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

Optimal. Leaf size=37 \[ \frac {1}{4} \log \left (1-2 x^2\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right )-\frac {1}{2} \sin ^{-1}(x) \]

[Out]

-1/2*arcsin(x)-1/2*arctanh(x/(-x^2+1)^(1/2))+1/4*ln(-2*x^2+1)

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6742, 260, 402, 216, 377, 207} \[ \frac {1}{4} \log \left (1-2 x^2\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right )-\frac {1}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x - Sqrt[1 - x^2])^(-1),x]

[Out]

-ArcSin[x]/2 - ArcTanh[x/Sqrt[1 - x^2]]/2 + Log[1 - 2*x^2]/4

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1}{x-\sqrt {1-x^2}} \, dx &=\int \left (\frac {x}{-1+2 x^2}+\frac {\sqrt {1-x^2}}{-1+2 x^2}\right ) \, dx\\ &=\int \frac {x}{-1+2 x^2} \, dx+\int \frac {\sqrt {1-x^2}}{-1+2 x^2} \, dx\\ &=\frac {1}{4} \log \left (1-2 x^2\right )-\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx+\frac {1}{2} \int \frac {1}{\sqrt {1-x^2} \left (-1+2 x^2\right )} \, dx\\ &=-\frac {1}{2} \sin ^{-1}(x)+\frac {1}{4} \log \left (1-2 x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right )\\ &=-\frac {1}{2} \sin ^{-1}(x)-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right )+\frac {1}{4} \log \left (1-2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 1.00 \[ \frac {1}{4} \log \left (1-2 x^2\right )-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right )-\frac {1}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(x - Sqrt[1 - x^2])^(-1),x]

[Out]

-1/2*ArcSin[x] - ArcTanh[x/Sqrt[1 - x^2]]/2 + Log[1 - 2*x^2]/4

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fricas [B]  time = 0.44, size = 84, normalized size = 2.27 \[ \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) + \frac {1}{4} \, \log \left (2 \, x^{2} - 1\right ) + \frac {1}{4} \, \log \left (-\frac {x^{2} + \sqrt {-x^{2} + 1} {\left (x + 1\right )} - x - 1}{x^{2}}\right ) - \frac {1}{4} \, \log \left (-\frac {x^{2} - \sqrt {-x^{2} + 1} {\left (x - 1\right )} + x - 1}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan((sqrt(-x^2 + 1) - 1)/x) + 1/4*log(2*x^2 - 1) + 1/4*log(-(x^2 + sqrt(-x^2 + 1)*(x + 1) - x - 1)/x^2) - 1
/4*log(-(x^2 - sqrt(-x^2 + 1)*(x - 1) + x - 1)/x^2)

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giac [B]  time = 0.44, size = 140, normalized size = 3.78 \[ -\frac {1}{4} \, \pi \mathrm {sgn}\relax (x) - \frac {1}{2} \, \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) + \frac {1}{4} \, \log \left ({\left | x + \frac {1}{2} \, \sqrt {2} \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - \frac {1}{2} \, \sqrt {2} \right |}\right ) - \frac {1}{4} \, \log \left ({\left | -\frac {x}{\sqrt {-x^{2} + 1} - 1} + \frac {\sqrt {-x^{2} + 1} - 1}{x} + 2 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | -\frac {x}{\sqrt {-x^{2} + 1} - 1} + \frac {\sqrt {-x^{2} + 1} - 1}{x} - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*pi*sgn(x) - 1/2*arctan(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1)) + 1/4*log(abs(x + 1/
2*sqrt(2))) + 1/4*log(abs(x - 1/2*sqrt(2))) - 1/4*log(abs(-x/(sqrt(-x^2 + 1) - 1) + (sqrt(-x^2 + 1) - 1)/x + 2
)) + 1/4*log(abs(-x/(sqrt(-x^2 + 1) - 1) + (sqrt(-x^2 + 1) - 1)/x - 2))

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maple [B]  time = 0.05, size = 175, normalized size = 4.73 \[ -\frac {\arctanh \left (\frac {\left (1-\left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}\right ) \sqrt {2}}{\sqrt {-4 \left (x -\frac {\sqrt {2}}{2}\right )^{2}-4 \left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}\right )}{4}+\frac {\arctanh \left (\frac {\left (\left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+1\right ) \sqrt {2}}{\sqrt {-4 \left (x +\frac {\sqrt {2}}{2}\right )^{2}+4 \left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}\right )}{4}-\frac {\arcsin \relax (x )}{2}+\frac {\ln \left (2 x^{2}-1\right )}{4}-\frac {\sqrt {2}\, \sqrt {-4 \left (x +\frac {\sqrt {2}}{2}\right )^{2}+4 \left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}{8}+\frac {\sqrt {2}\, \sqrt {-4 \left (x -\frac {\sqrt {2}}{2}\right )^{2}-4 \left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*ln(2*x^2-1)-1/8*2^(1/2)*(-4*(x+1/2*2^(1/2))^2+4*(x+1/2*2^(1/2))*2^(1/2)+2)^(1/2)-1/2*arcsin(x)+1/4*arctanh
(((x+1/2*2^(1/2))*2^(1/2)+1)*2^(1/2)/(-4*(x+1/2*2^(1/2))^2+4*(x+1/2*2^(1/2))*2^(1/2)+2)^(1/2))+1/8*2^(1/2)*(-4
*(x-1/2*2^(1/2))^2-4*(x-1/2*2^(1/2))*2^(1/2)+2)^(1/2)-1/4*arctanh((1-(x-1/2*2^(1/2))*2^(1/2))*2^(1/2)/(-4*(x-1
/2*2^(1/2))^2-4*(x-1/2*2^(1/2))*2^(1/2)+2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x - sqrt(-x^2 + 1)), x)

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mupad [B]  time = 0.14, size = 105, normalized size = 2.84 \[ \frac {\ln \left (x-\frac {\sqrt {2}}{2}\right )}{4}+\frac {\ln \left (x+\frac {\sqrt {2}}{2}\right )}{4}-\frac {\ln \left (\frac {\sqrt {2}\,\left (\frac {\sqrt {2}\,x}{2}-1\right )\,1{}\mathrm {i}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\frac {\sqrt {2}}{2}}\right )}{4}+\frac {\ln \left (\frac {\sqrt {2}\,\left (\frac {\sqrt {2}\,x}{2}+1\right )\,1{}\mathrm {i}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\frac {\sqrt {2}}{2}}\right )}{4}-\frac {\mathrm {asin}\relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x - (1 - x^2)^(1/2)),x)

[Out]

log(x - 2^(1/2)/2)/4 + log(x + 2^(1/2)/2)/4 - log((2^(1/2)*((2^(1/2)*x)/2 - 1)*1i - (1 - x^2)^(1/2)*1i)/(x - 2
^(1/2)/2))/4 + log((2^(1/2)*((2^(1/2)*x)/2 + 1)*1i + (1 - x^2)^(1/2)*1i)/(x + 2^(1/2)/2))/4 - asin(x)/2

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sympy [A]  time = 0.17, size = 17, normalized size = 0.46 \[ \frac {\log {\left (x - \sqrt {1 - x^{2}} \right )}}{2} - \frac {\operatorname {asin}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - sqrt(1 - x**2))/2 - asin(x)/2

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