\(\int \frac {x}{(2+3 x^4)^2} \, dx\) [700]

Optimal result

Integrand size = 11, antiderivative size = 38 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {x^2}{8 \left (2+3 x^4\right )}+\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}} \]

[Out]

1/8*x^2/(3*x^4+2)+1/48*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {281, 205, 209} \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}}+\frac {x^2}{8 \left (3 x^4+2\right )} \]

[In]

Int[x/(2 + 3*x^4)^2,x]

[Out]

x^2/(8*(2 + 3*x^4)) + ArcTan[Sqrt[3/2]*x^2]/(8*Sqrt[6])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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}{\left (2+3 x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{8 \left (2+3 x^4\right )}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{2+3 x^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{8 \left (2+3 x^4\right )}+\frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {x^2}{8 \left (2+3 x^4\right )}+\frac {\arctan \left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}} \]

[In]

Integrate[x/(2 + 3*x^4)^2,x]

[Out]

x^2/(8*(2 + 3*x^4)) + ArcTan[Sqrt[3/2]*x^2]/(8*Sqrt[6])

Maple [A] (verified)

Time = 3.92 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {x^{2}}{24 x^{4}+16}+\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{48}\)	\(28\)
default	\(\frac {x^{2}}{24 x^{4}+16}+\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{48}\)	\(30\)
meijerg	\(\frac {\sqrt {6}\, \left (\frac {\sqrt {6}\, x^{2}}{3 x^{4}+2}+\arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )\right )}{48}\)	\(35\)

[In]

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

[Out]

1/24*x^2/(x^4+2/3)+1/48*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {\sqrt {6} {\left (3 \, x^{4} + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + 6 \, x^{2}}{48 \, {\left (3 \, x^{4} + 2\right )}} \]

[In]

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

[Out]

1/48*(sqrt(6)*(3*x^4 + 2)*arctan(1/2*sqrt(6)*x^2) + 6*x^2)/(3*x^4 + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {x^{2}}{24 x^{4} + 16} + \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{48} \]

[In]

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

[Out]

x**2/(24*x**4 + 16) + sqrt(6)*atan(sqrt(6)*x**2/2)/48

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {1}{48} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + \frac {x^{2}}{8 \, {\left (3 \, x^{4} + 2\right )}} \]

[In]

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

[Out]

1/48*sqrt(6)*arctan(1/2*sqrt(6)*x^2) + 1/8*x^2/(3*x^4 + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {1}{48} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + \frac {x^{2}}{8 \, {\left (3 \, x^{4} + 2\right )}} \]

[In]

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

[Out]

1/48*sqrt(6)*arctan(1/2*sqrt(6)*x^2) + 1/8*x^2/(3*x^4 + 2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\left (2+3 x^4\right )^2} \, dx=\frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{48}+\frac {x^2}{24\,\left (x^4+\frac {2}{3}\right )} \]

[In]

int(x/(3*x^4 + 2)^2,x)

[Out]

(6^(1/2)*atan((6^(1/2)*x^2)/2))/48 + x^2/(24*(x^4 + 2/3))