3.1161 \(\int \sqrt{3-x} \sqrt{-2+x} \, dx\)

Optimal. Leaf size=51 \[ -\frac{1}{2} \sqrt{x-2} (3-x)^{3/2}+\frac{1}{4} \sqrt{x-2} \sqrt{3-x}-\frac{1}{8} \sin ^{-1}(5-2 x) \]

[Out]

(Sqrt[3 - x]*Sqrt[-2 + x])/4 - ((3 - x)^(3/2)*Sqrt[-2 + x])/2 - ArcSin[5 - 2*x]/8

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Rubi [A]  time = 0.0097638, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {50, 53, 619, 216} \[ -\frac{1}{2} \sqrt{x-2} (3-x)^{3/2}+\frac{1}{4} \sqrt{x-2} \sqrt{3-x}-\frac{1}{8} \sin ^{-1}(5-2 x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[3 - x]*Sqrt[-2 + x],x]

[Out]

(Sqrt[3 - x]*Sqrt[-2 + x])/4 - ((3 - x)^(3/2)*Sqrt[-2 + x])/2 - ArcSin[5 - 2*x]/8

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 53

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

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 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]

Rubi steps

\begin{align*} \int \sqrt{3-x} \sqrt{-2+x} \, dx &=-\frac{1}{2} (3-x)^{3/2} \sqrt{-2+x}+\frac{1}{4} \int \frac{\sqrt{3-x}}{\sqrt{-2+x}} \, dx\\ &=\frac{1}{4} \sqrt{3-x} \sqrt{-2+x}-\frac{1}{2} (3-x)^{3/2} \sqrt{-2+x}+\frac{1}{8} \int \frac{1}{\sqrt{3-x} \sqrt{-2+x}} \, dx\\ &=\frac{1}{4} \sqrt{3-x} \sqrt{-2+x}-\frac{1}{2} (3-x)^{3/2} \sqrt{-2+x}+\frac{1}{8} \int \frac{1}{\sqrt{-6+5 x-x^2}} \, dx\\ &=\frac{1}{4} \sqrt{3-x} \sqrt{-2+x}-\frac{1}{2} (3-x)^{3/2} \sqrt{-2+x}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,5-2 x\right )\\ &=\frac{1}{4} \sqrt{3-x} \sqrt{-2+x}-\frac{1}{2} (3-x)^{3/2} \sqrt{-2+x}-\frac{1}{8} \sin ^{-1}(5-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0226581, size = 69, normalized size = 1.35 \[ \frac{\sqrt{-x^2+5 x-6} \left (\sqrt{x-2} \left (2 x^2-11 x+15\right )+\sqrt{3-x} \sin ^{-1}\left (\sqrt{3-x}\right )\right )}{4 (x-3) \sqrt{x-2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[3 - x]*Sqrt[-2 + x],x]

[Out]

(Sqrt[-6 + 5*x - x^2]*(Sqrt[-2 + x]*(15 - 11*x + 2*x^2) + Sqrt[3 - x]*ArcSin[Sqrt[3 - x]]))/(4*(-3 + x)*Sqrt[-
2 + x])

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Maple [A]  time = 0.004, size = 61, normalized size = 1.2 \begin{align*} -{\frac{1}{2} \left ( 3-x \right ) ^{{\frac{3}{2}}}\sqrt{-2+x}}+{\frac{1}{4}\sqrt{3-x}\sqrt{-2+x}}+{\frac{\arcsin \left ( 2\,x-5 \right ) }{8}\sqrt{ \left ( -2+x \right ) \left ( 3-x \right ) }{\frac{1}{\sqrt{3-x}}}{\frac{1}{\sqrt{-2+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3-x)^(1/2)*(-2+x)^(1/2),x)

[Out]

-1/2*(3-x)^(3/2)*(-2+x)^(1/2)+1/4*(3-x)^(1/2)*(-2+x)^(1/2)+1/8*((-2+x)*(3-x))^(1/2)/(-2+x)^(1/2)/(3-x)^(1/2)*a
rcsin(2*x-5)

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Maxima [A]  time = 1.423, size = 51, normalized size = 1. \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} + 5 \, x - 6} x - \frac{5}{4} \, \sqrt{-x^{2} + 5 \, x - 6} + \frac{1}{8} \, \arcsin \left (2 \, x - 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-x)^(1/2)*(-2+x)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-x^2 + 5*x - 6)*x - 5/4*sqrt(-x^2 + 5*x - 6) + 1/8*arcsin(2*x - 5)

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Fricas [A]  time = 1.49894, size = 147, normalized size = 2.88 \begin{align*} \frac{1}{4} \,{\left (2 \, x - 5\right )} \sqrt{x - 2} \sqrt{-x + 3} - \frac{1}{8} \, \arctan \left (\frac{{\left (2 \, x - 5\right )} \sqrt{x - 2} \sqrt{-x + 3}}{2 \,{\left (x^{2} - 5 \, x + 6\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-x)^(1/2)*(-2+x)^(1/2),x, algorithm="fricas")

[Out]

1/4*(2*x - 5)*sqrt(x - 2)*sqrt(-x + 3) - 1/8*arctan(1/2*(2*x - 5)*sqrt(x - 2)*sqrt(-x + 3)/(x^2 - 5*x + 6))

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Sympy [A]  time = 3.0069, size = 124, normalized size = 2.43 \begin{align*} \begin{cases} - \frac{i \operatorname{acosh}{\left (\sqrt{x - 2} \right )}}{4} + \frac{i \left (x - 2\right )^{\frac{5}{2}}}{2 \sqrt{x - 3}} - \frac{3 i \left (x - 2\right )^{\frac{3}{2}}}{4 \sqrt{x - 3}} + \frac{i \sqrt{x - 2}}{4 \sqrt{x - 3}} & \text{for}\: \left |{x - 2}\right | > 1 \\\frac{\operatorname{asin}{\left (\sqrt{x - 2} \right )}}{4} - \frac{\left (x - 2\right )^{\frac{5}{2}}}{2 \sqrt{3 - x}} + \frac{3 \left (x - 2\right )^{\frac{3}{2}}}{4 \sqrt{3 - x}} - \frac{\sqrt{x - 2}}{4 \sqrt{3 - x}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-x)**(1/2)*(-2+x)**(1/2),x)

[Out]

Piecewise((-I*acosh(sqrt(x - 2))/4 + I*(x - 2)**(5/2)/(2*sqrt(x - 3)) - 3*I*(x - 2)**(3/2)/(4*sqrt(x - 3)) + I
*sqrt(x - 2)/(4*sqrt(x - 3)), Abs(x - 2) > 1), (asin(sqrt(x - 2))/4 - (x - 2)**(5/2)/(2*sqrt(3 - x)) + 3*(x -
2)**(3/2)/(4*sqrt(3 - x)) - sqrt(x - 2)/(4*sqrt(3 - x)), True))

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Giac [A]  time = 1.09726, size = 38, normalized size = 0.75 \begin{align*} \frac{1}{4} \,{\left (2 \, x - 5\right )} \sqrt{x - 2} \sqrt{-x + 3} + \frac{1}{4} \, \arcsin \left (\sqrt{x - 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3-x)^(1/2)*(-2+x)^(1/2),x, algorithm="giac")

[Out]

1/4*(2*x - 5)*sqrt(x - 2)*sqrt(-x + 3) + 1/4*arcsin(sqrt(x - 2))