3.30.97 \(\int \frac {e^{-e^x-2 x} (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} (e^{3 x} x^2+e^{2 x} (-x+x^2)))}{x^3} \, dx\) [2997]

3.30.97.1 Optimal result

Integrand size = 74, antiderivative size = 34 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{-2 x}+\frac {e^{e^x+x}-\frac {5 e^{-e^x}}{x}}{x}+\log (3) \]

output

(exp(exp(x)+x)-5/x/exp(exp(x)))/x+ln(3)+25/exp(x)^2
 

3.30.97.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{-2 x}-\frac {5 e^{-e^x}}{x^2}+\frac {e^{e^x+x}}{x} \]

input

Integrate[(E^(-E^x - 2*x)*(10*E^(2*x) + 5*E^(3*x)*x - 50*E^E^x*x^3 + E^(2* 
E^x + x)*(E^(3*x)*x^2 + E^(2*x)*(-x + x^2))))/x^3,x]
 

output

25/E^(2*x) - 5/(E^E^x*x^2) + E^(E^x + x)/x
 

3.30.97.3 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.

\(\displaystyle \int \frac {e^{-2 x-e^x} \left (-50 e^{e^x} x^3+e^{x+2 e^x} \left (e^{3 x} x^2+e^{2 x} \left (x^2-x\right )\right )+5 e^{3 x} x+10 e^{2 x}\right )}{x^3} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {10 e^{-e^x}}{x^3}+\frac {e^{x-e^x} \left (e^{2 e^x} x-e^{2 e^x}+5\right )}{x^2}-50 e^{-2 x}+\frac {e^{-2 x-e^x+2 \left (2 x+e^x\right )}}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 10 \int \frac {e^{-e^x}}{x^3}dx+5 \int \frac {e^{x-e^x}}{x^2}dx-\int \frac {e^{x+e^x}}{x^2}dx+\int \frac {e^{x+e^x}}{x}dx+\int \frac {e^{2 x+e^x}}{x}dx+25 e^{-2 x}\)

input

Int[(E^(-E^x - 2*x)*(10*E^(2*x) + 5*E^(3*x)*x - 50*E^E^x*x^3 + E^(2*E^x + 
x)*(E^(3*x)*x^2 + E^(2*x)*(-x + x^2))))/x^3,x]
 

output

$Aborted
 

3.30.97.3.1 Defintions of rubi rules used

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] 
]
 

3.30.97.4 Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79

method	result	size
risch	\(25 \,{\mathrm e}^{-2 x}+\frac {{\mathrm e}^{{\mathrm e}^{x}+x}}{x}-\frac {5 \,{\mathrm e}^{-{\mathrm e}^{x}}}{x^{2}}\)	\(27\)
parallelrisch	\(\frac {\left ({\mathrm e}^{2 x} {\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{{\mathrm e}^{x}+x} x +25 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-{\mathrm e}^{x}} {\mathrm e}^{-2 x}}{x^{2}}\)	\(43\)

input

int(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^3*exp( 
exp(x))+5*x*exp(x)^3+10*exp(x)^2)/x^3/exp(x)^2/exp(exp(x)),x,method=_RETUR 
NVERBOSE)
 

output

25*exp(-2*x)+1/x*exp(exp(x)+x)-5/x^2*exp(-exp(x))
 

3.30.97.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\frac {{\left (25 \, x^{2} e^{\left (e^{x}\right )} + x e^{\left (3 \, x + 2 \, e^{x}\right )} - 5 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{2}} \]

input

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^ 
3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)^2)/x^3/exp(x)^2/exp(exp(x)),x, algori 
thm=\
 

output

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

3.30.97.6 Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{- 2 x} + \frac {x^{2} e^{x} e^{e^{x}} - 5 x e^{- e^{x}}}{x^{3}} \]

input

integrate(((x**2*exp(x)**3+(x**2-x)*exp(x)**2)*exp(exp(x))*exp(exp(x)+x)-5 
0*x**3*exp(exp(x))+5*x*exp(x)**3+10*exp(x)**2)/x**3/exp(x)**2/exp(exp(x)), 
x)
 

output

25*exp(-2*x) + (x**2*exp(x)*exp(exp(x)) - 5*x*exp(-exp(x)))/x**3
 

3.30.97.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\frac {x e^{\left (x + e^{x}\right )} - 5 \, e^{\left (-e^{x}\right )}}{x^{2}} + 25 \, e^{\left (-2 \, x\right )} \]

input

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^ 
3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)^2)/x^3/exp(x)^2/exp(exp(x)),x, algori 
thm=\
 

output

(x*e^(x + e^x) - 5*e^(-e^x))/x^2 + 25*e^(-2*x)
 

3.30.97.8 Giac [F]

\[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\int { -\frac {{\left (50 \, x^{3} e^{\left (e^{x}\right )} - 5 \, x e^{\left (3 \, x\right )} - {\left (x^{2} e^{\left (3 \, x\right )} + {\left (x^{2} - x\right )} e^{\left (2 \, x\right )}\right )} e^{\left (x + 2 \, e^{x}\right )} - 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{3}} \,d x } \]

input

integrate(((x^2*exp(x)^3+(x^2-x)*exp(x)^2)*exp(exp(x))*exp(exp(x)+x)-50*x^ 
3*exp(exp(x))+5*x*exp(x)^3+10*exp(x)^2)/x^3/exp(x)^2/exp(exp(x)),x, algori 
thm=\
 

output

integrate(-(50*x^3*e^(e^x) - 5*x*e^(3*x) - (x^2*e^(3*x) + (x^2 - x)*e^(2*x 
))*e^(x + 2*e^x) - 10*e^(2*x))*e^(-2*x - e^x)/x^3, x)
 

3.30.97.9 Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25\,{\mathrm {e}}^{-2\,x}+\frac {{\mathrm {e}}^{x+{\mathrm {e}}^x}}{x}-\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{x^2} \]

input

int((exp(-2*x)*exp(-exp(x))*(10*exp(2*x) + 5*x*exp(3*x) - 50*x^3*exp(exp(x 
)) + exp(x + exp(x))*exp(exp(x))*(x^2*exp(3*x) - exp(2*x)*(x - x^2))))/x^3 
,x)
 

output

25*exp(-2*x) + exp(x + exp(x))/x - (5*exp(-exp(x)))/x^2