Integrand size = 59, antiderivative size = 23 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log (x) \log \left (2+\frac {\log (2 x)}{4 e}\right )}{9 x^2} \] Output:
1/9*ln(x)*ln(1/4*ln(2*x)/exp(1)+2)/x^2
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log (x) \left (-1+\log \left (\frac {1}{4} (8 e+\log (2 x))\right )\right )}{9 x^2} \] Input:
Integrate[(Log[x] + (8*E - 16*E*Log[x] + (1 - 2*Log[x])*Log[2*x])*Log[(8*E + Log[2*x])/(4*E)])/(72*E*x^3 + 9*x^3*Log[2*x]),x]
Output:
(Log[x]*(-1 + Log[(8*E + Log[2*x])/4]))/(9*x^2)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log (x)+(-16 e \log (x)+(1-2 \log (x)) \log (2 x)+8 e) \log \left (\frac {\log (2 x)+8 e}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {\log (x)+(-16 e \log (x)+(1-2 \log (x)) \log (2 x)+8 e) \log \left (\frac {\log (2 x)+8 e}{4 e}\right )}{x^3 (9 \log (2 x)+72 e)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\log (x)+(-16 e \log (x)+(1-2 \log (x)) \log (2 x)+8 e) \log \left (\frac {\log (2 x)+8 e}{4 e}\right )}{9 x^3 (\log (2 x)+8 e)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {\log (x)+(-16 e \log (x)+(1-2 \log (x)) \log (2 x)+8 e) \log \left (\frac {\log (2 x)+8 e}{4 e}\right )}{x^3 (\log (2 x)+8 e)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (\frac {\log (x)}{x^3 (\log (2 x)+8 e)}+\frac {(2 \log (x)-1) (-\log (\log (2 x)+8 e)+\log (4)+1)}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (\int \frac {\log (\log (2 x)+8 e)}{x^3}dx-2 \int \frac {\log (x) \log (\log (2 x)+8 e)}{x^3}dx+4 e^{16 e} \log (x) \operatorname {ExpIntegralEi}(-2 (\log (2 x)+8 e))-4 e^{16 e} (\log (2 x)+8 e) \operatorname {ExpIntegralEi}(-2 (\log (2 x)+8 e))-\frac {1}{2 x^2}-\frac {(1+\log (4)) \log (x)}{x^2}\right )\) |
Input:
Int[(Log[x] + (8*E - 16*E*Log[x] + (1 - 2*Log[x])*Log[2*x])*Log[(8*E + Log [2*x])/(4*E)])/(72*E*x^3 + 9*x^3*Log[2*x]),x]
Output:
$Aborted
Time = 1.94 (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 | \(\frac {-2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}-1}{18 x^{2}}+\frac {8 \,{\mathrm e} \,{\mathrm e}^{2 \ln \left (2\right )+16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right )}{9}+\frac {\ln \left (2\right ) {\mathrm e}^{2 \ln \left (2\right )+16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right )}{9}-\frac {8 \,{\mathrm e}}{9 x^{2}}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{9 x^{2}}-\frac {\ln \left (2\right )}{9 x^{2}}-\frac {\ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{18 x^{2}}-\frac {2 \left (-\frac {1}{4}-\frac {\ln \left (x \right )}{2}\right ) \ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{9 x^{2}}+\frac {1}{18 x^{2}}-\frac {4 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right ) \ln \left (2\right )}{9}-\frac {32 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right ) {\mathrm e}}{9}\) | \(315\) |
parts | \(\frac {-2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}-1}{18 x^{2}}+\frac {8 \,{\mathrm e} \,{\mathrm e}^{2 \ln \left (2\right )+16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right )}{9}+\frac {\ln \left (2\right ) {\mathrm e}^{2 \ln \left (2\right )+16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right )}{9}-\frac {8 \,{\mathrm e}}{9 x^{2}}-\frac {2 \ln \left (2\right ) \ln \left (x \right )}{9 x^{2}}-\frac {\ln \left (2\right )}{9 x^{2}}-\frac {\ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{18 x^{2}}-\frac {2 \left (-\frac {1}{4}-\frac {\ln \left (x \right )}{2}\right ) \ln \left (\ln \left (2\right )+\ln \left (x \right )+8 \,{\mathrm e}\right )}{9 x^{2}}+\frac {1}{18 x^{2}}-\frac {4 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right ) \ln \left (2\right )}{9}-\frac {32 \,{\mathrm e}^{16 \,{\mathrm e}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+2 \ln \left (2\right )+16 \,{\mathrm e}\right ) {\mathrm e}}{9}\) | \(315\) |
Input:
int((((-2*ln(x)+1)*ln(2*x)-16*exp(1)*ln(x)+8*exp(1))*ln(1/4*(ln(2*x)+8*exp (1))/exp(1))+ln(x))/(9*x^3*ln(2*x)+72*x^3*exp(1)),x,method=_RETURNVERBOSE)
Output:
1/9/x^2*ln(x)*ln(1/4*(ln(2)+ln(x)+8*exp(1))*exp(-1))
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log \left (\frac {1}{4} \, {\left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right )} e^{\left (-1\right )}\right ) \log \left (x\right )}{9 \, x^{2}} \] Input:
integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log (2*x)+8*exp(1))/exp(1))+log(x))/(9*x^3*log(2*x)+72*x^3*exp(1)),x, algorith m="fricas")
Output:
1/9*log(1/4*(8*e + log(2) + log(x))*e^(-1))*log(x)/x^2
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {\frac {\log {\left (x \right )}}{4} + \frac {\log {\left (2 \right )}}{4} + 2 e}{e} \right )}}{9 x^{2}} \] Input:
integrate((((-2*ln(x)+1)*ln(2*x)-16*exp(1)*ln(x)+8*exp(1))*ln(1/4*(ln(2*x) +8*exp(1))/exp(1))+ln(x))/(9*x**3*ln(2*x)+72*x**3*exp(1)),x)
Output:
log(x)*log((log(x)/4 + log(2)/4 + 2*E)*exp(-1))/(9*x**2)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=-\frac {{\left (2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) - \log \left (x\right ) \log \left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right )}{9 \, x^{2}} \] Input:
integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log (2*x)+8*exp(1))/exp(1))+log(x))/(9*x^3*log(2*x)+72*x^3*exp(1)),x, algorith m="maxima")
Output:
-1/9*((2*log(2) + 1)*log(x) - log(x)*log(8*e + log(2) + log(x)))/x^2
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=-\frac {2 \, \log \left (2\right ) \log \left (x\right ) - \log \left (x\right ) \log \left (8 \, e + \log \left (2\right ) + \log \left (x\right )\right ) + \log \left (x\right )}{9 \, x^{2}} \] Input:
integrate((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log (2*x)+8*exp(1))/exp(1))+log(x))/(9*x^3*log(2*x)+72*x^3*exp(1)),x, algorith m="giac")
Output:
-1/9*(2*log(2)*log(x) - log(x)*log(8*e + log(2) + log(x)) + log(x))/x^2
Time = 3.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\ln \left (x\right )\,\left (\ln \left (\frac {\ln \left (2\,x\right )}{4}+2\,\mathrm {e}\right )-1\right )}{9\,x^2} \] Input:
int((log(x) - log(exp(-1)*(log(2*x)/4 + 2*exp(1)))*(log(2*x)*(2*log(x) - 1 ) - 8*exp(1) + 16*exp(1)*log(x)))/(9*x^3*log(2*x) + 72*x^3*exp(1)),x)
Output:
(log(x)*(log(log(2*x)/4 + 2*exp(1)) - 1))/(9*x^2)
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x)+(8 e-16 e \log (x)+(1-2 \log (x)) \log (2 x)) \log \left (\frac {8 e+\log (2 x)}{4 e}\right )}{72 e x^3+9 x^3 \log (2 x)} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (2 x \right )+8 e}{4 e}\right ) \mathrm {log}\left (x \right )}{9 x^{2}} \] Input:
int((((-2*log(x)+1)*log(2*x)-16*exp(1)*log(x)+8*exp(1))*log(1/4*(log(2*x)+ 8*exp(1))/exp(1))+log(x))/(9*x^3*log(2*x)+72*x^3*exp(1)),x)
Output:
(log((log(2*x) + 8*e)/(4*e))*log(x))/(9*x**2)