\(\int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x (\frac {1}{12 x+4 x^2})^{e^{-3+x}}} (\frac {1}{12 x+4 x^2})^{e^{-3+x}} (e^{2 x} (-6-4 x)+e^{3+x} (6 x+2 x^2)+e^{2 x} (6 x+2 x^2) \log (\frac {5}{12 x+4 x^2}))}{3 x+x^2} \, dx\) [196]

Optimal result

Integrand size = 123, antiderivative size = 41 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{-8+2^{1-2 e^{-3+x}} 5^{e^{-3+x}} e^x \left (\frac {1}{x (3+x)}\right )^{e^{-3+x}}} \] Output:

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

Mathematica [F]

\[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx \] Input:

Integrate[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^ 
E^(-3 + x))*((12*x + 4*x^2)^(-1))^E^(-3 + x)*(E^(2*x)*(-6 - 4*x) + E^(3 + 
x)*(6*x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x + x^ 
2),x]
 

Output:

Integrate[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^ 
E^(-3 + x))*((12*x + 4*x^2)^(-1))^E^(-3 + x)*(E^(2*x)*(-6 - 4*x) + E^(3 + 
x)*(6*x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x + x^ 
2), 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.

\(\displaystyle \int \frac {5^{e^{x-3}} \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}-11\right ) \left (e^{x+3} \left (2 x^2+6 x\right )+e^{2 x} \left (2 x^2+6 x\right ) \log \left (\frac {5}{4 x^2+12 x}\right )+e^{2 x} (-4 x-6)\right )}{x^2+3 x} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {5^{e^{x-3}} \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}-11\right ) \left (e^{x+3} \left (2 x^2+6 x\right )+e^{2 x} \left (2 x^2+6 x\right ) \log \left (\frac {5}{4 x^2+12 x}\right )+e^{2 x} (-4 x-6)\right )}{x (x+3)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {5^{e^{x-3}} \left (\frac {1}{x (4 x+12)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}-11\right ) \left (e^{x+3} \left (2 x^2+6 x\right )+e^{2 x} \left (2 x^2+6 x\right ) \log \left (\frac {5}{4 x^2+12 x}\right )+e^{2 x} (-4 x-6)\right )}{x (x+3)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (2\ 5^{e^{x-3}} \left (\frac {1}{x (4 x+12)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+x-8\right )+\frac {2\ 5^{e^{x-3}} \left (\frac {1}{x (4 x+12)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+2 x-11\right ) \left (x^2 \log \left (\frac {5}{4 x (x+3)}\right )-2 x+3 x \log \left (\frac {5}{4 x (x+3)}\right )-3\right )}{x (x+3)}\right )dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (2\ 5^{e^{x-3}} \left (\frac {1}{x (4 x+12)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+x-8\right )+\frac {2^{1-2 e^{x-3}} 5^{e^{x-3}} \left (\frac {1}{x (x+3)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+2 x-11\right ) \left (x^2 \log \left (\frac {5}{4 x (x+3)}\right )-2 x+3 x \log \left (\frac {5}{4 x (x+3)}\right )-3\right )}{x (x+3)}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (2\ 5^{e^{x-3}} \left (\frac {1}{x (4 x+12)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+x-8\right )+\frac {2^{1-2 e^{x-3}} 5^{e^{x-3}} \left (\frac {1}{x (x+3)}\right )^{e^{x-3}} \exp \left (2\ 5^{e^{x-3}} e^x \left (\frac {1}{4 x^2+12 x}\right )^{e^{x-3}}+2 x-11\right ) \left (x^2 \log \left (\frac {5}{4 x (x+3)}\right )-2 x+3 x \log \left (\frac {5}{4 x (x+3)}\right )-3\right )}{x (x+3)}\right )dx\)

Input:

Int[(5^E^(-3 + x)*E^(-11 + 2*5^E^(-3 + x)*E^x*((12*x + 4*x^2)^(-1))^E^(-3 
+ x))*((12*x + 4*x^2)^(-1))^E^(-3 + x)*(E^(2*x)*(-6 - 4*x) + E^(3 + x)*(6* 
x + 2*x^2) + E^(2*x)*(6*x + 2*x^2)*Log[5/(12*x + 4*x^2)]))/(3*x + x^2),x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.70 (sec) , antiderivative size = 164, normalized size of antiderivative = 4.00

\[{\mathrm e}^{2 \,{\mathrm e}^{x -\ln \left (x \right ) {\mathrm e}^{-3+x}+{\mathrm e}^{-3+x} \ln \left (5\right )-2 \,{\mathrm e}^{-3+x} \ln \left (2\right )-{\mathrm e}^{-3+x} \ln \left (3+x \right )} {\mathrm e}^{-\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}} {\mathrm e}^{\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right )^{2} \operatorname {csgn}\left (\frac {i}{3+x}\right )}{2}} {\mathrm e}^{-\frac {i \pi \,{\mathrm e}^{-3+x} \operatorname {csgn}\left (\frac {i}{\left (3+x \right ) x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i}{3+x}\right )}{2}}-8}\]

Input:

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

Output:

exp(2*exp(x-ln(x)*exp(-3+x)+exp(-3+x)*ln(5)-2*exp(-3+x)*ln(2)-exp(-3+x)*ln 
(3+x))*exp(-1/2*I*Pi*exp(-3+x)*csgn(I/(3+x)/x)^3)*exp(1/2*I*Pi*exp(-3+x)*c 
sgn(I/(3+x)/x)^2*csgn(I/x))*exp(1/2*I*Pi*exp(-3+x)*csgn(I/(3+x)/x)^2*csgn( 
I/(3+x)))*exp(-1/2*I*Pi*exp(-3+x)*csgn(I/(3+x)/x)*csgn(I/x)*csgn(I/(3+x))) 
-8)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{\left ({\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (x + 3\right )} - 11 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} \] Input:

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

Output:

e^((2*(5/4/(x^2 + 3*x))^e^(x - 3)*e^(x + 3) - 11*e^3)*e^(-3) + 3)
 

Sympy [A] (verification not implemented)

Time = 25.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{2 e^{x} e^{\frac {e^{x} \log {\left (\frac {5}{4 x^{2} + 12 x} \right )}}{e^{3}}} - 8} \] Input:

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

Output:

exp(2*exp(x)*exp(exp(-3)*exp(x)*log(5/(4*x**2 + 12*x))) - 8)
 

Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=e^{\left (2 \, e^{\left (e^{\left (x - 3\right )} \log \left (5\right ) - 2 \, e^{\left (x - 3\right )} \log \left (2\right ) - e^{\left (x - 3\right )} \log \left (x + 3\right ) - e^{\left (x - 3\right )} \log \left (x\right ) + x\right )} - 8\right )} \] Input:

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

Output:

e^(2*e^(e^(x - 3)*log(5) - 2*e^(x - 3)*log(2) - e^(x - 3)*log(x + 3) - e^( 
x - 3)*log(x) + x) - 8)
 

Giac [F]

\[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\int { \frac {2 \, {\left ({\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right ) - {\left (2 \, x + 3\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 3 \, x\right )} e^{\left (x + 3\right )}\right )} \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{\left (2 \, \left (\frac {5}{4 \, {\left (x^{2} + 3 \, x\right )}}\right )^{e^{\left (x - 3\right )}} e^{x} - 11\right )}}{x^{2} + 3 \, x} \,d x } \] Input:

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

Output:

integrate(2*((x^2 + 3*x)*e^(2*x)*log(5/4/(x^2 + 3*x)) - (2*x + 3)*e^(2*x) 
+ (x^2 + 3*x)*e^(x + 3))*(5/4/(x^2 + 3*x))^e^(x - 3)*e^(2*(5/4/(x^2 + 3*x) 
)^e^(x - 3)*e^x - 11)/(x^2 + 3*x), x)
 

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx={\mathrm {e}}^{-8}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x\,{\left (\frac {5}{4\,x^2+12\,x}\right )}^{{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}} \] Input:

int((exp(-3)*exp(exp(-3)*exp(x)*log(5/(12*x + 4*x^2)))*exp(2*exp(exp(-3)*e 
xp(x)*log(5/(12*x + 4*x^2)))*exp(x) - 8)*(exp(3)*exp(x)*(6*x + 2*x^2) - ex 
p(2*x)*(4*x + 6) + exp(2*x)*log(5/(12*x + 4*x^2))*(6*x + 2*x^2)))/(3*x + x 
^2),x)
 

Output:

exp(-8)*exp(2*exp(x)*(5/(12*x + 4*x^2))^(exp(-3)*exp(x)))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {5^{e^{-3+x}} e^{-11+2\ 5^{e^{-3+x}} e^x \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}}} \left (\frac {1}{12 x+4 x^2}\right )^{e^{-3+x}} \left (e^{2 x} (-6-4 x)+e^{3+x} \left (6 x+2 x^2\right )+e^{2 x} \left (6 x+2 x^2\right ) \log \left (\frac {5}{12 x+4 x^2}\right )\right )}{3 x+x^2} \, dx=\frac {e^{2 e^{\frac {e^{x} \mathrm {log}\left (\frac {5}{4 x^{2}+12 x}\right )+e^{3} x}{e^{3}}}}}{e^{8}} \] Input:

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

Output:

e**(2*e**((e**x*log(5/(4*x**2 + 12*x)) + e**3*x)/e**3))/e**8