\(\int \frac {1}{\sqrt {16-x^4}} \, dx\) [971]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {1}{2} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {227} \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {1}{2} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[In]

Int[1/Sqrt[16 - x^4],x]

[Out]

EllipticF[ArcSin[x/2], -1]/2

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {1}{2} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[In]

Integrate[1/Sqrt[16 - x^4],x]

[Out]

EllipticF[ArcSin[x/2], -1]/2

Maple [C] (verified)

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

Time = 4.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25

method	result	size
meijerg	\(\frac {x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {x^{4}}{16}\right )}{4}\)	\(15\)
default	\(\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{2 \sqrt {-x^{4}+16}}\)	\(34\)
elliptic	\(\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{2 \sqrt {-x^{4}+16}}\)	\(34\)

[In]

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

[Out]

1/4*x*hypergeom([1/4,1/2],[5/4],1/16*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {1}{2} \, F(\arcsin \left (\frac {1}{2} \, x\right )\,|\,-1) \]

[In]

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

[Out]

1/2*elliptic_f(arcsin(1/2*x), -1)

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. 31 vs. \(2 (5) = 10\).

Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.58 \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), x**4*exp_polar(2*I*pi)/16)/(16*gamma(5/4))

Maxima [F]

\[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\int { \frac {1}{\sqrt {-x^{4} + 16}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(-x^4 + 16), x)

Giac [F]

\[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\int { \frac {1}{\sqrt {-x^{4} + 16}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(-x^4 + 16), x)

Mupad [B] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sqrt {16-x^4}} \, dx=\frac {x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ \frac {x^4}{16}\right )}{4} \]

[In]

int(1/(16 - x^4)^(1/2),x)

[Out]

(x*hypergeom([1/4, 1/2], 5/4, x^4/16))/4