3.4.29 \(\int \frac {x^9}{1+x^4+x^8} \, dx\) [329]

Optimal. Leaf size=54 \[ \frac {x^2}{2}+\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

[Out]

1/2*x^2+1/6*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)-1/6*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1373, 1136, 1175, 632, 210} \begin {gather*} \frac {\text {ArcTan}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\text {ArcTan}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(1 + x^4 + x^8),x]

[Out]

x^2/2 + ArcTan[(1 - 2*x^2)/Sqrt[3]]/(2*Sqrt[3]) - ArcTan[(1 + 2*x^2)/Sqrt[3]]/(2*Sqrt[3])

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 1136

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

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1373

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^9}{1+x^4+x^8} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^4}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=\frac {x^2}{2}-\frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=\frac {x^2}{2}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac {x^2}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=\frac {x^2}{2}+\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.11, size = 98, normalized size = 1.81 \begin {gather*} \frac {x^2}{2}-\frac {\left (i+\sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x^2\right )}{2 \sqrt {6+6 i \sqrt {3}}}-\frac {\left (-i+\sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (i+\sqrt {3}\right ) x^2\right )}{2 \sqrt {6-6 i \sqrt {3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(1 + x^4 + x^8),x]

[Out]

x^2/2 - ((I + Sqrt[3])*ArcTan[((-I + Sqrt[3])*x^2)/2])/(2*Sqrt[6 + (6*I)*Sqrt[3]]) - ((-I + Sqrt[3])*ArcTan[((
I + Sqrt[3])*x^2)/2])/(2*Sqrt[6 - (6*I)*Sqrt[3]])

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 43, normalized size = 0.80

method	result	size
default	\(\frac {x^{2}}{2}-\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{6}\)	\(43\)
risch	\(\frac {x^{2}}{2}-\frac {\sqrt {3}\, \arctan \left (\frac {x^{6} \sqrt {3}}{3}+\frac {2 x^{2} \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {x^{2} \sqrt {3}}{3}\right )}{6}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2-1/6*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)-1/6*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 42, normalized size = 0.78 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/2*x^2 - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1))

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 40, normalized size = 0.74 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x^{2}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{6} + 2 \, x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/2*x^2 - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*x^2) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(x^6 + 2*x^2))

________________________________________________________________________________________

Sympy [A]
time = 0.05, size = 51, normalized size = 0.94 \begin {gather*} \frac {x^{2}}{2} + \frac {\sqrt {3} \left (- 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{2}}{3} \right )} - 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{6}}{3} + \frac {2 \sqrt {3} x^{2}}{3} \right )}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(x**8+x**4+1),x)

[Out]

x**2/2 + sqrt(3)*(-2*atan(sqrt(3)*x**2/3) - 2*atan(sqrt(3)*x**6/3 + 2*sqrt(3)*x**2/3))/12

________________________________________________________________________________________

Giac [A]
time = 6.34, size = 42, normalized size = 0.78 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/2*x^2 - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1))

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 43, normalized size = 0.80 \begin {gather*} \frac {x^2}{2}-\frac {\sqrt {3}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x^6}{3}+\frac {2\,\sqrt {3}\,x^2}{3}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x^2}{3}\right )\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^4 + x^8 + 1),x)

[Out]

x^2/2 - (3^(1/2)*(2*atan((2*3^(1/2)*x^2)/3 + (3^(1/2)*x^6)/3) + 2*atan((3^(1/2)*x^2)/3)))/12

________________________________________________________________________________________