\(\int \frac {1}{3} e^{\frac {1}{9} (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} (6 e+6 e^{3 x}))} (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} (6+30 e^x x)+e^{e^{5 e^x} x} (6 e^{3 x}+e^{5 e^x} (2 e+10 e^{1+x} x+e^{3 x} (2+10 e^x x)))) \, dx\) [567]

Optimal result

Integrand size = 163, antiderivative size = 27 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{\left (e^{e^{5 e^x} x}+\frac {1}{3} \left (e+e^{3 x}\right )\right )^2} \] Output:

exp((exp(x*exp(5*exp(x)))+1/3*exp(1)+1/3*exp(3*x))^2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{\frac {1}{9} \left (e+e^{3 x}+3 e^{e^{5 e^x} x}\right )^2} \] Input:

Integrate[(E^((E^2 + E^(6*x) + 9*E^(2*E^(5*E^x)*x) + 2*E^(1 + 3*x) + E^(E^ 
(5*E^x)*x)*(6*E + 6*E^(3*x)))/9)*(2*E^(6*x) + 2*E^(1 + 3*x) + E^(5*E^x + 2 
*E^(5*E^x)*x)*(6 + 30*E^x*x) + E^(E^(5*E^x)*x)*(6*E^(3*x) + E^(5*E^x)*(2*E 
 + 10*E^(1 + x)*x + E^(3*x)*(2 + 10*E^x*x)))))/3,x]
 

Output:

E^((E + E^(3*x) + 3*E^(E^(5*E^x)*x))^2/9)
 

Rubi [A] (verified)

Time = 3.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {27, 27, 7292, 7257}

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.

\(\displaystyle \int \frac {1}{3} \left (e^{2 e^{5 e^x} x+5 e^x} \left (30 e^x x+6\right )+2 e^{6 x}+2 e^{3 x+1}+e^{e^{5 e^x} x} \left (e^{5 e^x} \left (10 e^{x+1} x+e^{3 x} \left (10 e^x x+2\right )+2 e\right )+6 e^{3 x}\right )\right ) \exp \left (\frac {1}{9} \left (e^{e^{5 e^x} x} \left (6 e^{3 x}+6 e\right )+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{3 x+1}+e^2\right )\right ) \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int 2 \exp \left (\frac {1}{9} \left (6 e^{e^{5 e^x} x} \left (e+e^{3 x}\right )+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{3 x+1}+e^2\right )\right ) \left (3 e^{2 e^{5 e^x} x+5 e^x} \left (5 e^x x+1\right )+e^{6 x}+e^{3 x+1}+e^{e^{5 e^x} x} \left (e^{5 e^x} \left (5 e^{x+1} x+e^{3 x} \left (5 e^x x+1\right )+e\right )+3 e^{3 x}\right )\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} \int \exp \left (\frac {1}{9} \left (6 e^{e^{5 e^x} x} \left (e+e^{3 x}\right )+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{3 x+1}+e^2\right )\right ) \left (3 e^{2 e^{5 e^x} x+5 e^x} \left (5 e^x x+1\right )+e^{6 x}+e^{3 x+1}+e^{e^{5 e^x} x} \left (e^{5 e^x} \left (5 e^{x+1} x+e^{3 x} \left (5 e^x x+1\right )+e\right )+3 e^{3 x}\right )\right )dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {2}{3} \int e^{\frac {1}{9} \left (e+e^{3 x}+3 e^{e^{5 e^x} x}\right )^2} \left (e+e^{3 x}+3 e^{e^{5 e^x} x}\right ) \left (5 e^{e^{5 e^x} x+x+5 e^x} x+e^{3 x}+e^{e^{5 e^x} x+5 e^x}\right )dx\)

\(\Big \downarrow \) 7257

\(\displaystyle e^{\frac {1}{9} \left (e^{3 x}+3 e^{e^{5 e^x} x}+e\right )^2}\)

Input:

Int[(E^((E^2 + E^(6*x) + 9*E^(2*E^(5*E^x)*x) + 2*E^(1 + 3*x) + E^(E^(5*E^x 
)*x)*(6*E + 6*E^(3*x)))/9)*(2*E^(6*x) + 2*E^(1 + 3*x) + E^(5*E^x + 2*E^(5* 
E^x)*x)*(6 + 30*E^x*x) + E^(E^(5*E^x)*x)*(6*E^(3*x) + E^(5*E^x)*(2*E + 10* 
E^(1 + x)*x + E^(3*x)*(2 + 10*E^x*x)))))/3,x]
 

Output:

E^((E + E^(3*x) + 3*E^(E^(5*E^x)*x))^2/9)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7257

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim 
p[q*(F^v/Log[F]), x] /;  !FalseQ[q]] /; FreeQ[F, x]
 

rule 7292

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(53\) vs. \(2(22)=44\).

Time = 78.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{5 \,{\mathrm e}^{x}}}+\frac {2 \,{\mathrm e}^{x \left ({\mathrm e}^{5 \,{\mathrm e}^{x}}+3\right )}}{3}+\frac {2 \,{\mathrm e}^{x \,{\mathrm e}^{5 \,{\mathrm e}^{x}}+1}}{3}+\frac {{\mathrm e}^{6 x}}{9}+\frac {2 \,{\mathrm e}^{1+3 x}}{9}+\frac {{\mathrm e}^{2}}{9}}\)	\(54\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{5 \,{\mathrm e}^{x}}}+\frac {\left (6 \,{\mathrm e}^{3 x}+6 \,{\mathrm e}\right ) {\mathrm e}^{x \,{\mathrm e}^{5 \,{\mathrm e}^{x}}}}{9}+\frac {{\mathrm e}^{6 x}}{9}+\frac {2 \,{\mathrm e} \,{\mathrm e}^{3 x}}{9}+\frac {{\mathrm e}^{2}}{9}}\)	\(56\)

Input:

int(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))^2+(((10*exp(x) 
*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x))*exp( 
x*exp(5*exp(x)))+2*exp(3*x)^2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*exp(x)))^ 
2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)^2+2/9*exp(1) 
*exp(3*x)+1/9*exp(1)^2),x,method=_RETURNVERBOSE)
 

Output:

exp(exp(2*x*exp(5*exp(x)))+2/3*exp(x*(exp(5*exp(x))+3))+2/3*exp(x*exp(5*ex 
p(x))+1)+1/9*exp(6*x)+2/9*exp(1+3*x)+1/9*exp(2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{\left (\frac {1}{9} \, {\left (6 \, {\left (e^{7} + e^{\left (3 \, x + 6\right )}\right )} e^{\left (x e^{\left (5 \, e^{x}\right )}\right )} + e^{8} + 9 \, e^{\left (2 \, x e^{\left (5 \, e^{x}\right )} + 6\right )} + e^{\left (6 \, x + 6\right )} + 2 \, e^{\left (3 \, x + 7\right )}\right )} e^{\left (-6\right )}\right )} \] Input:

integrate(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))^2+(((10* 
exp(x)*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x) 
)*exp(x*exp(5*exp(x)))+2*exp(3*x)^2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*exp 
(x)))^2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)^2+2/9* 
exp(1)*exp(3*x)+1/9*exp(1)^2),x, algorithm="fricas")
 

Output:

e^(1/9*(6*(e^7 + e^(3*x + 6))*e^(x*e^(5*e^x)) + e^8 + 9*e^(2*x*e^(5*e^x) + 
 6) + e^(6*x + 6) + 2*e^(3*x + 7))*e^(-6))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).

Time = 5.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{\left (\frac {2 e^{3 x}}{3} + \frac {2 e}{3}\right ) e^{x e^{5 e^{x}}} + \frac {e^{6 x}}{9} + \frac {2 e e^{3 x}}{9} + e^{2 x e^{5 e^{x}}} + \frac {e^{2}}{9}} \] Input:

integrate(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))**2+(((10 
*exp(x)*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x 
))*exp(x*exp(5*exp(x)))+2*exp(3*x)**2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*e 
xp(x)))**2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)**2+ 
2/9*exp(1)*exp(3*x)+1/9*exp(1)**2),x)
 

Output:

exp((2*exp(3*x)/3 + 2*E/3)*exp(x*exp(5*exp(x))) + exp(6*x)/9 + 2*E*exp(3*x 
)/9 + exp(2*x*exp(5*exp(x))) + exp(2)/9)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (22) = 44\).

Time = 0.37 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{\left (\frac {1}{9} \, e^{2} + e^{\left (2 \, x e^{\left (5 \, e^{x}\right )}\right )} + \frac {2}{3} \, e^{\left (x e^{\left (5 \, e^{x}\right )} + 3 \, x\right )} + \frac {2}{3} \, e^{\left (x e^{\left (5 \, e^{x}\right )} + 1\right )} + \frac {1}{9} \, e^{\left (6 \, x\right )} + \frac {2}{9} \, e^{\left (3 \, x + 1\right )}\right )} \] Input:

integrate(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))^2+(((10* 
exp(x)*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x) 
)*exp(x*exp(5*exp(x)))+2*exp(3*x)^2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*exp 
(x)))^2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)^2+2/9* 
exp(1)*exp(3*x)+1/9*exp(1)^2),x, algorithm="maxima")
 

Output:

e^(1/9*e^2 + e^(2*x*e^(5*e^x)) + 2/3*e^(x*e^(5*e^x) + 3*x) + 2/3*e^(x*e^(5 
*e^x) + 1) + 1/9*e^(6*x) + 2/9*e^(3*x + 1))
 

Giac [F]

\[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=\int { \frac {2}{3} \, {\left (3 \, {\left (5 \, x e^{x} + 1\right )} e^{\left (2 \, x e^{\left (5 \, e^{x}\right )} + 5 \, e^{x}\right )} + {\left ({\left ({\left (5 \, x e^{x} + 1\right )} e^{\left (3 \, x\right )} + 5 \, x e^{\left (x + 1\right )} + e\right )} e^{\left (5 \, e^{x}\right )} + 3 \, e^{\left (3 \, x\right )}\right )} e^{\left (x e^{\left (5 \, e^{x}\right )}\right )} + e^{\left (6 \, x\right )} + e^{\left (3 \, x + 1\right )}\right )} e^{\left (\frac {2}{3} \, {\left (e + e^{\left (3 \, x\right )}\right )} e^{\left (x e^{\left (5 \, e^{x}\right )}\right )} + \frac {1}{9} \, e^{2} + e^{\left (2 \, x e^{\left (5 \, e^{x}\right )}\right )} + \frac {1}{9} \, e^{\left (6 \, x\right )} + \frac {2}{9} \, e^{\left (3 \, x + 1\right )}\right )} \,d x } \] Input:

integrate(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))^2+(((10* 
exp(x)*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x) 
)*exp(x*exp(5*exp(x)))+2*exp(3*x)^2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*exp 
(x)))^2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)^2+2/9* 
exp(1)*exp(3*x)+1/9*exp(1)^2),x, algorithm="giac")
 

Output:

integrate(2/3*(3*(5*x*e^x + 1)*e^(2*x*e^(5*e^x) + 5*e^x) + (((5*x*e^x + 1) 
*e^(3*x) + 5*x*e^(x + 1) + e)*e^(5*e^x) + 3*e^(3*x))*e^(x*e^(5*e^x)) + e^( 
6*x) + e^(3*x + 1))*e^(2/3*(e + e^(3*x))*e^(x*e^(5*e^x)) + 1/9*e^2 + e^(2* 
x*e^(5*e^x)) + 1/9*e^(6*x) + 2/9*e^(3*x + 1)), x)
 

Mupad [B] (verification not implemented)

Time = 3.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{6\,x}}{9}}\,{\mathrm {e}}^{\frac {2\,\mathrm {e}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}}}{3}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{9}}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}}}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{3\,x}\,\mathrm {e}}{9}}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}}}{3}} \] Input:

int((exp(exp(6*x)/9 + exp(2)/9 + exp(2*x*exp(5*exp(x))) + (2*exp(3*x)*exp( 
1))/9 + (exp(x*exp(5*exp(x)))*(6*exp(3*x) + 6*exp(1)))/9)*(2*exp(6*x) + 2* 
exp(3*x)*exp(1) + exp(x*exp(5*exp(x)))*(6*exp(3*x) + exp(5*exp(x))*(2*exp( 
1) + exp(3*x)*(10*x*exp(x) + 2) + 10*x*exp(1)*exp(x))) + exp(5*exp(x))*exp 
(2*x*exp(5*exp(x)))*(30*x*exp(x) + 6)))/3,x)
 

Output:

exp(exp(6*x)/9)*exp((2*exp(1)*exp(x*exp(5*exp(x))))/3)*exp(exp(2)/9)*exp(e 
xp(2*x*exp(5*exp(x))))*exp((2*exp(3*x)*exp(1))/9)*exp((2*exp(3*x)*exp(x*ex 
p(5*exp(x))))/3)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {1}{3} e^{\frac {1}{9} \left (e^2+e^{6 x}+9 e^{2 e^{5 e^x} x}+2 e^{1+3 x}+e^{e^{5 e^x} x} \left (6 e+6 e^{3 x}\right )\right )} \left (2 e^{6 x}+2 e^{1+3 x}+e^{5 e^x+2 e^{5 e^x} x} \left (6+30 e^x x\right )+e^{e^{5 e^x} x} \left (6 e^{3 x}+e^{5 e^x} \left (2 e+10 e^{1+x} x+e^{3 x} \left (2+10 e^x x\right )\right )\right )\right ) \, dx=e^{e^{2 e^{5 e^{x}} x}+\frac {2 e^{e^{5 e^{x}} x +3 x}}{3}+\frac {2 e^{e^{5 e^{x}} x} e}{3}+\frac {e^{6 x}}{9}+\frac {2 e^{3 x} e}{9}+\frac {e^{2}}{9}} \] Input:

int(1/3*((30*exp(x)*x+6)*exp(5*exp(x))*exp(x*exp(5*exp(x)))^2+(((10*exp(x) 
*x+2)*exp(3*x)+10*x*exp(1)*exp(x)+2*exp(1))*exp(5*exp(x))+6*exp(3*x))*exp( 
x*exp(5*exp(x)))+2*exp(3*x)^2+2*exp(1)*exp(3*x))*exp(exp(x*exp(5*exp(x)))^ 
2+1/9*(6*exp(3*x)+6*exp(1))*exp(x*exp(5*exp(x)))+1/9*exp(3*x)^2+2/9*exp(1) 
*exp(3*x)+1/9*exp(1)^2),x)
 

Output:

e**((9*e**(2*e**(5*e**x)*x) + 6*e**(e**(5*e**x)*x + 3*x) + 6*e**(e**(5*e** 
x)*x)*e + e**(6*x) + 2*e**(3*x)*e + e**2)/9)