\(\int \frac {-8 x-2 x \log (4)+2 \log (\frac {3}{x})+(4 x+x \log (4)) \log ^2(\frac {3}{x})+(-2+\log ^2(\frac {3}{x})) \log (\frac {1}{6} e^{-4 x-x \log (4)} (-2+\log ^2(\frac {3}{x})))}{(-2 x+x \log ^2(\frac {3}{x})) \log (\frac {1}{6} e^{-4 x-x \log (4)} (-2+\log ^2(\frac {3}{x})))} \, dx\) [762]

Optimal result

Integrand size = 116, antiderivative size = 29 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=\log \left (\frac {x}{\log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}\right ) \] Output:

ln(x/ln(1/6*(ln(3/x)^2-2)/exp(x*(4+2*ln(2)))))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=\log (x)-\log \left (-4 x-x \log (4)+x (4+\log (4))+\log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )\right ) \] Input:

Integrate[(-8*x - 2*x*Log[4] + 2*Log[3/x] + (4*x + x*Log[4])*Log[3/x]^2 + 
(-2 + Log[3/x]^2)*Log[(E^(-4*x - x*Log[4])*(-2 + Log[3/x]^2))/6])/((-2*x + 
 x*Log[3/x]^2)*Log[(E^(-4*x - x*Log[4])*(-2 + Log[3/x]^2))/6]),x]
 

Output:

Log[x] - Log[-4*x - x*Log[4] + x*(4 + Log[4]) + Log[(-2 + Log[3/x]^2)/(6*E 
^(x*(4 + Log[4])))]]
 

Rubi [F]

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.

\(\displaystyle \int \frac {-8 x+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )+2 \log \left (\frac {3}{x}\right )-2 x \log (4)}{\left (x \log ^2\left (\frac {3}{x}\right )-2 x\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )+2 \log \left (\frac {3}{x}\right )+x (-8-2 \log (4))}{\left (x \log ^2\left (\frac {3}{x}\right )-2 x\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx\)

\(\Big \downarrow \) 3041

\(\displaystyle \int \frac {(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )+2 \log \left (\frac {3}{x}\right )+x (-8-2 \log (4))}{x \left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left ((4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )\right )-\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{x (-\log (4))-4 x} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )-2 \log \left (\frac {3}{x}\right )-x (-8-2 \log (4))}{x \left (2-\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {1}{x}+\frac {-4 x \left (1+\frac {\log (2)}{2}\right ) \log ^2\left (\frac {3}{x}\right )-2 \log \left (\frac {3}{x}\right )+8 x \left (1+\frac {\log (2)}{2}\right )}{x \left (2-\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 (2+\log (2)) \int \frac {1}{\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx+2 \int \frac {\log \left (\frac {3}{x}\right )}{x \left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx+2 (2+\log (2)) \int \frac {\log ^2\left (\frac {3}{x}\right )}{\left (\log ^2\left (\frac {3}{x}\right )-2\right ) \log \left (\frac {1}{6} e^{-x (4+\log (4))} \left (\log ^2\left (\frac {3}{x}\right )-2\right )\right )}dx+\log (x)\)

Input:

Int[(-8*x - 2*x*Log[4] + 2*Log[3/x] + (4*x + x*Log[4])*Log[3/x]^2 + (-2 + 
Log[3/x]^2)*Log[(E^(-4*x - x*Log[4])*(-2 + Log[3/x]^2))/6])/((-2*x + x*Log 
[3/x]^2)*Log[(E^(-4*x - x*Log[4])*(-2 + Log[3/x]^2))/6]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 6.92 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31

method	result	size
parallelrisch	\(-\ln \left (\ln \left (\frac {\left (\ln \left (\frac {3}{x}\right )^{2}-2\right ) {\mathrm e}^{-2 x \ln \left (2\right )-4 x}}{6}\right )\right )-\ln \left (\frac {3}{x}\right )\)	\(38\)
risch	\(\ln \left (x \right )-\ln \left (\ln \left (4^{x} {\mathrm e}^{4 x}\right )+\frac {i \left (2 \pi \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x} \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right ) \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x}\right ) \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x} \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right )+\pi \,\operatorname {csgn}\left (i \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right ) \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x} \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right )^{2}-\pi \,\operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x}\right ) \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x} \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right )^{2}+\pi \operatorname {csgn}\left (i 4^{-x} {\mathrm e}^{-4 x} \left (-8-8 \ln \left (3\right ) \ln \left (x \right )+4 \ln \left (x \right )^{2}+4 \ln \left (3\right )^{2}\right )\right )^{3}-2 i \ln \left (3\right )-6 i \ln \left (2\right )+2 i \ln \left (8+8 \ln \left (3\right ) \ln \left (x \right )-4 \ln \left (x \right )^{2}-4 \ln \left (3\right )^{2}\right )-2 \pi \right )}{2}\right )\)	\(317\)

Input:

int(((ln(3/x)^2-2)*ln(1/6*(ln(3/x)^2-2)/exp(2*x*ln(2)+4*x))+(2*x*ln(2)+4*x 
)*ln(3/x)^2+2*ln(3/x)-4*x*ln(2)-8*x)/(x*ln(3/x)^2-2*x)/ln(1/6*(ln(3/x)^2-2 
)/exp(2*x*ln(2)+4*x)),x,method=_RETURNVERBOSE)
 

Output:

-ln(ln(1/6*(ln(3/x)^2-2)/exp(2*x*ln(2)+4*x)))-ln(3/x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=-\log \left (\frac {3}{x}\right ) - \log \left (\log \left (\frac {1}{6} \, e^{\left (-2 \, x \log \left (2\right ) - 4 \, x\right )} \log \left (\frac {3}{x}\right )^{2} - \frac {1}{3} \, e^{\left (-2 \, x \log \left (2\right ) - 4 \, x\right )}\right )\right ) \] Input:

integrate(((log(3/x)^2-2)*log(1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x))+(2*x 
*log(2)+4*x)*log(3/x)^2+2*log(3/x)-4*x*log(2)-8*x)/(x*log(3/x)^2-2*x)/log( 
1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x)),x, algorithm="fricas")
 

Output:

-log(3/x) - log(log(1/6*e^(-2*x*log(2) - 4*x)*log(3/x)^2 - 1/3*e^(-2*x*log 
(2) - 4*x)))
 

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=\log {\left (x \right )} - \log {\left (\log {\left (\left (\frac {\log {\left (\frac {3}{x} \right )}^{2}}{6} - \frac {1}{3}\right ) e^{- 4 x - 2 x \log {\left (2 \right )}} \right )} \right )} \] Input:

integrate(((ln(3/x)**2-2)*ln(1/6*(ln(3/x)**2-2)/exp(2*x*ln(2)+4*x))+(2*x*l 
n(2)+4*x)*ln(3/x)**2+2*ln(3/x)-4*x*ln(2)-8*x)/(x*ln(3/x)**2-2*x)/ln(1/6*(l 
n(3/x)**2-2)/exp(2*x*ln(2)+4*x)),x)
 

Output:

log(x) - log(log((log(3/x)**2/6 - 1/3)*exp(-4*x - 2*x*log(2))))
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=-\log \left (-2 \, x {\left (\log \left (2\right ) + 2\right )} - \log \left (3\right ) - \log \left (2\right ) + \log \left (\log \left (3\right )^{2} - 2 \, \log \left (3\right ) \log \left (x\right ) + \log \left (x\right )^{2} - 2\right )\right ) + \log \left (x\right ) \] Input:

integrate(((log(3/x)^2-2)*log(1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x))+(2*x 
*log(2)+4*x)*log(3/x)^2+2*log(3/x)-4*x*log(2)-8*x)/(x*log(3/x)^2-2*x)/log( 
1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x)),x, algorithm="maxima")
 

Output:

-log(-2*x*(log(2) + 2) - log(3) - log(2) + log(log(3)^2 - 2*log(3)*log(x) 
+ log(x)^2 - 2)) + log(x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=-\log \left (2 \, x \log \left (2\right ) + 4 \, x + \log \left (3\right ) + \log \left (2\right ) - \log \left (\log \left (3\right )^{2} - 2 \, \log \left (3\right ) \log \left (x\right ) + \log \left (x\right )^{2} - 2\right )\right ) + \log \left (x\right ) \] Input:

integrate(((log(3/x)^2-2)*log(1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x))+(2*x 
*log(2)+4*x)*log(3/x)^2+2*log(3/x)-4*x*log(2)-8*x)/(x*log(3/x)^2-2*x)/log( 
1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x)),x, algorithm="giac")
 

Output:

-log(2*x*log(2) + 4*x + log(3) + log(2) - log(log(3)^2 - 2*log(3)*log(x) + 
 log(x)^2 - 2)) + log(x)
 

Mupad [B] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=\ln \left (x\right )-\ln \left (\ln \left (\frac {{\mathrm {e}}^{-4\,x}\,\left (\frac {{\ln \left (\frac {3}{x}\right )}^2}{6}-\frac {1}{3}\right )}{2^{2\,x}}\right )\right ) \] Input:

int(-(2*log(3/x) - 8*x + log(exp(- 4*x - 2*x*log(2))*(log(3/x)^2/6 - 1/3)) 
*(log(3/x)^2 - 2) - 4*x*log(2) + log(3/x)^2*(4*x + 2*x*log(2)))/(log(exp(- 
 4*x - 2*x*log(2))*(log(3/x)^2/6 - 1/3))*(2*x - x*log(3/x)^2)),x)
 

Output:

log(x) - log(log((exp(-4*x)*(log(3/x)^2/6 - 1/3))/2^(2*x)))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {-8 x-2 x \log (4)+2 \log \left (\frac {3}{x}\right )+(4 x+x \log (4)) \log ^2\left (\frac {3}{x}\right )+\left (-2+\log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )}{\left (-2 x+x \log ^2\left (\frac {3}{x}\right )\right ) \log \left (\frac {1}{6} e^{-4 x-x \log (4)} \left (-2+\log ^2\left (\frac {3}{x}\right )\right )\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {3}{x}\right )^{2}-2}{6 e^{4 x} 2^{2 x}}\right )\right )+\mathrm {log}\left (x \right ) \] Input:

int(((log(3/x)^2-2)*log(1/6*(log(3/x)^2-2)/exp(2*x*log(2)+4*x))+(2*x*log(2 
)+4*x)*log(3/x)^2+2*log(3/x)-4*x*log(2)-8*x)/(x*log(3/x)^2-2*x)/log(1/6*(l 
og(3/x)^2-2)/exp(2*x*log(2)+4*x)),x)
 

Output:

 - log(log((log(3/x)**2 - 2)/(6*e**(4*x)*2**(2*x)))) + log(x)