3.332 \(\int \frac {x^3}{1+x^4+x^8} \, dx\)

Optimal. Leaf size=23 \[ \frac {\tan ^{-1}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

[Out]

1/6*arctan(1/3*(2*x^4+1)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1352, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + x^4 + x^8),x]

[Out]

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

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 1352

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {2 x^4+1}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x^4 + x^8),x]

[Out]

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

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fricas [A]  time = 0.78, size = 18, normalized size = 0.78 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1))

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giac [A]  time = 0.36, size = 18, normalized size = 0.78 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1))

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maple [A]  time = 0.00, size = 19, normalized size = 0.83 \[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}+1\right ) \sqrt {3}}{3}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^8+x^4+1),x)

[Out]

1/6*3^(1/2)*arctan(1/3*(2*x^4+1)*3^(1/2))

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maxima [A]  time = 2.47, size = 18, normalized size = 0.78 \[ \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^4 + 1))

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mupad [B]  time = 1.30, size = 17, normalized size = 0.74 \[ \frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,\left (\frac {2\,x^4}{3}+\frac {1}{3}\right )\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^4 + x^8 + 1),x)

[Out]

(3^(1/2)*atan(3^(1/2)*((2*x^4)/3 + 1/3)))/6

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sympy [A]  time = 0.12, size = 26, normalized size = 1.13 \[ \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(x**8+x**4+1),x)

[Out]

sqrt(3)*atan(2*sqrt(3)*x**4/3 + sqrt(3)/3)/6

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