3.873 \(\int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx\)

Optimal. Leaf size=28 \[ \frac {2 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}} \]

[Out]

2/3*2^(1/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*EllipticE(sin(1/2*arcsin(1
/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.00, antiderivative size = 28, 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 = {228} \[ \frac {2 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(2 - 3*x^2)^(-1/4),x]

[Out]

(2*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/Sqrt[3]

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx &=\frac {2 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.86 \[ \frac {x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^2}{2}\right )}{\sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - 3*x^2)^(-1/4),x]

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{3 \, x^{2} - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(-3*x^2 + 2)^(3/4)/(3*x^2 - 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-3*x^2 + 2)^(-1/4), x)

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maple [C]  time = 0.28, size = 18, normalized size = 0.64 \[ \frac {2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-3*x^2 + 2)^(-1/4), x)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.57 \[ \frac {2^{3/4}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ \frac {3\,x^2}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2 - 3*x^2)^(1/4),x)

[Out]

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

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sympy [C]  time = 0.69, size = 27, normalized size = 0.96 \[ \frac {2^{\frac {3}{4}} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{2 i \pi }}{2}} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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