\(\int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\) [1150]

Optimal result

Integrand size = 23, antiderivative size = 67 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {x}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}+\frac {2 x}{3 a^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {40, 39} \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {2 x}{3 a^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {x}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}} \]

[In]

Int[1/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]

[Out]

x/(3*a^2*c*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)) + (2*x)/(3*a^4*c^2*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]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(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] && ILtQ[m + 3/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}+\frac {2 \int \frac {1}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx}{3 a^2 c} \\ & = \frac {x}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}+\frac {2 x}{3 a^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {3 a^2 x-2 b^2 x^3}{3 a^4 c (c (a-b x))^{3/2} (a+b x)^{3/2}} \]

[In]

Integrate[1/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]

[Out]

(3*a^2*x - 2*b^2*x^3)/(3*a^4*c*(c*(a - b*x))^(3/2)*(a + b*x)^(3/2))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {\left (-b x +a \right ) x \left (-2 b^{2} x^{2}+3 a^{2}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} a^{4} \left (-b c x +a c \right )^{\frac {5}{2}}}\)	\(45\)
default	\(-\frac {1}{3 b a c \left (b x +a \right )^{\frac {3}{2}} \left (-b c x +a c \right )^{\frac {3}{2}}}+\frac {-\frac {1}{b a c \sqrt {b x +a}\, \left (-b c x +a c \right )^{\frac {3}{2}}}+\frac {\frac {2 \sqrt {b x +a}}{3 b a c \left (-b c x +a c \right )^{\frac {3}{2}}}+\frac {2 \sqrt {b x +a}}{3 b \,a^{2} c^{2} \sqrt {-b c x +a c}}}{a}}{a}\)	\(129\)

[In]

int(1/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=-\frac {{\left (2 \, b^{2} x^{3} - 3 \, a^{2} x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{3 \, {\left (a^{4} b^{4} c^{3} x^{4} - 2 \, a^{6} b^{2} c^{3} x^{2} + a^{8} c^{3}\right )}} \]

[In]

integrate(1/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*b^2*x^3 - 3*a^2*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a)/(a^4*b^4*c^3*x^4 - 2*a^6*b^2*c^3*x^2 + a^8*c^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.54 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {1}{2}, \frac {5}{2}, 3 \\\frac {5}{4}, \frac {7}{4}, 2, \frac {5}{2}, 3 & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{3 \pi ^{\frac {3}{2}} a^{4} b c^{\frac {5}{2}}} + \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, 0, \frac {1}{2}, \frac {3}{4}, \frac {5}{4}, 1 & \\\frac {3}{4}, \frac {5}{4} & - \frac {1}{2}, 0, 2, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{3 \pi ^{\frac {3}{2}} a^{4} b c^{\frac {5}{2}}} \]

[In]

integrate(1/(b*x+a)**(5/2)/(-b*c*x+a*c)**(5/2),x)

[Out]

I*meijerg(((5/4, 7/4, 1), (1/2, 5/2, 3)), ((5/4, 7/4, 2, 5/2, 3), (0,)), a**2/(b**2*x**2))/(3*pi**(3/2)*a**4*b
*c**(5/2)) + meijerg(((-1/2, 0, 1/2, 3/4, 5/4, 1), ()), ((3/4, 5/4), (-1/2, 0, 2, 0)), a**2*exp_polar(-2*I*pi)
/(b**2*x**2))/(3*pi**(3/2)*a**4*b*c**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {x}{3 \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{2} c} + \frac {2 \, x}{3 \, \sqrt {-b^{2} c x^{2} + a^{2} c} a^{4} c^{2}} \]

[In]

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

[Out]

1/3*x/((-b^2*c*x^2 + a^2*c)^(3/2)*a^2*c) + 2/3*x/(sqrt(-b^2*c*x^2 + a^2*c)*a^4*c^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (55) = 110\).

Time = 0.33 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.97 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=-\frac {\frac {\sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a} {\left (\frac {4 \, {\left (b x + a\right )}}{a^{4} c} - \frac {9}{a^{3} c}\right )}}{{\left ({\left (b x + a\right )} c - 2 \, a c\right )}^{2}} + \frac {4 \, {\left (3 \, {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} - 18 \, a {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} c + 16 \, a^{2} c^{2}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} - 2 \, a c\right )}^{3} a^{3} \sqrt {-c} c}}{12 \, b} \]

[In]

integrate(1/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="giac")

[Out]

-1/12*(sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)*(4*(b*x + a)/(a^4*c) - 9/(a^3*c))/((b*x + a)*c - 2*a*c)^2 + 4*
(3*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 - 18*a*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c
+ 2*a*c))^2*c + 16*a^2*c^2)/(((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^2 - 2*a*c)^3*a^3*sqrt(-c)*
c))/b

Mupad [B] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=-\frac {3\,a^2\,x\,\sqrt {a\,c-b\,c\,x}-2\,b^2\,x^3\,\sqrt {a\,c-b\,c\,x}}{{\left (a\,c-b\,c\,x\right )}^2\,\left (3\,a^4\,\left (a\,c-b\,c\,x\right )-6\,a^5\,c\right )\,\sqrt {a+b\,x}} \]

[In]

int(1/((a*c - b*c*x)^(5/2)*(a + b*x)^(5/2)),x)

[Out]

-(3*a^2*x*(a*c - b*c*x)^(1/2) - 2*b^2*x^3*(a*c - b*c*x)^(1/2))/((a*c - b*c*x)^2*(3*a^4*(a*c - b*c*x) - 6*a^5*c
)*(a + b*x)^(1/2))