3.1147 \(\int \sqrt{a+b x} \sqrt{a c-b c x} \, dx\)

Optimal. Leaf size=68 \[ \frac{a^2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{b}+\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x} \]

[Out]

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

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Rubi [A]  time = 0.0265255, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {38, 63, 217, 203} \[ \frac{a^2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{b}+\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*Sqrt[a*c - b*c*x],x]

[Out]

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

Rule 38

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 \sqrt{a+b x} \sqrt{a c-b c x} \, dx &=\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{2} \left (a^2 c\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{a c-b c x}} \, dx\\ &=\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a c-c x^2}} \, dx,x,\sqrt{a+b x}\right )}{b}\\ &=\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{1+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{b}\\ &=\frac{1}{2} x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{a^2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.118363, size = 95, normalized size = 1.4 \[ \frac{c \left (a^2 b x-2 a^{5/2} \sqrt{a-b x} \sqrt{\frac{b x}{a}+1} \sin ^{-1}\left (\frac{\sqrt{a-b x}}{\sqrt{2} \sqrt{a}}\right )-b^3 x^3\right )}{2 b \sqrt{a+b x} \sqrt{c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]*Sqrt[a*c - b*c*x],x]

[Out]

(c*(a^2*b*x - b^3*x^3 - 2*a^(5/2)*Sqrt[a - b*x]*Sqrt[1 + (b*x)/a]*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]))/(2
*b*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])

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Maple [B]  time = 0.004, size = 127, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,bc}\sqrt{bx+a} \left ( -bcx+ac \right ) ^{{\frac{3}{2}}}}+{\frac{a}{2\,b}\sqrt{bx+a}\sqrt{-bcx+ac}}+{\frac{{a}^{2}c}{2}\sqrt{ \left ( bx+a \right ) \left ( -bcx+ac \right ) }\arctan \left ({x\sqrt{{b}^{2}c}{\frac{1}{\sqrt{-{b}^{2}c{x}^{2}+{a}^{2}c}}}} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{-bcx+ac}}}{\frac{1}{\sqrt{{b}^{2}c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66942, size = 373, normalized size = 5.49 \begin{align*} \left [\frac{a^{2} \sqrt{-c} \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt{-b c x + a c} \sqrt{b x + a} b \sqrt{-c} x - a^{2} c\right ) + 2 \, \sqrt{-b c x + a c} \sqrt{b x + a} b x}{4 \, b}, -\frac{a^{2} \sqrt{c} \arctan \left (\frac{\sqrt{-b c x + a c} \sqrt{b x + a} b \sqrt{c} x}{b^{2} c x^{2} - a^{2} c}\right ) - \sqrt{-b c x + a c} \sqrt{b x + a} b x}{2 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(a^2*sqrt(-c)*log(2*b^2*c*x^2 + 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c) + 2*sqrt(-b*c*x
+ a*c)*sqrt(b*x + a)*b*x)/b, -1/2*(a^2*sqrt(c)*arctan(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2
- a^2*c)) - sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*x)/b]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(-a + b*x))*sqrt(a + b*x), x)

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

[Out]

Timed out