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

Optimal. Leaf size=224 \[ \frac {4 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {2}{15} \left (-3 x^2-2\right )^{3/4} x-\frac {8 \sqrt [4]{-3 x^2-2} x}{15 \left (\sqrt {-3 x^2-2}+\sqrt {2}\right )}-\frac {8 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x} \]

[Out]

-2/15*x*(-3*x^2-2)^(3/4)-8/15*x*(-3*x^2-2)^(1/4)/(2^(1/2)+(-3*x^2-2)^(1/2))-8/45*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)+4/45*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 = 0.09, antiderivative size = 224, normalized size of antiderivative = 1.00, 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 = {321, 230, 305, 220, 1196} \[ -\frac {2}{15} \left (-3 x^2-2\right )^{3/4} x-\frac {8 \sqrt [4]{-3 x^2-2} x}{15 \left (\sqrt {-3 x^2-2}+\sqrt {2}\right )}+\frac {4 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {8 \sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(-2 - 3*x^2)^(1/4),x]

[Out]

(-2*x*(-2 - 3*x^2)^(3/4))/15 - (8*x*(-2 - 3*x^2)^(1/4))/(15*(Sqrt[2] + Sqrt[-2 - 3*x^2])) - (8*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])/(15*Sqrt[3]*x) + (4*2^(1/4)*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])/(15*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 321

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

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 {x^2}{\sqrt [4]{-2-3 x^2}} \, dx &=-\frac {2}{15} x \left (-2-3 x^2\right )^{3/4}-\frac {4}{15} \int \frac {1}{\sqrt [4]{-2-3 x^2}} \, dx\\ &=-\frac {2}{15} x \left (-2-3 x^2\right )^{3/4}+\frac {\left (4 \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 )}{15 x}\\ &=-\frac {2}{15} x \left (-2-3 x^2\right )^{3/4}+\frac {\left (8 \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 )}{15 \sqrt {3} x}-\frac {\left (8 \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 )}{15 \sqrt {3} x}\\ &=-\frac {2}{15} x \left (-2-3 x^2\right )^{3/4}-\frac {8 x \sqrt [4]{-2-3 x^2}}{15 \left (\sqrt {2}+\sqrt {-2-3 x^2}\right )}-\frac {8 \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 )}{15 \sqrt {3} x}+\frac {4 \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 )}{15 \sqrt {3} x}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 58, normalized size = 0.26 \[ \frac {2 x \left (-2^{3/4} \sqrt [4]{3 x^2+2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {3 x^2}{2}\right )+3 x^2+2\right )}{15 \sqrt [4]{-3 x^2-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(-2 - 3*x^2)^(1/4),x]

[Out]

(2*x*(2 + 3*x^2 - 2^(3/4)*(2 + 3*x^2)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (-3*x^2)/2]))/(15*(-2 - 3*x^2)^(1
/4))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \frac {45 \, x {\rm integral}\left (\frac {16 \, {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{45 \, {\left (3 \, x^{4} + 2 \, x^{2}\right )}}, x\right ) - 2 \, {\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} - 2\right )}^{\frac {3}{4}}}{45 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(45*x*integral(16/45*(-3*x^2 - 2)^(3/4)/(3*x^4 + 2*x^2), x) - 2*(3*x^2 - 4)*(-3*x^2 - 2)^(3/4))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-3*x^2 - 2)^(1/4), x)

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maple [C]  time = 0.27, size = 41, normalized size = 0.18 \[ \frac {2 \left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{15}+\frac {2 \left (3 x^{2}+2\right ) x}{15 \left (-3 x^{2}-2\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x*(3*x^2+2)/(-3*x^2-2)^(1/4)+2/15*(-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 \[ \int \frac {x^{2}}{{\left (-3 \, x^{2} - 2\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(-3*x^2 - 2)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(- 3*x^2 - 2)^(1/4),x)

[Out]

int(x^2/(- 3*x^2 - 2)^(1/4), x)

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sympy [C]  time = 0.72, size = 34, normalized size = 0.15 \[ \frac {2^{\frac {3}{4}} x^{3} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(3/4)*x**3*exp(-I*pi/4)*hyper((1/4, 3/2), (5/2,), 3*x**2*exp_polar(I*pi)/2)/6

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