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

Optimal. Leaf size=63 \[ \frac {3 x}{2 \sqrt [4]{3 x^2+2}}-\frac {\left (3 x^2+2\right )^{3/4}}{2 x}-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2^{3/4}} \]

[Out]

3/2*x/(3*x^2+2)^(1/4)-1/2*(3*x^2+2)^(3/4)/x-1/2*2^(1/4)*(cos(1/2*arctan(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcta
n(1/2*x*6^(1/2)))*EllipticE(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 = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {325, 227, 196} \[ \frac {3 x}{2 \sqrt [4]{3 x^2+2}}-\frac {\left (3 x^2+2\right )^{3/4}}{2 x}-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(2 + 3*x^2)^(1/4)),x]

[Out]

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

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]

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 \sqrt [4]{2+3 x^2}} \, dx &=-\frac {\left (2+3 x^2\right )^{3/4}}{2 x}+\frac {3}{4} \int \frac {1}{\sqrt [4]{2+3 x^2}} \, dx\\ &=\frac {3 x}{2 \sqrt [4]{2+3 x^2}}-\frac {\left (2+3 x^2\right )^{3/4}}{2 x}-\frac {3}{2} \int \frac {1}{\left (2+3 x^2\right )^{5/4}} \, dx\\ &=\frac {3 x}{2 \sqrt [4]{2+3 x^2}}-\frac {\left (2+3 x^2\right )^{3/4}}{2 x}-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.43 \[ -\frac {\, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};-\frac {3 x^2}{2}\right )}{\sqrt [4]{2} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} + 2\right )}^{\frac {3}{4}}}{3 \, x^{4} + 2 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 33, normalized size = 0.52 \[ \frac {3 \,2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{8}-\frac {\left (3 x^{2}+2\right )^{\frac {3}{4}}}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*x^2+2)^(3/4)/x+3/8*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 \[ \int \frac {1}{{\left (3 \, x^{2} + 2\right )}^{\frac {1}{4}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.02, size = 36, normalized size = 0.57 \[ -\frac {2\,3^{3/4}\,{\left (\frac {2}{x^2}+3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {2}{3\,x^2}\right )}{9\,x\,{\left (3\,x^2+2\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(3*x^2 + 2)^(1/4)),x)

[Out]

-(2*3^(3/4)*(2/x^2 + 3)^(1/4)*hypergeom([1/4, 3/4], 7/4, -2/(3*x^2)))/(9*x*(3*x^2 + 2)^(1/4))

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sympy [C]  time = 0.74, size = 29, normalized size = 0.46 \[ - \frac {2^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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