3.259 \(\int \sqrt{(b-x) (-a+x)} \, dx\)

Optimal. Leaf size=71 \[ -\frac{1}{4} (a+b-2 x) \sqrt{x (a+b)-a b-x^2}-\frac{1}{8} (a-b)^2 \tan ^{-1}\left (\frac{a+b-2 x}{2 \sqrt{x (a+b)-a b-x^2}}\right ) \]

[Out]

-((a + b - 2*x)*Sqrt[-(a*b) + (a + b)*x - x^2])/4 - ((a - b)^2*ArcTan[(a + b - 2*x)/(2*Sqrt[-(a*b) + (a + b)*x
- x^2])])/8

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Rubi [A]  time = 0.0254128, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {1981, 612, 621, 204} \[ -\frac{1}{4} (a+b-2 x) \sqrt{x (a+b)-a b-x^2}-\frac{1}{8} (a-b)^2 \tan ^{-1}\left (\frac{a+b-2 x}{2 \sqrt{x (a+b)-a b-x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(b - x)*(-a + x)],x]

[Out]

-((a + b - 2*x)*Sqrt[-(a*b) + (a + b)*x - x^2])/4 - ((a - b)^2*ArcTan[(a + b - 2*x)/(2*Sqrt[-(a*b) + (a + b)*x
- x^2])])/8

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rule 612

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

Rule 621

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

Rule 204

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

Rubi steps

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

Mathematica [A]  time = 0.157026, size = 106, normalized size = 1.49 \[ \frac{(a-x) \left ((a-b)^{5/2} \sqrt{b-x} \sqrt{\frac{a-x}{a-b}} \sinh ^{-1}\left (\frac{\sqrt{b-x}}{\sqrt{a-b}}\right )-(a-x) (b-x) (a+b-2 x)\right )}{4 (x-a) \sqrt{(a-x) (x-b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(b - x)*(-a + x)],x]

[Out]

((a - x)*(-((a + b - 2*x)*(a - x)*(b - x)) + (a - b)^(5/2)*Sqrt[(a - x)/(a - b)]*Sqrt[b - x]*ArcSinh[Sqrt[b -
x]/Sqrt[a - b]]))/(4*(-a + x)*Sqrt[(a - x)*(-b + x)])

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Maple [A]  time = 0.013, size = 122, normalized size = 1.7 \begin{align*} -{\frac{a+b-2\,x}{4}\sqrt{-ab+ \left ( a+b \right ) x-{x}^{2}}}-{\frac{ab}{4}\arctan \left ({ \left ( x-{\frac{b}{2}}-{\frac{a}{2}} \right ){\frac{1}{\sqrt{-ab+ \left ( a+b \right ) x-{x}^{2}}}}} \right ) }+{\frac{{a}^{2}}{8}\arctan \left ({ \left ( x-{\frac{b}{2}}-{\frac{a}{2}} \right ){\frac{1}{\sqrt{-ab+ \left ( a+b \right ) x-{x}^{2}}}}} \right ) }+{\frac{{b}^{2}}{8}\arctan \left ({ \left ( x-{\frac{b}{2}}-{\frac{a}{2}} \right ){\frac{1}{\sqrt{-ab+ \left ( a+b \right ) x-{x}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a+b-2*x)*(-a*b+(a+b)*x-x^2)^(1/2)-1/4*arctan((x-1/2*b-1/2*a)/(-a*b+(a+b)*x-x^2)^(1/2))*a*b+1/8*arctan((x
-1/2*b-1/2*a)/(-a*b+(a+b)*x-x^2)^(1/2))*a^2+1/8*arctan((x-1/2*b-1/2*a)/(-a*b+(a+b)*x-x^2)^(1/2))*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27196, size = 209, normalized size = 2.94 \begin{align*} -\frac{1}{8} \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (-\frac{\sqrt{-a b +{\left (a + b\right )} x - x^{2}}{\left (a + b - 2 \, x\right )}}{2 \,{\left (a b -{\left (a + b\right )} x + x^{2}\right )}}\right ) - \frac{1}{4} \, \sqrt{-a b +{\left (a + b\right )} x - x^{2}}{\left (a + b - 2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a^2 - 2*a*b + b^2)*arctan(-1/2*sqrt(-a*b + (a + b)*x - x^2)*(a + b - 2*x)/(a*b - (a + b)*x + x^2)) - 1/4
*sqrt(-a*b + (a + b)*x - x^2)*(a + b - 2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (- a + x\right ) \left (b - x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((-a + x)*(b - x)), x)

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Giac [A]  time = 1.20003, size = 82, normalized size = 1.15 \begin{align*} \frac{1}{8} \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \arcsin \left (\frac{a + b - 2 \, x}{a - b}\right ) \mathrm{sgn}\left (-a + b\right ) - \frac{1}{4} \, \sqrt{-a b + a x + b x - x^{2}}{\left (a + b - 2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2 - 2*a*b + b^2)*arcsin((a + b - 2*x)/(a - b))*sgn(-a + b) - 1/4*sqrt(-a*b + a*x + b*x - x^2)*(a + b -
2*x)