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

Optimal. Leaf size=24 \[ -\sqrt {2 x-x^2}-2 \sin ^{-1}(1-x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {640, 619, 216} \[ -\sqrt {2 x-x^2}-2 \sin ^{-1}(1-x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/Sqrt[2*x - x^2],x]

[Out]

-Sqrt[2*x - x^2] - 2*ArcSin[1 - x]

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 619

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

Rule 640

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 27, normalized size = 1.12 \[ -\sqrt {-((x-2) x)}-4 \sin ^{-1}\left (\sqrt {1-\frac {x}{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/Sqrt[2*x - x^2],x]

[Out]

-Sqrt[-((-2 + x)*x)] - 4*ArcSin[Sqrt[1 - x/2]]

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fricas [A]  time = 0.41, size = 32, normalized size = 1.33 \[ -\sqrt {-x^{2} + 2 \, x} - 4 \, \arctan \left (\frac {\sqrt {-x^{2} + 2 \, x}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.03, size = 20, normalized size = 0.83 \[ -\sqrt {-x^{2} + 2 \, x} + 2 \, \arcsin \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 2*x) + 2*arcsin(x - 1)

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maple [A]  time = 0.01, size = 21, normalized size = 0.88 \[ 2 \arcsin \left (x -1\right )-\sqrt {-x^{2}+2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.20, size = 22, normalized size = 0.92 \[ -\sqrt {-x^{2} + 2 \, x} - 2 \, \arcsin \left (-x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 2*x) - 2*arcsin(-x + 1)

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mupad [B]  time = 0.27, size = 20, normalized size = 0.83 \[ 2\,\mathrm {asin}\left (x-1\right )-\sqrt {2\,x-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)/(2*x - x^2)^(1/2),x)

[Out]

2*asin(x - 1) - (2*x - x^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 1)/sqrt(-x*(x - 2)), x)

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