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

Optimal. Leaf size=29 \[ \frac {1}{x+\frac {4 \left (3+\frac {16 \log ^2(3)}{(-x+x (1+\log (3)))^2}\right )}{x}} \]

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Rubi [A]  time = 0.18, antiderivative size = 37, normalized size of antiderivative = 1.28, number of steps used = 13, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {1594, 2075, 638, 618, 204} \begin {gather*} \frac {x+4}{2 \left (x^2+2 x+8\right )}-\frac {4-x}{2 \left (x^2-2 x+8\right )} \end {gather*}

Antiderivative was successfully verified.

[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 204

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

Rule 618

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 638

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)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), 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 1594

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 2075

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 {gather*} \begin {aligned} \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+\operatorname {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )+\operatorname {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 {\tan ^{-1}\left (\frac {-1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\frac {\tan ^{-1}\left (\frac {1+x}{\sqrt {7}}\right )}{2 \sqrt {7}}-\operatorname {Subst}\left (\int \frac {1}{-28-x^2} \, dx,x,-2+2 x\right )-\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.55 \begin {gather*} \frac {x^3}{64+12 x^2+x^4} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.92, size = 16, normalized size = 0.55 \begin {gather*} \frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.34, size = 16, normalized size = 0.55 \begin {gather*} \frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 17, normalized size = 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	\(\frac {x^{3}}{x^{4}+12 x^{2}+64}\)	\(17\)
default	\(-\frac {-x -4}{2 \left (x^{2}+2 x +8\right )}-\frac {-x +4}{2 \left (x^{2}-2 x +8\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.51, size = 16, normalized size = 0.55 \begin {gather*} \frac {x^{3}}{x^{4} + 12 \, x^{2} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.05, size = 16, normalized size = 0.55 \begin {gather*} \frac {x^3}{x^4+12\,x^2+64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.11, size = 12, normalized size = 0.41 \begin {gather*} \frac {x^{3}}{x^{4} + 12 x^{2} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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