3.1149 \(\int \frac{1}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac{x}{a^2 c \sqrt{a+b x} \sqrt{a c-b c x}} \]

[Out]

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

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Rubi [A]  time = 0.0038488, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {39} \[ \frac{x}{a^2 c \sqrt{a+b x} \sqrt{a c-b c x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]

[Out]

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

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx &=\frac{x}{a^2 c \sqrt{a+b x} \sqrt{a c-b c x}}\\ \end{align*}

Mathematica [A]  time = 0.0129921, size = 29, normalized size = 0.97 \[ \frac{x}{a^2 c \sqrt{a+b x} \sqrt{c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.002, size = 30, normalized size = 1. \begin{align*}{\frac{x \left ( -bx+a \right ) }{{a}^{2}}{\frac{1}{\sqrt{bx+a}}} \left ( -bcx+ac \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00213, size = 34, normalized size = 1.13 \begin{align*} \frac{x}{\sqrt{-b^{2} c x^{2} + a^{2} c} a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(sqrt(-b^2*c*x^2 + a^2*c)*a^2*c)

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Fricas [A]  time = 1.56822, size = 88, normalized size = 2.93 \begin{align*} -\frac{\sqrt{-b c x + a c} \sqrt{b x + a} x}{a^{2} b^{2} c^{2} x^{2} - a^{4} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="fricas")

[Out]

-sqrt(-b*c*x + a*c)*sqrt(b*x + a)*x/(a^2*b^2*c^2*x^2 - a^4*c^2)

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Sympy [C]  time = 5.54556, size = 94, normalized size = 3.13 \begin{align*} - \frac{i{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4}, 1 & \frac{1}{2}, \frac{3}{2}, 2 \\\frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2}, 2 & 0 \end{matrix} \middle |{\frac{a^{2}}{b^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} a^{2} b c^{\frac{3}{2}}} + \frac{{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{1}{2}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1 & \\\frac{1}{4}, \frac{3}{4} & - \frac{1}{2}, 0, 1, 0 \end{matrix} \middle |{\frac{a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} a^{2} b c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*meijerg(((3/4, 5/4, 1), (1/2, 3/2, 2)), ((3/4, 1, 5/4, 3/2, 2), (0,)), a**2/(b**2*x**2))/(2*pi**(3/2)*a**2*
b*c**(3/2)) + meijerg(((-1/2, 0, 1/4, 1/2, 3/4, 1), ()), ((1/4, 3/4), (-1/2, 0, 1, 0)), a**2*exp_polar(-2*I*pi
)/(b**2*x**2))/(2*pi**(3/2)*a**2*b*c**(3/2))

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Giac [B]  time = 1.11973, size = 155, normalized size = 5.17 \begin{align*} \frac{2 \, \sqrt{-c} c}{{\left (2 \, a c^{2} -{\left (\sqrt{-b c x + a c} \sqrt{-c} - \sqrt{2 \, a c^{2} +{\left (b c x - a c\right )} c}\right )}^{2}\right )} a b{\left | c \right |}} - \frac{\sqrt{-b c x + a c}}{2 \, \sqrt{2 \, a c^{2} +{\left (b c x - a c\right )} c} a^{2} b{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="giac")

[Out]

2*sqrt(-c)*c/((2*a*c^2 - (sqrt(-b*c*x + a*c)*sqrt(-c) - sqrt(2*a*c^2 + (b*c*x - a*c)*c))^2)*a*b*abs(c)) - 1/2*
sqrt(-b*c*x + a*c)/(sqrt(2*a*c^2 + (b*c*x - a*c)*c)*a^2*b*abs(c))