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

Optimal. Leaf size=22 \[ \frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}} \]

[Out]

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

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Rubi [A]  time = 0.0130876, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {12, 275, 203} \[ \frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x)/(2 + 3*x^4),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0090014, size = 22, normalized size = 1. \[ \frac{b \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right )}{2 \sqrt{6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x)/(2 + 3*x^4),x]

[Out]

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

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Maple [A]  time = 0.04, size = 16, normalized size = 0.7 \begin{align*}{\frac{b\sqrt{6}}{12}\arctan \left ({\frac{{x}^{2}\sqrt{6}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42433, size = 20, normalized size = 0.91 \begin{align*} \frac{1}{12} \, \sqrt{6} b \arctan \left (\frac{1}{2} \, \sqrt{6} x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.49792, size = 54, normalized size = 2.45 \begin{align*} \frac{1}{12} \, \sqrt{6} b \arctan \left (\frac{1}{2} \, \sqrt{6} x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.089657, size = 19, normalized size = 0.86 \begin{align*} \frac{\sqrt{6} b \operatorname{atan}{\left (\frac{\sqrt{6} x^{2}}{2} \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08175, size = 20, normalized size = 0.91 \begin{align*} \frac{1}{12} \, \sqrt{6} b \arctan \left (\frac{1}{2} \, \sqrt{6} x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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