3.850 \(\int \sqrt {\frac {1}{1-x^2}} \, dx\)

Optimal. Leaf size=27 \[ \sqrt {\frac {1}{1-x^2}} \sqrt {1-x^2} \sin ^{-1}(x) \]

[Out]

arcsin(x)*(1/(-x^2+1))^(1/2)*(-x^2+1)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6720, 216} \[ \sqrt {\frac {1}{1-x^2}} \sqrt {1-x^2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 - x^2)^(-1)],x]

[Out]

Sqrt[(1 - x^2)^(-1)]*Sqrt[1 - x^2]*ArcSin[x]

Rule 216

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

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \sqrt {\frac {1}{1-x^2}} \sqrt {1-x^2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 - x^2)^(-1)],x]

[Out]

Sqrt[(1 - x^2)^(-1)]*Sqrt[1 - x^2]*ArcSin[x]

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fricas [A]  time = 0.45, size = 26, normalized size = 0.96 \[ 2 \, \arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {-\frac {1}{x^{2} - 1}} + 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 10, normalized size = 0.37 \[ -\arcsin \relax (x) \mathrm {sgn}\left (x^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(x)*sgn(x^2 - 1)

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maple [A]  time = 0.01, size = 30, normalized size = 1.11 \[ \sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/(x^2-1))^(1/2)*(x^2-1)^(1/2)*ln(x+(x^2-1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-1/(x^2 - 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \sqrt {-\frac {1}{x^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1/(x^2 - 1))^(1/2),x)

[Out]

int((-1/(x^2 - 1))^(1/2), x)

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sympy [A]  time = 1.01, size = 7, normalized size = 0.26 \[ \begin {cases} \operatorname {asin}{\relax (x )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((asin(x), (x > -1) & (x < 1)))

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