3.4.11 \(\int \frac {\sqrt {2 x-x^2}}{2-2 x} \, dx\) [311]

Optimal. Leaf size=36 \[ -\frac {1}{2} \sqrt {2 x-x^2}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {699, 702, 213} \begin {gather*} \frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x-x^2}\right )-\frac {1}{2} \sqrt {2 x-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2*x - x^2]/(2 - 2*x),x]

[Out]

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

Rule 213

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

Rule 699

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1))), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 702

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

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Mathematica [A]
time = 0.06, size = 49, normalized size = 1.36 \begin {gather*} \frac {1}{2} \sqrt {-((-2+x) x)} \left (-1+\frac {2 \tan ^{-1}\left (1+\sqrt {-2+x} \sqrt {x}-x\right )}{\sqrt {-2+x} \sqrt {x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.45, size = 29, normalized size = 0.81

method	result	size
default	\(-\frac {\sqrt {-\left (x -1\right )^{2}+1}}{2}+\frac {\arctanh \left (\frac {1}{\sqrt {-\left (x -1\right )^{2}+1}}\right )}{2}\)	\(29\)
risch	\(\frac {x \left (x -2\right )}{2 \sqrt {-x \left (x -2\right )}}+\frac {\arctanh \left (\frac {1}{\sqrt {-\left (x -1\right )^{2}+1}}\right )}{2}\)	\(30\)
trager	\(-\frac {\sqrt {-x^{2}+2 x}}{2}-\frac {\ln \left (\frac {\sqrt {-x^{2}+2 x}-1}{x -1}\right )}{2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 45, normalized size = 1.25 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} + \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac {2}{{\left | x - 1 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).
time = 1.11, size = 57, normalized size = 1.58 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} + \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-x^2 + 2*x) + 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt {- x^{2} + 2 x}}{x - 1}\, dx}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.49, size = 40, normalized size = 1.11 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 2 \, x} - \frac {1}{2} \, \log \left (-\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {\sqrt {2\,x-x^2}}{2\,x-2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - x^2)^(1/2)/(2*x - 2),x)

[Out]

-int((2*x - x^2)^(1/2)/(2*x - 2), x)

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