3.7.91 \(\int \frac {x}{2+3 x^4} \, dx\) [691]

Optimal. Leaf size=21 \[ \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \]

[Out]

1/12*arctan(1/2*x^2*6^(1/2))*6^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 209} \begin {gather*} \frac {\text {ArcTan}\left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(2 + 3*x^4),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 281

Int[(x_)^(m_.)*((a_) + (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))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(2 + 3*x^4),x]

[Out]

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

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Maple [A]
time = 0.14, size = 15, normalized size = 0.71

method	result	size
default	\(\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{12}\)	\(15\)
risch	\(\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{12}\)	\(15\)
meijerg	\(\frac {\sqrt {6}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )}{12}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(3*x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/12*arctan(1/2*x^2*6^(1/2))*6^(1/2)

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Maxima [A]
time = 0.50, size = 14, normalized size = 0.67 \begin {gather*} \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x^4+2),x, algorithm="maxima")

[Out]

1/12*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

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Fricas [A]
time = 0.35, size = 14, normalized size = 0.67 \begin {gather*} \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x^4+2),x, algorithm="fricas")

[Out]

1/12*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

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Sympy [A]
time = 0.03, size = 17, normalized size = 0.81 \begin {gather*} \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x**4+2),x)

[Out]

sqrt(6)*atan(sqrt(6)*x**2/2)/12

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Giac [A]
time = 0.50, size = 14, normalized size = 0.67 \begin {gather*} \frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(3*x^4+2),x, algorithm="giac")

[Out]

1/12*sqrt(6)*arctan(1/2*sqrt(6)*x^2)

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.67 \begin {gather*} \frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(3*x^4 + 2),x)

[Out]

(6^(1/2)*atan((6^(1/2)*x^2)/2))/12

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