\(\int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} (e^x (-6 x-6 x^2+2 x \log (9))+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))) \log (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))))}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx\) [1292]

Optimal result

Integrand size = 137, antiderivative size = 20 \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=\left (x+\frac {1}{2} \log (-3-3 x+\log (9))\right )^{e^{e^x}} \] Output:

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

Mathematica [F]

\[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=\int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \int 2^{1-e^{e^x}} e^{e^x} (2 x+\log (-3 x+\log (9)-3))^{-1+e^{e^x}}dx+3 \int \frac {2^{-e^{e^x}} e^{e^x} (2 x+\log (-3 x+\log (9)-3))^{-1+e^{e^x}}}{3 x-\log (9)+3}dx+\int 2^{-e^{e^x}} e^{x+e^x} (2 x+\log (-3 x+\log (9)-3))^{e^{e^x}} \log \left (\frac {1}{2} (2 x+\log (-3 x+\log (9)-3))\right )dx\)

Input:

Int[((2*x + Log[-3 - 3*x + Log[9]])^E^E^x*(E^E^x*(-9 - 6*x + 2*Log[9]) + E 
^E^x*(E^x*(-6*x - 6*x^2 + 2*x*Log[9]) + E^x*(-3 - 3*x + Log[9])*Log[-3 - 3 
*x + Log[9]])*Log[(2*x + Log[-3 - 3*x + Log[9]])/2]))/(2^E^E^x*(-6*x - 6*x 
^2 + 2*x*Log[9] + (-3 - 3*x + Log[9])*Log[-3 - 3*x + Log[9]])),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

\[\left (\frac {\ln \left (2 \ln \left (3\right )-3 x -3\right )}{2}+x \right )^{{\mathrm e}^{{\mathrm e}^{x}}}\]

Input:

int((((2*ln(3)-3*x-3)*exp(x)*ln(2*ln(3)-3*x-3)+(4*x*ln(3)-6*x^2-6*x)*exp(x 
))*exp(exp(x))*ln(1/2*ln(2*ln(3)-3*x-3)+x)+(4*ln(3)-6*x-9)*exp(exp(x)))*ex 
p(exp(exp(x))*ln(1/2*ln(2*ln(3)-3*x-3)+x))/((2*ln(3)-3*x-3)*ln(2*ln(3)-3*x 
-3)+4*x*ln(3)-6*x^2-6*x),x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx={\left (x + \frac {1}{2} \, \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right )\right )}^{e^{\left (e^{x}\right )}} \] Input:

integrate((((2*log(3)-3*x-3)*exp(x)*log(2*log(3)-3*x-3)+(4*x*log(3)-6*x^2- 
6*x)*exp(x))*exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x)+(4*log(3)-6*x-9)*e 
xp(exp(x)))*exp(exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x))/((2*log(3)-3*x 
-3)*log(2*log(3)-3*x-3)+4*x*log(3)-6*x^2-6*x),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=\text {Timed out} \] Input:

integrate((((2*ln(3)-3*x-3)*exp(x)*ln(2*ln(3)-3*x-3)+(4*x*ln(3)-6*x**2-6*x 
)*exp(x))*exp(exp(x))*ln(1/2*ln(2*ln(3)-3*x-3)+x)+(4*ln(3)-6*x-9)*exp(exp( 
x)))*exp(exp(exp(x))*ln(1/2*ln(2*ln(3)-3*x-3)+x))/((2*ln(3)-3*x-3)*ln(2*ln 
(3)-3*x-3)+4*x*ln(3)-6*x**2-6*x),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=e^{\left (-e^{\left (e^{x}\right )} \log \left (2\right ) + e^{\left (e^{x}\right )} \log \left (2 \, x + \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right )\right )\right )} \] Input:

integrate((((2*log(3)-3*x-3)*exp(x)*log(2*log(3)-3*x-3)+(4*x*log(3)-6*x^2- 
6*x)*exp(x))*exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x)+(4*log(3)-6*x-9)*e 
xp(exp(x)))*exp(exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x))/((2*log(3)-3*x 
-3)*log(2*log(3)-3*x-3)+4*x*log(3)-6*x^2-6*x),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=\int { \frac {{\left ({\left ({\left (3 \, x - 2 \, \log \left (3\right ) + 3\right )} e^{x} \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right ) + 2 \, {\left (3 \, x^{2} - 2 \, x \log \left (3\right ) + 3 \, x\right )} e^{x}\right )} e^{\left (e^{x}\right )} \log \left (x + \frac {1}{2} \, \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right )\right ) + {\left (6 \, x - 4 \, \log \left (3\right ) + 9\right )} e^{\left (e^{x}\right )}\right )} {\left (x + \frac {1}{2} \, \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right )\right )}^{e^{\left (e^{x}\right )}}}{6 \, x^{2} - 4 \, x \log \left (3\right ) + {\left (3 \, x - 2 \, \log \left (3\right ) + 3\right )} \log \left (-3 \, x + 2 \, \log \left (3\right ) - 3\right ) + 6 \, x} \,d x } \] Input:

integrate((((2*log(3)-3*x-3)*exp(x)*log(2*log(3)-3*x-3)+(4*x*log(3)-6*x^2- 
6*x)*exp(x))*exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x)+(4*log(3)-6*x-9)*e 
xp(exp(x)))*exp(exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x))/((2*log(3)-3*x 
-3)*log(2*log(3)-3*x-3)+4*x*log(3)-6*x^2-6*x),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 8.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx={\left (x+\frac {\ln \left (\ln \left (9\right )-3\,x-3\right )}{2}\right )}^{{\mathrm {e}}^{{\mathrm {e}}^x}} \] Input:

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

Output:

(x + log(log(9) - 3*x - 3)/2)^exp(exp(x))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {2^{-e^{e^x}} (2 x+\log (-3-3 x+\log (9)))^{e^{e^x}} \left (e^{e^x} (-9-6 x+2 \log (9))+e^{e^x} \left (e^x \left (-6 x-6 x^2+2 x \log (9)\right )+e^x (-3-3 x+\log (9)) \log (-3-3 x+\log (9))\right ) \log \left (\frac {1}{2} (2 x+\log (-3-3 x+\log (9)))\right )\right )}{-6 x-6 x^2+2 x \log (9)+(-3-3 x+\log (9)) \log (-3-3 x+\log (9))} \, dx=e^{e^{e^{x}} \mathrm {log}\left (\frac {\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )-3 x -3\right )}{2}+x \right )} \] Input:

int((((2*log(3)-3*x-3)*exp(x)*log(2*log(3)-3*x-3)+(4*x*log(3)-6*x^2-6*x)*e 
xp(x))*exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x)+(4*log(3)-6*x-9)*exp(exp 
(x)))*exp(exp(exp(x))*log(1/2*log(2*log(3)-3*x-3)+x))/((2*log(3)-3*x-3)*lo 
g(2*log(3)-3*x-3)+4*x*log(3)-6*x^2-6*x),x)
 

Output:

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