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]

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

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 {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*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \[ \frac {b \tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right )}{2 \sqrt {6}} \]

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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fricas [A]  time = 0.78, size = 15, normalized size = 0.68 \[ \frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

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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giac [A]  time = 0.17, size = 15, normalized size = 0.68 \[ \frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

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)

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maple [A]  time = 0.04, size = 16, normalized size = 0.73 \[ \frac {\sqrt {6}\, b \arctan \left (\frac {\sqrt {6}\, x^{2}}{2}\right )}{12} \]

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

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maxima [A]  time = 2.88, size = 15, normalized size = 0.68 \[ \frac {1}{12} \, \sqrt {6} b \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) \]

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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mupad [B]  time = 4.77, size = 15, normalized size = 0.68 \[ \frac {\sqrt {6}\,b\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x)/(3*x^4 + 2),x)

[Out]

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

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sympy [A]  time = 0.13, size = 19, normalized size = 0.86 \[ \frac {\sqrt {6} b \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{12} \]

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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