3.9.50 \(\int \sqrt {\frac {1}{1-x^2}} \, dx\) [850]

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 = 0.00, 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 = {1972, 222} \begin {gather*} \sqrt {\frac {1}{1-x^2}} \sqrt {1-x^2} \text {ArcSin}(x) \end {gather*}

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 222

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 1972

Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(a + b*x^n)
^(p*q)], Int[u*(a + b*x^n)^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]

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 = 0.02, size = 39, normalized size = 1.44 \begin {gather*} \sqrt {\frac {1}{1-x^2}} \sqrt {1-x^2} \tan ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 - x^2)^(-1)],x]

[Out]

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

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Maple [A]
time = 0.52, size = 30, normalized size = 1.11

method	result	size
meijerg	\(\arcsin \left (x \right ) \sqrt {\frac {1}{-x^{2}+1}}\, \sqrt {-x^{2}+1}\)	\(24\)
default	\(\sqrt {-\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right )\)	\(30\)
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}}\, x^{2}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}}+x \right )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/(-x^2+1))^(1/2),x,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.43, size = 26, normalized size = 0.96 \begin {gather*} 2 \, \arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {-\frac {1}{x^{2} - 1}} + 1}{x}\right ) \end {gather*}

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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Sympy [A]
time = 0.86, size = 7, normalized size = 0.26 \begin {gather*} \begin {cases} \operatorname {asin}{\left (x \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \end {gather*}

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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Giac [A]
time = 2.17, size = 10, normalized size = 0.37 \begin {gather*} -\arcsin \left (x\right ) \mathrm {sgn}\left (x^{2} - 1\right ) \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \sqrt {-\frac {1}{x^2-1}} \,d x \end {gather*}

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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