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

Optimal. Leaf size=16 \[ \sin ^{-1}(x)-\sqrt {1-x^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {665, 216} \[ \sin ^{-1}(x)-\sqrt {1-x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^2]/(1 - x),x]

[Out]

-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 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 44, normalized size = 2.75 \[ \frac {x^2+2 \sqrt {1-x^2} \sin ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )-1}{\sqrt {1-x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - x^2]/(1 - x),x]

[Out]

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

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fricas [B]  time = 0.96, size = 30, normalized size = 1.88 \[ -\sqrt {-x^{2} + 1} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 14, normalized size = 0.88 \[ -\sqrt {-x^{2} + 1} + \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 1) + arcsin(x)

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maple [A]  time = 0.04, size = 20, normalized size = 1.25 \[ \arcsin \relax (x )-\sqrt {-2 x -\left (x -1\right )^{2}+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.92, size = 14, normalized size = 0.88 \[ -\sqrt {-x^{2} + 1} + \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 1) + arcsin(x)

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mupad [B]  time = 0.02, size = 14, normalized size = 0.88 \[ \mathrm {asin}\relax (x)-\sqrt {1-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1 - x^2)^(1/2)/(x - 1),x)

[Out]

asin(x) - (1 - x^2)^(1/2)

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sympy [A]  time = 2.40, size = 17, normalized size = 1.06 \[ - \begin {cases} \sqrt {1 - x^{2}} - \operatorname {asin}{\relax (x )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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