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

Optimal. Leaf size=30 \[ 5 \left (10+\frac {\frac {4}{5}-x^2}{8 (-2+x) \left (1+e^4+x\right )}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 51, normalized size of antiderivative = 1.70, number of steps used = 4, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {1680, 12, 1814, 8} \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[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 1680

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 1814

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 {gather*} \begin {aligned} \text {integral} &=\operatorname {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} \operatorname {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 {\operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 34, normalized size = 1.13 \begin {gather*} \frac {-6+5 e^4 (-2+x)-5 x}{8 \left (-2+e^4 (-2+x)-x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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 + E^4*(-2 + x) - x + x^2))

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 30, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (x - 2\right )} e^{4} - 5 \, x - 6}{8 \, {\left (x^{2} + {\left (x - 2\right )} e^{4} - x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/8*((-16*exp(8)+16*exp(4)^2)/(exp(8)^2+
12*exp(8)*exp(4)+18*exp(8)+36*exp(4)^2+108*exp(4)+81)*ln(sageVARx^2+2*sageVARx*exp(4)+2*sageVARx+exp(8)+2*exp(
4)+1)+(-5*exp(8)^2*ex

________________________________________________________________________________________

maple [A]  time = 0.12, size = 34, normalized size = 1.13

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

Verification of antiderivative is not currently implemented for this CAS.

[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)/(x-2)/(exp(2)^2+x+1)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 32, normalized size = 1.07 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 2.21, size = 28, normalized size = 0.93 \begin {gather*} -\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.68, size = 34, normalized size = 1.13 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________