3.700 \(\int \frac {x}{(2+3 x^4)^2} \, dx\)

Optimal. Leaf size=38 \[ \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}}+\frac {x^2}{8 \left (3 x^4+2\right )} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {275, 199, 203} \[ \frac {x^2}{8 \left (3 x^4+2\right )}+\frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}} \]

Antiderivative was successfully verified.

[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 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

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

Rule 275

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}{\left (2+3 x^4\right )^2} \, dx &=\frac {1}{2} \operatorname {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} \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{8 \sqrt {6}}+\frac {x^2}{8 \left (3 x^4+2\right )} \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.68, size = 37, normalized size = 0.97 \[ \frac {\sqrt {6} {\left (3 \, x^{4} + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + 6 \, x^{2}}{48 \, {\left (3 \, x^{4} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 29, normalized size = 0.76 \[ \frac {1}{48} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + \frac {x^{2}}{8 \, {\left (3 \, x^{4} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 30, normalized size = 0.79 \[ \frac {x^{2}}{24 x^{4}+16}+\frac {\sqrt {6}\, \arctan \left (\frac {\sqrt {6}\, x^{2}}{2}\right )}{48} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.91, size = 29, normalized size = 0.76 \[ \frac {1}{48} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) + \frac {x^{2}}{8 \, {\left (3 \, x^{4} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.04, size = 28, normalized size = 0.74 \[ \frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{48}+\frac {x^2}{24\,\left (x^4+\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.17, size = 27, normalized size = 0.71 \[ \frac {x^{2}}{24 x^{4} + 16} + \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{48} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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