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

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Rubi [A]  time = 0.0069099, antiderivative size = 43, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 0.0033747, 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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Maple [C]  time = 0.01, size = 18, normalized size = 0.4 \begin{align*}{\frac{{2}^{{\frac{3}{4}}}x}{2}{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,-{\frac{3\,{x}^{2}}{2}})}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (3 \, x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{{\left (3 \, x^{2} + 2\right )}^{\frac{1}{4}}}, x\right ) \end{align*}

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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Sympy [C]  time = 0.58462, size = 26, normalized size = 0.6 \begin{align*} \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} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (3 \, x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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)