Optimal. Leaf size=23 \[ \frac {\log (x) \log \left (2+\frac {\log (2 x)}{4 e}\right )}{9 x^2} \]
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Rubi [F] time = 1.01, 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 {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{x^3 (72 e+9 \log (2 x))} \, dx\\ &=\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{9 x^3 (8 e+\log (2 x))} \, dx\\ &=\frac {1}{9} \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{x^3 (8 e+\log (2 x))} \, dx\\ &=\frac {1}{9} \int \frac {\frac {\log (x)}{8 e+\log (2 x)}+(-1+2 \log (x)) (1+\log (4)-\log (8 e+\log (2 x)))}{x^3} \, dx\\ &=\frac {1}{9} \int \left (\frac {-8 e (1+\log (4))+(1+16 e (1+\log (4))) \log (x)-(1+\log (4)) \log (2 x)+2 (1+\log (4)) \log (x) \log (2 x)}{x^3 (8 e+\log (2 x))}-\frac {(-1+2 \log (x)) \log (8 e+\log (2 x))}{x^3}\right ) \, dx\\ &=\frac {1}{9} \int \frac {-8 e (1+\log (4))+(1+16 e (1+\log (4))) \log (x)-(1+\log (4)) \log (2 x)+2 (1+\log (4)) \log (x) \log (2 x)}{x^3 (8 e+\log (2 x))} \, dx-\frac {1}{9} \int \frac {(-1+2 \log (x)) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {1}{9} \int \frac {-1-\log (4)+\frac {\log (x) (1+16 e (1+\log (4))+2 (1+\log (4)) \log (2 x))}{8 e+\log (2 x)}}{x^3} \, dx-\frac {1}{9} \int \left (-\frac {\log (8 e+\log (2 x))}{x^3}+\frac {2 \log (x) \log (8 e+\log (2 x))}{x^3}\right ) \, dx\\ &=\frac {1}{9} \int \left (\frac {(1+\log (4)) (-1+2 \log (x))}{x^3}+\frac {\log (x)}{x^3 (8 e+\log (2 x))}\right ) \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {1}{9} \int \frac {\log (x)}{x^3 (8 e+\log (2 x))} \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx+\frac {1}{9} (1+\log (4)) \int \frac {-1+2 \log (x)}{x^3} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}-\frac {1}{9} \int \frac {4 e^{16 e} \text {Ei}(-2 (8 e+\log (2 x)))}{x} \, dx+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx-\frac {1}{9} \left (4 e^{16 e}\right ) \int \frac {\text {Ei}(-2 (8 e+\log (2 x)))}{x} \, dx\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx-\frac {1}{9} \left (4 e^{16 e}\right ) \operatorname {Subst}(\int \text {Ei}(-2 (8 e+x)) \, dx,x,\log (2 x))\\ &=\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx+\frac {1}{9} \left (2 e^{16 e}\right ) \operatorname {Subst}(\int \text {Ei}(x) \, dx,x,-16 e-2 \log (2 x))\\ &=-\frac {1}{18 x^2}+\frac {4}{9} e^{16 e} \text {Ei}(-2 (8 e+\log (2 x))) \log (x)-\frac {(1+\log (4)) \log (x)}{9 x^2}-\frac {4}{9} e^{16 e} \text {Ei}(-16 e-2 \log (2 x)) (8 e+\log (2 x))+\frac {1}{9} \int \frac {\log (8 e+\log (2 x))}{x^3} \, dx-\frac {2}{9} \int \frac {\log (x) \log (8 e+\log (2 x))}{x^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.11, size = 24, normalized size = 1.04 \begin {gather*} \frac {\log (x) \left (-1+\log \left (\frac {1}{4} (8 e+\log (2 x))\right )\right )}{9 x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 21, normalized size = 0.91 \begin {gather*} \frac {\log \left (\frac {1}{4} \, {\left (8 \, e + \log \relax (2) + \log \relax (x)\right )} e^{\left (-1\right )}\right ) \log \relax (x)}{9 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 28, normalized size = 1.22 \begin {gather*} -\frac {2 \, \log \relax (2) \log \relax (x) - \log \relax (x) \log \left (8 \, e + \log \relax (2) + \log \relax (x)\right ) + \log \relax (x)}{9 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 22, normalized size = 0.96
method | result | size |
risch | \(\frac {\ln \relax (x ) \ln \left (\frac {\left (\ln \relax (2)+\ln \relax (x )+8 \,{\mathrm e}\right ) {\mathrm e}^{-1}}{4}\right )}{9 x^{2}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 29, normalized size = 1.26 \begin {gather*} -\frac {{\left (2 \, \log \relax (2) + 1\right )} \log \relax (x) - \log \relax (x) \log \left (8 \, e + \log \relax (2) + \log \relax (x)\right )}{9 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.98, size = 21, normalized size = 0.91 \begin {gather*} \frac {\ln \relax (x)\,\left (\ln \left (\frac {\ln \left (2\,x\right )}{4}+2\,\mathrm {e}\right )-1\right )}{9\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.48, size = 27, normalized size = 1.17 \begin {gather*} \frac {\log {\relax (x )} \log {\left (\frac {\frac {\log {\relax (x )}}{4} + \frac {\log {\relax (2 )}}{4} + 2 e}{e} \right )}}{9 x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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