\(\int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e (24-24 x^2)+e^4 (-8-6 e+24 x^2)+e^2 (32 x+24 e x-32 x^3)} \, dx\) [4633]

Optimal result

Integrand size = 91, antiderivative size = 22 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=\frac {x^2}{-4-3 e+\left (-e^2+2 x\right )^2} \]

[Out]

x^2/((2*x-exp(2))^2-4-3*exp(1))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6, 1607, 1694, 790} \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=-\frac {4 e^2 x-e^4+3 e+4}{4 \left (-4 \left (x-\frac {e^2}{2}\right )^2+3 e+4\right )} \]

[In]

Int[(-8*x - 6*E*x + 2*E^4*x - 4*E^2*x^2)/(16 + 9*E^2 + E^8 - 8*E^6*x - 32*x^2 + 16*x^4 + E*(24 - 24*x^2) + E^4
*(-8 - 6*E + 24*x^2) + E^2*(32*x + 24*E*x - 32*x^3)),x]

[Out]

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

Rule 6

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

Rule 790

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] /; FreeQ[{a, c, d, e, f, g, p}, x] &
& EqQ[a*e*g - c*d*f*(2*p + 3), 0] && NeQ[p, -1]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(-8-6 e) x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx \\ & = \int \frac {\left (-8-6 e+2 e^4\right ) x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx \\ & = \int \frac {x \left (-8-6 e+2 e^4-4 e^2 x\right )}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx \\ & = \text {Subst}\left (\int \frac {\left (e^2+2 x\right ) \left (-4-3 e-2 e^2 x\right )}{\left (4+3 e-4 x^2\right )^2} \, dx,x,-\frac {e^2}{2}+x\right ) \\ & = -\frac {4+3 e-e^4+4 e^2 x}{4 \left (4+3 e-4 \left (-\frac {e^2}{2}+x\right )^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=\frac {4+3 e-e^4+4 e^2 x}{4 \left (-4-3 e+e^4-4 e^2 x+4 x^2\right )} \]

[In]

Integrate[(-8*x - 6*E*x + 2*E^4*x - 4*E^2*x^2)/(16 + 9*E^2 + E^8 - 8*E^6*x - 32*x^2 + 16*x^4 + E*(24 - 24*x^2)
+ E^4*(-8 - 6*E + 24*x^2) + E^2*(32*x + 24*E*x - 32*x^3)),x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64

method	result	size
risch	\(\frac {{\mathrm e}^{2} x -\frac {{\mathrm e}^{4}}{4}+\frac {3 \,{\mathrm e}}{4}+1}{{\mathrm e}^{4}-4 \,{\mathrm e}^{2} x +4 x^{2}-3 \,{\mathrm e}-4}\)	\(36\)
gosper	\(-\frac {{\mathrm e}^{4}-4 \,{\mathrm e}^{2} x -3 \,{\mathrm e}-4}{4 \left ({\mathrm e}^{4}-4 \,{\mathrm e}^{2} x +4 x^{2}-3 \,{\mathrm e}-4\right )}\)	\(40\)
norman	\(\frac {{\mathrm e}^{2} x -\frac {{\mathrm e}^{4}}{4}+\frac {3 \,{\mathrm e}}{4}+1}{{\mathrm e}^{4}-4 \,{\mathrm e}^{2} x +4 x^{2}-3 \,{\mathrm e}-4}\)	\(40\)
parallelrisch	\(\frac {-{\mathrm e}^{4}+4 \,{\mathrm e}^{2} x +3 \,{\mathrm e}+4}{4 \,{\mathrm e}^{4}-16 \,{\mathrm e}^{2} x +16 x^{2}-12 \,{\mathrm e}-16}\)	\(42\)
default	\(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16 \textit {\_Z}^{4}-32 \textit {\_Z}^{3} {\mathrm e}^{2}+\left (24 \,{\mathrm e}^{4}-24 \,{\mathrm e}-32\right ) \textit {\_Z}^{2}+\left (-8 \,{\mathrm e}^{6}+24 \,{\mathrm e}^{3}+32 \,{\mathrm e}^{2}\right ) \textit {\_Z} +{\mathrm e}^{8}+16-6 \,{\mathrm e}^{5}-8 \,{\mathrm e}^{4}+9 \,{\mathrm e}^{2}+24 \,{\mathrm e}\right )}{\sum }\frac {\left (2 \textit {\_R}^{2} {\mathrm e}^{2}+\left (-{\mathrm e}^{4}+4+3 \,{\mathrm e}\right ) \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{-{\mathrm e}^{6}+6 \textit {\_R} \,{\mathrm e}^{4}-12 \textit {\_R}^{2} {\mathrm e}^{2}+8 \textit {\_R}^{3}+3 \,{\mathrm e}^{3}-6 \textit {\_R} \,{\mathrm e}+4 \,{\mathrm e}^{2}-8 \textit {\_R}}\right )}{4}\)	\(135\)

[In]

int((2*x*exp(2)^2-4*x^2*exp(2)-6*x*exp(1)-8*x)/(exp(2)^4-8*x*exp(2)^3+(-6*exp(1)+24*x^2-8)*exp(2)^2+(24*x*exp(
1)-32*x^3+32*x)*exp(2)+9*exp(1)^2+(-24*x^2+24)*exp(1)+16*x^4-32*x^2+16),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=\frac {4 \, x e^{2} - e^{4} + 3 \, e + 4}{4 \, {\left (4 \, x^{2} - 4 \, x e^{2} + e^{4} - 3 \, e - 4\right )}} \]

[In]

integrate((2*x*exp(2)^2-4*x^2*exp(2)-6*x*exp(1)-8*x)/(exp(2)^4-8*x*exp(2)^3+(-6*exp(1)+24*x^2-8)*exp(2)^2+(24*
x*exp(1)-32*x^3+32*x)*exp(2)+9*exp(1)^2+(-24*x^2+24)*exp(1)+16*x^4-32*x^2+16),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.61 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=- \frac {- 4 x e^{2} - 3 e - 4 + e^{4}}{16 x^{2} - 16 x e^{2} - 12 e - 16 + 4 e^{4}} \]

[In]

integrate((2*x*exp(2)**2-4*x**2*exp(2)-6*x*exp(1)-8*x)/(exp(2)**4-8*x*exp(2)**3+(-6*exp(1)+24*x**2-8)*exp(2)**
2+(24*x*exp(1)-32*x**3+32*x)*exp(2)+9*exp(1)**2+(-24*x**2+24)*exp(1)+16*x**4-32*x**2+16),x)

[Out]

-(-4*x*exp(2) - 3*E - 4 + exp(4))/(16*x**2 - 16*x*exp(2) - 12*E - 16 + 4*exp(4))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=\frac {4 \, x e^{2} - e^{4} + 3 \, e + 4}{4 \, {\left (4 \, x^{2} - 4 \, x e^{2} + e^{4} - 3 \, e - 4\right )}} \]

[In]

integrate((2*x*exp(2)^2-4*x^2*exp(2)-6*x*exp(1)-8*x)/(exp(2)^4-8*x*exp(2)^3+(-6*exp(1)+24*x^2-8)*exp(2)^2+(24*
x*exp(1)-32*x^3+32*x)*exp(2)+9*exp(1)^2+(-24*x^2+24)*exp(1)+16*x^4-32*x^2+16),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=\frac {4 \, x e^{2} - e^{4} + 3 \, e + 4}{4 \, {\left (4 \, x^{2} - 4 \, x e^{2} + e^{4} - 3 \, e - 4\right )}} \]

[In]

integrate((2*x*exp(2)^2-4*x^2*exp(2)-6*x*exp(1)-8*x)/(exp(2)^4-8*x*exp(2)^3+(-6*exp(1)+24*x^2-8)*exp(2)^2+(24*
x*exp(1)-32*x^3+32*x)*exp(2)+9*exp(1)^2+(-24*x^2+24)*exp(1)+16*x^4-32*x^2+16),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {-8 x-6 e x+2 e^4 x-4 e^2 x^2}{16+9 e^2+e^8-8 e^6 x-32 x^2+16 x^4+e \left (24-24 x^2\right )+e^4 \left (-8-6 e+24 x^2\right )+e^2 \left (32 x+24 e x-32 x^3\right )} \, dx=-\frac {\frac {3\,\mathrm {e}}{4}-\frac {{\mathrm {e}}^4}{4}+x\,{\mathrm {e}}^2+1}{-4\,x^2+4\,{\mathrm {e}}^2\,x+3\,\mathrm {e}-{\mathrm {e}}^4+4} \]

[In]

int(-(8*x + 6*x*exp(1) - 2*x*exp(4) + 4*x^2*exp(2))/(9*exp(2) + exp(8) - exp(4)*(6*exp(1) - 24*x^2 + 8) - 8*x*
exp(6) - exp(1)*(24*x^2 - 24) + exp(2)*(32*x + 24*x*exp(1) - 32*x^3) - 32*x^2 + 16*x^4 + 16),x)

[Out]

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