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

Optimal. Leaf size=25 \[ \sqrt{1-x^2} \sqrt{\frac{1}{x^2-1}} \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.014449, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6720, 217, 206} \[ \sqrt{\frac{1}{x^2-1}} \sqrt{x^2-1} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-1 + x^2)^(-1)],x]

[Out]

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

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])

Rule 217

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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\\ &=\left (\sqrt{\frac{1}{-1+x^2}} \sqrt{-1+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-1+x^2}}\right )\\ &=\sqrt{\frac{1}{-1+x^2}} \sqrt{-1+x^2} \tanh ^{-1}\left (\frac{x}{\sqrt{-1+x^2}}\right )\\ \end{align*}

Mathematica [B]  time = 0.0198312, size = 56, normalized size = 2.24 \[ \frac{1}{2} \sqrt{\frac{1}{x^2-1}} \sqrt{x^2-1} \left (\log \left (\frac{x}{\sqrt{x^2-1}}+1\right )-\log \left (1-\frac{x}{\sqrt{x^2-1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(-1 + x^2)^(-1)],x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 1.1 \begin{align*} \sqrt{ \left ({x}^{2}-1 \right ) ^{-1}}\sqrt{{x}^{2}-1}\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \end{align*}

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 [A]  time = 1.2255, size = 19, normalized size = 0.76 \begin{align*} \log \left (2 \, x + 2 \, \sqrt{x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*x + 2*sqrt(x^2 - 1))

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Fricas [A]  time = 1.42682, size = 35, normalized size = 1.4 \begin{align*} -\log \left (-x + \sqrt{x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-x + sqrt(x^2 - 1))

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Sympy [A]  time = 1.18868, size = 15, normalized size = 0.6 \begin{align*} \begin{cases} \log{\left (x + \sqrt{x^{2} - 1} \right )} & \text{for}\: x > -1 \wedge x < 1 \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(x + sqrt(x**2 - 1)), (x > -1) & (x < 1)))

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Giac [A]  time = 1.15246, size = 20, normalized size = 0.8 \begin{align*} -\log \left ({\left | -x + \sqrt{x^{2} - 1} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-x + sqrt(x^2 - 1)))