3.9.19 \(\int \frac {x^7}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=136 \[ \frac {2}{891} \left (2-3 x^2\right )^{11/4}-\frac {16}{567} \left (2-3 x^2\right )^{7/4}+\frac {56}{243} \left (2-3 x^2\right )^{3/4}+\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt {2-3 x^2}+\sqrt {2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {440, 261, 266, 43, 439} \begin {gather*} \frac {2}{891} \left (2-3 x^2\right )^{11/4}-\frac {16}{567} \left (2-3 x^2\right )^{7/4}+\frac {56}{243} \left (2-3 x^2\right )^{3/4}+\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt {2-3 x^2}+\sqrt {2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(56*(2 - 3*x^2)^(3/4))/243 - (16*(2 - 3*x^2)^(7/4))/567 + (2*(2 - 3*x^2)^(11/4))/891 + (32*2^(1/4)*ArcTan[(Sqr
t[2] - Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))])/81 + (32*2^(1/4)*ArcTanh[(Sqrt[2] + Sqrt[2 - 3*x^2])/(2^
(3/4)*(2 - 3*x^2)^(1/4))])/81

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 439

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

Rule 440

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(1/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

Rubi steps

\begin {align*} \int \frac {x^7}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac {16 x}{27 \sqrt [4]{2-3 x^2}}-\frac {4 x^3}{9 \sqrt [4]{2-3 x^2}}-\frac {x^5}{3 \sqrt [4]{2-3 x^2}}+\frac {64 x}{27 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {x^5}{\sqrt [4]{2-3 x^2}} \, dx\right )-\frac {4}{9} \int \frac {x^3}{\sqrt [4]{2-3 x^2}} \, dx-\frac {16}{27} \int \frac {x}{\sqrt [4]{2-3 x^2}} \, dx+\frac {64}{27} \int \frac {x}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\\ &=\frac {32}{243} \left (2-3 x^2\right )^{3/4}+\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{2-3 x}} \, dx,x,x^2\right )-\frac {2}{9} \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{2-3 x}} \, dx,x,x^2\right )\\ &=\frac {32}{243} \left (2-3 x^2\right )^{3/4}+\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \left (\frac {4}{9 \sqrt [4]{2-3 x}}-\frac {4}{9} (2-3 x)^{3/4}+\frac {1}{9} (2-3 x)^{7/4}\right ) \, dx,x,x^2\right )-\frac {2}{9} \operatorname {Subst}\left (\int \left (\frac {2}{3 \sqrt [4]{2-3 x}}-\frac {1}{3} (2-3 x)^{3/4}\right ) \, dx,x,x^2\right )\\ &=\frac {56}{243} \left (2-3 x^2\right )^{3/4}-\frac {16}{567} \left (2-3 x^2\right )^{7/4}+\frac {2}{891} \left (2-3 x^2\right )^{11/4}+\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 46, normalized size = 0.34 \begin {gather*} \frac {2 \left (2-3 x^2\right )^{3/4} \left (-2464 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {3 x^2}{2}-1\right )+189 x^4+540 x^2+1712\right )}{18711} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(2*(2 - 3*x^2)^(3/4)*(1712 + 540*x^2 + 189*x^4 - 2464*Hypergeometric2F1[3/4, 1, 7/4, -1 + (3*x^2)/2]))/18711

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IntegrateAlgebraic [A]  time = 0.18, size = 124, normalized size = 0.91 \begin {gather*} -\frac {32}{81} \sqrt [4]{2} \tan ^{-1}\left (\frac {\frac {\sqrt {2-3 x^2}}{2^{3/4}}-\frac {1}{\sqrt [4]{2}}}{\sqrt [4]{2-3 x^2}}\right )+\frac {32}{81} \sqrt [4]{2} \tanh ^{-1}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{2-3 x^2}}{\sqrt {2} \sqrt {2-3 x^2}+2}\right )+\frac {2 \left (2-3 x^2\right )^{3/4} \left (189 x^4+540 x^2+1712\right )}{18711} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(2*(2 - 3*x^2)^(3/4)*(1712 + 540*x^2 + 189*x^4))/18711 - (32*2^(1/4)*ArcTan[(-2^(-1/4) + Sqrt[2 - 3*x^2]/2^(3/
4))/(2 - 3*x^2)^(1/4)])/81 + (32*2^(1/4)*ArcTanh[(2*2^(1/4)*(2 - 3*x^2)^(1/4))/(2 + Sqrt[2]*Sqrt[2 - 3*x^2])])
/81

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fricas [B]  time = 1.28, size = 253, normalized size = 1.86 \begin {gather*} \frac {2}{18711} \, {\left (189 \, x^{4} + 540 \, x^{2} + 1712\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} + \frac {32}{81} \cdot 8^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{4} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4 \, \sqrt {2} + 4 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) + \frac {32}{81} \cdot 8^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{8} \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-4 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 16 \, \sqrt {2} + 16 \, \sqrt {-3 \, x^{2} + 2}} - \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {8}{81} \cdot 8^{\frac {1}{4}} \sqrt {2} \log \left (4 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 16 \, \sqrt {2} + 16 \, \sqrt {-3 \, x^{2} + 2}\right ) - \frac {8}{81} \cdot 8^{\frac {1}{4}} \sqrt {2} \log \left (-4 \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 16 \, \sqrt {2} + 16 \, \sqrt {-3 \, x^{2} + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="fricas")

[Out]

2/18711*(189*x^4 + 540*x^2 + 1712)*(-3*x^2 + 2)^(3/4) + 32/81*8^(1/4)*sqrt(2)*arctan(1/4*8^(1/4)*sqrt(2)*sqrt(
8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4)
- 1) + 32/81*8^(1/4)*sqrt(2)*arctan(1/8*8^(1/4)*sqrt(2)*sqrt(-4*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 16*sqrt(2
) + 16*sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 1) + 8/81*8^(1/4)*sqrt(2)*log(4*8^(3/4)*sq
rt(2)*(-3*x^2 + 2)^(1/4) + 16*sqrt(2) + 16*sqrt(-3*x^2 + 2)) - 8/81*8^(1/4)*sqrt(2)*log(-4*8^(3/4)*sqrt(2)*(-3
*x^2 + 2)^(1/4) + 16*sqrt(2) + 16*sqrt(-3*x^2 + 2))

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giac [A]  time = 0.45, size = 160, normalized size = 1.18 \begin {gather*} \frac {2}{891} \, {\left (3 \, x^{2} - 2\right )}^{2} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} - \frac {16}{567} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {7}{4}} - \frac {32}{81} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {32}{81} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {16}{81} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {16}{81} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {56}{243} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="giac")

[Out]

2/891*(3*x^2 - 2)^2*(-3*x^2 + 2)^(3/4) - 16/567*(-3*x^2 + 2)^(7/4) - 32/81*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4)
+ 2*(-3*x^2 + 2)^(1/4))) - 32/81*2^(1/4)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(-3*x^2 + 2)^(1/4))) + 16/81*2^(1/4
)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 16/81*2^(1/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4)
+ sqrt(2) + sqrt(-3*x^2 + 2)) + 56/243*(-3*x^2 + 2)^(3/4)

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maple [C]  time = 2.85, size = 217, normalized size = 1.60 \begin {gather*} \frac {16 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+8\right )^{2}\right ) \ln \left (\frac {\left (-3 x^{2}+2\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+8\right )^{2}-6 x^{2}+2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}+8\right )^{2}-4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+8\right )^{2}\right )}{3 x^{2}-4}\right )}{81}-\frac {16 \RootOf \left (\textit {\_Z}^{4}+8\right ) \ln \left (\frac {\left (-3 x^{2}+2\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}+8\right )^{3}-6 x^{2}-2 \sqrt {-3 x^{2}+2}\, \RootOf \left (\textit {\_Z}^{4}+8\right )^{2}+4 \left (-3 x^{2}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+8\right )}{3 x^{2}-4}\right )}{81}-\frac {2 \left (189 x^{4}+540 x^{2}+1712\right ) \left (3 x^{2}-2\right )}{18711 \left (-3 x^{2}+2\right )^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)

[Out]

-2/18711*(189*x^4+540*x^2+1712)*(3*x^2-2)/(-3*x^2+2)^(1/4)-16/81*RootOf(_Z^4+8)*ln((RootOf(_Z^4+8)^3*(-3*x^2+2
)^(3/4)-2*RootOf(_Z^4+8)^2*(-3*x^2+2)^(1/2)+4*RootOf(_Z^4+8)*(-3*x^2+2)^(1/4)-6*x^2)/(3*x^2-4))+16/81*RootOf(_
Z^2+RootOf(_Z^4+8)^2)*ln((RootOf(_Z^2+RootOf(_Z^4+8)^2)*RootOf(_Z^4+8)^2*(-3*x^2+2)^(3/4)+2*RootOf(_Z^4+8)^2*(
-3*x^2+2)^(1/2)-4*RootOf(_Z^2+RootOf(_Z^4+8)^2)*(-3*x^2+2)^(1/4)-6*x^2)/(3*x^2-4))

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maxima [A]  time = 1.96, size = 151, normalized size = 1.11 \begin {gather*} \frac {2}{891} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {11}{4}} - \frac {16}{567} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {7}{4}} - \frac {32}{81} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {32}{81} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {16}{81} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {16}{81} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {56}{243} \, {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="maxima")

[Out]

2/891*(-3*x^2 + 2)^(11/4) - 16/567*(-3*x^2 + 2)^(7/4) - 32/81*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2
+ 2)^(1/4))) - 32/81*2^(1/4)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(-3*x^2 + 2)^(1/4))) + 16/81*2^(1/4)*log(2^(3/4)
*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 16/81*2^(1/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) +
sqrt(-3*x^2 + 2)) + 56/243*(-3*x^2 + 2)^(3/4)

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mupad [B]  time = 0.91, size = 82, normalized size = 0.60 \begin {gather*} \frac {56\,{\left (2-3\,x^2\right )}^{3/4}}{243}-\frac {16\,{\left (2-3\,x^2\right )}^{7/4}}{567}+\frac {2\,{\left (2-3\,x^2\right )}^{11/4}}{891}+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {32}{81}+\frac {32}{81}{}\mathrm {i}\right )+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {32}{81}-\frac {32}{81}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^7/((2 - 3*x^2)^(1/4)*(3*x^2 - 4)),x)

[Out]

(56*(2 - 3*x^2)^(3/4))/243 - 2^(1/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 + 1i/2))*(32/81 + 32i/81) - 2^(1/4)*a
tan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 - 1i/2))*(32/81 - 32i/81) - (16*(2 - 3*x^2)^(7/4))/567 + (2*(2 - 3*x^2)^(11
/4))/891

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x^{7}}{3 x^{2} \sqrt [4]{2 - 3 x^{2}} - 4 \sqrt [4]{2 - 3 x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(-3*x**2+2)**(1/4)/(-3*x**2+4),x)

[Out]

-Integral(x**7/(3*x**2*(2 - 3*x**2)**(1/4) - 4*(2 - 3*x**2)**(1/4)), x)

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