\(\int \frac {x^4}{\sqrt {16-x^4}} \, dx\) [969]

Optimal result

Integrand size = 15, antiderivative size = 29 \[ \int \frac {x^4}{\sqrt {16-x^4}} \, dx=-\frac {1}{3} x \sqrt {16-x^4}+\frac {8}{3} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[Out]

8/3*EllipticF(1/2*x,I)-1/3*x*(-x^4+16)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {327, 227} \[ \int \frac {x^4}{\sqrt {16-x^4}} \, dx=\frac {8}{3} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right )-\frac {1}{3} x \sqrt {16-x^4} \]

[In]

Int[x^4/Sqrt[16 - x^4],x]

[Out]

-1/3*(x*Sqrt[16 - x^4]) + (8*EllipticF[ArcSin[x/2], -1])/3

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 327

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*} \text {integral}& = -\frac {1}{3} x \sqrt {16-x^4}+\frac {16}{3} \int \frac {1}{\sqrt {16-x^4}} \, dx \\ & = -\frac {1}{3} x \sqrt {16-x^4}+\frac {8}{3} 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 = 36, normalized size of antiderivative = 1.24 \[ \int \frac {x^4}{\sqrt {16-x^4}} \, dx=-\frac {1}{3} x \left (\sqrt {16-x^4}-4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {x^4}{16}\right )\right ) \]

[In]

Integrate[x^4/Sqrt[16 - x^4],x]

[Out]

-1/3*(x*(Sqrt[16 - x^4] - 4*Hypergeometric2F1[1/4, 1/2, 5/4, x^4/16]))

Maple [C] (verified)

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

Time = 4.48 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59

method	result	size
meijerg	\(\frac {x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};\frac {x^{4}}{16}\right )}{20}\)	\(17\)
default	\(-\frac {x \sqrt {-x^{4}+16}}{3}+\frac {8 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{3 \sqrt {-x^{4}+16}}\)	\(47\)
elliptic	\(-\frac {x \sqrt {-x^{4}+16}}{3}+\frac {8 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{3 \sqrt {-x^{4}+16}}\)	\(47\)
risch	\(\frac {x \left (x^{4}-16\right )}{3 \sqrt {-x^{4}+16}}+\frac {8 \sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{3 \sqrt {-x^{4}+16}}\)	\(52\)

[In]

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

[Out]

1/20*x^5*hypergeom([1/2,5/4],[9/4],1/16*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {x^4}{\sqrt {16-x^4}} \, dx=-\frac {1}{3} \, \sqrt {-x^{4} + 16} x + \frac {8}{3} i \, F(\arcsin \left (\frac {2}{x}\right )\,|\,-1) \]

[In]

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

[Out]

-1/3*sqrt(-x^4 + 16)*x + 8/3*I*elliptic_f(arcsin(2/x), -1)

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {x^4}{\sqrt {16-x^4}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), x**4*exp_polar(2*I*pi)/16)/(16*gamma(9/4))

Maxima [F]

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

[In]

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

[Out]

integrate(x^4/sqrt(-x^4 + 16), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^4/sqrt(-x^4 + 16), x)

Mupad [F(-1)]

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

[In]

int(x^4/(16 - x^4)^(1/2),x)

[Out]

int(x^4/(16 - x^4)^(1/2), x)