3.15.61 \(\int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} (e^x (-125-25 x+80 x^2-20 x^3) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} (e^x (60 x-24 x^2)-60 e^x x \log (x)+e^x (375+75 x-240 x^2+60 x^3) \log ^2(x)))}{(125-100 x+20 x^2) \log ^2(x)} \, dx\) [1461]

3.15.61.1 Optimal result

Integrand size = 133, antiderivative size = 32 \[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=e^{e^x \left (-x+3 e^{\frac {4}{5 \left (2-\frac {5}{x}\right ) \log (x)}} x\right )} \]

output

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

3.15.61.2 Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=e^{e^x \left (-1+3 e^{\frac {4 x}{5 (-5+2 x) \log (x)}}\right ) x} \]

input

Integrate[(E^(-(E^x*x) + 3*E^(x + (4*x)/((-25 + 10*x)*Log[x]))*x)*(E^x*(-1 
25 - 25*x + 80*x^2 - 20*x^3)*Log[x]^2 + E^((4*x)/((-25 + 10*x)*Log[x]))*(E 
^x*(60*x - 24*x^2) - 60*E^x*x*Log[x] + E^x*(375 + 75*x - 240*x^2 + 60*x^3) 
*Log[x]^2)))/((125 - 100*x + 20*x^2)*Log[x]^2),x]
 

output

E^(E^x*(-1 + 3*E^((4*x)/(5*(-5 + 2*x)*Log[x])))*x)
 

3.15.61.3 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.

\(\displaystyle \int \frac {e^{3 x e^{x+\frac {4 x}{(10 x-25) \log (x)}}-e^x x} \left (e^x \left (-20 x^3+80 x^2-25 x-125\right ) \log ^2(x)+e^{\frac {4 x}{(10 x-25) \log (x)}} \left (e^x \left (60 x-24 x^2\right )+e^x \left (60 x^3-240 x^2+75 x+375\right ) \log ^2(x)-60 e^x x \log (x)\right )\right )}{\left (20 x^2-100 x+125\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 80 \int -\frac {\exp \left (3 e^{x-\frac {4 x}{5 (5-2 x) \log (x)}} x-e^x x\right ) \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-3 e^{-\frac {4 x}{5 (5-2 x) \log (x)}} \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-20 e^x x \log (x)+4 e^x \left (5 x-2 x^2\right )\right )\right )}{400 (5-2 x)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{5} \int \frac {\exp \left (3 e^{x-\frac {4 x}{5 (5-2 x) \log (x)}} x-e^x x\right ) \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-3 e^{-\frac {4 x}{5 (5-2 x) \log (x)}} \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-20 e^x x \log (x)+4 e^x \left (5 x-2 x^2\right )\right )\right )}{(5-2 x)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {1}{5} \int \frac {\exp \left (e^x \left (-1+3 e^{\frac {4 x}{5 (2 x-5) \log (x)}}\right ) x\right ) \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-3 e^{-\frac {4 x}{5 (5-2 x) \log (x)}} \left (5 e^x \left (4 x^3-16 x^2+5 x+25\right ) \log ^2(x)-20 e^x x \log (x)+4 e^x \left (5 x-2 x^2\right )\right )\right )}{(5-2 x)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{5} \int \left (5 \exp \left (e^x \left (-1+3 e^{\frac {4 x}{5 (2 x-5) \log (x)}}\right ) x+x\right ) (x+1)-\frac {3 \exp \left (e^x \left (-1+3 e^{\frac {4 x}{5 (2 x-5) \log (x)}}\right ) x+\frac {4 x}{5 (2 x-5) \log (x)}+x\right ) \left (20 \log ^2(x) x^3-80 \log ^2(x) x^2-8 x^2+25 \log ^2(x) x-20 \log (x) x+20 x+125 \log ^2(x)\right )}{(2 x-5)^2 \log ^2(x)}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle -\frac {1}{5} \int \left (5 \exp \left (e^x \left (-1+3 e^{\frac {4 x}{5 (2 x-5) \log (x)}}\right ) x+x\right ) (x+1)-\frac {3 \exp \left (e^x \left (-1+3 e^{\frac {4 x}{5 (2 x-5) \log (x)}}\right ) x+\frac {4 x}{5 (2 x-5) \log (x)}+x\right ) \left (20 \log ^2(x) x^3-80 \log ^2(x) x^2-8 x^2+25 \log ^2(x) x-20 \log (x) x+20 x+125 \log ^2(x)\right )}{(2 x-5)^2 \log ^2(x)}\right )dx\)

input

Int[(E^(-(E^x*x) + 3*E^(x + (4*x)/((-25 + 10*x)*Log[x]))*x)*(E^x*(-125 - 2 
5*x + 80*x^2 - 20*x^3)*Log[x]^2 + E^((4*x)/((-25 + 10*x)*Log[x]))*(E^x*(60 
*x - 24*x^2) - 60*E^x*x*Log[x] + E^x*(375 + 75*x - 240*x^2 + 60*x^3)*Log[x 
]^2)))/((125 - 100*x + 20*x^2)*Log[x]^2),x]
 

output

$Aborted
 

3.15.61.3.1 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 7277

Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> 
 Simp[1/(4^p*c^p)   Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} 
, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] &&  !AlgebraicFu 
nctionQ[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] 
]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

3.15.61.4 Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12

\[{\mathrm e}^{-x \left (-3 \,{\mathrm e}^{\frac {x \left (10 x \ln \left (x \right )-25 \ln \left (x \right )+4\right )}{5 \left (-5+2 x \right ) \ln \left (x \right )}}+{\mathrm e}^{x}\right )}\]

input

int((((60*x^3-240*x^2+75*x+375)*exp(x)*ln(x)^2-60*x*exp(x)*ln(x)+(-24*x^2+ 
60*x)*exp(x))*exp(4*x/(10*x-25)/ln(x))+(-20*x^3+80*x^2-25*x-125)*exp(x)*ln 
(x)^2)*exp(3*x*exp(x)*exp(4*x/(10*x-25)/ln(x))-exp(x)*x)/(20*x^2-100*x+125 
)/ln(x)^2,x)
 

output

exp(-x*(-3*exp(1/5*x*(10*x*ln(x)-25*ln(x)+4)/(-5+2*x)/ln(x))+exp(x)))
 

3.15.61.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=e^{\left (-x e^{x} + 3 \, x e^{\left (\frac {5 \, {\left (2 \, x^{2} - 5 \, x\right )} \log \left (x\right ) + 4 \, x}{5 \, {\left (2 \, x - 5\right )} \log \left (x\right )}\right )}\right )} \]

input

integrate((((60*x^3-240*x^2+75*x+375)*exp(x)*log(x)^2-60*x*exp(x)*log(x)+( 
-24*x^2+60*x)*exp(x))*exp(4*x/(10*x-25)/log(x))+(-20*x^3+80*x^2-25*x-125)* 
exp(x)*log(x)^2)*exp(3*x*exp(x)*exp(4*x/(10*x-25)/log(x))-exp(x)*x)/(20*x^ 
2-100*x+125)/log(x)^2,x, algorithm=\
 

output

e^(-x*e^x + 3*x*e^(1/5*(5*(2*x^2 - 5*x)*log(x) + 4*x)/((2*x - 5)*log(x))))
 

3.15.61.6 Sympy [A] (verification not implemented)

Time = 37.78 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=e^{3 x e^{x} e^{\frac {4 x}{\left (10 x - 25\right ) \log {\left (x \right )}}} - x e^{x}} \]

input

integrate((((60*x**3-240*x**2+75*x+375)*exp(x)*ln(x)**2-60*x*exp(x)*ln(x)+ 
(-24*x**2+60*x)*exp(x))*exp(4*x/(10*x-25)/ln(x))+(-20*x**3+80*x**2-25*x-12 
5)*exp(x)*ln(x)**2)*exp(3*x*exp(x)*exp(4*x/(10*x-25)/ln(x))-exp(x)*x)/(20* 
x**2-100*x+125)/ln(x)**2,x)
 

output

exp(3*x*exp(x)*exp(4*x/((10*x - 25)*log(x))) - x*exp(x))
 

3.15.61.7 Maxima [F]

\[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=\int { -\frac {{\left (5 \, {\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} \log \left (x\right )^{2} - 3 \, {\left (5 \, {\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} \log \left (x\right )^{2} - 20 \, x e^{x} \log \left (x\right ) - 4 \, {\left (2 \, x^{2} - 5 \, x\right )} e^{x}\right )} e^{\left (\frac {4 \, x}{5 \, {\left (2 \, x - 5\right )} \log \left (x\right )}\right )}\right )} e^{\left (3 \, x e^{\left (x + \frac {4 \, x}{5 \, {\left (2 \, x - 5\right )} \log \left (x\right )}\right )} - x e^{x}\right )}}{5 \, {\left (4 \, x^{2} - 20 \, x + 25\right )} \log \left (x\right )^{2}} \,d x } \]

input

integrate((((60*x^3-240*x^2+75*x+375)*exp(x)*log(x)^2-60*x*exp(x)*log(x)+( 
-24*x^2+60*x)*exp(x))*exp(4*x/(10*x-25)/log(x))+(-20*x^3+80*x^2-25*x-125)* 
exp(x)*log(x)^2)*exp(3*x*exp(x)*exp(4*x/(10*x-25)/log(x))-exp(x)*x)/(20*x^ 
2-100*x+125)/log(x)^2,x, algorithm=\
 

output

e^(3*x*e^(x + 2/5/log(x) + 2/((2*x - 5)*log(x))) - x*e^x) - 1/5*integrate( 
0, x)
 

3.15.61.8 Giac [F]

\[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx=\int { -\frac {{\left (5 \, {\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} \log \left (x\right )^{2} - 3 \, {\left (5 \, {\left (4 \, x^{3} - 16 \, x^{2} + 5 \, x + 25\right )} e^{x} \log \left (x\right )^{2} - 20 \, x e^{x} \log \left (x\right ) - 4 \, {\left (2 \, x^{2} - 5 \, x\right )} e^{x}\right )} e^{\left (\frac {4 \, x}{5 \, {\left (2 \, x - 5\right )} \log \left (x\right )}\right )}\right )} e^{\left (3 \, x e^{\left (x + \frac {4 \, x}{5 \, {\left (2 \, x - 5\right )} \log \left (x\right )}\right )} - x e^{x}\right )}}{5 \, {\left (4 \, x^{2} - 20 \, x + 25\right )} \log \left (x\right )^{2}} \,d x } \]

input

integrate((((60*x^3-240*x^2+75*x+375)*exp(x)*log(x)^2-60*x*exp(x)*log(x)+( 
-24*x^2+60*x)*exp(x))*exp(4*x/(10*x-25)/log(x))+(-20*x^3+80*x^2-25*x-125)* 
exp(x)*log(x)^2)*exp(3*x*exp(x)*exp(4*x/(10*x-25)/log(x))-exp(x)*x)/(20*x^ 
2-100*x+125)/log(x)^2,x, algorithm=\
 

output

undef
 

3.15.61.9 Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x x+3 e^{x+\frac {4 x}{(-25+10 x) \log (x)}} x} \left (e^x \left (-125-25 x+80 x^2-20 x^3\right ) \log ^2(x)+e^{\frac {4 x}{(-25+10 x) \log (x)}} \left (e^x \left (60 x-24 x^2\right )-60 e^x x \log (x)+e^x \left (375+75 x-240 x^2+60 x^3\right ) \log ^2(x)\right )\right )}{\left (125-100 x+20 x^2\right ) \log ^2(x)} \, dx={\mathrm {e}}^{3\,x\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {4\,x}{25\,\ln \left (x\right )-10\,x\,\ln \left (x\right )}}}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^x} \]

input

int((exp(3*x*exp((4*x)/(log(x)*(10*x - 25)))*exp(x) - x*exp(x))*(exp((4*x) 
/(log(x)*(10*x - 25)))*(exp(x)*(60*x - 24*x^2) + exp(x)*log(x)^2*(75*x - 2 
40*x^2 + 60*x^3 + 375) - 60*x*exp(x)*log(x)) - exp(x)*log(x)^2*(25*x - 80* 
x^2 + 20*x^3 + 125)))/(log(x)^2*(20*x^2 - 100*x + 125)),x)
 

output

exp(3*x*exp(x)*exp(-(4*x)/(25*log(x) - 10*x*log(x))))*exp(-x*exp(x))