3.14.40 \(\int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} (324-4 e^3+e^{2+x} (-81 x+e^3 x))}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx\) [1340]

3.14.40.1 Optimal result

Integrand size = 84, antiderivative size = 27 \[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=\frac {\left (-81+e^3\right ) x}{-2-e^{\frac {e^{2+x}}{4}}+2 x} \]

output

(exp(3)-81)/(2*x-2-exp(1/4*exp(2+x)))*x
 

3.14.40.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=-\frac {\left (-81+e^3\right ) x}{2+e^{\frac {e^{2+x}}{4}}-2 x} \]

input

Integrate[(648 - 8*E^3 + E^(E^(2 + x)/4)*(324 - 4*E^3 + E^(2 + x)*(-81*x + 
 E^3*x)))/(16 + 4*E^(E^(2 + x)/2) + E^(E^(2 + x)/4)*(16 - 16*x) - 32*x + 1 
6*x^2),x]
 

output

-(((-81 + E^3)*x)/(2 + E^(E^(2 + x)/4) - 2*x))
 

3.14.40.3 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.

\(\displaystyle \int \frac {e^{\frac {e^{x+2}}{4}} \left (e^{x+2} \left (e^3 x-81 x\right )-4 e^3+324\right )-8 e^3+648}{16 x^2-32 x+4 e^{\frac {e^{x+2}}{2}}+e^{\frac {e^{x+2}}{4}} (16-16 x)+16} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (81-e^3\right ) \left (-e^{x+\frac {e^{x+2}}{4}+2} x+4 e^{\frac {e^{x+2}}{4}}+8\right )}{4 \left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (81-e^3\right ) \int \frac {-e^{x+\frac {e^{x+2}}{4}+2} x+4 e^{\frac {e^{x+2}}{4}}+8}{\left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{4} \left (81-e^3\right ) \int \left (\frac {4 \left (2+e^{\frac {e^{x+2}}{4}}\right )}{\left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}-\frac {e^{\frac {1}{4} \left (4 x+e^{x+2}+8\right )} x}{\left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (81-e^3\right ) \left (4 \int \frac {1}{-2 x+e^{\frac {e^{x+2}}{4}}+2}dx+8 \int \frac {x}{\left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}dx-\int \frac {e^{\frac {1}{4} \left (4 x+e^{x+2}+8\right )} x}{\left (-2 x+e^{\frac {e^{x+2}}{4}}+2\right )^2}dx\right )\)

input

Int[(648 - 8*E^3 + E^(E^(2 + x)/4)*(324 - 4*E^3 + E^(2 + x)*(-81*x + E^3*x 
)))/(16 + 4*E^(E^(2 + x)/2) + E^(E^(2 + x)/4)*(16 - 16*x) - 32*x + 16*x^2) 
,x]
 

output

$Aborted
 

3.14.40.3.1 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 2009

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

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] 
]
 

3.14.40.4 Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left ({\mathrm e}^{3}-81\right ) x}{2 x -2-{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}}\)	\(23\)
parallelrisch	\(\frac {4 x \,{\mathrm e}^{3}-324 x}{8 x -8-4 \,{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}}\)	\(28\)
norman	\(\frac {\left (-\frac {81}{2}+\frac {{\mathrm e}^{3}}{2}\right ) {\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}-81+{\mathrm e}^{3}}{2 x -2-{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{4}}}\)	\(36\)

input

int((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp(3)+64 
8)/(4*exp(1/4*exp(2+x))^2+(-16*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+16),x,m 
ethod=_RETURNVERBOSE)
 

output

(exp(3)-81)/(2*x-2-exp(1/4*exp(2+x)))*x
 

3.14.40.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=\frac {x e^{3} - 81 \, x}{2 \, x - e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 2} \]

input

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp 
(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+1 
6),x, algorithm=\
 

output

(x*e^3 - 81*x)/(2*x - e^(1/4*e^(x + 2)) - 2)
 

3.14.40.6 Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=\frac {- x e^{3} + 81 x}{- 2 x + e^{\frac {e^{x + 2}}{4}} + 2} \]

input

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp 
(3)+648)/(4*exp(1/4*exp(2+x))**2+(-16*x+16)*exp(1/4*exp(2+x))+16*x**2-32*x 
+16),x)
 

output

(-x*exp(3) + 81*x)/(-2*x + exp(exp(x + 2)/4) + 2)
 

3.14.40.7 Maxima [F]

\[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=\int { \frac {{\left ({\left (x e^{3} - 81 \, x\right )} e^{\left (x + 2\right )} - 4 \, e^{3} + 324\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, e^{3} + 648}{4 \, {\left (4 \, x^{2} - 4 \, {\left (x - 1\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, x + e^{\left (\frac {1}{2} \, e^{\left (x + 2\right )}\right )} + 4\right )}} \,d x } \]

input

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp 
(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+1 
6),x, algorithm=\
 

output

1/4*integrate((((x*e^3 - 81*x)*e^(x + 2) - 4*e^3 + 324)*e^(1/4*e^(x + 2)) 
- 8*e^3 + 648)/(4*x^2 - 4*(x - 1)*e^(1/4*e^(x + 2)) - 8*x + e^(1/2*e^(x + 
2)) + 4), x)
 

3.14.40.8 Giac [F]

\[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=\int { \frac {{\left ({\left (x e^{3} - 81 \, x\right )} e^{\left (x + 2\right )} - 4 \, e^{3} + 324\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, e^{3} + 648}{4 \, {\left (4 \, x^{2} - 4 \, {\left (x - 1\right )} e^{\left (\frac {1}{4} \, e^{\left (x + 2\right )}\right )} - 8 \, x + e^{\left (\frac {1}{2} \, e^{\left (x + 2\right )}\right )} + 4\right )}} \,d x } \]

input

integrate((((x*exp(3)-81*x)*exp(2+x)-4*exp(3)+324)*exp(1/4*exp(2+x))-8*exp 
(3)+648)/(4*exp(1/4*exp(2+x))^2+(-16*x+16)*exp(1/4*exp(2+x))+16*x^2-32*x+1 
6),x, algorithm=\
 

output

integrate(1/4*(((x*e^3 - 81*x)*e^(x + 2) - 4*e^3 + 324)*e^(1/4*e^(x + 2)) 
- 8*e^3 + 648)/(4*x^2 - 4*(x - 1)*e^(1/4*e^(x + 2)) - 8*x + e^(1/2*e^(x + 
2)) + 4), x)
 

3.14.40.9 Mupad [F(-1)]

Timed out. \[ \int \frac {648-8 e^3+e^{\frac {e^{2+x}}{4}} \left (324-4 e^3+e^{2+x} \left (-81 x+e^3 x\right )\right )}{16+4 e^{\frac {e^{2+x}}{2}}+e^{\frac {e^{2+x}}{4}} (16-16 x)-32 x+16 x^2} \, dx=-\int \frac {8\,{\mathrm {e}}^3+{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{4}}\,\left (4\,{\mathrm {e}}^3+{\mathrm {e}}^{x+2}\,\left (81\,x-x\,{\mathrm {e}}^3\right )-324\right )-648}{4\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{2}}-32\,x-{\mathrm {e}}^{\frac {{\mathrm {e}}^{x+2}}{4}}\,\left (16\,x-16\right )+16\,x^2+16} \,d x \]

input

int(-(8*exp(3) + exp(exp(x + 2)/4)*(4*exp(3) + exp(x + 2)*(81*x - x*exp(3) 
) - 324) - 648)/(4*exp(exp(x + 2)/2) - 32*x - exp(exp(x + 2)/4)*(16*x - 16 
) + 16*x^2 + 16),x)
 

output

-int((8*exp(3) + exp(exp(x + 2)/4)*(4*exp(3) + exp(x + 2)*(81*x - x*exp(3) 
) - 324) - 648)/(4*exp(exp(x + 2)/2) - 32*x - exp(exp(x + 2)/4)*(16*x - 16 
) + 16*x^2 + 16), x)