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

Optimal. Leaf size=43 \[ \frac {2 x}{\sqrt [4]{3 x^2+2}}-\frac {2 \sqrt [4]{2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}} \]

[Out]

2*x/(3*x^2+2)^(1/4)-2/3*2^(1/4)*(cos(1/2*arctan(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arctan(1/2*x*6^(1/2)))*Ellipt
icE(sin(1/2*arctan(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {227, 196} \[ \frac {2 x}{\sqrt [4]{3 x^2+2}}-\frac {2 \sqrt [4]{2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x^2)^(-1/4),x]

[Out]

(2*x)/(2 + 3*x^2)^(1/4) - (2*2^(1/4)*EllipticE[ArcTan[Sqrt[3/2]*x]/2, 2])/Sqrt[3]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{2+3 x^2}} \, dx &=\frac {2 x}{\sqrt [4]{2+3 x^2}}-2 \int \frac {1}{\left (2+3 x^2\right )^{5/4}} \, dx\\ &=\frac {2 x}{\sqrt [4]{2+3 x^2}}-\frac {2 \sqrt [4]{2} E\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 = 0.00, size = 24, normalized size = 0.56 \[ \frac {x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^2}{2}\right )}{\sqrt [4]{2}} \]

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 = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{{\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}}}, x\right ) \]

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)^(-1/4), x)

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

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 = 0.27, size = 18, normalized size = 0.42 \[ \frac {2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2} \]

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 = 0.37 \[ \frac {8^{1/4}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {3\,x^2}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x^2 + 2)^(1/4),x)

[Out]

(8^(1/4)*x*hypergeom([1/4, 1/2], 3/2, -(3*x^2)/2))/2

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sympy [C]  time = 0.67, size = 26, normalized size = 0.60 \[ \frac {2^{\frac {3}{4}} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2} \]

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(I*pi)/2)/2

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