3.1146 \(\int (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx\)

Optimal. Leaf size=102 \[ \frac{3 a^4 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{4 b}+\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2} \]

[Out]

(3*a^2*c*x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/8 + (x*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/4 + (3*a^4*c^(3/2)*Arc
Tan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[c*(a - b*x)]])/(4*b)

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Rubi [A]  time = 0.0361188, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {38, 63, 217, 203} \[ \frac{3 a^4 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{4 b}+\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2),x]

[Out]

(3*a^2*c*x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/8 + (x*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/4 + (3*a^4*c^(3/2)*Arc
Tan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[c*(a - b*x)]])/(4*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 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx &=\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac{1}{4} \left (3 a^2 c\right ) \int \sqrt{a+b x} \sqrt{a c-b c x} \, dx\\ &=\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac{1}{8} \left (3 a^4 c^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{a c-b c x}} \, dx\\ &=\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac{\left (3 a^4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a c-c x^2}} \, dx,x,\sqrt{a+b x}\right )}{4 b}\\ &=\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac{\left (3 a^4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{4 b}\\ &=\frac{3}{8} a^2 c x \sqrt{a+b x} \sqrt{a c-b c x}+\frac{1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac{3 a^4 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{c (a-b x)}}\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.122282, size = 109, normalized size = 1.07 \[ \frac{c^2 \left (-7 a^2 b^3 x^3+5 a^4 b x-6 a^{9/2} \sqrt{a-b x} \sqrt{\frac{b x}{a}+1} \sin ^{-1}\left (\frac{\sqrt{a-b x}}{\sqrt{2} \sqrt{a}}\right )+2 b^5 x^5\right )}{8 b \sqrt{a+b x} \sqrt{c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2),x]

[Out]

(c^2*(5*a^4*b*x - 7*a^2*b^3*x^3 + 2*b^5*x^5 - 6*a^(9/2)*Sqrt[a - b*x]*Sqrt[1 + (b*x)/a]*ArcSin[Sqrt[a - b*x]/(
Sqrt[2]*Sqrt[a])]))/(8*b*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])

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Maple [B]  time = 0.004, size = 185, normalized size = 1.8 \begin{align*} -{\frac{1}{4\,bc} \left ( bx+a \right ) ^{{\frac{3}{2}}} \left ( -bcx+ac \right ) ^{{\frac{5}{2}}}}-{\frac{a}{4\,bc}\sqrt{bx+a} \left ( -bcx+ac \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}}{8\,b} \left ( -bcx+ac \right ) ^{{\frac{3}{2}}}\sqrt{bx+a}}+{\frac{3\,{a}^{3}c}{8\,b}\sqrt{bx+a}\sqrt{-bcx+ac}}+{\frac{3\,{a}^{4}{c}^{2}}{8}\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)^(3/2)*(-b*c*x+a*c)^(3/2),x)

[Out]

-1/4/b/c*(b*x+a)^(3/2)*(-b*c*x+a*c)^(5/2)-1/4*a/b/c*(b*x+a)^(1/2)*(-b*c*x+a*c)^(5/2)+1/8*a^2/b*(-b*c*x+a*c)^(3
/2)*(b*x+a)^(1/2)+3/8*a^3*c/b*(-b*c*x+a*c)^(1/2)*(b*x+a)^(1/2)+3/8*a^4*c^2*((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)^(3/2)*(-b*c*x+a*c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62313, size = 447, normalized size = 4.38 \begin{align*} \left [\frac{3 \, a^{4} \sqrt{-c} 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 \,{\left (2 \, b^{3} c x^{3} - 5 \, a^{2} b c x\right )} \sqrt{-b c x + a c} \sqrt{b x + a}}{16 \, b}, -\frac{3 \, a^{4} c^{\frac{3}{2}} \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 ) +{\left (2 \, b^{3} c x^{3} - 5 \, a^{2} b c x\right )} \sqrt{-b c x + a c} \sqrt{b x + a}}{8 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*a^4*sqrt(-c)*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*(2*b^3*
c*x^3 - 5*a^2*b*c*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b, -1/8*(3*a^4*c^(3/2)*arctan(sqrt(-b*c*x + a*c)*sqrt(b
*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c)) + (2*b^3*c*x^3 - 5*a^2*b*c*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- c \left (- a + b x\right )\right )^{\frac{3}{2}} \left (a + b x\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(-b*c*x+a*c)**(3/2),x)

[Out]

Integral((-c*(-a + b*x))**(3/2)*(a + b*x)**(3/2), 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)^(3/2)*(-b*c*x+a*c)^(3/2),x, algorithm="giac")

[Out]

Timed out