3.10.72 \(\int \frac {1}{x^2 \sqrt {16-x^4}} \, dx\) [972]

Optimal. Leaf size=43 \[ -\frac {\sqrt {16-x^4}}{16 x}-\frac {1}{8} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {1}{8} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {331, 313, 227, 1195, 21, 435} \begin {gather*} \frac {1}{8} F\left (\left .\text {ArcSin}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{8} E\left (\left .\text {ArcSin}\left (\frac {x}{2}\right )\right |-1\right )-\frac {\sqrt {16-x^4}}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[16 - x^4]),x]

[Out]

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

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 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]

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*} \int \frac {1}{x^2 \sqrt {16-x^4}} \, dx &=-\frac {\sqrt {16-x^4}}{16 x}-\frac {1}{16} \int \frac {x^2}{\sqrt {16-x^4}} \, dx\\ &=-\frac {\sqrt {16-x^4}}{16 x}+\frac {1}{4} \int \frac {1}{\sqrt {16-x^4}} \, dx-\frac {1}{4} \int \frac {1+\frac {x^2}{4}}{\sqrt {16-x^4}} \, dx\\ &=-\frac {\sqrt {16-x^4}}{16 x}+\frac {1}{8} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{4} \int \frac {1+\frac {x^2}{4}}{\sqrt {4-x^2} \sqrt {4+x^2}} \, dx\\ &=-\frac {\sqrt {16-x^4}}{16 x}+\frac {1}{8} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{16} \int \frac {\sqrt {4+x^2}}{\sqrt {4-x^2}} \, dx\\ &=-\frac {\sqrt {16-x^4}}{16 x}-\frac {1}{8} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )+\frac {1}{8} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 24, normalized size = 0.56 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\frac {x^4}{16}\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[16 - x^4]),x]

[Out]

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

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Maple [A]
time = 0.17, size = 58, normalized size = 1.35

method	result	size
meijerg	\(-\frac {\hypergeom \left (\left [-\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{4}\right ], \frac {x^{4}}{16}\right )}{4 x}\)	\(17\)
default	\(-\frac {\sqrt {-x^{4}+16}}{16 x}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{8 \sqrt {-x^{4}+16}}\)	\(58\)
elliptic	\(-\frac {\sqrt {-x^{4}+16}}{16 x}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{8 \sqrt {-x^{4}+16}}\)	\(58\)
risch	\(\frac {x^{4}-16}{16 x \sqrt {-x^{4}+16}}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (\frac {x}{2}, i\right )-\EllipticE \left (\frac {x}{2}, i\right )\right )}{8 \sqrt {-x^{4}+16}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-x^4+16)^(1/2)/x+1/8*(-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
))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^2), x)

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Fricas [A]
time = 0.07, size = 34, normalized size = 0.79 \begin {gather*} -\frac {x E(\arcsin \left (\frac {1}{2} \, x\right )\,|\,-1) - x F(\arcsin \left (\frac {1}{2} \, x\right )\,|\,-1) + 2 \, \sqrt {-x^{4} + 16}}{32 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(x*elliptic_e(arcsin(1/2*x), -1) - x*elliptic_f(arcsin(1/2*x), -1) + 2*sqrt(-x^4 + 16))/x

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Sympy [A]
time = 0.35, size = 34, normalized size = 0.79 \begin {gather*} \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-x^4 + 16)*x^2), x)

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Mupad [B]
time = 1.25, size = 33, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {1-\frac {16}{x^4}}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ \frac {16}{x^4}\right )}{3\,x\,\sqrt {16-x^4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(16 - x^4)^(1/2)),x)

[Out]

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

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