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

Optimal. Leaf size=223 \[ -\frac{\sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac{1}{2}\right )}{2\ 2^{3/4} x}+\frac{3 \sqrt [4]{-3 x^2-2} x}{2 \left (\sqrt{-3 x^2-2}+\sqrt{2}\right )}+\frac{\left (-3 x^2-2\right )^{3/4}}{2 x}+\frac{\sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{2^{3/4} x} \]

[Out]

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

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Rubi [A]  time = 0.092718, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {325, 230, 305, 220, 1196} \[ \frac{3 \sqrt [4]{-3 x^2-2} x}{2 \left (\sqrt{-3 x^2-2}+\sqrt{2}\right )}+\frac{\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac{\sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{2\ 2^{3/4} x}+\frac{\sqrt{3} \sqrt{-\frac{x^2}{\left (\sqrt{-3 x^2-2}+\sqrt{2}\right )^2}} \left (\sqrt{-3 x^2-2}+\sqrt{2}\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac{1}{2}\right )}{2^{3/4} x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(-2 - 3*x^2)^(1/4)),x]

[Out]

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

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 230

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[x^2/Sqrt[1 - x^4/a
], x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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

Mathematica [C]  time = 0.0074077, size = 46, normalized size = 0.21 \[ -\frac{\sqrt [4]{\frac{3 x^2}{2}+1} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{1}{2};-\frac{3 x^2}{2}\right )}{x \sqrt [4]{-3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.019, size = 43, normalized size = 0.2 \begin{align*} -{\frac{3\,{x}^{2}+2}{2\,x}{\frac{1}{\sqrt [4]{-3\,{x}^{2}-2}}}}-{\frac{3\, \left ( -1 \right ) ^{3/4}x{2}^{3/4}}{8}{\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/x^2/(-3*x^2-2)^(1/4),x)

[Out]

-1/2*(3*x^2+2)/x/(-3*x^2-2)^(1/4)-3/8*(-1)^(3/4)*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}} 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)^(1/4),x, algorithm="maxima")

[Out]

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

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

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]

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

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Sympy [C]  time = 0.669019, size = 36, normalized size = 0.16 \begin{align*} \frac{2^{\frac{3}{4}} e^{\frac{3 i \pi }{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} \end{align*}

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)*exp(3*I*pi/4)*hyper((-1/2, 1/4), (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{1}{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)^(1/4),x, algorithm="giac")

[Out]

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