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

Optimal. Leaf size=202 \[ -\frac {\sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{\sqrt {3} x}+\frac {2 \sqrt [4]{-3 x^2-2} x}{\sqrt {-3 x^2-2}+\sqrt {2}}+\frac {2 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {3} x} \]

[Out]

2*x*(-3*x^2-2)^(1/4)/(2^(1/2)+(-3*x^2-2)^(1/2))+2/3*2^(1/4)*(cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))^2)^(1
/2)/cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))*EllipticE(sin(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1
/2))*(2^(1/2)+(-3*x^2-2)^(1/2))*(-x^2/(2^(1/2)+(-3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)-1/3*2^(1/4)*(cos(2*arctan(
1/2*(-3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)))*EllipticF(sin(2*arctan(1/2
*(-3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(-3*x^2-2)^(1/2))*(-x^2/(2^(1/2)+(-3*x^2-2)^(1/2))^2)^(1/2)/
x*3^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {230, 305, 220, 1196} \[ \frac {2 \sqrt [4]{-3 x^2-2} x}{\sqrt {-3 x^2-2}+\sqrt {2}}-\frac {\sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {3} x}+\frac {2 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {3} x} \]

Antiderivative was successfully verified.

[In]

Int[(-2 - 3*x^2)^(-1/4),x]

[Out]

(2*x*(-2 - 3*x^2)^(1/4))/(Sqrt[2] + Sqrt[-2 - 3*x^2]) + (2*2^(1/4)*Sqrt[-(x^2/(Sqrt[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])/(Sqrt[3]*x) - (2^(1/4)*Sqr
t[-(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])/(Sqrt[3]*x)

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 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 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}{\sqrt [4]{-2-3 x^2}} \, dx &=-\frac {\left (\sqrt {\frac {2}{3}} \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 )}{x}\\ &=-\frac {\left (2 \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 )}{\sqrt {3} x}+\frac {\left (2 \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 )}{\sqrt {3} x}\\ &=\frac {2 x \sqrt [4]{-2-3 x^2}}{\sqrt {2}+\sqrt {-2-3 x^2}}+\frac {2 \sqrt [4]{2} \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 )}{\sqrt {3} x}-\frac {\sqrt [4]{2} \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 )}{\sqrt {3} x}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 43, normalized size = 0.21 \[ \frac {x \sqrt [4]{\frac {3 x^2}{2}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^2}{2}\right )}{\sqrt [4]{-3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 3*x^2)^(-1/4),x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \frac {3 \, x {\rm integral}\left (-\frac {4 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{3 \, {\left (3 \, x^{4} + 2 \, x^{2}\right )}}, x\right ) - 2 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{3 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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maple [C]  time = 0.27, size = 21, normalized size = 0.10 \[ -\frac {\left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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mupad [B]  time = 4.88, size = 34, normalized size = 0.17 \[ \frac {2^{3/4}\,x\,{\left (3\,x^2+2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ -\frac {3\,x^2}{2}\right )}{2\,{\left (-3\,x^2-2\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(- 3*x^2 - 2)^(1/4),x)

[Out]

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

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sympy [C]  time = 0.69, size = 32, normalized size = 0.16 \[ \frac {2^{\frac {3}{4}} x e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2} \]

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

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