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

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

[Out]

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

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Rubi [A]  time = 0.0198747, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, 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 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 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 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 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]

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 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]

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*}

Mathematica [C]  time = 0.0027426, 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]

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

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Maple [A]  time = 0.009, size = 58, normalized size = 1.4 \begin{align*} -{\frac{1}{16\,x}\sqrt{-{x}^{4}+16}}+{\frac{1}{8}\sqrt{-{x}^{2}+4}\sqrt{{x}^{2}+4} \left ({\it EllipticF} \left ({\frac{x}{2}},i \right ) -{\it EllipticE} \left ({\frac{x}{2}},i \right ) \right ){\frac{1}{\sqrt{-{x}^{4}+16}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 16} x^{2}}\,{d x} \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-x^{4} + 16}}{x^{6} - 16 \, x^{2}}, x\right ) \end{align*}

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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Sympy [A]  time = 0.677896, size = 34, normalized size = 0.79 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{4} + 16} x^{2}}\,{d x} \end{align*}

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)