left (-\frac {x}{2}\right )\right )\right )}^2}{x^2\,\ln \left (-\frac {x}{2}\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.51, size = 41, normalized size = 1.64 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\left (5 \log {\left (- \frac {x}{2} \right )} \right )}^{2} - \frac {25 \log {\relax (3 )}^{2}}{x e^{8}} \right )} - 2 \log {\left (\log {\left (5 \log {\left (- \frac {x}{2} \right )} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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