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

Optimal. Leaf size=53 \[ \frac{3}{2} \text{EllipticF}\left (\sin ^{-1}(x),-1\right )-\frac{3 \sqrt{1-x^4}}{2 x}+\frac{1}{2 x \sqrt{1-x^4}}-\frac{3}{2} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

[Out]

1/(2*x*Sqrt[1 - x^4]) - (3*Sqrt[1 - x^4])/(2*x) - (3*EllipticE[ArcSin[x], -1])/2 + (3*EllipticF[ArcSin[x], -1]
)/2

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Rubi [A]  time = 0.0204178, antiderivative size = 53, 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 = {290, 325, 307, 221, 1181, 424} \[ -\frac{3 \sqrt{1-x^4}}{2 x}+\frac{1}{2 x \sqrt{1-x^4}}+\frac{3}{2} F\left (\left .\sin ^{-1}(x)\right |-1\right )-\frac{3}{2} E\left (\left .\sin ^{-1}(x)\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 - x^4)^(3/2)),x]

[Out]

1/(2*x*Sqrt[1 - x^4]) - (3*Sqrt[1 - x^4])/(2*x) - (3*EllipticE[ArcSin[x], -1])/2 + (3*EllipticF[ArcSin[x], -1]
)/2

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

Mathematica [C]  time = 0.0028323, size = 18, normalized size = 0.34 \[ -\frac{\, _2F_1\left (-\frac{1}{4},\frac{3}{2};\frac{3}{4};x^4\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(1 - x^4)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.01, size = 68, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}}{2}{\frac{1}{\sqrt{-{x}^{4}+1}}}}-{\frac{1}{x}\sqrt{-{x}^{4}+1}}+{\frac{3\,{\it EllipticF} \left ( x,i \right ) -3\,{\it EllipticE} \left ( x,i \right ) }{2}\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^2/(-x^4+1)^(3/2),x)

[Out]

1/2*x^3/(-x^4+1)^(1/2)-(-x^4+1)^(1/2)/x+3/2*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/2)*(EllipticF(x,I)-Ellipt
icE(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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((-x^4 + 1)^(3/2)*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} + 1}}{x^{10} - 2 \, x^{6} + x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.972371, size = 32, normalized size = 0.6 \begin{align*} \frac{\Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{3}{2} \\ \frac{3}{4} \end{matrix}\middle |{x^{4} e^{2 i \pi }} \right )}}{4 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+1)**(3/2),x)

[Out]

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

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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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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