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

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

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Rubi [A]  time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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) \end {gather*}

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 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]

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*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.35 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.06, size = 78, normalized size = 1.53 \begin {gather*} \frac {\frac {\sqrt {3-x}}{\sqrt {x-2}}-\frac {(3-x)^{3/2}}{(x-2)^{3/2}}}{4 \left (\frac {3-x}{x-2}+1\right )^2}-\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt {3-x}}{\sqrt {x-2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[3 - x]*Sqrt[-2 + x],x]

[Out]

(-((3 - x)^(3/2)/(-2 + x)^(3/2)) + Sqrt[3 - x]/Sqrt[-2 + x])/(4*(1 + (3 - x)/(-2 + x))^2) - ArcTan[Sqrt[3 - x]
/Sqrt[-2 + x]]/4

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fricas [A]  time = 1.29, size = 52, normalized size = 1.02 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.02, size = 42, normalized size = 0.82 \begin {gather*} \frac {1}{4} \, {\left (2 \, x + 3\right )} \sqrt {x - 2} \sqrt {-x + 3} - 2 \, \sqrt {x - 2} \sqrt {-x + 3} + \frac {1}{4} \, \arcsin \left (\sqrt {x - 2}\right ) \end {gather*}

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

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maple [A]  time = 0.01, size = 61, normalized size = 1.20 \begin {gather*} \frac {\sqrt {\left (x -2\right ) \left (-x +3\right )}\, \arcsin \left (2 x -5\right )}{8 \sqrt {x -2}\, \sqrt {-x +3}}-\frac {\left (-x +3\right )^{\frac {3}{2}} \sqrt {x -2}}{2}+\frac {\sqrt {-x +3}\, \sqrt {x -2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 38, normalized size = 0.75 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.21, size = 41, normalized size = 0.80 \begin {gather*} \left (\frac {x}{2}-\frac {5}{4}\right )\,\sqrt {x-2}\,\sqrt {3-x}-\frac {\ln \left (x-\frac {5}{2}-\sqrt {x-2}\,\sqrt {3-x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 2)^(1/2)*(3 - x)^(1/2),x)

[Out]

(x/2 - 5/4)*(x - 2)^(1/2)*(3 - x)^(1/2) - (log(x - (x - 2)^(1/2)*(3 - x)^(1/2)*1i - 5/2)*1i)/8

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sympy [A]  time = 3.01, size = 124, normalized size = 2.43 \begin {gather*} \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 {gather*}

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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