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

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

[Out]

-1/2*(3*x^2+2)^(1/4)/x-1/4*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)))*Ell
ipticF(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 = 49, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}}}{3 \, x^{4} + 2 \, x^{2}}, x\right ) \]

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

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)

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maple [C]  time = 0.27, size = 33, normalized size = 0.67 \[ -\frac {3 \,2^{\frac {1}{4}} x \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{8}-\frac {\left (3 x^{2}+2\right )^{\frac {1}{4}}}{2 x} \]

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

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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mupad [B]  time = 4.99, size = 36, normalized size = 0.73 \[ -\frac {2\,3^{1/4}\,{\left (\frac {2}{x^2}+3\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ -\frac {2}{3\,x^2}\right )}{15\,x\,{\left (3\,x^2+2\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(3*x^2 + 2)^(3/4)),x)

[Out]

-(2*3^(1/4)*(2/x^2 + 3)^(3/4)*hypergeom([3/4, 5/4], 9/4, -2/(3*x^2)))/(15*x*(3*x^2 + 2)^(3/4))

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sympy [C]  time = 0.82, size = 29, normalized size = 0.59 \[ - \frac {\sqrt [4]{2} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2 x} \]

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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