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

Optimal. Leaf size=14 \[ \sqrt {1-x^2}+\sin ^{-1}(x) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {665, 216} \begin {gather*} \sqrt {1-x^2}+\sin ^{-1}(x) \end {gather*}

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 = 30, normalized size = 2.14 \begin {gather*} \sqrt {1-x^2}-2 \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - x^2]/(1 + x),x]

[Out]

Sqrt[1 - x^2] - 2*ArcSin[Sqrt[1 - x]/Sqrt[2]]

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IntegrateAlgebraic [B]  time = 0.13, size = 32, normalized size = 2.29 \begin {gather*} \sqrt {1-x^2}-2 \tan ^{-1}\left (\frac {\sqrt {1-x^2}}{x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 28, normalized size = 2.00 \begin {gather*} \sqrt {-x^{2} + 1} - 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \end {gather*}

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 = 0.18, size = 12, normalized size = 0.86 \begin {gather*} \sqrt {-x^{2} + 1} + \arcsin \relax (x) \end {gather*}

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 = 0.05, size = 18, normalized size = 1.29 \begin {gather*} \arcsin \relax (x )+\sqrt {2 x -\left (x +1\right )^{2}+2} \end {gather*}

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 = 2.96, size = 12, normalized size = 0.86 \begin {gather*} \sqrt {-x^{2} + 1} + \arcsin \relax (x) \end {gather*}

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 = 0.48, size = 12, normalized size = 0.86 \begin {gather*} \mathrm {asin}\relax (x)+\sqrt {1-x^2} \end {gather*}

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 = 2.18, size = 15, normalized size = 1.07 \begin {gather*} \begin {cases} \sqrt {1 - x^{2}} + \operatorname {asin}{\relax (x )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \end {gather*}

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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