3.336 \(\int \frac{x+x^5}{(1+2 x^2+2 x^4)^3} \, dx\)

Optimal. Leaf size=59 \[ \frac{2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}+\frac{4 x^2+3}{16 \left (2 x^4+2 x^2+1\right )^2}+\tan ^{-1}\left (2 x^2+1\right ) \]

[Out]

(3 + 4*x^2)/(16*(1 + 2*x^2 + 2*x^4)^2) + (1 + 2*x^2)/(2*(1 + 2*x^2 + 2*x^4)) + ArcTan[1 + 2*x^2]

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Rubi [A]  time = 0.0642882, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1593, 1663, 1660, 12, 614, 617, 204} \[ \frac{2 x^2+1}{2 \left (2 x^4+2 x^2+1\right )}+\frac{4 x^2+3}{16 \left (2 x^4+2 x^2+1\right )^2}+\tan ^{-1}\left (2 x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]

[Out]

(3 + 4*x^2)/(16*(1 + 2*x^2 + 2*x^4)^2) + (1 + 2*x^2)/(2*(1 + 2*x^2 + 2*x^4)) + ArcTan[1 + 2*x^2]

Rule 1593

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 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

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

Rule 614

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int \frac{x+x^5}{\left (1+2 x^2+2 x^4\right )^3} \, dx &=\int \frac{x \left (1+x^4\right )}{\left (1+2 x^2+2 x^4\right )^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1+x^2}{\left (1+2 x+2 x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac{3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{16}{\left (1+2 x+2 x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\operatorname{Subst}\left (\int \frac{1}{\left (1+2 x+2 x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac{1+2 x^2}{2 \left (1+2 x^2+2 x^4\right )}+\operatorname{Subst}\left (\int \frac{1}{1+2 x+2 x^2} \, dx,x,x^2\right )\\ &=\frac{3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac{1+2 x^2}{2 \left (1+2 x^2+2 x^4\right )}-\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+2 x^2\right )\\ &=\frac{3+4 x^2}{16 \left (1+2 x^2+2 x^4\right )^2}+\frac{1+2 x^2}{2 \left (1+2 x^2+2 x^4\right )}+\tan ^{-1}\left (1+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.02139, size = 44, normalized size = 0.75 \[ \frac{32 x^6+48 x^4+36 x^2+11}{16 \left (2 x^4+2 x^2+1\right )^2}+\tan ^{-1}\left (2 x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x + x^5)/(1 + 2*x^2 + 2*x^4)^3,x]

[Out]

(11 + 36*x^2 + 48*x^4 + 32*x^6)/(16*(1 + 2*x^2 + 2*x^4)^2) + ArcTan[1 + 2*x^2]

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Maple [A]  time = 0.009, size = 41, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ \left ( 2\,{x}^{4}+2\,{x}^{2}+1 \right ) ^{2}} \left ({x}^{6}+3/2\,{x}^{4}+{\frac{9\,{x}^{2}}{8}}+{\frac{11}{32}} \right ) }+\arctan \left ( 2\,{x}^{2}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5+x)/(2*x^4+2*x^2+1)^3,x)

[Out]

2*(x^6+3/2*x^4+9/8*x^2+11/32)/(2*x^4+2*x^2+1)^2+arctan(2*x^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 11}{16 \,{\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )}} + 2 \, \int \frac{x}{2 \, x^{4} + 2 \, x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="maxima")

[Out]

1/16*(32*x^6 + 48*x^4 + 36*x^2 + 11)/(4*x^8 + 8*x^6 + 8*x^4 + 4*x^2 + 1) + 2*integrate(x/(2*x^4 + 2*x^2 + 1),
x)

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Fricas [A]  time = 1.54041, size = 180, normalized size = 3.05 \begin{align*} \frac{32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 16 \,{\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (2 \, x^{2} + 1\right ) + 11}{16 \,{\left (4 \, x^{8} + 8 \, x^{6} + 8 \, x^{4} + 4 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="fricas")

[Out]

1/16*(32*x^6 + 48*x^4 + 36*x^2 + 16*(4*x^8 + 8*x^6 + 8*x^4 + 4*x^2 + 1)*arctan(2*x^2 + 1) + 11)/(4*x^8 + 8*x^6
+ 8*x^4 + 4*x^2 + 1)

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Sympy [A]  time = 0.182194, size = 46, normalized size = 0.78 \begin{align*} \frac{32 x^{6} + 48 x^{4} + 36 x^{2} + 11}{64 x^{8} + 128 x^{6} + 128 x^{4} + 64 x^{2} + 16} + \operatorname{atan}{\left (2 x^{2} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5+x)/(2*x**4+2*x**2+1)**3,x)

[Out]

(32*x**6 + 48*x**4 + 36*x**2 + 11)/(64*x**8 + 128*x**6 + 128*x**4 + 64*x**2 + 16) + atan(2*x**2 + 1)

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Giac [A]  time = 1.10982, size = 57, normalized size = 0.97 \begin{align*} \frac{32 \, x^{6} + 48 \, x^{4} + 36 \, x^{2} + 11}{16 \,{\left (2 \, x^{4} + 2 \, x^{2} + 1\right )}^{2}} + \arctan \left (2 \, x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+x)/(2*x^4+2*x^2+1)^3,x, algorithm="giac")

[Out]

1/16*(32*x^6 + 48*x^4 + 36*x^2 + 11)/(2*x^4 + 2*x^2 + 1)^2 + arctan(2*x^2 + 1)