3.1138 \(\int (a+a x)^{3/2} (c-c x)^{3/2} \, dx\)

Optimal. Leaf size=96 \[ \frac{3}{4} a^{3/2} c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a x+a}}{\sqrt{a} \sqrt{c-c x}}\right )+\frac{3}{8} a c x \sqrt{a x+a} \sqrt{c-c x}+\frac{1}{4} x (a x+a)^{3/2} (c-c x)^{3/2} \]

[Out]

(3*a*c*x*Sqrt[a + a*x]*Sqrt[c - c*x])/8 + (x*(a + a*x)^(3/2)*(c - c*x)^(3/2))/4 + (3*a^(3/2)*c^(3/2)*ArcTan[(S
qrt[c]*Sqrt[a + a*x])/(Sqrt[a]*Sqrt[c - c*x])])/4

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Rubi [A]  time = 0.0365369, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {38, 63, 217, 203} \[ \frac{3}{4} a^{3/2} c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a x+a}}{\sqrt{a} \sqrt{c-c x}}\right )+\frac{3}{8} a c x \sqrt{a x+a} \sqrt{c-c x}+\frac{1}{4} x (a x+a)^{3/2} (c-c x)^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*c*x*Sqrt[a + a*x]*Sqrt[c - c*x])/8 + (x*(a + a*x)^(3/2)*(c - c*x)^(3/2))/4 + (3*a^(3/2)*c^(3/2)*ArcTan[(S
qrt[c]*Sqrt[a + a*x])/(Sqrt[a]*Sqrt[c - c*x])])/4

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

Mathematica [A]  time = 0.0788277, size = 104, normalized size = 1.08 \[ \frac{\sqrt{c} (a (x+1))^{3/2} \sqrt{c-c x} \left (\sqrt{c} x \sqrt{x+1} \left (-2 x^3+2 x^2+5 x-5\right )+6 \sqrt{c-c x} \sin ^{-1}\left (\frac{\sqrt{c-c x}}{\sqrt{2} \sqrt{c}}\right )\right )}{8 (x-1) (x+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(a*(1 + x))^(3/2)*Sqrt[c - c*x]*(Sqrt[c]*x*Sqrt[1 + x]*(-5 + 5*x + 2*x^2 - 2*x^3) + 6*Sqrt[c - c*x]*A
rcSin[Sqrt[c - c*x]/(Sqrt[2]*Sqrt[c])]))/(8*(-1 + x)*(1 + x)^(3/2))

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Maple [B]  time = 0.004, size = 143, normalized size = 1.5 \begin{align*} -{\frac{1}{4\,c} \left ( ax+a \right ) ^{{\frac{3}{2}}} \left ( -cx+c \right ) ^{{\frac{5}{2}}}}-{\frac{a}{4\,c}\sqrt{ax+a} \left ( -cx+c \right ) ^{{\frac{5}{2}}}}+{\frac{a}{8}\sqrt{ax+a} \left ( -cx+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,ac}{8}\sqrt{ax+a}\sqrt{-cx+c}}+{\frac{3\,{a}^{2}{c}^{2}}{8}\sqrt{ \left ( -cx+c \right ) \left ( ax+a \right ) }\arctan \left ({x\sqrt{ac}{\frac{1}{\sqrt{-ac{x}^{2}+ac}}}} \right ){\frac{1}{\sqrt{ax+a}}}{\frac{1}{\sqrt{-cx+c}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+a)^(3/2)*(-c*x+c)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63984, size = 393, normalized size = 4.09 \begin{align*} \left [\frac{3}{16} \, \sqrt{-a c} a c \log \left (2 \, a c x^{2} + 2 \, \sqrt{-a c} \sqrt{a x + a} \sqrt{-c x + c} x - a c\right ) - \frac{1}{8} \,{\left (2 \, a c x^{3} - 5 \, a c x\right )} \sqrt{a x + a} \sqrt{-c x + c}, -\frac{3}{8} \, \sqrt{a c} a c \arctan \left (\frac{\sqrt{a c} \sqrt{a x + a} \sqrt{-c x + c} x}{a c x^{2} - a c}\right ) - \frac{1}{8} \,{\left (2 \, a c x^{3} - 5 \, a c x\right )} \sqrt{a x + a} \sqrt{-c x + c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (x + 1\right )\right )^{\frac{3}{2}} \left (- c \left (x - 1\right )\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(x + 1))**(3/2)*(-c*(x - 1))**(3/2), x)

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Giac [B]  time = 1.24884, size = 277, normalized size = 2.89 \begin{align*} -\frac{{\left (\frac{2 \, a^{3} c \log \left ({\left | -\sqrt{-a c} \sqrt{a x + a} + \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt{-a c}} - \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \sqrt{a x + a} a x\right )} c{\left | a \right |}}{2 \, a^{2}} + \frac{{\left (\frac{2 \, a^{3} c \log \left ({\left | -\sqrt{-a c} \sqrt{a x + a} + \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt{-a c}} - \sqrt{-{\left (a x + a\right )} a c + 2 \, a^{2} c}{\left ({\left (a x + a\right )}{\left (2 \,{\left (a x + a\right )}{\left (\frac{a x + a}{a^{2}} - \frac{3}{a}\right )} + 5\right )} - a\right )} \sqrt{a x + a}\right )} c{\left | a \right |}}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^3*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c + 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x + a
)*a*c + 2*a^2*c)*sqrt(a*x + a)*a*x)*c*abs(a)/a^2 + 1/8*(2*a^3*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x
+ a)*a*c + 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x + a)*a*c + 2*a^2*c)*((a*x + a)*(2*(a*x + a)*((a*x + a)/a^2 - 3/
a) + 5) - a)*sqrt(a*x + a))*c*abs(a)/a^2