\(\int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 (-4 e^4+4 e^8-28 x^2-8 x^3)+e^{10} (-e^4+e^8-23 x^2-2 x^3)}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} (24 x^2+2 x^3)+e^5 (32 x^2+8 x^3)} \, dx\) [2504]

Optimal result

Integrand size = 160, antiderivative size = 34 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x+\frac {\frac {e^4-e^8-x}{x}+x}{\left (2+e^5\right )^2+x} \] Output:

(x+(exp(4)-exp(4)^2-x)/x)/((exp(5)+2)^2+x)-x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.59 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-\frac {-e^4+e^8+4 e^5 x (1+x)+e^{10} x (1+x)+x \left (5+4 x+x^2\right )}{x \left (4+4 e^5+e^{10}+x\right )} \] Input:

Integrate[(E^4*(-4 - 2*x) - 11*x^2 - 8*E^15*x^2 - E^20*x^2 - 8*x^3 - x^4 + 
 E^8*(4 + 2*x) + E^5*(-4*E^4 + 4*E^8 - 28*x^2 - 8*x^3) + E^10*(-E^4 + E^8 
- 23*x^2 - 2*x^3))/(16*x^2 + 8*E^15*x^2 + E^20*x^2 + 8*x^3 + x^4 + E^10*(2 
4*x^2 + 2*x^3) + E^5*(32*x^2 + 8*x^3)),x]
 

Output:

-((-E^4 + E^8 + 4*E^5*x*(1 + x) + E^10*x*(1 + x) + x*(5 + 4*x + x^2))/(x*( 
4 + 4*E^5 + E^10 + x)))
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(34)=68\).

Time = 0.47 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.15, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6, 6, 6, 6, 2026, 2007, 2123, 2009}

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.

\(\displaystyle \int \frac {-x^4-8 x^3-e^{20} x^2-8 e^{15} x^2-11 x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}\right ) x^2-e^{20} x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+\left (16+8 e^{15}\right ) x^2+e^{20} x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+\left (16+8 e^{15}+e^{20}\right ) x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^2 \left (x^2+2 \left (2+e^5\right )^2 x+\left (2+e^5\right )^4\right )}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^2 \left (x+\left (2+e^5\right )^2\right )^2}dx\)

\(\Big \downarrow \) 2123

\(\displaystyle \int \left (\frac {e^4 \left (e^4-1\right )}{\left (2+e^5\right )^2 x^2}+\frac {20+e^4+36 e^5-e^8+25 e^{10}+8 e^{15}+e^{20}}{\left (2+e^5\right )^2 \left (x+e^{10}+4 e^5+4\right )^2}-1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -x-\frac {20+e^4+36 e^5-e^8+25 e^{10}+8 e^{15}+e^{20}}{\left (2+e^5\right )^2 \left (x+\left (2+e^5\right )^2\right )}+\frac {e^4 \left (1-e^4\right )}{\left (2+e^5\right )^2 x}\)

Input:

Int[(E^4*(-4 - 2*x) - 11*x^2 - 8*E^15*x^2 - E^20*x^2 - 8*x^3 - x^4 + E^8*( 
4 + 2*x) + E^5*(-4*E^4 + 4*E^8 - 28*x^2 - 8*x^3) + E^10*(-E^4 + E^8 - 23*x 
^2 - 2*x^3))/(16*x^2 + 8*E^15*x^2 + E^20*x^2 + 8*x^3 + x^4 + E^10*(24*x^2 
+ 2*x^3) + E^5*(32*x^2 + 8*x^3)),x]
 

Output:

(E^4*(1 - E^4))/((2 + E^5)^2*x) - x - (20 + E^4 + 36*E^5 - E^8 + 25*E^10 + 
 8*E^15 + E^20)/((2 + E^5)^2*((2 + E^5)^2 + x))
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 2007

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex 
pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol 
yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2009

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

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15

method	result	size
risch	\(-x +\frac {\left (-4 \,{\mathrm e}^{5}-{\mathrm e}^{10}-5\right ) x -{\mathrm e}^{8}+{\mathrm e}^{4}}{\left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right ) x}\)	\(39\)
norman	\(\frac {\left (28 \,{\mathrm e}^{5}+23 \,{\mathrm e}^{10}+{\mathrm e}^{20}+8 \,{\mathrm e}^{15}+11\right ) x -x^{3}-{\mathrm e}^{8}+{\mathrm e}^{4}}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\)	\(56\)
gosper	\(\frac {{\mathrm e}^{20} x +8 x \,{\mathrm e}^{15}+23 x \,{\mathrm e}^{10}-x^{3}+28 x \,{\mathrm e}^{5}-{\mathrm e}^{8}+{\mathrm e}^{4}+11 x}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\)	\(60\)
parallelrisch	\(\frac {{\mathrm e}^{20} x +8 x \,{\mathrm e}^{15}+23 x \,{\mathrm e}^{10}-x^{3}+28 x \,{\mathrm e}^{5}-{\mathrm e}^{8}+{\mathrm e}^{4}+11 x}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\)	\(60\)

Input:

int((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp(5)^2+ 
(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)*exp(4) 
-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5)^2+(8 
*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x,method=_RETURNVERBOSE)
 

Output:

-x+((-4*exp(5)-exp(10)-5)*x-exp(8)+exp(4))/(exp(10)+4*exp(5)+x+4)/x
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-\frac {x^{3} + 4 \, x^{2} + {\left (x^{2} + x\right )} e^{10} + 4 \, {\left (x^{2} + x\right )} e^{5} + 5 \, x + e^{8} - e^{4}}{x^{2} + x e^{10} + 4 \, x e^{5} + 4 \, x} \] Input:

integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp 
(5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* 
exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 
)^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm="fricas")
 

Output:

-(x^3 + 4*x^2 + (x^2 + x)*e^10 + 4*(x^2 + x)*e^5 + 5*x + e^8 - e^4)/(x^2 + 
 x*e^10 + 4*x*e^5 + 4*x)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 3.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=- x - \frac {x \left (5 + 4 e^{5} + e^{10}\right ) - e^{4} + e^{8}}{x^{2} + x \left (4 + 4 e^{5} + e^{10}\right )} \] Input:

integrate((-x**2*exp(5)**4-8*x**2*exp(5)**3+(exp(4)**2-exp(4)-2*x**3-23*x* 
*2)*exp(5)**2+(4*exp(4)**2-4*exp(4)-8*x**3-28*x**2)*exp(5)+(4+2*x)*exp(4)* 
*2+(-2*x-4)*exp(4)-x**4-8*x**3-11*x**2)/(x**2*exp(5)**4+8*x**2*exp(5)**3+( 
2*x**3+24*x**2)*exp(5)**2+(8*x**3+32*x**2)*exp(5)+x**4+8*x**3+16*x**2),x)
 

Output:

-x - (x*(5 + 4*exp(5) + exp(10)) - exp(4) + exp(8))/(x**2 + x*(4 + 4*exp(5 
) + exp(10)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x - \frac {x {\left (e^{10} + 4 \, e^{5} + 5\right )} + e^{8} - e^{4}}{x^{2} + x {\left (e^{10} + 4 \, e^{5} + 4\right )}} \] Input:

integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp 
(5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* 
exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 
)^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm="maxima")
 

Output:

-x - (x*(e^10 + 4*e^5 + 5) + e^8 - e^4)/(x^2 + x*(e^10 + 4*e^5 + 4))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x - \frac {x e^{10} + 4 \, x e^{5} + 5 \, x + e^{8} - e^{4}}{x^{2} + x e^{10} + 4 \, x e^{5} + 4 \, x} \] Input:

integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp 
(5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* 
exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 
)^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm="giac")
 

Output:

-x - (x*e^10 + 4*x*e^5 + 5*x + e^8 - e^4)/(x^2 + x*e^10 + 4*x*e^5 + 4*x)
 

Mupad [B] (verification not implemented)

Time = 2.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x-\frac {{\mathrm {e}}^8-{\mathrm {e}}^4+x\,\left (4\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+5\right )}{x\,\left (x+4\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+4\right )} \] Input:

int(-(exp(5)*(4*exp(4) - 4*exp(8) + 28*x^2 + 8*x^3) + 8*x^2*exp(15) + x^2* 
exp(20) + exp(10)*(exp(4) - exp(8) + 23*x^2 + 2*x^3) + 11*x^2 + 8*x^3 + x^ 
4 + exp(4)*(2*x + 4) - exp(8)*(2*x + 4))/(exp(10)*(24*x^2 + 2*x^3) + exp(5 
)*(32*x^2 + 8*x^3) + 8*x^2*exp(15) + x^2*exp(20) + 16*x^2 + 8*x^3 + x^4),x 
)
 

Output:

- x - (exp(8) - exp(4) + x*(4*exp(5) + exp(10) + 5))/(x*(x + 4*exp(5) + ex 
p(10) + 4))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 3.74 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=\frac {-e^{20} x^{2}-e^{18}-8 e^{15} x^{2}+e^{14}-4 e^{13}-e^{10} x^{3}-23 e^{10} x^{2}+4 e^{9}-4 e^{8}-4 e^{5} x^{3}-28 e^{5} x^{2}+4 e^{4}-4 x^{3}-11 x^{2}}{x \left (e^{20}+8 e^{15}+e^{10} x +24 e^{10}+4 e^{5} x +32 e^{5}+4 x +16\right )} \] Input:

int((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp(5)^2+ 
(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)*exp(4) 
-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5)^2+(8 
*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x)
 

Output:

( - e**20*x**2 - e**18 - 8*e**15*x**2 + e**14 - 4*e**13 - e**10*x**3 - 23* 
e**10*x**2 + 4*e**9 - 4*e**8 - 4*e**5*x**3 - 28*e**5*x**2 + 4*e**4 - 4*x** 
3 - 11*x**2)/(x*(e**20 + 8*e**15 + e**10*x + 24*e**10 + 4*e**5*x + 32*e**5 
 + 4*x + 16))