\(\int \frac {1}{x^4 \sqrt {16-x^4}} \, dx\) [973]

Optimal result

Integrand size = 15, antiderivative size = 31 \[ \int \frac {1}{x^4 \sqrt {16-x^4}} \, dx=-\frac {\sqrt {16-x^4}}{48 x^3}+\frac {1}{96} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right ) \]

[Out]

1/96*EllipticF(1/2*x,I)-1/48*(-x^4+16)^(1/2)/x^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, 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 = {331, 227} \[ \int \frac {1}{x^4 \sqrt {16-x^4}} \, dx=\frac {1}{96} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{2}\right ),-1\right )-\frac {\sqrt {16-x^4}}{48 x^3} \]

[In]

Int[1/(x^4*Sqrt[16 - x^4]),x]

[Out]

-1/48*Sqrt[16 - x^4]/x^3 + EllipticF[ArcSin[x/2], -1]/96

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {16-x^4}}{48 x^3}+\frac {1}{48} \int \frac {1}{\sqrt {16-x^4}} \, dx \\ & = -\frac {\sqrt {16-x^4}}{48 x^3}+\frac {1}{96} 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 = 0.77 \[ \int \frac {1}{x^4 \sqrt {16-x^4}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\frac {x^4}{16}\right )}{12 x^3} \]

[In]

Integrate[1/(x^4*Sqrt[16 - x^4]),x]

[Out]

-1/12*Hypergeometric2F1[-3/4, 1/2, 1/4, x^4/16]/x^3

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 = 0.55

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};\frac {x^{4}}{16}\right )}{12 x^{3}}\)	\(17\)
default	\(-\frac {\sqrt {-x^{4}+16}}{48 x^{3}}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{96 \sqrt {-x^{4}+16}}\)	\(49\)
elliptic	\(-\frac {\sqrt {-x^{4}+16}}{48 x^{3}}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{96 \sqrt {-x^{4}+16}}\)	\(49\)
risch	\(\frac {x^{4}-16}{48 x^{3} \sqrt {-x^{4}+16}}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, F\left (\frac {x}{2}, i\right )}{96 \sqrt {-x^{4}+16}}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^4 \sqrt {16-x^4}} \, dx=\frac {x^{3} F(\arcsin \left (\frac {1}{2} \, x\right )\,|\,-1) - 2 \, \sqrt {-x^{4} + 16}}{96 \, x^{3}} \]

[In]

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

[Out]

1/96*(x^3*elliptic_f(arcsin(1/2*x), -1) - 2*sqrt(-x^4 + 16))/x^3

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \sqrt {16-x^4}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^4), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^4), x)

Mupad [F(-1)]

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

[In]

int(1/(x^4*(16 - x^4)^(1/2)),x)

[Out]

int(1/(x^4*(16 - x^4)^(1/2)), x)