3.45 \(\int \frac{x^2}{13-6 x^3+x^6} \, dx\)

Optimal. Leaf size=14 \[ \frac{1}{6} \tan ^{-1}\left (\frac{1}{2} \left (x^3-3\right )\right ) \]

[Out]

ArcTan[(-3 + x^3)/2]/6

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Rubi [A]  time = 0.0159955, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1352, 618, 204} \[ \frac{1}{6} \tan ^{-1}\left (\frac{1}{2} \left (x^3-3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(13 - 6*x^3 + x^6),x]

[Out]

ArcTan[(-3 + x^3)/2]/6

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]

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 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^2}{13-6 x^3+x^6} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{13-6 x+x^2} \, dx,x,x^3\right )\\ &=-\left (\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{-16-x^2} \, dx,x,2 \left (-3+x^3\right )\right )\right )\\ &=\frac{1}{6} \tan ^{-1}\left (\frac{1}{2} \left (-3+x^3\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0044996, size = 14, normalized size = 1. \[ \frac{1}{6} \tan ^{-1}\left (\frac{1}{2} \left (x^3-3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(13 - 6*x^3 + x^6),x]

[Out]

ArcTan[(-3 + x^3)/2]/6

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Maple [A]  time = 0.003, size = 11, normalized size = 0.8 \begin{align*}{\frac{1}{6}\arctan \left ({\frac{{x}^{3}}{2}}-{\frac{3}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^6-6*x^3+13),x)

[Out]

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

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Maxima [A]  time = 1.43972, size = 14, normalized size = 1. \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{2} \, x^{3} - \frac{3}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-6*x^3+13),x, algorithm="maxima")

[Out]

1/6*arctan(1/2*x^3 - 3/2)

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Fricas [A]  time = 1.84042, size = 36, normalized size = 2.57 \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{2} \, x^{3} - \frac{3}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-6*x^3+13),x, algorithm="fricas")

[Out]

1/6*arctan(1/2*x^3 - 3/2)

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Sympy [A]  time = 0.105266, size = 10, normalized size = 0.71 \begin{align*} \frac{\operatorname{atan}{\left (\frac{x^{3}}{2} - \frac{3}{2} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**6-6*x**3+13),x)

[Out]

atan(x**3/2 - 3/2)/6

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Giac [A]  time = 1.07075, size = 14, normalized size = 1. \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{2} \, x^{3} - \frac{3}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-6*x^3+13),x, algorithm="giac")

[Out]

1/6*arctan(1/2*x^3 - 3/2)