\(\int \frac {x^2}{\sqrt {16-x^4}} \, dx\) [970]

Optimal result

Integrand size = 15, antiderivative size = 21 \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=2 E\left (\left .\arcsin \left (\frac {x}{2}\right )\right |-1\right )-2 \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {313, 227, 1195, 21, 435} \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=2 E\left (\left .\arcsin \left (\frac {x}{2}\right )\right |-1\right )-2 \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[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 227

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 313

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 435

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

Rule 1195

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*} \text {integral}& = -\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\frac {1}{12} x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {x^4}{16}\right ) \]

[In]

Integrate[x^2/Sqrt[16 - x^4],x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
meijerg	\(\frac {x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {x^{4}}{16}\right )}{12}\)	\(17\)
default	\(-\frac {2 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (F\left (\frac {x}{2}, i\right )-E\left (\frac {x}{2}, i\right )\right )}{\sqrt {-x^{4}+16}}\)	\(43\)
elliptic	\(-\frac {2 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (F\left (\frac {x}{2}, i\right )-E\left (\frac {x}{2}, i\right )\right )}{\sqrt {-x^{4}+16}}\)	\(43\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (15) = 30\).

Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\frac {-8 i \, x E(\arcsin \left (\frac {2}{x}\right )\,|\,-1) + 8 i \, x F(\arcsin \left (\frac {2}{x}\right )\,|\,-1) - \sqrt {-x^{4} + 16}}{x} \]

[In]

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

[Out]

(-8*I*x*elliptic_e(arcsin(2/x), -1) + 8*I*x*elliptic_f(arcsin(2/x), -1) - sqrt(-x^4 + 16))/x

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (12) = 24\).

Time = 0.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {7}{4}\right )} \]

[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))

Maxima [F]

\[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {-x^{4} + 16}} \,d x } \]

[In]

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

[Out]

integrate(x^2/sqrt(-x^4 + 16), x)

Giac [F]

\[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {-x^{4} + 16}} \,d x } \]

[In]

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

[Out]

integrate(x^2/sqrt(-x^4 + 16), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {16-x^4}} \, dx=\int \frac {x^2}{\sqrt {16-x^4}} \,d x \]

[In]

int(x^2/(16 - x^4)^(1/2),x)

[Out]

int(x^2/(16 - x^4)^(1/2), x)