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

Optimal. Leaf size=47 \[ \frac {8 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}-\frac {2}{15} x \left (2-3 x^2\right )^{3/4} \]

[Out]

-2/15*x*(-3*x^2+2)^(3/4)+8/45*2^(1/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*
EllipticE(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {321, 228} \[ \frac {8 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}-\frac {2}{15} x \left (2-3 x^2\right )^{3/4} \]

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*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/(15*Sqrt[3])

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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]

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 {8 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.87 \[ -\frac {2}{15} x \left (\left (2-3 x^2\right )^{3/4}-2^{3/4} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^2}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(2 - 3*x^2)^(1/4),x]

[Out]

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

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x^{2}}{3 \, x^{2} - 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(-3*x^2 + 2)^(3/4)*x^2/(3*x^2 - 2), 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 = 0.29, size = 38, normalized size = 0.81 \[ \frac {2 \,2^{\frac {3}{4}} x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{15}+\frac {2 \left (3 x^{2}-2\right ) x}{15 \left (-3 x^{2}+2\right )^{\frac {1}{4}}} \]

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*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 = -0.02 \[ \int \frac {x^2}{{\left (2-3\,x^2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(2 - 3*x^2)^(1/4),x)

[Out]

int(x^2/(2 - 3*x^2)^(1/4), x)

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sympy [C]  time = 0.72, size = 29, normalized size = 0.62 \[ \frac {2^{\frac {3}{4}} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{2 i \pi }}{2}} \right )}}{6} \]

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

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