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

Optimal. Leaf size=18 \[ \frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {688, 207} \[ \frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((2 - 2*x)*Sqrt[2*x - x^2]),x]

[Out]

ArcTanh[Sqrt[2*x - x^2]]/2

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 688

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

Rubi steps

\begin {align*} \int \frac {1}{(2-2 x) \sqrt {2 x-x^2}} \, dx &=-\left (4 \operatorname {Subst}\left (\int \frac {1}{-8+8 x^2} \, dx,x,\sqrt {2 x-x^2}\right )\right )\\ &=\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right )\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 38, normalized size = 2.11 \[ -\frac {\sqrt {x-2} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt {x-2}}{\sqrt {x}}\right )}{\sqrt {-((x-2) x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.79, size = 44, normalized size = 2.44 \[ \frac {1}{2} \, \log \left (\frac {x + \sqrt {-x^{2} + 2 \, x}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - \sqrt {-x^{2} + 2 \, x}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 26, normalized size = 1.44 \[ -\frac {1}{2} \, \log \left (-\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 15, normalized size = 0.83 \[ \frac {\arctanh \left (\frac {1}{\sqrt {-\left (x -1\right )^{2}+1}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.32, size = 31, normalized size = 1.72 \[ \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac {2}{{\left | x - 1 \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ -\int \frac {1}{\left (2\,x-2\right )\,\sqrt {2\,x-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(x*sqrt(-x**2 + 2*x) - sqrt(-x**2 + 2*x)), x)/2

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