3.6.92 \(\int \frac {-4+4 \log (x)+(-e^x+e^x \log (x)) \log (3 x)+(-6 e^x x+e^x \log (x)+(-6 e^x x^2+e^x x \log (x)) \log (3 x)) \log (\frac {-6 x+\log (x)}{x})}{-96 x^2+16 x \log (x)+(-48 e^x x^2+8 e^x x \log (x)) \log (3 x)+(-6 e^{2 x} x^2+e^{2 x} x \log (x)) \log ^2(3 x)} \, dx\) [592]

3.6.92.1 Optimal result

Integrand size = 132, antiderivative size = 25 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=-20+\frac {\log \left (-6+\frac {\log (x)}{x}\right )}{-4-e^x \log (3 x)} \]

output

ln(ln(x)/x-6)/(-4-ln(3*x)*exp(x))-20
 

3.6.92.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=-\frac {\log \left (-6+\frac {\log (x)}{x}\right )}{4+e^x \log (3 x)} \]

input

Integrate[(-4 + 4*Log[x] + (-E^x + E^x*Log[x])*Log[3*x] + (-6*E^x*x + E^x* 
Log[x] + (-6*E^x*x^2 + E^x*x*Log[x])*Log[3*x])*Log[(-6*x + Log[x])/x])/(-9 
6*x^2 + 16*x*Log[x] + (-48*E^x*x^2 + 8*E^x*x*Log[x])*Log[3*x] + (-6*E^(2*x 
)*x^2 + E^(2*x)*x*Log[x])*Log[3*x]^2),x]
 

output

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

3.6.92.3 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.

\(\displaystyle \int \frac {\left (\left (e^x x \log (x)-6 e^x x^2\right ) \log (3 x)-6 e^x x+e^x \log (x)\right ) \log \left (\frac {\log (x)-6 x}{x}\right )+4 \log (x)+\left (e^x \log (x)-e^x\right ) \log (3 x)-4}{-96 x^2+\left (e^{2 x} x \log (x)-6 e^{2 x} x^2\right ) \log ^2(3 x)+\left (8 e^x x \log (x)-48 e^x x^2\right ) \log (3 x)+16 x \log (x)} \, dx\)

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[(-4 + 4*Log[x] + (-E^x + E^x*Log[x])*Log[3*x] + (-6*E^x*x + E^x*Log[x] 
 + (-6*E^x*x^2 + E^x*x*Log[x])*Log[3*x])*Log[(-6*x + Log[x])/x])/(-96*x^2 
+ 16*x*Log[x] + (-48*E^x*x^2 + 8*E^x*x*Log[x])*Log[3*x] + (-6*E^(2*x)*x^2 
+ E^(2*x)*x*Log[x])*Log[3*x]^2),x]
 

output

$Aborted
 

3.6.92.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.6.92.4 Maple [A] (verified)

Time = 32.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(-\frac {\ln \left (\frac {\ln \left (x \right )-6 x}{x}\right )}{\ln \left (3 x \right ) {\mathrm e}^{x}+4}\)	\(25\)
risch	\(-\frac {2 \ln \left (-\frac {\ln \left (x \right )}{6}+x \right )}{8+2 \ln \left (3\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (x \right )}-\frac {-2 i \pi \operatorname {csgn}\left (\frac {i \left (\frac {\ln \left (x \right )}{6}-x \right )}{x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (\frac {\ln \left (x \right )}{6}-x \right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {\ln \left (x \right )}{6}-x \right )}{x}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (\frac {\ln \left (x \right )}{6}-x \right )\right ) \operatorname {csgn}\left (\frac {i \left (\frac {\ln \left (x \right )}{6}-x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-i \pi \operatorname {csgn}\left (\frac {i \left (\frac {\ln \left (x \right )}{6}-x \right )}{x}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i \left (\frac {\ln \left (x \right )}{6}-x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i \pi +2 \ln \left (3\right )+2 \ln \left (2\right )-2 \ln \left (x \right )}{8+2 \ln \left (3\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (x \right )}\)	\(203\)

input

int((((x*exp(x)*ln(x)-6*exp(x)*x^2)*ln(3*x)+exp(x)*ln(x)-6*exp(x)*x)*ln((l 
n(x)-6*x)/x)+(exp(x)*ln(x)-exp(x))*ln(3*x)+4*ln(x)-4)/((x*exp(x)^2*ln(x)-6 
*exp(x)^2*x^2)*ln(3*x)^2+(8*x*exp(x)*ln(x)-48*exp(x)*x^2)*ln(3*x)+16*x*ln( 
x)-96*x^2),x,method=_RETURNVERBOSE)
 

output

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

3.6.92.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=-\frac {\log \left (-\frac {6 \, x - \log \left (x\right )}{x}\right )}{e^{x} \log \left (3\right ) + e^{x} \log \left (x\right ) + 4} \]

input

integrate((((x*exp(x)*log(x)-6*exp(x)*x^2)*log(3*x)+exp(x)*log(x)-6*exp(x) 
*x)*log((log(x)-6*x)/x)+(exp(x)*log(x)-exp(x))*log(3*x)+4*log(x)-4)/((x*ex 
p(x)^2*log(x)-6*exp(x)^2*x^2)*log(3*x)^2+(8*x*exp(x)*log(x)-48*exp(x)*x^2) 
*log(3*x)+16*x*log(x)-96*x^2),x, algorithm=\
 

output

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

3.6.92.6 Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=- \frac {\log {\left (\frac {- 6 x + \log {\left (x \right )}}{x} \right )}}{\left (\log {\left (x \right )} + \log {\left (3 \right )}\right ) e^{x} + 4} \]

input

integrate((((x*exp(x)*ln(x)-6*exp(x)*x**2)*ln(3*x)+exp(x)*ln(x)-6*exp(x)*x 
)*ln((ln(x)-6*x)/x)+(exp(x)*ln(x)-exp(x))*ln(3*x)+4*ln(x)-4)/((x*exp(x)**2 
*ln(x)-6*exp(x)**2*x**2)*ln(3*x)**2+(8*x*exp(x)*ln(x)-48*exp(x)*x**2)*ln(3 
*x)+16*x*ln(x)-96*x**2),x)
 

output

-log((-6*x + log(x))/x)/((log(x) + log(3))*exp(x) + 4)
 

3.6.92.7 Maxima [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=\frac {\log \left (x\right ) - \log \left (-6 \, x + \log \left (x\right )\right )}{{\left (\log \left (3\right ) + \log \left (x\right )\right )} e^{x} + 4} \]

input

integrate((((x*exp(x)*log(x)-6*exp(x)*x^2)*log(3*x)+exp(x)*log(x)-6*exp(x) 
*x)*log((log(x)-6*x)/x)+(exp(x)*log(x)-exp(x))*log(3*x)+4*log(x)-4)/((x*ex 
p(x)^2*log(x)-6*exp(x)^2*x^2)*log(3*x)^2+(8*x*exp(x)*log(x)-48*exp(x)*x^2) 
*log(3*x)+16*x*log(x)-96*x^2),x, algorithm=\
 

output

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

3.6.92.8 Giac [A] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=\frac {\log \left (x\right ) - \log \left (-6 \, x + \log \left (x\right )\right )}{e^{x} \log \left (3\right ) + e^{x} \log \left (x\right ) + 4} \]

input

integrate((((x*exp(x)*log(x)-6*exp(x)*x^2)*log(3*x)+exp(x)*log(x)-6*exp(x) 
*x)*log((log(x)-6*x)/x)+(exp(x)*log(x)-exp(x))*log(3*x)+4*log(x)-4)/((x*ex 
p(x)^2*log(x)-6*exp(x)^2*x^2)*log(3*x)^2+(8*x*exp(x)*log(x)-48*exp(x)*x^2) 
*log(3*x)+16*x*log(x)-96*x^2),x, algorithm=\
 

output

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

3.6.92.9 Mupad [B] (verification not implemented)

Time = 12.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-4+4 \log (x)+\left (-e^x+e^x \log (x)\right ) \log (3 x)+\left (-6 e^x x+e^x \log (x)+\left (-6 e^x x^2+e^x x \log (x)\right ) \log (3 x)\right ) \log \left (\frac {-6 x+\log (x)}{x}\right )}{-96 x^2+16 x \log (x)+\left (-48 e^x x^2+8 e^x x \log (x)\right ) \log (3 x)+\left (-6 e^{2 x} x^2+e^{2 x} x \log (x)\right ) \log ^2(3 x)} \, dx=-\frac {\ln \left (-\frac {6\,x-\ln \left (x\right )}{x}\right )}{\ln \left (3\,x\right )\,{\mathrm {e}}^x+4} \]

input

int((log(3*x)*(exp(x) - exp(x)*log(x)) - 4*log(x) + log(-(6*x - log(x))/x) 
*(log(3*x)*(6*x^2*exp(x) - x*exp(x)*log(x)) - exp(x)*log(x) + 6*x*exp(x)) 
+ 4)/(log(3*x)*(48*x^2*exp(x) - 8*x*exp(x)*log(x)) - 16*x*log(x) + log(3*x 
)^2*(6*x^2*exp(2*x) - x*exp(2*x)*log(x)) + 96*x^2),x)
 

output

-log(-(6*x - log(x))/x)/(log(3*x)*exp(x) + 4)