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

Optimal. Leaf size=29 \[ -\frac {\sqrt {\frac {1}{x}+1} \sqrt {x} \sin ^{-1}(1-2 x)}{\sqrt {x+1}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1448, 26, 53, 619, 216} \[ -\frac {\sqrt {\frac {1}{x}+1} \sqrt {x} \sin ^{-1}(1-2 x)}{\sqrt {x+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^(-1)]/Sqrt[1 - x^2],x]

[Out]

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

Rule 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 53

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

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 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 1448

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Dist[(e^IntPart[q]*(d + e*x
^mn)^FracPart[q])/(x^(mn*FracPart[q])*(1 + d/(x^mn*e))^FracPart[q]), Int[x^(mn*q)*(1 + d/(x^mn*e))^q*(a + c*x^
n2)^p, x], x] /; FreeQ[{a, c, d, e, mn, p, q}, x] && EqQ[n2, -2*mn] &&  !IntegerQ[p] &&  !IntegerQ[q] && PosQ[
n2]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 41, normalized size = 1.41 \[ -\tan ^{-1}\left (\frac {\sqrt {\frac {x+1}{x}} (2 x-1) \sqrt {1-x^2}}{2 \left (x^2-1\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 34, normalized size = 1.17 \[ -\arctan \left (\frac {2 \, \sqrt {-x^{2} + 1} x \sqrt {\frac {x + 1}{x}}}{2 \, x^{2} + x - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(1/x + 1)/sqrt(-x^2 + 1), x)

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maple [A]  time = 0.02, size = 40, normalized size = 1.38 \[ \frac {\sqrt {\frac {x +1}{x}}\, \sqrt {-x^{2}+1}\, x \arcsin \left (2 x -1\right )}{\left (x +1\right ) \sqrt {-\left (x -1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(1/x + 1)/sqrt(-x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {\frac {1}{x}+1}}{\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {1 + \frac {1}{x}}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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