3.972 \(\int \frac {1}{x^2 \sqrt {16-x^4}} \, dx\)

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

[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 = 0.02, 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 = {325, 307, 221, 1181, 21, 424} \[ -\frac {\sqrt {16-x^4}}{16 x}+\frac {1}{8} F\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right )-\frac {1}{8} E\left (\left .\sin ^{-1}\left (\frac {x}{2}\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[16 - x^4]),x]

[Out]

-Sqrt[16 - x^4]/(16*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 221

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 307

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 325

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 424

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

Rule 1181

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]  time = 0.00, size = 24, normalized size = 0.56 \[ -\frac {\, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\frac {x^4}{16}\right )}{4 x} \]

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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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + 16}}{x^{6} - 16 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + 16)/(x^6 - 16*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-x^{4} + 16} x^{2}}\,{d x} \]

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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maple [A]  time = 0.01, size = 58, normalized size = 1.35 \[ -\frac {\sqrt {-x^{4}+16}}{16 x}+\frac {\sqrt {-x^{2}+4}\, \sqrt {x^{2}+4}\, \left (-\EllipticE \left (\frac {x}{2}, i\right )+\EllipticF \left (\frac {x}{2}, i\right )\right )}{8 \sqrt {-x^{4}+16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \[ \int \frac {1}{\sqrt {-x^{4} + 16} x^{2}}\,{d x} \]

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

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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sympy [A]  time = 0.98, size = 34, normalized size = 0.79 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{16}} \right )}}{16 x \Gamma \left (\frac {3}{4}\right )} \]

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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