\(\int \frac {4+12 x+5 x^2+e^4 (-4+20 x-5 x^2)}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 (32-32 x+8 x^2)+e^4 (64-48 x^2+16 x^3)} \, dx\) [3593]

Optimal result

Integrand size = 77, antiderivative size = 30 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=5 \left (10+\frac {\frac {4}{5}-x^2}{8 (-2+x) \left (1+e^4+x\right )}\right ) \]

[Out]

5*(4/5-x^2)/(-2+x)/(8*x+8*exp(2)^2+8)+50

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {1694, 12, 1828, 8} \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {5 \left (1-e^4\right ) x+2 \left (3+5 e^4\right )}{2 \left (\left (3+e^4\right )^2-4 \left (x+\frac {1}{32} \left (16 e^4-16\right )\right )^2\right )} \]

[In]

Int[(4 + 12*x + 5*x^2 + E^4*(-4 + 20*x - 5*x^2))/(32 + 32*x - 24*x^2 - 16*x^3 + 8*x^4 + E^8*(32 - 32*x + 8*x^2
) + E^4*(64 - 48*x^2 + 16*x^3)),x]

[Out]

(2*(3 + 5*E^4) + 5*(1 - E^4)*x)/(2*((3 + E^4)^2 - 4*((-16 + 16*E^4)/32 + x)^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {5 \left (1-e^4\right ) \left (3+e^4\right )^2+4 \left (17+10 e^4+5 e^8\right ) x+20 \left (1-e^4\right ) x^2}{2 \left (9+6 e^4+e^8-4 x^2\right )^2} \, dx,x,\frac {1}{32} \left (-16+16 e^4\right )+x\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {5 \left (1-e^4\right ) \left (3+e^4\right )^2+4 \left (17+10 e^4+5 e^8\right ) x+20 \left (1-e^4\right ) x^2}{\left (9+6 e^4+e^8-4 x^2\right )^2} \, dx,x,\frac {1}{32} \left (-16+16 e^4\right )+x\right ) \\ & = \frac {2 \left (3+5 e^4\right )+5 \left (1-e^4\right ) x}{2 \left (\left (3+e^4\right )^2-\left (1-e^4-2 x\right )^2\right )}-\frac {\text {Subst}\left (\int 0 \, dx,x,\frac {1}{32} \left (-16+16 e^4\right )+x\right )}{4 \left (3+e^4\right )^2} \\ & = \frac {2 \left (3+5 e^4\right )+5 \left (1-e^4\right ) x}{2 \left (\left (3+e^4\right )^2-\left (1-e^4-2 x\right )^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {-6+5 e^4 (-2+x)-5 x}{8 (-2+x) \left (1+e^4+x\right )} \]

[In]

Integrate[(4 + 12*x + 5*x^2 + E^4*(-4 + 20*x - 5*x^2))/(32 + 32*x - 24*x^2 - 16*x^3 + 8*x^4 + E^8*(32 - 32*x +
8*x^2) + E^4*(64 - 48*x^2 + 16*x^3)),x]

[Out]

(-6 + 5*E^4*(-2 + x) - 5*x)/(8*(-2 + x)*(1 + E^4 + x))

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13

method	result	size
norman	\(\frac {\left (\frac {5 \,{\mathrm e}^{4}}{8}-\frac {5}{8}\right ) x -\frac {5 \,{\mathrm e}^{4}}{4}-\frac {3}{4}}{\left (-2+x \right ) \left ({\mathrm e}^{4}+x +1\right )}\)	\(34\)
risch	\(\frac {\left (\frac {5 \,{\mathrm e}^{4}}{8}-\frac {5}{8}\right ) x -\frac {5 \,{\mathrm e}^{4}}{4}-\frac {3}{4}}{x \,{\mathrm e}^{4}-2 \,{\mathrm e}^{4}+x^{2}-x -2}\)	\(34\)
gosper	\(\frac {5 x \,{\mathrm e}^{4}-10 \,{\mathrm e}^{4}-5 x -6}{8 x \,{\mathrm e}^{4}-16 \,{\mathrm e}^{4}+8 x^{2}-8 x -16}\)	\(43\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{4}-10 \,{\mathrm e}^{4}-5 x -6}{8 x \,{\mathrm e}^{4}-16 \,{\mathrm e}^{4}+8 x^{2}-8 x -16}\)	\(43\)

[In]

int(((-5*x^2+20*x-4)*exp(2)^2+5*x^2+12*x+4)/((8*x^2-32*x+32)*exp(2)^4+(16*x^3-48*x^2+64)*exp(2)^2+8*x^4-16*x^3
-24*x^2+32*x+32),x,method=_RETURNVERBOSE)

[Out]

((5/8*exp(2)^2-5/8)*x-5/4*exp(2)^2-3/4)/(-2+x)/(exp(2)^2+x+1)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {5 \, {\left (x - 2\right )} e^{4} - 5 \, x - 6}{8 \, {\left (x^{2} + {\left (x - 2\right )} e^{4} - x - 2\right )}} \]

[In]

integrate(((-5*x^2+20*x-4)*exp(2)^2+5*x^2+12*x+4)/((8*x^2-32*x+32)*exp(2)^4+(16*x^3-48*x^2+64)*exp(2)^2+8*x^4-
16*x^3-24*x^2+32*x+32),x, algorithm="fricas")

[Out]

1/8*(5*(x - 2)*e^4 - 5*x - 6)/(x^2 + (x - 2)*e^4 - x - 2)

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {x \left (-5 + 5 e^{4}\right ) - 10 e^{4} - 6}{8 x^{2} + x \left (-8 + 8 e^{4}\right ) - 16 e^{4} - 16} \]

[In]

integrate(((-5*x**2+20*x-4)*exp(2)**2+5*x**2+12*x+4)/((8*x**2-32*x+32)*exp(2)**4+(16*x**3-48*x**2+64)*exp(2)**
2+8*x**4-16*x**3-24*x**2+32*x+32),x)

[Out]

(x*(-5 + 5*exp(4)) - 10*exp(4) - 6)/(8*x**2 + x*(-8 + 8*exp(4)) - 16*exp(4) - 16)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {5 \, x {\left (e^{4} - 1\right )} - 10 \, e^{4} - 6}{8 \, {\left (x^{2} + x {\left (e^{4} - 1\right )} - 2 \, e^{4} - 2\right )}} \]

[In]

integrate(((-5*x^2+20*x-4)*exp(2)^2+5*x^2+12*x+4)/((8*x^2-32*x+32)*exp(2)^4+(16*x^3-48*x^2+64)*exp(2)^2+8*x^4-
16*x^3-24*x^2+32*x+32),x, algorithm="maxima")

[Out]

1/8*(5*x*(e^4 - 1) - 10*e^4 - 6)/(x^2 + x*(e^4 - 1) - 2*e^4 - 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=\frac {5 \, x e^{4} - 5 \, x - 10 \, e^{4} - 6}{8 \, {\left (x^{2} + x e^{4} - x - 2 \, e^{4} - 2\right )}} \]

[In]

integrate(((-5*x^2+20*x-4)*exp(2)^2+5*x^2+12*x+4)/((8*x^2-32*x+32)*exp(2)^4+(16*x^3-48*x^2+64)*exp(2)^2+8*x^4-
16*x^3-24*x^2+32*x+32),x, algorithm="giac")

[Out]

1/8*(5*x*e^4 - 5*x - 10*e^4 - 6)/(x^2 + x*e^4 - x - 2*e^4 - 2)

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {4+12 x+5 x^2+e^4 \left (-4+20 x-5 x^2\right )}{32+32 x-24 x^2-16 x^3+8 x^4+e^8 \left (32-32 x+8 x^2\right )+e^4 \left (64-48 x^2+16 x^3\right )} \, dx=-\frac {5\,x+10\,{\mathrm {e}}^4-5\,x\,{\mathrm {e}}^4+6}{8\,\left (x-2\right )\,\left (x+{\mathrm {e}}^4+1\right )} \]

[In]

int((12*x - exp(4)*(5*x^2 - 20*x + 4) + 5*x^2 + 4)/(32*x + exp(8)*(8*x^2 - 32*x + 32) + exp(4)*(16*x^3 - 48*x^
2 + 64) - 24*x^2 - 16*x^3 + 8*x^4 + 32),x)

[Out]

-(5*x + 10*exp(4) - 5*x*exp(4) + 6)/(8*(x - 2)*(x + exp(4) + 1))