3.970 \(\int \frac {x^2}{\sqrt {16-x^4}} \, dx\)

Optimal. Leaf size=21 \[ 2 E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-2 F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 21, 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} \[ 2 E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-2 F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[16 - x^4],x]

[Out]

2*EllipticE[ArcSin[x/2], -1] - 2*EllipticF[ArcSin[x/2], -1]

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 {16-x^4}} \, dx &=-\left (4 \int \frac {1}{\sqrt {16-x^4}} \, dx\right )+4 \int \frac {1+\frac {x^2}{4}}{\sqrt {16-x^4}} \, dx\\ &=-2 F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+4 \int \frac {1+\frac {x^2}{4}}{\sqrt {4-x^2} \sqrt {4+x^2}} \, dx\\ &=-2 F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\int \frac {\sqrt {4+x^2}}{\sqrt {4-x^2}} \, dx\\ &=2 E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-2 F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.00, size = 24, normalized size = 1.14 \[ \frac {1}{12} x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {x^4}{16}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[16 - x^4],x]

[Out]

(x^3*Hypergeometric2F1[1/2, 3/4, 7/4, x^4/16])/12

________________________________________________________________________________________

fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 16} x^{2}}{x^{4} - 16}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^4+16)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-x^4 + 16)*x^2/(x^4 - 16), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-x^{4} + 16}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^4+16)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(-x^4 + 16), x)

________________________________________________________________________________________

maple [B]  time = 0.01, size = 43, normalized size = 2.05 \[ -\frac {2 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (-\EllipticE \left (\frac {x}{2}, i\right )+\EllipticF \left (\frac {x}{2}, i\right )\right )}{\sqrt {-x^{4}+16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-x^4+16)^(1/2),x)

[Out]

-2*(-x^2+4)^(1/2)*(x^2+4)^(1/2)/(-x^4+16)^(1/2)*(EllipticF(1/2*x,I)-EllipticE(1/2*x,I))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-x^{4} + 16}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^4+16)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(-x^4 + 16), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {x^2}{\sqrt {16-x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(16 - x^4)^(1/2),x)

[Out]

int(x^2/(16 - x^4)^(1/2), x)

________________________________________________________________________________________

sympy [B]  time = 1.31, size = 32, normalized size = 1.52 \[ \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-x**4+16)**(1/2),x)

[Out]

x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**4*exp_polar(2*I*pi)/16)/(16*gamma(7/4))

________________________________________________________________________________________