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

Optimal. Leaf size=49 \[ -\frac{\sqrt{3} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right ),2\right )}{2 \sqrt [4]{2}}-\frac{\sqrt [4]{3 x^2+2}}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.0083085, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {325, 231} \[ -\frac{\sqrt [4]{3 x^2+2}}{2 x}-\frac{\sqrt{3} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right |2\right )}{2 \sqrt [4]{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(2 + 3*x^2)^(3/4)),x]

[Out]

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.0047161, size = 27, normalized size = 0.55 \[ -\frac{\, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{1}{2};-\frac{3 x^2}{2}\right )}{2^{3/4} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.021, size = 33, normalized size = 0.7 \begin{align*} -{\frac{1}{2\,x}\sqrt [4]{3\,{x}^{2}+2}}-{\frac{3\,\sqrt [4]{2}x}{8}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{3}{4}};\,{\frac{3}{2}};\,-{\frac{3\,{x}^{2}}{2}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*x^2+2)^(1/4)/x-3/8*2^(1/4)*x*hypergeom([1/2,3/4],[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{3}{4}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.768287, size = 29, normalized size = 0.59 \begin{align*} - \frac{\sqrt [4]{2}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{1}{2} \end{matrix}\middle |{\frac{3 x^{2} e^{i \pi }}{2}} \right )}}{2 x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{3}{4}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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