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

Optimal. Leaf size=48 \[ \frac {\sqrt [4]{3} E\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}-\frac {\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}} \]

[Out]

1/2*3^(1/4)*EllipticE(1/3*2^(1/4)*3^(3/4)*x,I)*2^(1/4)-1/2*3^(1/4)*EllipticF(1/3*2^(1/4)*3^(3/4)*x,I)*2^(1/4)

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Rubi [A]  time = 0.03, antiderivative size = 48, 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 = {307, 221, 1181, 21, 424} \[ \frac {\sqrt [4]{3} E\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}-\frac {\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[3 - 2*x^4],x]

[Out]

(3^(1/4)*EllipticE[ArcSin[(2/3)^(1/4)*x], -1])/2^(3/4) - (3^(1/4)*EllipticF[ArcSin[(2/3)^(1/4)*x], -1])/2^(3/4
)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 221

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

Rule 307

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1181

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

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {3-2 x^4}} \, dx &=-\left (\sqrt {\frac {3}{2}} \int \frac {1}{\sqrt {3-2 x^4}} \, dx\right )+\sqrt {\frac {3}{2}} \int \frac {1+\sqrt {\frac {2}{3}} x^2}{\sqrt {3-2 x^4}} \, dx\\ &=-\frac {\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}+\sqrt {3} \int \frac {1+\sqrt {\frac {2}{3}} x^2}{\sqrt {\sqrt {6}-2 x^2} \sqrt {\sqrt {6}+2 x^2}} \, dx\\ &=-\frac {\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}+\frac {\int \frac {\sqrt {\sqrt {6}+2 x^2}}{\sqrt {\sqrt {6}-2 x^2}} \, dx}{\sqrt {2}}\\ &=\frac {\sqrt [4]{3} E\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}-\frac {\sqrt [4]{3} F\left (\left .\sin ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )\right |-1\right )}{2^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.60 \[ \frac {x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {2 x^4}{3}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[3 - 2*x^4],x]

[Out]

(x^3*Hypergeometric2F1[1/2, 3/4, 7/4, (2*x^4)/3])/(3*Sqrt[3])

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-2 \, x^{4} + 3} x^{2}}{2 \, x^{4} - 3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-2 \, x^{4} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-2*x^4 + 3), x)

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maple [A]  time = 0.04, size = 69, normalized size = 1.44 \[ -\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {-3 \sqrt {6}\, x^{2}+9}\, \sqrt {3 \sqrt {6}\, x^{2}+9}\, \left (-\EllipticE \left (\frac {\sqrt {3}\, 6^{\frac {1}{4}} x}{3}, i\right )+\EllipticF \left (\frac {\sqrt {3}\, 6^{\frac {1}{4}} x}{3}, i\right )\right )}{18 \sqrt {-2 x^{4}+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*3^(1/2)*6^(1/4)*(9-3*6^(1/2)*x^2)^(1/2)*(9+3*6^(1/2)*x^2)^(1/2)/(-2*x^4+3)^(1/2)*(EllipticF(1/3*x*3^(1/2
)*6^(1/4),I)-EllipticE(1/3*x*3^(1/2)*6^(1/4),I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-2 \, x^{4} + 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-2*x^4 + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\sqrt {3-2\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(3 - 2*x^4)^(1/2),x)

[Out]

int(x^2/(3 - 2*x^4)^(1/2), x)

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sympy [A]  time = 1.28, size = 39, normalized size = 0.81 \[ \frac {\sqrt {3} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {2 x^{4} e^{2 i \pi }}{3}} \right )}}{12 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(3)*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), 2*x**4*exp_polar(2*I*pi)/3)/(12*gamma(7/4))

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