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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 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 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 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 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*}

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Mathematica [C]  time = 0.01, 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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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} - 1}}{x^{6} - x^{2}}, x\right ) \]

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

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)

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maple [C]  time = 0.01, size = 56, normalized size = 0.40 \[ \frac {\sqrt {x^{4}-1}}{x}+\frac {i \sqrt {x^{2}+1}\, \sqrt {-x^{2}+1}\, \left (-\EllipticE \left (i x , i\right )+\EllipticF \left (i x , i\right )\right )}{\sqrt {x^{4}-1}} \]

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

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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mupad [B]  time = 1.21, size = 18, normalized size = 0.13 \[ -\frac {\sqrt {\frac {1}{x^4}}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ \frac {1}{x^4}\right )}{3\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(x^4 - 1)^(1/2)),x)

[Out]

-((1/x^4)^(1/2)*hypergeom([1/2, 3/4], 7/4, 1/x^4))/(3*x)

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sympy [C]  time = 0.97, size = 29, normalized size = 0.21 \[ - \frac {i \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

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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