\(\int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} (-2-7 x+6 x^2)+(3 e^4-3 x+e^{2 x} x) \log (\frac {1}{3} (15 e^4-15 x+5 e^{2 x} x))}{12 x^2-6 x^3+e^4 (-12 x+6 x^2)+e^{2 x} (-4 x^2+2 x^3)+(6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} (-2 x+x^2)) \log (\frac {1}{3} (15 e^4-15 x+5 e^{2 x} x))} \, dx\) [2289]

Optimal result

Integrand size = 172, antiderivative size = 31 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log \left ((4-2 x) \left (2 x+\log \left (5 \left (e^4-x+\frac {1}{3} e^{2 x} x\right )\right )\right )\right ) \] Output:

ln((2*x+ln(5/3*x*exp(2*x)+5*exp(4)-5*x))*(4-2*x))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log (2-x)+\log \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right ) \] Input:

Integrate[(6 + 9*x - 12*x^2 + E^4*(-12 + 12*x) + E^(2*x)*(-2 - 7*x + 6*x^2 
) + (3*E^4 - 3*x + E^(2*x)*x)*Log[(15*E^4 - 15*x + 5*E^(2*x)*x)/3])/(12*x^ 
2 - 6*x^3 + E^4*(-12*x + 6*x^2) + E^(2*x)*(-4*x^2 + 2*x^3) + (6*x - 3*x^2 
+ E^4*(-6 + 3*x) + E^(2*x)*(-2*x + x^2))*Log[(15*E^4 - 15*x + 5*E^(2*x)*x) 
/3]),x]
 

Output:

Log[2 - x] + Log[2*x + Log[5*E^4 - 5*x + (5*E^(2*x)*x)/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.

\(\displaystyle \int \frac {-12 x^2+e^{2 x} \left (6 x^2-7 x-2\right )+9 x+e^4 (12 x-12)+\left (e^{2 x} x-3 x+3 e^4\right ) \log \left (\frac {1}{3} \left (5 e^{2 x} x-15 x+15 e^4\right )\right )+6}{-6 x^3+12 x^2+e^4 \left (6 x^2-12 x\right )+\left (-3 x^2+e^{2 x} \left (x^2-2 x\right )+6 x+e^4 (3 x-6)\right ) \log \left (\frac {1}{3} \left (5 e^{2 x} x-15 x+15 e^4\right )\right )+e^{2 x} \left (2 x^3-4 x^2\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {12 x^2-e^{2 x} \left (6 x^2-7 x-2\right )-9 x-e^4 (12 x-12)-\left (e^{2 x} x-3 x+3 e^4\right ) \log \left (\frac {1}{3} \left (5 e^{2 x} x-15 x+15 e^4\right )\right )-6}{(2-x) \left (e^{2 x} x-3 x+3 e^4\right ) \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {3 \left (2 x^2-2 e^4 x-e^4\right )}{x \left (e^{2 x} x-3 x+3 e^4\right ) \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}+\frac {6 x^2-7 x+x \log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )-2}{(x-2) x \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \int \frac {1}{2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )}dx+\int \frac {1}{x \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}dx-6 e^4 \int \frac {1}{\left (e^{2 x} x-3 x+3 e^4\right ) \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}dx-3 e^4 \int \frac {1}{x \left (e^{2 x} x-3 x+3 e^4\right ) \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}dx+6 \int \frac {x}{\left (e^{2 x} x-3 x+3 e^4\right ) \left (2 x+\log \left (\frac {5}{3} e^{2 x} x-5 x+5 e^4\right )\right )}dx+\log (2-x)\)

Input:

Int[(6 + 9*x - 12*x^2 + E^4*(-12 + 12*x) + E^(2*x)*(-2 - 7*x + 6*x^2) + (3 
*E^4 - 3*x + E^(2*x)*x)*Log[(15*E^4 - 15*x + 5*E^(2*x)*x)/3])/(12*x^2 - 6* 
x^3 + E^4*(-12*x + 6*x^2) + E^(2*x)*(-4*x^2 + 2*x^3) + (6*x - 3*x^2 + E^4* 
(-6 + 3*x) + E^(2*x)*(-2*x + x^2))*Log[(15*E^4 - 15*x + 5*E^(2*x)*x)/3]),x 
]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method	result	size
norman	\(\ln \left (-2+x \right )+\ln \left (2 x +\ln \left (\frac {5 x \,{\mathrm e}^{2 x}}{3}+5 \,{\mathrm e}^{4}-5 x \right )\right )\)	\(27\)
risch	\(\ln \left (-2+x \right )+\ln \left (2 x +\ln \left (\frac {5 x \,{\mathrm e}^{2 x}}{3}+5 \,{\mathrm e}^{4}-5 x \right )\right )\)	\(27\)
parallelrisch	\(\ln \left (-2+x \right )+\ln \left (x +\frac {\ln \left (\frac {5 x \,{\mathrm e}^{2 x}}{3}+5 \,{\mathrm e}^{4}-5 x \right )}{2}\right )\)	\(27\)

Input:

int(((x*exp(2*x)+3*exp(4)-3*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x- 
2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(-6+3*x)*e 
xp(4)-3*x^2+6*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp(2*x)+(6 
*x^2-12*x)*exp(4)-6*x^3+12*x^2),x,method=_RETURNVERBOSE)
 

Output:

ln(-2+x)+ln(2*x+ln(5/3*x*exp(2*x)+5*exp(4)-5*x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log \left (2 \, x + \log \left (\frac {5}{3} \, x e^{\left (2 \, x\right )} - 5 \, x + 5 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \] Input:

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x 
^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(-6 
+3*x)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp 
(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="fricas")
 

Output:

log(2*x + log(5/3*x*e^(2*x) - 5*x + 5*e^4)) + log(x - 2)
 

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log {\left (x - 2 \right )} + \log {\left (2 x + \log {\left (\frac {5 x e^{2 x}}{3} - 5 x + 5 e^{4} \right )} \right )} \] Input:

integrate(((x*exp(2*x)+3*exp(4)-3*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x* 
*2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-12*x**2+9*x+6)/(((x**2-2*x)*exp(2*x)+( 
-6+3*x)*exp(4)-3*x**2+6*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x**3-4*x**2) 
*exp(2*x)+(6*x**2-12*x)*exp(4)-6*x**3+12*x**2),x)
 

Output:

log(x - 2) + log(2*x + log(5*x*exp(2*x)/3 - 5*x + 5*exp(4)))
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log \left (2 \, x + \log \left (5\right ) - \log \left (3\right ) + \log \left (x e^{\left (2 \, x\right )} - 3 \, x + 3 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \] Input:

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x 
^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(-6 
+3*x)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp 
(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="maxima")
 

Output:

log(2*x + log(5) - log(3) + log(x*e^(2*x) - 3*x + 3*e^4)) + log(x - 2)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\log \left (2 \, x + \log \left (\frac {5}{3} \, x e^{\left (2 \, x\right )} - 5 \, x + 5 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \] Input:

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x 
^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(-6 
+3*x)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp 
(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="giac")
 

Output:

log(2*x + log(5/3*x*e^(2*x) - 5*x + 5*e^4)) + log(x - 2)
 

Mupad [B] (verification not implemented)

Time = 2.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\ln \left (2\,x+\ln \left (5\,{\mathrm {e}}^4-5\,x+\frac {5\,x\,{\mathrm {e}}^{2\,x}}{3}\right )\right )+\ln \left (x-2\right ) \] Input:

int(-(9*x - exp(2*x)*(7*x - 6*x^2 + 2) - 12*x^2 + log(5*exp(4) - 5*x + (5* 
x*exp(2*x))/3)*(3*exp(4) - 3*x + x*exp(2*x)) + exp(4)*(12*x - 12) + 6)/(ex 
p(4)*(12*x - 6*x^2) - log(5*exp(4) - 5*x + (5*x*exp(2*x))/3)*(6*x - exp(2* 
x)*(2*x - x^2) - 3*x^2 + exp(4)*(3*x - 6)) + exp(2*x)*(4*x^2 - 2*x^3) - 12 
*x^2 + 6*x^3),x)
 

Output:

log(2*x + log(5*exp(4) - 5*x + (5*x*exp(2*x))/3)) + log(x - 2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {5 e^{2 x} x}{3}+5 e^{4}-5 x \right )+2 x \right )+\mathrm {log}\left (x -2\right ) \] Input:

int(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x 
-2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(-6+3*x)* 
exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp(2*x)+ 
(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x)
 

Output:

log(log((5*e**(2*x)*x + 15*e**4 - 15*x)/3) + 2*x) + log(x - 2)