\(\int x \sqrt {\frac {-a+x}{b-x}} \, dx\) [219]

Optimal result

Integrand size = 19, antiderivative size = 92 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=\frac {1}{4} (a-5 b) (b-x) \sqrt {\frac {-a+x}{b-x}}+\frac {1}{2} (b-x)^2 \sqrt {\frac {-a+x}{b-x}}-\frac {1}{4} (a-b) (a+3 b) \arctan \left (\sqrt {\frac {-a+x}{b-x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1980, 466, 393, 209} \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=-\frac {1}{4} (a-b) (a+3 b) \arctan \left (\sqrt {-\frac {a-x}{b-x}}\right )+\frac {1}{2} (b-x)^2 \sqrt {-\frac {a-x}{b-x}}+\frac {1}{4} (a-5 b) (b-x) \sqrt {-\frac {a-x}{b-x}} \]

[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 209

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

Rule 393

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 466

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 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\left ((2 (a-b)) \text {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) \text {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)) \text {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) \arctan \left (\sqrt {-\frac {a-x}{b-x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=\frac {\sqrt {\frac {-a+x}{b-x}} \left ((a-3 b-2 x) (b-x) \sqrt {-a+x}+\left (-a^2-2 a b+3 b^2\right ) \sqrt {b-x} \arctan \left (\frac {\sqrt {-a+x}}{\sqrt {b-x}}\right )\right )}{4 \sqrt {-a+x}} \]

[In]

Integrate[x*Sqrt[(-a + x)/(b - x)],x]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53

method	result	size
risch	\(\frac {\left (a -3 b -2 x \right ) \left (b -x \right ) \sqrt {-\frac {a -x}{b -x}}\, \sqrt {-\left (b -x \right ) \left (a -x \right )}}{4 \sqrt {-\left (-b +x \right ) \left (-a +x \right )}}+\frac {\left (\frac {1}{4} a b -\frac {3}{8} b^{2}+\frac {1}{8} a^{2}\right ) \arctan \left (\frac {x -\frac {b}{2}-\frac {a}{2}}{\sqrt {-a b +\left (a +b \right ) x -x^{2}}}\right ) \sqrt {-\frac {a -x}{b -x}}\, \sqrt {-\left (b -x \right ) \left (a -x \right )}}{a -x}\)	\(141\)
default	\(\frac {\sqrt {-\frac {a -x}{b -x}}\, \left (b -x \right ) \left (\arctan \left (\frac {a +b -2 x}{2 \sqrt {-a b +a x +b x -x^{2}}}\right ) a^{2}+2 b \arctan \left (\frac {a +b -2 x}{2 \sqrt {-a b +a x +b x -x^{2}}}\right ) a -3 \arctan \left (\frac {a +b -2 x}{2 \sqrt {-a b +a x +b x -x^{2}}}\right ) b^{2}+2 \sqrt {-a b +a x +b x -x^{2}}\, a -6 \sqrt {-a b +a x +b x -x^{2}}\, b -4 \sqrt {-a b +a x +b x -x^{2}}\, x \right )}{8 \sqrt {-\left (b -x \right ) \left (a -x \right )}}\)	\(196\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=-\frac {1}{4} \, {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \arctan \left (\sqrt {-\frac {a - x}{b - x}}\right ) + \frac {1}{4} \, {\left (a b - 3 \, b^{2} - {\left (a - b\right )} x + 2 \, x^{2}\right )} \sqrt {-\frac {a - x}{b - x}} \]

[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))

Sympy [F]

\[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=\int x \sqrt {\frac {- a + x}{b - x}}\, dx \]

[In]

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

[Out]

Integral(x*sqrt((-a + x)/(b - x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=-\frac {1}{4} \, {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \arctan \left (\sqrt {-\frac {a - x}{b - x}}\right ) - \frac {{\left (a^{2} - 6 \, a b + 5 \, b^{2}\right )} \left (-\frac {a - x}{b - x}\right )^{\frac {3}{2}} - {\left (a^{2} + 2 \, a b - 3 \, b^{2}\right )} \sqrt {-\frac {a - x}{b - x}}}{4 \, {\left (\frac {{\left (a - x\right )}^{2}}{{\left (b - x\right )}^{2}} - \frac {2 \, {\left (a - x\right )}}{b - x} + 1\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=\frac {1}{8} \, {\left (a^{2} \mathrm {sgn}\left (-b + x\right ) + 2 \, a b \mathrm {sgn}\left (-b + x\right ) - 3 \, b^{2} \mathrm {sgn}\left (-b + x\right )\right )} \arcsin \left (\frac {a + b - 2 \, x}{a - b}\right ) \mathrm {sgn}\left (-a + b\right ) - \frac {1}{4} \, \sqrt {-a b + a x + b x - x^{2}} {\left (a \mathrm {sgn}\left (-b + x\right ) - 3 \, b \mathrm {sgn}\left (-b + x\right ) - 2 \, x \mathrm {sgn}\left (-b + x\right )\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.52 \[ \int x \sqrt {\frac {-a+x}{b-x}} \, dx=-\frac {\sqrt {-\frac {a-x}{b-x}}\,\left (\frac {a^2\,1{}\mathrm {i}}{4}+\frac {a\,b\,1{}\mathrm {i}}{2}-\frac {b^2\,3{}\mathrm {i}}{4}\right )\,1{}\mathrm {i}-{\left (-\frac {a-x}{b-x}\right )}^{3/2}\,\left (\frac {a^2\,1{}\mathrm {i}}{4}-\frac {a\,b\,3{}\mathrm {i}}{2}+\frac {b^2\,5{}\mathrm {i}}{4}\right )\,1{}\mathrm {i}}{\frac {{\left (a-x\right )}^2}{{\left (b-x\right )}^2}-\frac {2\,\left (a-x\right )}{b-x}+1}-\frac {\mathrm {atan}\left (\sqrt {-\frac {a-x}{b-x}}\right )\,\left (a-b\right )\,\left (a+3\,b\right )}{4} \]

[In]

int(x*(-(a - x)/(b - x))^(1/2),x)

[Out]

- ((-(a - x)/(b - x))^(1/2)*((a*b*1i)/2 + (a^2*1i)/4 - (b^2*3i)/4)*1i - (-(a - x)/(b - x))^(3/2)*((a^2*1i)/4 -
(a*b*3i)/2 + (b^2*5i)/4)*1i)/((a - x)^2/(b - x)^2 - (2*(a - x))/(b - x) + 1) - (atan((-(a - x)/(b - x))^(1/2)
)*(a - b)*(a + 3*b))/4