Integrand size = 59, antiderivative size = 23 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log (x) \log \left (2+\frac {\log (2 x)}{4 e}\right )}{9 x^2} \]
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\[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx \]
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Rubi steps \begin{align*} \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 ) \text {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 ) \text {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{align*}
Time = 0.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log (x) \left (-1+\log \left (\frac {1}{4} (8 e+\log (2 x))\right )\right )}{9 x^2} \]
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Time = 6.63 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {\ln \left (x \right ) \ln \left (\frac {\left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right ) {\mathrm e}^{-1}}{4}\right )}{9 x^{2}}\) | \(22\) |
parallelrisch | \(\frac {\ln \left (x \right ) \ln \left (\frac {\left (\ln \left (2 x \right )+8 \,{\mathrm e}\right ) {\mathrm e}^{-1}}{4}\right )}{9 x^{2}}\) | \(24\) |
default | \(\frac {-2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )-1}{18 x^{2}}+\frac {8 \,{\mathrm e} \,{\mathrm e}^{16 \,{\mathrm e}+2 \ln \left (2\right )} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right )}{9}+\frac {\ln \left (2\right ) {\mathrm e}^{16 \,{\mathrm e}+2 \ln \left (2\right )} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right )}{9}-\frac {8 \,{\mathrm e}}{9 x^{2}}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{9 x^{2}}-\frac {\ln \left (2\right )}{9 x^{2}}-\frac {\ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{18 x^{2}}-\frac {2 \left (-\frac {1}{4}-\frac {\ln \left (x \right )}{2}\right ) \ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{9 x^{2}}+\frac {1}{18 x^{2}}-\frac {32 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right ) {\mathrm e}}{9}-\frac {4 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right ) \ln \left (2\right )}{9}\) | \(315\) |
parts | \(\frac {-2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )-1}{18 x^{2}}+\frac {8 \,{\mathrm e} \,{\mathrm e}^{16 \,{\mathrm e}+2 \ln \left (2\right )} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right )}{9}+\frac {\ln \left (2\right ) {\mathrm e}^{16 \,{\mathrm e}+2 \ln \left (2\right )} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right )}{9}-\frac {8 \,{\mathrm e}}{9 x^{2}}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{9 x^{2}}-\frac {\ln \left (2\right )}{9 x^{2}}-\frac {\ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{18 x^{2}}-\frac {2 \left (-\frac {1}{4}-\frac {\ln \left (x \right )}{2}\right ) \ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{9 x^{2}}+\frac {1}{18 x^{2}}-\frac {32 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right ) {\mathrm e}}{9}-\frac {4 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {Ei}_{1}\left (2 \ln \left (x \right )+16 \,{\mathrm e}+2 \ln \left (2\right )\right ) \ln \left (2\right )}{9}\) | \(315\) |
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log \left (\frac {1}{4} \, {\left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right )} e^{\left (-1\right )}\right ) \log \left (x\right )}{9 \, x^{2}} \]
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {\frac {\log {\left (x \right )}}{4} + \frac {\log {\left (2 \right )}}{4} + 2 e}{e} \right )}}{9 x^{2}} \]
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Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=-\frac {{\left (2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) - \log \left (x\right ) \log \left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right )}{9 \, x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=-\frac {2 \, \log \left (2\right ) \log \left (x\right ) - \log \left (x\right ) \log \left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right ) + \log \left (x\right )}{9 \, x^{2}} \]
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Time = 9.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\ln \left (x\right )\,\left (\ln \left (\frac {\ln \left (2\,x\right )}{4}+2\,\mathrm {e}\right )-1\right )}{9\,x^2} \]
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