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]

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

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Rubi [A]  time = 0.01, 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 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])

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 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\\ &=\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*}

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Mathematica [B]  time = 0.03, 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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fricas [A]  time = 0.41, size = 14, normalized size = 0.56 \[ -\log \left (-x + \sqrt {x^{2} - 1}\right ) \]

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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giac [A]  time = 0.36, size = 15, normalized size = 0.60 \[ -\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \]

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)))

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maple [A]  time = 0.01, size = 28, normalized size = 1.12 \[ \sqrt {\frac {1}{x^{2}-1}}\, \sqrt {x^{2}-1}\, \ln \left (x +\sqrt {x^{2}-1}\right ) \]

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 [A]  time = 0.88, size = 14, normalized size = 0.56 \[ \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \]

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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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 = 1.52, size = 15, normalized size = 0.60 \[ \begin {cases} \log {\left (x + \sqrt {x^{2} - 1} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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