\(\int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\) [139]

Optimal result

Integrand size = 26, antiderivative size = 146 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 x^3}{3 b^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {5 x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^6 c^3}+\frac {5 a^2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^7 c^{5/2}} \] Output:

1/3*x^5/b^2/c/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2)-5/3*x^3/b^4/c^2/(b*x+a)^(1/ 
2)/(-b*c*x+a*c)^(1/2)-5/2*x*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)/b^6/c^3+5*a^2 
*arctan(c^(1/2)*(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2))/b^7/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.85 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {15 a^4 b x-20 a^2 b^3 x^3+3 b^5 x^5-30 a^2 \sqrt {a-b x} \sqrt {a+b x} \left (a^2-b^2 x^2\right ) \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{6 b^7 c^2 \sqrt {c (a-b x)} (-a+b x) (a+b x)^{3/2}} \] Input:

Integrate[x^6/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

(15*a^4*b*x - 20*a^2*b^3*x^3 + 3*b^5*x^5 - 30*a^2*Sqrt[a - b*x]*Sqrt[a + b 
*x]*(a^2 - b^2*x^2)*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]])/(6*b^7*c^2*Sqrt[c 
*(a - b*x)]*(-a + b*x)*(a + b*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {109, 27, 35, 109, 27, 35, 101, 25, 27, 45, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\)

\(\Big \downarrow \) 109

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {\int \frac {5 a c x^4 (a-b x)}{(a+b x)^{3/2} (a c-b c x)^{5/2}}dx}{3 a b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \int \frac {x^4 (a-b x)}{(a+b x)^{3/2} (a c-b c x)^{5/2}}dx}{3 b^2}\)

\(\Big \downarrow \) 35

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \int \frac {x^4}{(a+b x)^{3/2} (a c-b c x)^{3/2}}dx}{3 b^2 c}\)

\(\Big \downarrow \) 109

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\int \frac {3 a c x^2 (a-b x)}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{a b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \int \frac {x^2 (a-b x)}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{b^2}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 35

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \int \frac {x^2}{\sqrt {a+b x} \sqrt {a c-b c x}}dx}{b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \left (-\frac {\int -\frac {a^2 c}{\sqrt {a+b x} \sqrt {a c-b c x}}dx}{2 b^2 c}-\frac {x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\right )}{b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \left (\frac {\int \frac {a^2 c}{\sqrt {a+b x} \sqrt {a c-b c x}}dx}{2 b^2 c}-\frac {x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\right )}{b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \left (\frac {a^2 \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}}dx}{2 b^2}-\frac {x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\right )}{b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 45

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \left (\frac {a^2 \int \frac {1}{\frac {c (a+b x) b}{a c-b c x}+b}d\frac {\sqrt {a+b x}}{\sqrt {a c-b c x}}}{b^2}-\frac {x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\right )}{b^2 c}\right )}{3 b^2 c}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^5}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {5 \left (\frac {x^3}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {3 \left (\frac {a^2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^3 \sqrt {c}}-\frac {x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\right )}{b^2 c}\right )}{3 b^2 c}\)

Input:

Int[x^6/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 35

Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} 
, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] &&  !(IntegerQ[n] && SimplerQ[a + 
b*x, c + d*x])
 

rule 45

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

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(259\) vs. \(2(120)=240\).

Time = 0.33 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.78

method	result	size
default	\(\frac {\sqrt {c \left (-b x +a \right )}\, \left (15 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{2} b^{4} c \,x^{4}-3 b^{4} x^{5} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}-30 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{4} b^{2} c \,x^{2}+20 a^{2} b^{2} x^{3} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}+15 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{6} c -15 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}\, a^{4} x \right )}{6 c^{3} \left (-b x +a \right )^{2} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, b^{6} \left (b x +a \right )^{\frac {3}{2}}}\)	\(260\)
risch	\(-\frac {x \sqrt {b x +a}\, \left (-b x +a \right )}{2 b^{6} \sqrt {-c \left (b x -a \right )}\, c^{2}}+\frac {\left (\frac {5 a^{2} \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right )}{2 b^{6} \sqrt {b^{2} c}}+\frac {7 a^{2} \sqrt {-b^{2} c \left (x -\frac {a}{b}\right )^{2}-2 a b c \left (x -\frac {a}{b}\right )}}{6 b^{8} c \left (x -\frac {a}{b}\right )}+\frac {7 a^{2} \sqrt {-b^{2} c \left (x +\frac {a}{b}\right )^{2}+2 a b c \left (x +\frac {a}{b}\right )}}{6 b^{8} c \left (x +\frac {a}{b}\right )}+\frac {a^{3} \sqrt {-b^{2} c \left (x -\frac {a}{b}\right )^{2}-2 a b c \left (x -\frac {a}{b}\right )}}{12 b^{9} c \left (x -\frac {a}{b}\right )^{2}}-\frac {a^{3} \sqrt {-b^{2} c \left (x +\frac {a}{b}\right )^{2}+2 a b c \left (x +\frac {a}{b}\right )}}{12 b^{9} c \left (x +\frac {a}{b}\right )^{2}}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}}{\sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}\, c^{2}}\)	\(325\)

Input:

int(x^6/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(c*(-b*x+a))^(1/2)/c^3*(15*arctan((b^2*c)^(1/2)*x/(c*(-b^2*x^2+a^2))^( 
1/2))*a^2*b^4*c*x^4-3*b^4*x^5*(b^2*c)^(1/2)*(c*(-b^2*x^2+a^2))^(1/2)-30*ar 
ctan((b^2*c)^(1/2)*x/(c*(-b^2*x^2+a^2))^(1/2))*a^4*b^2*c*x^2+20*a^2*b^2*x^ 
3*(b^2*c)^(1/2)*(c*(-b^2*x^2+a^2))^(1/2)+15*arctan((b^2*c)^(1/2)*x/(c*(-b^ 
2*x^2+a^2))^(1/2))*a^6*c-15*(c*(-b^2*x^2+a^2))^(1/2)*(b^2*c)^(1/2)*a^4*x)/ 
(-b*x+a)^2/(b^2*c)^(1/2)/(c*(-b^2*x^2+a^2))^(1/2)/b^6/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.21 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\left [-\frac {15 \, {\left (a^{2} b^{4} x^{4} - 2 \, a^{4} b^{2} x^{2} + a^{6}\right )} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) + 2 \, {\left (3 \, b^{5} x^{5} - 20 \, a^{2} b^{3} x^{3} + 15 \, a^{4} b x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{12 \, {\left (b^{11} c^{3} x^{4} - 2 \, a^{2} b^{9} c^{3} x^{2} + a^{4} b^{7} c^{3}\right )}}, -\frac {15 \, {\left (a^{2} b^{4} x^{4} - 2 \, a^{4} b^{2} x^{2} + a^{6}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (3 \, b^{5} x^{5} - 20 \, a^{2} b^{3} x^{3} + 15 \, a^{4} b x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{6 \, {\left (b^{11} c^{3} x^{4} - 2 \, a^{2} b^{9} c^{3} x^{2} + a^{4} b^{7} c^{3}\right )}}\right ] \] Input:

integrate(x^6/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(15*(a^2*b^4*x^4 - 2*a^4*b^2*x^2 + a^6)*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*(3*b^5*x^5 - 
20*a^2*b^3*x^3 + 15*a^4*b*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^11*c^3*x 
^4 - 2*a^2*b^9*c^3*x^2 + a^4*b^7*c^3), -1/6*(15*(a^2*b^4*x^4 - 2*a^4*b^2*x 
^2 + a^6)*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)) + (3*b^5*x^5 - 20*a^2*b^3*x^3 + 15*a^4*b*x)*sqrt(-b*c*x + 
 a*c)*sqrt(b*x + a))/(b^11*c^3*x^4 - 2*a^2*b^9*c^3*x^2 + a^4*b^7*c^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(x**6/(b*x+a)**(5/2)/(-b*c*x+a*c)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {5 \, a^{2} x {\left (\frac {3 \, x^{2}}{{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2} c} - \frac {2 \, a^{2}}{{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{4} c}\right )}}{6 \, b^{2}} - \frac {x^{5}}{2 \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2} c} - \frac {5 \, a^{2} x}{6 \, \sqrt {-b^{2} c x^{2} + a^{2} c} b^{6} c^{2}} + \frac {5 \, a^{2} \arcsin \left (\frac {b x}{a}\right )}{2 \, b^{7} c^{\frac {5}{2}}} \] Input:

integrate(x^6/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="maxima")
 

Output:

5/6*a^2*x*(3*x^2/((-b^2*c*x^2 + a^2*c)^(3/2)*b^2*c) - 2*a^2/((-b^2*c*x^2 + 
 a^2*c)^(3/2)*b^4*c))/b^2 - 1/2*x^5/((-b^2*c*x^2 + a^2*c)^(3/2)*b^2*c) - 5 
/6*a^2*x/(sqrt(-b^2*c*x^2 + a^2*c)*b^6*c^2) + 5/2*a^2*arcsin(b*x/a)/(b^7*c 
^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (120) = 240\).

Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.00 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=-\frac {\frac {{\left (2 \, {\left (b x + a\right )} {\left (3 \, {\left (b x + a\right )} {\left (\frac {b x + a}{b^{6} c} - \frac {5 \, a}{b^{6} c}\right )} + \frac {17 \, a^{2}}{b^{6} c}\right )} + \frac {3 \, a^{3}}{b^{6} c}\right )} \sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} c - 2 \, a c\right )}^{2}} + \frac {30 \, a^{2} \log \left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}\right )}{b^{6} \sqrt {-c} c^{2}} - \frac {4 \, {\left (15 \, a^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} - 54 \, a^{4} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} c + 56 \, a^{5} c^{2}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} - 2 \, a c\right )}^{3} b^{6} \sqrt {-c} c}}{12 \, b} \] Input:

integrate(x^6/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="giac")
 

Output:

-1/12*((2*(b*x + a)*(3*(b*x + a)*((b*x + a)/(b^6*c) - 5*a/(b^6*c)) + 17*a^ 
2/(b^6*c)) + 3*a^3/(b^6*c))*sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)/((b*x 
 + a)*c - 2*a*c)^2 + 30*a^2*log((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)* 
c + 2*a*c))^2)/(b^6*sqrt(-c)*c^2) - 4*(15*a^3*(sqrt(b*x + a)*sqrt(-c) - sq 
rt(-(b*x + a)*c + 2*a*c))^4 - 54*a^4*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x 
+ a)*c + 2*a*c))^2*c + 56*a^5*c^2)/(((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x 
+ a)*c + 2*a*c))^2 - 2*a*c)^3*b^6*sqrt(-c)*c))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\int \frac {x^6}{{\left (a\,c-b\,c\,x\right )}^{5/2}\,{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:

int(x^6/((a*c - b*c*x)^(5/2)*(a + b*x)^(5/2)),x)
                                                                                    
                                                                                    
 

Output:

int(x^6/((a*c - b*c*x)^(5/2)*(a + b*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.99 \[ \int \frac {x^6}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-30 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{4}+30 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{2} b^{2} x^{2}-15 a^{4} b x +20 a^{2} b^{3} x^{3}-3 b^{5} x^{5}\right )}{6 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{7} c^{3} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:

int(x^6/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x)
 

Output:

(sqrt(c)*( - 30*sqrt(a + b*x)*sqrt(a - b*x)*asin(sqrt(a - b*x)/(sqrt(a)*sq 
rt(2)))*a**4 + 30*sqrt(a + b*x)*sqrt(a - b*x)*asin(sqrt(a - b*x)/(sqrt(a)* 
sqrt(2)))*a**2*b**2*x**2 - 15*a**4*b*x + 20*a**2*b**3*x**3 - 3*b**5*x**5)) 
/(6*sqrt(a + b*x)*sqrt(a - b*x)*b**7*c**3*(a**2 - b**2*x**2))