3.980 \(\int \sqrt{(4-x) x} \, dx\)

Optimal. Leaf size=33 \[ -\frac{1}{2} \sqrt{4 x-x^2} (2-x)-2 \sin ^{-1}\left (1-\frac{x}{2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0116904, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {1979, 612, 619, 216} \[ -\frac{1}{2} \sqrt{4 x-x^2} (2-x)-2 \sin ^{-1}\left (1-\frac{x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(4 - x)*x],x]

[Out]

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

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[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 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{(4-x) x} \, dx &=\int \sqrt{4 x-x^2} \, dx\\ &=-\frac{1}{2} (2-x) \sqrt{4 x-x^2}+2 \int \frac{1}{\sqrt{4 x-x^2}} \, dx\\ &=-\frac{1}{2} (2-x) \sqrt{4 x-x^2}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{16}}} \, dx,x,4-2 x\right )\\ &=-\frac{1}{2} (2-x) \sqrt{4 x-x^2}-2 \sin ^{-1}\left (1-\frac{x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0365776, size = 32, normalized size = 0.97 \[ \frac{1}{2} (x-2) \sqrt{-(x-4) x}-4 \sin ^{-1}\left (\sqrt{1-\frac{x}{4}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(4 - x)*x],x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 0.9 \begin{align*} -{\frac{4-2\,x}{4}\sqrt{-{x}^{2}+4\,x}}+2\,\arcsin \left ( x/2-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4-x)*x)^(1/2),x)

[Out]

-1/4*(4-2*x)*(-x^2+4*x)^(1/2)+2*arcsin(1/2*x-1)

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Maxima [A]  time = 1.53334, size = 49, normalized size = 1.48 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} + 4 \, x} x - \sqrt{-x^{2} + 4 \, x} - 2 \, \arcsin \left (-\frac{1}{2} \, x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-x)*x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44239, size = 82, normalized size = 2.48 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} + 4 \, x}{\left (x - 2\right )} - 4 \, \arctan \left (\frac{\sqrt{-x^{2} + 4 \, x}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-x)*x)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x \left (4 - x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-x)*x)**(1/2),x)

[Out]

Integral(sqrt(x*(4 - x)), x)

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Giac [A]  time = 1.15543, size = 34, normalized size = 1.03 \begin{align*} \frac{1}{2} \, \sqrt{-x^{2} + 4 \, x}{\left (x - 2\right )} + 2 \, \arcsin \left (\frac{1}{2} \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4-x)*x)^(1/2),x, algorithm="giac")

[Out]

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