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]

ArcTanh[Sqrt[2*x - x^2]]/2

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Rubi [A]  time = 0.009766, antiderivative size = 18, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [B]  time = 0.0108599, 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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Maple [A]  time = 0.046, size = 15, normalized size = 0.8 \begin{align*}{\frac{1}{2}{\it Artanh} \left ({\frac{1}{\sqrt{- \left ( -1+x \right ) ^{2}+1}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16931, size = 42, normalized size = 2.33 \begin{align*} \frac{1}{2} \, \log \left (\frac{2 \, \sqrt{-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac{2}{{\left | x - 1 \right |}}\right ) \end{align*}

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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Fricas [B]  time = 1.99834, size = 97, normalized size = 5.39 \begin{align*} \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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{x \sqrt{- x^{2} + 2 x} - \sqrt{- x^{2} + 2 x}}\, dx}{2} \end{align*}

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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Giac [A]  time = 1.41763, size = 35, normalized size = 1.94 \begin{align*} -\frac{1}{2} \, \log \left (-\frac{2 \,{\left (\sqrt{-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \end{align*}

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))