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

Optimal. Leaf size=23 \[ -\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {627, 63, 206} \begin {gather*} -\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 + x]*Sqrt[1 - x^2]),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 627

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 45, normalized size = 1.96 \begin {gather*} \frac {\sqrt {2} \sqrt {x-1} \sqrt {x+1} \tan ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {2}}\right )}{\sqrt {1-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 38, normalized size = 1.65 \begin {gather*} -\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x+1}}{\sqrt {2 (x+1)-(x+1)^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.40, size = 45, normalized size = 1.96 \begin {gather*} \frac {1}{2} \, \sqrt {2} \log \left (-\frac {x^{2} + 2 \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {x + 1} - 2 \, x - 3}{x^{2} + 2 \, x + 1}\right ) \end {gather*}

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]

1/2*sqrt(2)*log(-(x^2 + 2*sqrt(2)*sqrt(-x^2 + 1)*sqrt(x + 1) - 2*x - 3)/(x^2 + 2*x + 1))

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giac [B]  time = 0.20, size = 37, normalized size = 1.61 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \log \left (\sqrt {2} + \sqrt {-x + 1}\right ) + \frac {1}{2} \, \sqrt {2} \log \left (\sqrt {2} - \sqrt {-x + 1}\right ) \end {gather*}

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]

-1/2*sqrt(2)*log(sqrt(2) + sqrt(-x + 1)) + 1/2*sqrt(2)*log(sqrt(2) - sqrt(-x + 1))

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maple [B]  time = 0.06, size = 40, normalized size = 1.74 \begin {gather*} -\frac {\sqrt {-x^{2}+1}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {-x +1}\, \sqrt {2}}{2}\right )}{\sqrt {x +1}\, \sqrt {-x +1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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(1/(sqrt(-x^2 + 1)*sqrt(x + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\sqrt {1-x^2}\,\sqrt {x+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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