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

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 209} \[ \int \frac {x}{2+3 x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \]

[In]

Integrate[x/(2 + 3*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 3.93 (sec) , antiderivative size = 15, normalized size of antiderivative = 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	\(\frac {\sqrt {6}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )}{12}\)	\(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{12} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {1}{12} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {x}{2+3 x^4} \, dx=\frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{12} \]

[In]

int(x/(3*x^4 + 2),x)

[Out]

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