\(\int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} (-2 e^x-x)}{(2 e^x+x)^2}+e^x (-28-32 x-8 x^2)+\frac {e^x (1+7 x+4 x^2+e^x (16+8 x))}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x (e^x (-16-8 x)-8 x-4 x^2)}{2 e^x+x}+e^x (32+32 x+8 x^2)} \, dx\) [1337]

Optimal result

Integrand size = 161, antiderivative size = 28 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=40-x-\frac {1}{4+2 x-\frac {e^x}{2 e^x+x}} \] Output:

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

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=-x-\frac {2}{7+4 x}+\frac {x}{(7+4 x) \left (2 x (2+x)+e^x (7+4 x)\right )} \] Input:

Integrate[(-14*x - 16*x^2 - 4*x^3 + (E^(2*x)*(-2*E^x - x))/(2*E^x + x)^2 + 
 E^x*(-28 - 32*x - 8*x^2) + (E^x*(1 + 7*x + 4*x^2 + E^x*(16 + 8*x)))/(2*E^ 
x + x))/(16*x + 16*x^2 + 4*x^3 + E^(2*x)/(2*E^x + x) + (E^x*(E^x*(-16 - 8* 
x) - 8*x - 4*x^2))/(2*E^x + x) + E^x*(32 + 32*x + 8*x^2)),x]
 

Output:

-x - 2/(7 + 4*x) + x/((7 + 4*x)*(2*x*(2 + x) + E^x*(7 + 4*x)))
 

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.

\(\displaystyle \int \frac {-4 x^3-16 x^2+e^x \left (-8 x^2-32 x-28\right )+\frac {e^x \left (4 x^2+7 x+e^x (8 x+16)+1\right )}{x+2 e^x}-14 x+\frac {e^{2 x} \left (-x-2 e^x\right )}{\left (x+2 e^x\right )^2}}{4 x^3+16 x^2+\frac {e^x \left (-4 x^2-8 x+e^x (-8 x-16)\right )}{x+2 e^x}+e^x \left (8 x^2+32 x+32\right )+16 x+\frac {e^{2 x}}{x+2 e^x}} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {-2 \left (2 x^2+8 x+7\right ) x^2-e^{2 x} \left (16 x^2+56 x+41\right )-e^x \left (16 x^3+60 x^2+49 x-1\right )}{\left (2 x (x+2)+e^x (4 x+7)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-16 x^2-56 x-41}{(4 x+7)^2}-\frac {4 x^2+11 x-7}{(4 x+7)^2 \left (2 x^2+4 e^x x+4 x+7 e^x\right )}+\frac {2 x \left (4 x^3+11 x^2-14\right )}{(4 x+7)^2 \left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {7}{32} \int \frac {1}{\left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}dx-\frac {3}{8} \int \frac {x}{\left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}dx+\frac {1}{2} \int \frac {x^2}{\left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}dx+\frac {49}{8} \int \frac {1}{(4 x+7)^2 \left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}dx+\frac {21}{32} \int \frac {1}{(4 x+7) \left (2 x^2+4 e^x x+4 x+7 e^x\right )^2}dx-\frac {1}{4} \int \frac {1}{2 x^2+4 e^x x+4 x+7 e^x}dx+14 \int \frac {1}{(4 x+7)^2 \left (2 x^2+4 e^x x+4 x+7 e^x\right )}dx+\frac {3}{4} \int \frac {1}{(4 x+7) \left (2 x^2+4 e^x x+4 x+7 e^x\right )}dx-x-\frac {2}{4 x+7}\)

Input:

Int[(-14*x - 16*x^2 - 4*x^3 + (E^(2*x)*(-2*E^x - x))/(2*E^x + x)^2 + E^x*( 
-28 - 32*x - 8*x^2) + (E^x*(1 + 7*x + 4*x^2 + E^x*(16 + 8*x)))/(2*E^x + x) 
)/(16*x + 16*x^2 + 4*x^3 + E^(2*x)/(2*E^x + x) + (E^x*(E^x*(-16 - 8*x) - 8 
*x - 4*x^2))/(2*E^x + x) + E^x*(32 + 32*x + 8*x^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46

method	result	size
risch	\(-x -\frac {1}{2 \left (\frac {7}{4}+x \right )}+\frac {x}{\left (7+4 x \right ) \left (2 x^{2}+4 \,{\mathrm e}^{x} x +4 x +7 \,{\mathrm e}^{x}\right )}\)	\(41\)
parallelrisch	\(\frac {-8 x^{2}+4 \,{\mathrm e}^{-\ln \left (2 \,{\mathrm e}^{x}+x \right )+x} x -4-16 x}{8 x -4 \,{\mathrm e}^{-\ln \left (2 \,{\mathrm e}^{x}+x \right )+x}+16}\)	\(49\)
norman	\(\frac {7 x^{2}+24 \,{\mathrm e}^{2 x}+{\mathrm e}^{x} x^{2}+2 x \,{\mathrm e}^{2 x}+26 \,{\mathrm e}^{x} x -2 x^{4}-8 \,{\mathrm e}^{x} x^{3}-8 \,{\mathrm e}^{2 x} x^{2}}{\left (2 \,{\mathrm e}^{x}+x \right ) \left (2 x^{2}+4 \,{\mathrm e}^{x} x +4 x +7 \,{\mathrm e}^{x}\right )}\)	\(81\)

Input:

int(((-2*exp(x)-x)*exp(-ln(2*exp(x)+x)+x)^2+((8*x+16)*exp(x)+4*x^2+7*x+1)* 
exp(-ln(2*exp(x)+x)+x)+(-8*x^2-32*x-28)*exp(x)-4*x^3-16*x^2-14*x)/((2*exp( 
x)+x)*exp(-ln(2*exp(x)+x)+x)^2+((-8*x-16)*exp(x)-4*x^2-8*x)*exp(-ln(2*exp( 
x)+x)+x)+(8*x^2+32*x+32)*exp(x)+4*x^3+16*x^2+16*x),x,method=_RETURNVERBOSE 
)
 

Output:

-x-1/2/(7/4+x)+x/(7+4*x)/(2*x^2+4*exp(x)*x+4*x+7*exp(x))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=-\frac {2 \, x^{3} + 4 \, x^{2} + {\left (4 \, x^{2} + 7 \, x + 2\right )} e^{x} + x}{2 \, x^{2} + {\left (4 \, x + 7\right )} e^{x} + 4 \, x} \] Input:

integrate(((-2*exp(x)-x)*exp(-log(2*exp(x)+x)+x)^2+((8*x+16)*exp(x)+4*x^2+ 
7*x+1)*exp(-log(2*exp(x)+x)+x)+(-8*x^2-32*x-28)*exp(x)-4*x^3-16*x^2-14*x)/ 
((2*exp(x)+x)*exp(-log(2*exp(x)+x)+x)^2+((-8*x-16)*exp(x)-4*x^2-8*x)*exp(- 
log(2*exp(x)+x)+x)+(8*x^2+32*x+32)*exp(x)+4*x^3+16*x^2+16*x),x, algorithm= 
"fricas")
 

Output:

-(2*x^3 + 4*x^2 + (4*x^2 + 7*x + 2)*e^x + x)/(2*x^2 + (4*x + 7)*e^x + 4*x)
 

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=- x + \frac {x}{8 x^{3} + 30 x^{2} + 28 x + \left (16 x^{2} + 56 x + 49\right ) e^{x}} - \frac {2}{4 x + 7} \] Input:

integrate(((-2*exp(x)-x)*exp(-ln(2*exp(x)+x)+x)**2+((8*x+16)*exp(x)+4*x**2 
+7*x+1)*exp(-ln(2*exp(x)+x)+x)+(-8*x**2-32*x-28)*exp(x)-4*x**3-16*x**2-14* 
x)/((2*exp(x)+x)*exp(-ln(2*exp(x)+x)+x)**2+((-8*x-16)*exp(x)-4*x**2-8*x)*e 
xp(-ln(2*exp(x)+x)+x)+(8*x**2+32*x+32)*exp(x)+4*x**3+16*x**2+16*x),x)
 

Output:

-x + x/(8*x**3 + 30*x**2 + 28*x + (16*x**2 + 56*x + 49)*exp(x)) - 2/(4*x + 
 7)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=-\frac {2 \, x^{3} + 4 \, x^{2} + {\left (4 \, x^{2} + 7 \, x + 2\right )} e^{x} + x}{2 \, x^{2} + {\left (4 \, x + 7\right )} e^{x} + 4 \, x} \] Input:

integrate(((-2*exp(x)-x)*exp(-log(2*exp(x)+x)+x)^2+((8*x+16)*exp(x)+4*x^2+ 
7*x+1)*exp(-log(2*exp(x)+x)+x)+(-8*x^2-32*x-28)*exp(x)-4*x^3-16*x^2-14*x)/ 
((2*exp(x)+x)*exp(-log(2*exp(x)+x)+x)^2+((-8*x-16)*exp(x)-4*x^2-8*x)*exp(- 
log(2*exp(x)+x)+x)+(8*x^2+32*x+32)*exp(x)+4*x^3+16*x^2+16*x),x, algorithm= 
"maxima")
 

Output:

-(2*x^3 + 4*x^2 + (4*x^2 + 7*x + 2)*e^x + x)/(2*x^2 + (4*x + 7)*e^x + 4*x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=-\frac {2 \, x^{3} + 4 \, x^{2} e^{x} + 4 \, x^{2} + 7 \, x e^{x} + x + 2 \, e^{x}}{2 \, x^{2} + 4 \, x e^{x} + 4 \, x + 7 \, e^{x}} \] Input:

integrate(((-2*exp(x)-x)*exp(-log(2*exp(x)+x)+x)^2+((8*x+16)*exp(x)+4*x^2+ 
7*x+1)*exp(-log(2*exp(x)+x)+x)+(-8*x^2-32*x-28)*exp(x)-4*x^3-16*x^2-14*x)/ 
((2*exp(x)+x)*exp(-log(2*exp(x)+x)+x)^2+((-8*x-16)*exp(x)-4*x^2-8*x)*exp(- 
log(2*exp(x)+x)+x)+(8*x^2+32*x+32)*exp(x)+4*x^3+16*x^2+16*x),x, algorithm= 
"giac")
 

Output:

-(2*x^3 + 4*x^2*e^x + 4*x^2 + 7*x*e^x + x + 2*e^x)/(2*x^2 + 4*x*e^x + 4*x 
+ 7*e^x)
 

Mupad [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=-x-\frac {x+2\,{\mathrm {e}}^x}{4\,x+7\,{\mathrm {e}}^x+4\,x\,{\mathrm {e}}^x+2\,x^2} \] Input:

int(-(14*x - exp(x - log(x + 2*exp(x)))*(7*x + exp(x)*(8*x + 16) + 4*x^2 + 
 1) + exp(x)*(32*x + 8*x^2 + 28) + 16*x^2 + 4*x^3 + exp(2*x - 2*log(x + 2* 
exp(x)))*(x + 2*exp(x)))/(16*x - exp(x - log(x + 2*exp(x)))*(8*x + exp(x)* 
(8*x + 16) + 4*x^2) + exp(x)*(32*x + 8*x^2 + 32) + 16*x^2 + 4*x^3 + exp(2* 
x - 2*log(x + 2*exp(x)))*(x + 2*exp(x))),x)
 

Output:

- x - (x + 2*exp(x))/(4*x + 7*exp(x) + 4*x*exp(x) + 2*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-14 x-16 x^2-4 x^3+\frac {e^{2 x} \left (-2 e^x-x\right )}{\left (2 e^x+x\right )^2}+e^x \left (-28-32 x-8 x^2\right )+\frac {e^x \left (1+7 x+4 x^2+e^x (16+8 x)\right )}{2 e^x+x}}{16 x+16 x^2+4 x^3+\frac {e^{2 x}}{2 e^x+x}+\frac {e^x \left (e^x (-16-8 x)-8 x-4 x^2\right )}{2 e^x+x}+e^x \left (32+32 x+8 x^2\right )} \, dx=\frac {x \left (-28 e^{x} x -41 e^{x}-14 x^{2}-24 x +1\right )}{28 e^{x} x +49 e^{x}+14 x^{2}+28 x} \] Input:

int(((-2*exp(x)-x)*exp(-log(2*exp(x)+x)+x)^2+((8*x+16)*exp(x)+4*x^2+7*x+1) 
*exp(-log(2*exp(x)+x)+x)+(-8*x^2-32*x-28)*exp(x)-4*x^3-16*x^2-14*x)/((2*ex 
p(x)+x)*exp(-log(2*exp(x)+x)+x)^2+((-8*x-16)*exp(x)-4*x^2-8*x)*exp(-log(2* 
exp(x)+x)+x)+(8*x^2+32*x+32)*exp(x)+4*x^3+16*x^2+16*x),x)
 

Output:

(x*( - 28*e**x*x - 41*e**x - 14*x**2 - 24*x + 1))/(7*(4*e**x*x + 7*e**x + 
2*x**2 + 4*x))