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

Optimal. Leaf size=92 \[ \frac{1}{2} (b-x)^2 \sqrt{\frac{x-a}{b-x}}+\frac{1}{4} (a-5 b) (b-x) \sqrt{\frac{x-a}{b-x}}-\frac{1}{4} (a-b) (a+3 b) \tan ^{-1}\left (\sqrt{\frac{x-a}{b-x}}\right ) \]

[Out]

((a - 5*b)*(b - x)*Sqrt[(-a + x)/(b - x)])/4 + ((b - x)^2*Sqrt[(-a + x)/(b - x)])/2 - ((a - b)*(a + 3*b)*ArcTa
n[Sqrt[(-a + x)/(b - x)]])/4

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Rubi [A]  time = 0.0589155, antiderivative size = 95, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1960, 455, 385, 203} \[ \frac{1}{2} (b-x)^2 \sqrt{-\frac{a-x}{b-x}}+\frac{1}{4} (a-5 b) (b-x) \sqrt{-\frac{a-x}{b-x}}-\frac{1}{4} (a-b) (a+3 b) \tan ^{-1}\left (\sqrt{-\frac{a-x}{b-x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[(-a + x)/(b - x)],x]

[Out]

((a - 5*b)*Sqrt[-((a - x)/(b - x))]*(b - x))/4 + (Sqrt[-((a - x)/(b - x))]*(b - x)^2)/2 - ((a - b)*(a + 3*b)*A
rcTan[Sqrt[-((a - x)/(b - x))]])/4

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 385

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

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.2566, size = 115, normalized size = 1.25 \[ \frac{\sqrt{\frac{x-a}{b-x}} \left ((b-x) (a-3 b-2 x) \sqrt{\frac{a-x}{a-b}}-\sqrt{a-b} (a+3 b) \sqrt{b-x} \sinh ^{-1}\left (\frac{\sqrt{b-x}}{\sqrt{a-b}}\right )\right )}{4 \sqrt{\frac{a-x}{a-b}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[(-a + x)/(b - x)],x]

[Out]

(Sqrt[(-a + x)/(b - x)]*((a - 3*b - 2*x)*Sqrt[(a - x)/(a - b)]*(b - x) - Sqrt[a - b]*(a + 3*b)*Sqrt[b - x]*Arc
Sinh[Sqrt[b - x]/Sqrt[a - b]]))/(4*Sqrt[(a - x)/(a - b)])

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Maple [B]  time = 0.019, size = 208, normalized size = 2.3 \begin{align*}{\frac{-b+x}{8}\sqrt{-{\frac{-a+x}{-b+x}}} \left ( \arctan \left ({\frac{-a+2\,x-b}{2}{\frac{1}{\sqrt{-ab+ax+bx-{x}^{2}}}}} \right ){a}^{2}+2\,b\arctan \left ( 1/2\,{\frac{-a+2\,x-b}{\sqrt{-ab+ax+bx-{x}^{2}}}} \right ) a-3\,\arctan \left ( 1/2\,{\frac{-a+2\,x-b}{\sqrt{-ab+ax+bx-{x}^{2}}}} \right ){b}^{2}+4\,\sqrt{-ab+ax+bx-{x}^{2}}x-2\,\sqrt{-ab+ax+bx-{x}^{2}}a+6\,\sqrt{-ab+ax+bx-{x}^{2}}b \right ){\frac{1}{\sqrt{- \left ( -b+x \right ) \left ( -a+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-(-a+x)/(-b+x))^(1/2)*(-b+x)*(arctan(1/2*(-a+2*x-b)/(-a*b+a*x+b*x-x^2)^(1/2))*a^2+2*b*arctan(1/2*(-a+2*x-
b)/(-a*b+a*x+b*x-x^2)^(1/2))*a-3*arctan(1/2*(-a+2*x-b)/(-a*b+a*x+b*x-x^2)^(1/2))*b^2+4*(-a*b+a*x+b*x-x^2)^(1/2
)*x-2*(-a*b+a*x+b*x-x^2)^(1/2)*a+6*(-a*b+a*x+b*x-x^2)^(1/2)*b)/(-(-b+x)*(-a+x))^(1/2)

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Maxima [A]  time = 1.42183, size = 176, normalized size = 1.91 \begin{align*} -\frac{1}{4} \,{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \arctan \left (\sqrt{-\frac{a - x}{b - x}}\right ) - \frac{{\left (a^{2} - 6 \, a b + 5 \, b^{2}\right )} \left (-\frac{a - x}{b - x}\right )^{\frac{3}{2}} -{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \sqrt{-\frac{a - x}{b - x}}}{4 \,{\left (\frac{{\left (a - x\right )}^{2}}{{\left (b - x\right )}^{2}} - \frac{2 \,{\left (a - x\right )}}{b - x} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.79824, size = 165, normalized size = 1.79 \begin{align*} -\frac{1}{4} \,{\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \arctan \left (\sqrt{-\frac{a - x}{b - x}}\right ) + \frac{1}{4} \,{\left (a b - 3 \, b^{2} -{\left (a - b\right )} x + 2 \, x^{2}\right )} \sqrt{-\frac{a - x}{b - x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2 + 2*a*b - 3*b^2)*arctan(sqrt(-(a - x)/(b - x))) + 1/4*(a*b - 3*b^2 - (a - b)*x + 2*x^2)*sqrt(-(a - x
)/(b - x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.09141, size = 139, normalized size = 1.51 \begin{align*} \frac{1}{8} \,{\left (a^{2} \mathrm{sgn}\left (-b + x\right ) + 2 \, a b \mathrm{sgn}\left (-b + x\right ) - 3 \, b^{2} \mathrm{sgn}\left (-b + x\right )\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 \mathrm{sgn}\left (-b + x\right ) - 3 \, b \mathrm{sgn}\left (-b + x\right ) - 2 \, x \mathrm{sgn}\left (-b + x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*sgn(-b + x) + 2*a*b*sgn(-b + x) - 3*b^2*sgn(-b + x))*arcsin((a + b - 2*x)/(a - b))*sgn(-a + b) - 1/4*
sqrt(-a*b + a*x + b*x - x^2)*(a*sgn(-b + x) - 3*b*sgn(-b + x) - 2*x*sgn(-b + x))