\(\int \frac {32 e^8+128 x^4+e^4 (24+128 x^2)}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 (24 x^3+64 x^5)} \, dx\) [2232]

Optimal result

Integrand size = 64, antiderivative size = 26 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4}{4 x^2+\frac {3 x}{\frac {e^4}{x}+2 x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4 \left (e^4+2 x^2\right )}{x^2 \left (3+4 e^4+8 x^2\right )} \] Input:

Integrate[(32*E^8 + 128*x^4 + E^4*(24 + 128*x^2))/(9*x^3 + 16*E^8*x^3 + 48 
*x^5 + 64*x^7 + E^4*(24*x^3 + 64*x^5)),x]
 

Output:

(-4*(E^4 + 2*x^2))/(x^2*(3 + 4*E^4 + 8*x^2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 2026, 1380, 27, 2089, 1578, 1195, 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.

\(\displaystyle \int \frac {128 x^4+e^4 \left (128 x^2+24\right )+32 e^8}{64 x^7+48 x^5+16 e^8 x^3+9 x^3+e^4 \left (64 x^5+24 x^3\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {128 x^4+e^4 \left (128 x^2+24\right )+32 e^8}{64 x^7+48 x^5+\left (9+16 e^8\right ) x^3+e^4 \left (64 x^5+24 x^3\right )}dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {128 x^4+e^4 \left (128 x^2+24\right )+32 e^8}{x^3 \left (64 x^4+16 \left (3+4 e^4\right ) x^2+\left (3+4 e^4\right )^2\right )}dx\)

\(\Big \downarrow \) 1380

\(\displaystyle 64 \int \frac {16 x^4+e^4 \left (16 x^2+3\right )+4 e^8}{8 x^3 \left (8 x^2+4 e^4+3\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 8 \int \frac {16 x^4+e^4 \left (16 x^2+3\right )+4 e^8}{x^3 \left (8 x^2+4 e^4+3\right )^2}dx\)

\(\Big \downarrow \) 2089

\(\displaystyle 8 \int \frac {16 x^4+16 e^4 x^2+e^4 \left (3+4 e^4\right )}{x^3 \left (8 x^2+4 e^4+3\right )^2}dx\)

\(\Big \downarrow \) 1578

\(\displaystyle 4 \int \frac {16 x^4+16 e^4 x^2+e^4 \left (3+4 e^4\right )}{x^4 \left (8 x^2+4 e^4+3\right )^2}dx^2\)

\(\Big \downarrow \) 1195

\(\displaystyle 4 \int \left (\frac {e^4}{\left (3+4 e^4\right ) x^4}+\frac {48}{\left (3+4 e^4\right ) \left (8 x^2+4 e^4+3\right )^2}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \left (-\frac {e^4}{\left (3+4 e^4\right ) x^2}-\frac {6}{\left (3+4 e^4\right ) \left (8 x^2+4 e^4+3\right )}\right )\)

Input:

Int[(32*E^8 + 128*x^4 + E^4*(24 + 128*x^2))/(9*x^3 + 16*E^8*x^3 + 48*x^5 + 
 64*x^7 + E^4*(24*x^3 + 64*x^5)),x]
 

Output:

4*(-(E^4/((3 + 4*E^4)*x^2)) - 6/((3 + 4*E^4)*(3 + 4*E^4 + 8*x^2)))
 

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 1380

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

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

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 2089

Int[(u_)^(p_.)*((f_.)*(x_))^(m_.)*(z_)^(q_.), x_Symbol] :> Int[(f*x)^m*Expa 
ndToSum[z, x]^q*ExpandToSum[u, x]^p, x] /; FreeQ[{f, m, p, q}, x] && Binomi 
alQ[z, x] && TrinomialQ[u, x] &&  !(BinomialMatchQ[z, x] && TrinomialMatchQ 
[u, x])
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
gosper	\(-\frac {4 \left (2 x^{2}+{\mathrm e}^{4}\right )}{x^{2} \left (8 x^{2}+4 \,{\mathrm e}^{4}+3\right )}\)	\(27\)
norman	\(\frac {-8 x^{2}-4 \,{\mathrm e}^{4}}{x^{2} \left (8 x^{2}+4 \,{\mathrm e}^{4}+3\right )}\)	\(28\)
risch	\(\frac {-8 x^{2}-4 \,{\mathrm e}^{4}}{x^{2} \left (8 x^{2}+4 \,{\mathrm e}^{4}+3\right )}\)	\(29\)
parallelrisch	\(\frac {-64 x^{2}-32 \,{\mathrm e}^{4}}{8 x^{2} \left (8 x^{2}+4 \,{\mathrm e}^{4}+3\right )}\)	\(29\)

Input:

int((32*exp(4)^2+(128*x^2+24)*exp(4)+128*x^4)/(16*x^3*exp(4)^2+(64*x^5+24* 
x^3)*exp(4)+64*x^7+48*x^5+9*x^3),x,method=_RETURNVERBOSE)
 

Output:

-4/x^2*(2*x^2+exp(4))/(8*x^2+4*exp(4)+3)
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4 \, {\left (2 \, x^{2} + e^{4}\right )}}{8 \, x^{4} + 4 \, x^{2} e^{4} + 3 \, x^{2}} \] Input:

integrate((32*exp(4)^2+(128*x^2+24)*exp(4)+128*x^4)/(16*x^3*exp(4)^2+(64*x 
^5+24*x^3)*exp(4)+64*x^7+48*x^5+9*x^3),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=\frac {- 8 x^{2} - 4 e^{4}}{8 x^{4} + x^{2} \cdot \left (3 + 4 e^{4}\right )} \] Input:

integrate((32*exp(4)**2+(128*x**2+24)*exp(4)+128*x**4)/(16*x**3*exp(4)**2+ 
(64*x**5+24*x**3)*exp(4)+64*x**7+48*x**5+9*x**3),x)
 

Output:

(-8*x**2 - 4*exp(4))/(8*x**4 + x**2*(3 + 4*exp(4)))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4 \, {\left (2 \, x^{2} + e^{4}\right )}}{8 \, x^{4} + x^{2} {\left (4 \, e^{4} + 3\right )}} \] Input:

integrate((32*exp(4)^2+(128*x^2+24)*exp(4)+128*x^4)/(16*x^3*exp(4)^2+(64*x 
^5+24*x^3)*exp(4)+64*x^7+48*x^5+9*x^3),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4 \, {\left (2 \, x^{2} + e^{4}\right )}}{8 \, x^{4} + 4 \, x^{2} e^{4} + 3 \, x^{2}} \] Input:

integrate((32*exp(4)^2+(128*x^2+24)*exp(4)+128*x^4)/(16*x^3*exp(4)^2+(64*x 
^5+24*x^3)*exp(4)+64*x^7+48*x^5+9*x^3),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=-\frac {4\,\left (2\,x^2+{\mathrm {e}}^4\right )}{x^2\,\left (8\,x^2+4\,{\mathrm {e}}^4+3\right )} \] Input:

int((32*exp(8) + exp(4)*(128*x^2 + 24) + 128*x^4)/(exp(4)*(24*x^3 + 64*x^5 
) + 16*x^3*exp(8) + 9*x^3 + 48*x^5 + 64*x^7),x)
 

Output:

-(4*(exp(4) + 2*x^2))/(x^2*(4*exp(4) + 8*x^2 + 3))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {32 e^8+128 x^4+e^4 \left (24+128 x^2\right )}{9 x^3+16 e^8 x^3+48 x^5+64 x^7+e^4 \left (24 x^3+64 x^5\right )} \, dx=\frac {-16 e^{8}-12 e^{4}+64 x^{4}}{x^{2} \left (16 e^{8}+32 e^{4} x^{2}+24 e^{4}+24 x^{2}+9\right )} \] Input:

int((32*exp(4)^2+(128*x^2+24)*exp(4)+128*x^4)/(16*x^3*exp(4)^2+(64*x^5+24* 
x^3)*exp(4)+64*x^7+48*x^5+9*x^3),x)
 

Output:

(4*( - 4*e**8 - 3*e**4 + 16*x**4))/(x**2*(16*e**8 + 32*e**4*x**2 + 24*e**4 
 + 24*x**2 + 9))