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

Optimal. Leaf size=30 \[ -\frac{1}{2} \tan ^{-1}\left (\frac{2-x^2}{2 \sqrt{-x^4+x^2-1}}\right ) \]

[Out]

-ArcTan[(2 - x^2)/(2*Sqrt[-1 + x^2 - x^4])]/2

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Rubi [A]  time = 0.0210465, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1114, 724, 204} \[ -\frac{1}{2} \tan ^{-1}\left (\frac{2-x^2}{2 \sqrt{-x^4+x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-1 + x^2 - x^4]),x]

[Out]

-ArcTan[(2 - x^2)/(2*Sqrt[-1 + x^2 - x^4])]/2

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \sqrt{-1+x^2-x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1+x-x^2}} \, dx,x,x^2\right )\\ &=-\operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,\frac{-2+x^2}{\sqrt{-1+x^2-x^4}}\right )\\ &=\frac{1}{2} \tan ^{-1}\left (\frac{-2+x^2}{2 \sqrt{-1+x^2-x^4}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0040188, size = 28, normalized size = 0.93 \[ \frac{1}{2} \tan ^{-1}\left (\frac{x^2-2}{2 \sqrt{-x^4+x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(-2 + x^2)/(2*Sqrt[-1 + x^2 - x^4])]/2

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Maple [A]  time = 0.011, size = 23, normalized size = 0.8 \begin{align*}{\frac{1}{2}\arctan \left ({\frac{{x}^{2}-2}{2}{\frac{1}{\sqrt{-{x}^{4}+{x}^{2}-1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(1/2*(x^2-2)/(-x^4+x^2-1)^(1/2))

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Maxima [C]  time = 1.44107, size = 23, normalized size = 0.77 \begin{align*} -\frac{1}{2} i \, \operatorname{arsinh}\left (-\frac{1}{3} \, \sqrt{3} + \frac{2 \, \sqrt{3}}{3 \, x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*arcsinh(-1/3*sqrt(3) + 2/3*sqrt(3)/x^2)

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Fricas [C]  time = 1.70981, size = 155, normalized size = 5.17 \begin{align*} \frac{1}{4} i \, \log \left (\frac{x^{2} + 2 i \, \sqrt{-x^{4} + x^{2} - 1} - 2}{2 \, x^{2}}\right ) - \frac{1}{4} i \, \log \left (\frac{x^{2} - 2 i \, \sqrt{-x^{4} + x^{2} - 1} - 2}{2 \, x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*log(1/2*(x^2 + 2*I*sqrt(-x^4 + x^2 - 1) - 2)/x^2) - 1/4*I*log(1/2*(x^2 - 2*I*sqrt(-x^4 + x^2 - 1) - 2)/x
^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*sqrt(-x**4 + x**2 - 1)), x)

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Giac [C]  time = 1.12593, size = 20, normalized size = 0.67 \begin{align*} \frac{1}{2} i \, \arcsin \left (\frac{1}{3} \, \sqrt{3}{\left (\frac{2 i}{x^{2}} - i\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*arcsin(1/3*sqrt(3)*(2*I/x^2 - I))