3.915 \(\int \frac{1}{x^6 (1-x^4)^{3/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac{21}{10} \text{EllipticF}\left (\sin ^{-1}(x),-1\right )-\frac{21 \sqrt{1-x^4}}{10 x}-\frac{7 \sqrt{1-x^4}}{10 x^5}+\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

1/(2*x^5*Sqrt[1 - x^4]) - (7*Sqrt[1 - x^4])/(10*x^5) - (21*Sqrt[1 - x^4])/(10*x) - (21*EllipticE[ArcSin[x], -1
])/10 + (21*EllipticF[ArcSin[x], -1])/10

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Rubi [A]  time = 0.0263981, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {290, 325, 307, 221, 1181, 424} \[ -\frac{21 \sqrt{1-x^4}}{10 x}-\frac{7 \sqrt{1-x^4}}{10 x^5}+\frac{1}{2 x^5 \sqrt{1-x^4}}+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(1 - x^4)^(3/2)),x]

[Out]

1/(2*x^5*Sqrt[1 - x^4]) - (7*Sqrt[1 - x^4])/(10*x^5) - (21*Sqrt[1 - x^4])/(10*x) - (21*EllipticE[ArcSin[x], -1
])/10 + (21*EllipticF[ArcSin[x], -1])/10

Rule 290

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

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 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^6 \left (1-x^4\right )^{3/2}} \, dx &=\frac{1}{2 x^5 \sqrt{1-x^4}}+\frac{7}{2} \int \frac{1}{x^6 \sqrt{1-x^4}} \, dx\\ &=\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{7 \sqrt{1-x^4}}{10 x^5}+\frac{21}{10} \int \frac{1}{x^2 \sqrt{1-x^4}} \, dx\\ &=\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{7 \sqrt{1-x^4}}{10 x^5}-\frac{21 \sqrt{1-x^4}}{10 x}-\frac{21}{10} \int \frac{x^2}{\sqrt{1-x^4}} \, dx\\ &=\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{7 \sqrt{1-x^4}}{10 x^5}-\frac{21 \sqrt{1-x^4}}{10 x}+\frac{21}{10} \int \frac{1}{\sqrt{1-x^4}} \, dx-\frac{21}{10} \int \frac{1+x^2}{\sqrt{1-x^4}} \, dx\\ &=\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{7 \sqrt{1-x^4}}{10 x^5}-\frac{21 \sqrt{1-x^4}}{10 x}+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{21}{10} \int \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} \, dx\\ &=\frac{1}{2 x^5 \sqrt{1-x^4}}-\frac{7 \sqrt{1-x^4}}{10 x^5}-\frac{21 \sqrt{1-x^4}}{10 x}-\frac{21}{10} E\left (\left .\sin ^{-1}(x)\right |-1\right )+\frac{21}{10} F\left (\left .\sin ^{-1}(x)\right |-1\right )\\ \end{align*}

Mathematica [C]  time = 0.0029563, size = 20, normalized size = 0.28 \[ -\frac{\, _2F_1\left (-\frac{5}{4},\frac{3}{2};-\frac{1}{4};x^4\right )}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(1 - x^4)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.061, size = 82, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}}{2}{\frac{1}{\sqrt{-{x}^{4}+1}}}}-{\frac{1}{5\,{x}^{5}}\sqrt{-{x}^{4}+1}}-{\frac{8}{5\,x}\sqrt{-{x}^{4}+1}}+{\frac{21\,{\it EllipticF} \left ( x,i \right ) -21\,{\it EllipticE} \left ( x,i \right ) }{10}\sqrt{-{x}^{2}+1}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{-{x}^{4}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(-x^4+1)^(3/2),x)

[Out]

1/2*x^3/(-x^4+1)^(1/2)-1/5*(-x^4+1)^(1/2)/x^5-8/5*(-x^4+1)^(1/2)/x+21/10*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)
^(1/2)*(EllipticF(x,I)-EllipticE(x,I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-x^{4} + 1\right )}^{\frac{3}{2}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^6), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-x^{4} + 1}}{x^{14} - 2 \, x^{10} + x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-x^4 + 1)/(x^14 - 2*x^10 + x^6), x)

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Sympy [A]  time = 1.4366, size = 37, normalized size = 0.52 \begin{align*} \frac{\Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{3}{2} \\ - \frac{1}{4} \end{matrix}\middle |{x^{4} e^{2 i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(-x**4+1)**(3/2),x)

[Out]

gamma(-5/4)*hyper((-5/4, 3/2), (-1/4,), x**4*exp_polar(2*I*pi))/(4*x**5*gamma(-1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-x^{4} + 1\right )}^{\frac{3}{2}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((-x^4 + 1)^(3/2)*x^6), x)