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

Optimal. Leaf size=140 \[ -\frac{\sqrt{x^2-1} \sqrt{x^2+1} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right ),\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4-1}}-\frac{x \left (x^2+1\right )}{\sqrt{x^4-1}}+\frac{\sqrt{x^4-1}}{x}+\frac{\sqrt{2} \sqrt{x^2-1} \sqrt{x^2+1} E\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{\sqrt{x^4-1}} \]

[Out]

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

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Rubi [A]  time = 0.0162141, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {325, 306, 222, 1185} \[ -\frac{x \left (x^2+1\right )}{\sqrt{x^4-1}}+\frac{\sqrt{x^4-1}}{x}-\frac{\sqrt{x^2-1} \sqrt{x^2+1} F\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{\sqrt{2} \sqrt{x^4-1}}+\frac{\sqrt{2} \sqrt{x^2-1} \sqrt{x^2+1} E\left (\sin ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{x^2-1}}\right )|\frac{1}{2}\right )}{\sqrt{x^4-1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[-1 + x^4]),x]

[Out]

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

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 306

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, Dist[1/q, 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] && LtQ[a, 0] && GtQ[b, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 1185

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Simp[(e*x*(q + c*x
^2))/(c*Sqrt[a + c*x^4]), x] - Simp[(Sqrt[2]*e*q*Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2)/q]*EllipticE[ArcSin[x/Sqrt[
(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[-a]*c*Sqrt[a + c*x^4]), x] /; EqQ[c*d + e*q, 0] && IntegerQ[q]] /; FreeQ[{a,
c, d, e}, x] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps

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

Mathematica [C]  time = 0.0052404, size = 38, normalized size = 0.27 \[ -\frac{\sqrt{1-x^4} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};x^4\right )}{x \sqrt{x^4-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[-1 + x^4]),x]

[Out]

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

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Maple [C]  time = 0.008, size = 56, normalized size = 0.4 \begin{align*}{\frac{1}{x}\sqrt{{x}^{4}-1}}+{i \left ({\it EllipticF} \left ( ix,i \right ) -{\it EllipticE} \left ( ix,i \right ) \right ) \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)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{4} - 1} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*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^{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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 0.79357, size = 29, normalized size = 0.21 \begin{align*} - \frac{i \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{x^{4}} \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)**(1/2),x)

[Out]

-I*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), x**4)/(4*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} - 1} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(x^4 - 1)*x^2), x)