\(\int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx\) [474]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 39, antiderivative size = 29 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {1}{x+\frac {4 \left (3+\frac {16 \log ^2(3)}{(-x+x (1+\log (3)))^2}\right )}{x}} \]

[Out]

1/(4*(3+16*ln(3)^2/((ln(3)+1)*x-x)^2)/x+x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1608, 2100, 652, 632, 210} \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x+4}{2 \left (x^2+2 x+8\right )}-\frac {4-x}{2 \left (x^2-2 x+8\right )} \]

[In]

Int[(192*x^2 + 12*x^4 - x^6)/(4096 + 1536*x^2 + 272*x^4 + 24*x^6 + x^8),x]

[Out]

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 1608

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

Rule 2100

Int[(P_)^(p_)*(Qm_), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Qm, x], x] /; QuadraticProdu
ctQ[PP, x]] /; PolyQ[Qm, x] && PolyQ[P, x] && ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (192+12 x^2-x^4\right )}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx \\ & = \int \left (\frac {4+3 x}{\left (8-2 x+x^2\right )^2}-\frac {1}{2 \left (8-2 x+x^2\right )}+\frac {4-3 x}{\left (8+2 x+x^2\right )^2}-\frac {1}{2 \left (8+2 x+x^2\right )}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{8-2 x+x^2} \, dx\right )-\frac {1}{2} \int \frac {1}{8+2 x+x^2} \, dx+\int \frac {4+3 x}{\left (8-2 x+x^2\right )^2} \, dx+\int \frac {4-3 x}{\left (8+2 x+x^2\right )^2} \, dx \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )}+\frac {1}{2} \int \frac {1}{8-2 x+x^2} \, dx+\frac {1}{2} \int \frac {1}{8+2 x+x^2} \, dx+\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )+\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,2+2 x\right ) \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )}-\frac {\arctan \left (\frac {-1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\frac {\arctan \left (\frac {1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )-\text {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,2+2 x\right ) \\ & = -\frac {4-x}{2 \left (8-2 x+x^2\right )}+\frac {4+x}{2 \left (8+2 x+x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^3}{64+12 x^2+x^4} \]

[In]

Integrate[(192*x^2 + 12*x^4 - x^6)/(4096 + 1536*x^2 + 272*x^4 + 24*x^6 + x^8),x]

[Out]

x^3/(64 + 12*x^2 + x^4)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59

method result size
gosper \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) \(17\)
norman \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) \(17\)
risch \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) \(17\)
parallelrisch \(\frac {x^{3}}{x^{4}+12 x^{2}+64}\) \(17\)
default \(-\frac {-x +4}{2 \left (x^{2}-2 x +8\right )}-\frac {-4-x}{2 \left (x^{2}+2 x +8\right )}\) \(36\)

[In]

int((-x^6+12*x^4+192*x^2)/(x^8+24*x^6+272*x^4+1536*x^2+4096),x,method=_RETURNVERBOSE)

[Out]

x^3/(x^4+12*x^2+64)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]

[In]

integrate((-x^6+12*x^4+192*x^2)/(x^8+24*x^6+272*x^4+1536*x^2+4096),x, algorithm="fricas")

[Out]

x^3/(x^4 + 12*x^2 + 64)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.41 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 x^{2} + 64} \]

[In]

integrate((-x**6+12*x**4+192*x**2)/(x**8+24*x**6+272*x**4+1536*x**2+4096),x)

[Out]

x**3/(x**4 + 12*x**2 + 64)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]

[In]

integrate((-x^6+12*x^4+192*x^2)/(x^8+24*x^6+272*x^4+1536*x^2+4096),x, algorithm="maxima")

[Out]

x^3/(x^4 + 12*x^2 + 64)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \]

[In]

integrate((-x^6+12*x^4+192*x^2)/(x^8+24*x^6+272*x^4+1536*x^2+4096),x, algorithm="giac")

[Out]

x^3/(x^4 + 12*x^2 + 64)

Mupad [B] (verification not implemented)

Time = 7.46 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.55 \[ \int \frac {192 x^2+12 x^4-x^6}{4096+1536 x^2+272 x^4+24 x^6+x^8} \, dx=\frac {x^3}{x^4+12\,x^2+64} \]

[In]

int((192*x^2 + 12*x^4 - x^6)/(1536*x^2 + 272*x^4 + 24*x^6 + x^8 + 4096),x)

[Out]

x^3/(12*x^2 + x^4 + 64)