3.26.65 \(\int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} (-1+e^{-x+e^{-x} x} (1-x) x)}{8 x^2} \, dx\) [2565]

3.26.65.1 Optimal result

Integrand size = 127, antiderivative size = 33 \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\frac {1}{4} e^{e^{\frac {1}{4} e^{\frac {e^{e^{e^{-x} x}}+3 x}{x}}}} \]

output

1/4*exp(exp(1/2*exp((3*x+exp(exp(exp(ln(x)-x))))/x-ln(2))))
 

3.26.65.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\frac {1}{4} e^{e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}}} \]

input

Integrate[(E^(E^(E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2) + E^(x/E^x) + E^( 
(E^E^(x/E^x) + 3*x - x*Log[2])/x)/2 + (E^E^(x/E^x) + 3*x - x*Log[2])/x)*(- 
1 + E^(-x + x/E^x)*(1 - x)*x))/(8*x^2),x]
 

output

E^E^(E^(3 + E^E^(x/E^x)/x)/4)/4
 

3.26.65.3 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.

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[(E^(E^(E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2) + E^(x/E^x) + E^((E^E^( 
x/E^x) + 3*x - x*Log[2])/x)/2 + (E^E^(x/E^x) + 3*x - x*Log[2])/x)*(-1 + E^ 
(-x + x/E^x)*(1 - x)*x))/(8*x^2),x]
 

output

$Aborted
 

3.26.65.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 2009

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

rule 7293

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

3.26.65.4 Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73

\[\frac {{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}}}+3 x}{x}}}{4}}}}{4}\]

input

int(1/8*((1-x)*exp(ln(x)-x)*exp(exp(ln(x)-x))-1)*exp(exp(exp(ln(x)-x)))*ex 
p((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(ln(x)-x 
)))-x*ln(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x)/ 
x)))/x^2,x)
 

output

1/4*exp(exp(1/4*exp((exp(exp(x*exp(-x)))+3*x)/x)))
 

3.26.65.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (29) = 58\).

Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 6.00 \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\frac {1}{4} \, e^{\left (\frac {{\left (x e^{\left (-x - \frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x} + \log \left (x\right )\right )} + 2 \, x e^{\left (-x + e^{\left (-x + \log \left (x\right )\right )} + \log \left (x\right )\right )} + 2 \, x e^{\left (-x + \frac {1}{2} \, e^{\left (-\frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x}\right )} + \log \left (x\right )\right )} - 2 \, {\left (x \log \left (2\right ) - 3 \, x\right )} e^{\left (-x + \log \left (x\right )\right )} + 2 \, e^{\left (-x + e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )} + \log \left (x\right )\right )}\right )} e^{\left (x - \log \left (x\right )\right )}}{2 \, x} + \frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x} - \frac {1}{2} \, e^{\left (-\frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x}\right )} - e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )} \]

input

integrate(1/8*((1-x)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x 
)-x)))*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp( 
exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x))) 
-x*log(2)+3*x)/x)))/x^2,x, algorithm=\
 

output

1/4*e^(1/2*(x*e^(-x - (x*log(2) - 3*x - e^(e^(e^(-x + log(x)))))/x + log(x 
)) + 2*x*e^(-x + e^(-x + log(x)) + log(x)) + 2*x*e^(-x + 1/2*e^(-(x*log(2) 
 - 3*x - e^(e^(e^(-x + log(x)))))/x) + log(x)) - 2*(x*log(2) - 3*x)*e^(-x 
+ log(x)) + 2*e^(-x + e^(e^(-x + log(x))) + log(x)))*e^(x - log(x))/x + (x 
*log(2) - 3*x - e^(e^(e^(-x + log(x)))))/x - 1/2*e^(-(x*log(2) - 3*x - e^( 
e^(e^(-x + log(x)))))/x) - e^(e^(-x + log(x))))
 

3.26.65.6 Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\text {Timed out} \]

input

integrate(1/8*((1-x)*exp(ln(x)-x)*exp(exp(ln(x)-x))-1)*exp(exp(exp(ln(x)-x 
)))*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(l 
n(x)-x)))-x*ln(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln(2) 
+3*x)/x)))/x**2,x)
 

output

Timed out
 

3.26.65.7 Maxima [A] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\frac {1}{4} \, e^{\left (e^{\left (\frac {1}{4} \, e^{\left (\frac {e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )}}{x} + 3\right )}\right )}\right )} \]

input

integrate(1/8*((1-x)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x 
)-x)))*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp( 
exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x))) 
-x*log(2)+3*x)/x)))/x^2,x, algorithm=\
 

output

1/4*e^(e^(1/4*e^(e^(e^(x*e^(-x)))/x + 3)))
 

3.26.65.8 Giac [F]

\[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\int { -\frac {{\left ({\left (x - 1\right )} e^{\left (-x + e^{\left (-x + \log \left (x\right )\right )} + \log \left (x\right )\right )} + 1\right )} e^{\left (-\frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x} + e^{\left (\frac {1}{2} \, e^{\left (-\frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x}\right )}\right )} + \frac {1}{2} \, e^{\left (-\frac {x \log \left (2\right ) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{x}\right )} + e^{\left (e^{\left (-x + \log \left (x\right )\right )}\right )}\right )}}{8 \, x^{2}} \,d x } \]

input

integrate(1/8*((1-x)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x 
)-x)))*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp( 
exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x))) 
-x*log(2)+3*x)/x)))/x^2,x, algorithm=\
 

output

integrate(-1/8*((x - 1)*e^(-x + e^(-x + log(x)) + log(x)) + 1)*e^(-(x*log( 
2) - 3*x - e^(e^(e^(-x + log(x)))))/x + e^(1/2*e^(-(x*log(2) - 3*x - e^(e^ 
(e^(-x + log(x)))))/x)) + 1/2*e^(-(x*log(2) - 3*x - e^(e^(e^(-x + log(x))) 
))/x) + e^(e^(-x + log(x))))/x^2, x)
 

3.26.65.9 Mupad [B] (verification not implemented)

Time = 12.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}}{x}}}{4}}}}{4} \]

input

int(-(exp(exp((3*x + exp(exp(exp(log(x) - x))) - x*log(2))/x)/2)*exp(exp(e 
xp(log(x) - x)))*exp((3*x + exp(exp(exp(log(x) - x))) - x*log(2))/x)*exp(e 
xp(exp((3*x + exp(exp(exp(log(x) - x))) - x*log(2))/x)/2))*(exp(log(x) - x 
)*exp(exp(log(x) - x))*(x - 1) + 1))/(8*x^2),x)
 

output

exp(exp((exp(3)*exp(exp(exp(x*exp(-x)))/x))/4))/4