3.880 \(\int \frac {1}{(2+3 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=27 \[ \frac {2^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right ),2\right )}{\sqrt {3}} \]

[Out]

1/3*2^(3/4)*(cos(1/2*arctan(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arctan(1/2*x*6^(1/2)))*EllipticF(sin(1/2*arctan(1
/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.00, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {231} \[ \frac {2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)^(-3/4),x]

[Out]

(2^(3/4)*EllipticF[ArcTan[Sqrt[3/2]*x]/2, 2])/Sqrt[3]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (2+3 x^2\right )^{3/4}} \, dx &=\frac {2^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 24, normalized size = 0.89 \[ \frac {x \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {3 x^2}{2}\right )}{2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x^2)^(-3/4),x]

[Out]

(x*Hypergeometric2F1[1/2, 3/4, 3/2, (-3*x^2)/2])/2^(3/4)

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+2)^(3/4),x, algorithm="fricas")

[Out]

integral((3*x^2 + 2)^(-3/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+2)^(3/4),x, algorithm="giac")

[Out]

integrate((3*x^2 + 2)^(-3/4), x)

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maple [C]  time = 0.27, size = 18, normalized size = 0.67 \[ \frac {2^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2+2)^(3/4),x)

[Out]

1/2*2^(1/4)*x*hypergeom([1/2,3/4],[3/2],-3/2*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x^2+2)^(3/4),x, algorithm="maxima")

[Out]

integrate((3*x^2 + 2)^(-3/4), x)

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mupad [B]  time = 0.08, size = 16, normalized size = 0.59 \[ \frac {2^{1/4}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {3}{2};\ -\frac {3\,x^2}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2 + 2)^(3/4),x)

[Out]

(2^(1/4)*x*hypergeom([1/2, 3/4], 3/2, -(3*x^2)/2))/2

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sympy [C]  time = 0.68, size = 26, normalized size = 0.96 \[ \frac {\sqrt [4]{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3*x**2+2)**(3/4),x)

[Out]

2**(1/4)*x*hyper((1/2, 3/4), (3/2,), 3*x**2*exp_polar(I*pi)/2)/2

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