3.292 \(\int \frac{x}{10+2 x^2+x^4} \, dx\)

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

[Out]

ArcTan[(1 + x^2)/3]/6

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Rubi [A]  time = 0.013418, antiderivative size = 14, normalized size of antiderivative = 1., 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 = {1107, 618, 204} \[ \frac{1}{6} \tan ^{-1}\left (\frac{1}{3} \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/(10 + 2*x^2 + x^4),x]

[Out]

ArcTan[(1 + x^2)/3]/6

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
 x, x^2], x] /; FreeQ[{a, b, c, p}, x]

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}{10+2 x^2+x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{10+2 x+x^2} \, dx,x,x^2\right )\\ &=-\operatorname{Subst}\left (\int \frac{1}{-36-x^2} \, dx,x,2 \left (1+x^2\right )\right )\\ &=\frac{1}{6} \tan ^{-1}\left (\frac{1}{3} \left (1+x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0043375, size = 14, normalized size = 1. \[ \frac{1}{6} \tan ^{-1}\left (\frac{1}{3} \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(10 + 2*x^2 + x^4),x]

[Out]

ArcTan[(1 + x^2)/3]/6

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Maple [A]  time = 0.002, size = 11, normalized size = 0.8 \begin{align*}{\frac{1}{6}\arctan \left ({\frac{{x}^{2}}{3}}+{\frac{1}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^4+2*x^2+10),x)

[Out]

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

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Maxima [A]  time = 1.41837, size = 14, normalized size = 1. \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{3} \, x^{2} + \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="maxima")

[Out]

1/6*arctan(1/3*x^2 + 1/3)

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Fricas [A]  time = 1.83923, size = 36, normalized size = 2.57 \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{3} \, x^{2} + \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="fricas")

[Out]

1/6*arctan(1/3*x^2 + 1/3)

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Sympy [A]  time = 0.095847, size = 10, normalized size = 0.71 \begin{align*} \frac{\operatorname{atan}{\left (\frac{x^{2}}{3} + \frac{1}{3} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x**4+2*x**2+10),x)

[Out]

atan(x**2/3 + 1/3)/6

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Giac [A]  time = 1.05179, size = 14, normalized size = 1. \begin{align*} \frac{1}{6} \, \arctan \left (\frac{1}{3} \, x^{2} + \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^4+2*x^2+10),x, algorithm="giac")

[Out]

1/6*arctan(1/3*x^2 + 1/3)