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]

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

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Rubi [A]  time = 0.01, antiderivative size = 14, 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 = {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 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 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]

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*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \frac {1}{6} \tan ^{-1}\left (\frac {1}{3} \left (x^2+1\right )\right ) \]

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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fricas [A]  time = 0.39, size = 10, normalized size = 0.71 \[ \frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

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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giac [A]  time = 1.65, size = 10, normalized size = 0.71 \[ \frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

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)

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maple [A]  time = 0.00, size = 11, normalized size = 0.79 \[ \frac {\arctan \left (\frac {x^{2}}{3}+\frac {1}{3}\right )}{6} \]

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 = 1.25, size = 10, normalized size = 0.71 \[ \frac {1}{6} \, \arctan \left (\frac {1}{3} \, x^{2} + \frac {1}{3}\right ) \]

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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mupad [B]  time = 0.19, size = 10, normalized size = 0.71 \[ \frac {\mathrm {atan}\left (\frac {x^2}{3}+\frac {1}{3}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(2*x^2 + x^4 + 10),x)

[Out]

atan(x^2/3 + 1/3)/6

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sympy [A]  time = 0.11, size = 10, normalized size = 0.71 \[ \frac {\operatorname {atan}{\left (\frac {x^{2}}{3} + \frac {1}{3} \right )}}{6} \]

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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