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

3.2.96.1 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
 

3.2.96.2 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]
 

3.2.96.3 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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 2009

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

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
 

3.2.96.3.1 Defintions of rubi rules used

rule 2009

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

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

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] 
]
 

3.2.96.4 Maple [C] (warning: unable to verify)

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

Time = 0.91 (sec) , antiderivative size = 164, normalized size of antiderivative = 4.00

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

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

3.2.96.5 Fricas [A] (verification not implemented)

Time = 0.26 (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=\
 

output

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

3.2.96.6 Sympy [A] (verification not implemented)

Time = 22.19 (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)
 

3.2.96.7 Maxima [A] (verification not implemented)

Time = 0.45 (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=\
 

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)
 

3.2.96.8 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=\
 

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)
 

3.2.96.9 Mupad [B] (verification not implemented)

Time = 8.45 (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)))