\(\int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) [414]

Optimal result

Integrand size = 22, antiderivative size = 154 \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 x^3}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {4 c x^2}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {16 c^3 \sqrt {a+b x}}{3 d (b c-a d)^3 (c+d x)^{3/2}}+\frac {16 c^2 (b c-3 a d) \sqrt {a+b x}}{3 d (b c-a d)^4 \sqrt {c+d x}} \] Output:

-2/3*x^3/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/2)-4*c*x^2/(-a*d+b*c)^2/(b*x+ 
a)^(1/2)/(d*x+c)^(3/2)-16/3*c^3*(b*x+a)^(1/2)/d/(-a*d+b*c)^3/(d*x+c)^(3/2) 
+16/3*c^2*(-3*a*d+b*c)*(b*x+a)^(1/2)/d/(-a*d+b*c)^4/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.60 \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (c^3-\frac {9 a c^2 (c+d x)}{a+b x}-\frac {9 a^2 c (c+d x)^2}{(a+b x)^2}+\frac {a^3 (c+d x)^3}{(a+b x)^3}\right )}{3 (b c-a d)^4 (c+d x)^{3/2}} \] Input:

Integrate[x^3/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(2*(a + b*x)^(3/2)*(c^3 - (9*a*c^2*(c + d*x))/(a + b*x) - (9*a^2*c*(c + d* 
x)^2)/(a + b*x)^2 + (a^3*(c + d*x)^3)/(a + b*x)^3))/(3*(b*c - a*d)^4*(c + 
d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {105, 100, 27, 87, 48}

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.

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {2 c \left (-\frac {\frac {2 \sqrt {a+b x} \left (3 a^2 d^2+b^2 c^2\right )}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {b \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}}{b^2 (b c-a d)}-\frac {2 a^2}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2 x^3}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {2 c \left (-\frac {\frac {2 \sqrt {a+b x} \left (3 a^2 d^2+b^2 c^2\right )}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {2 b \sqrt {a+b x} \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right )}{3 d \sqrt {c+d x} (b c-a d)^2}}{b^2 (b c-a d)}-\frac {2 a^2}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2 x^3}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[x^3/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(-2*x^3)/(3*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) + (2*c*((-2*a^2)/ 
(b^2*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((2*(b^2*c^2 + 3*a^2*d^2 
)*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - (2*b*(b^2*c^2 - 6*a*b 
*c*d - 3*a^2*d^2)*Sqrt[a + b*x])/(3*d*(b*c - a*d)^2*Sqrt[c + d*x]))/(b^2*( 
b*c - a*d))))/(b*c - a*d)
 

Defintions of rubi rules used

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 48

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

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

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

rule 105

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {2 \left (-x^{3} a^{3} d^{3}+9 x^{3} a^{2} b c \,d^{2}+9 x^{3} a \,b^{2} c^{2} d -x^{3} b^{3} c^{3}+6 a^{3} c \,d^{2} x^{2}+36 a^{2} b \,c^{2} d \,x^{2}+6 a \,b^{2} c^{3} x^{2}+24 a^{3} c^{2} d x +24 a^{2} b \,c^{3} x +16 c^{3} a^{3}\right )}{3 \left (a d -b c \right )^{4} \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}\)	\(141\)
gosper	\(-\frac {2 \left (-x^{3} a^{3} d^{3}+9 x^{3} a^{2} b c \,d^{2}+9 x^{3} a \,b^{2} c^{2} d -x^{3} b^{3} c^{3}+6 a^{3} c \,d^{2} x^{2}+36 a^{2} b \,c^{2} d \,x^{2}+6 a \,b^{2} c^{3} x^{2}+24 a^{3} c^{2} d x +24 a^{2} b \,c^{3} x +16 c^{3} a^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(182\)
orering	\(-\frac {2 \left (-x^{3} a^{3} d^{3}+9 x^{3} a^{2} b c \,d^{2}+9 x^{3} a \,b^{2} c^{2} d -x^{3} b^{3} c^{3}+6 a^{3} c \,d^{2} x^{2}+36 a^{2} b \,c^{2} d \,x^{2}+6 a \,b^{2} c^{3} x^{2}+24 a^{3} c^{2} d x +24 a^{2} b \,c^{3} x +16 c^{3} a^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(182\)

Input:

int(x^3/(b*x+a)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(-a^3*d^3*x^3+9*a^2*b*c*d^2*x^3+9*a*b^2*c^2*d*x^3-b^3*c^3*x^3+6*a^3*c 
*d^2*x^2+36*a^2*b*c^2*d*x^2+6*a*b^2*c^3*x^2+24*a^3*c^2*d*x+24*a^2*b*c^3*x+ 
16*a^3*c^3)/(a*d-b*c)^4/(b*x+a)^(3/2)/(d*x+c)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (132) = 264\).

Time = 0.48 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.91 \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (16 \, a^{3} c^{3} - {\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{3} + 6 \, {\left (a b^{2} c^{3} + 6 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{2} + 24 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} + {\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \, {\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} + {\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \] Input:

integrate(x^3/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

-2/3*(16*a^3*c^3 - (b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*x^3 
 + 6*(a*b^2*c^3 + 6*a^2*b*c^2*d + a^3*c*d^2)*x^2 + 24*(a^2*b*c^3 + a^3*c^2 
*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4* 
b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d 
^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d 
 - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d 
^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 
9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3* 
b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)
 

Sympy [F]

\[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{3}}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**3/(b*x+a)**(5/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(x**3/((a + b*x)**(5/2)*(c + d*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 771 vs. \(2 (132) = 264\).

Time = 0.38 (sec) , antiderivative size = 771, normalized size of antiderivative = 5.01 \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^3/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

2/3*sqrt(b*x + a)*((b^7*c^6*d*abs(b) - 12*a*b^6*c^5*d^2*abs(b) + 30*a^2*b^ 
5*c^4*d^3*abs(b) - 28*a^3*b^4*c^3*d^4*abs(b) + 9*a^4*b^3*c^2*d^5*abs(b))*( 
b*x + a)/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^ 
4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^ 
2*d^8) - 9*(a*b^7*c^6*d*abs(b) - 4*a^2*b^6*c^5*d^2*abs(b) + 6*a^3*b^5*c^4* 
d^3*abs(b) - 4*a^4*b^4*c^3*d^4*abs(b) + a^5*b^3*c^2*d^5*abs(b))/(b^9*c^7*d 
 - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5* 
c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8))/(b^2*c + (b 
*x + a)*b*d - a*b*d)^(3/2) - 4/3*(9*sqrt(b*d)*a^2*b^5*c^3 - 19*sqrt(b*d)*a 
^3*b^4*c^2*d + 11*sqrt(b*d)*a^4*b^3*c*d^2 - sqrt(b*d)*a^5*b^2*d^3 - 18*sqr 
t(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a 
^2*b^3*c^2 + 18*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d))^2*a^3*b^2*c*d + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr 
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b*c - 3*sqrt(b*d)*(sqrt(b*d)*sqrt( 
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*d)/((b^3*c^3*abs(b) 
- 3*a*b^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*(b^2*c - a 
*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^ 
3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {x^3}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(x^3/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x)
 

Output:

int(x^3/((a + b*x)^(5/2)*(c + d*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 2.46 (sec) , antiderivative size = 916, normalized size of antiderivative = 5.95 \[ \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^3/(b*x+a)^(5/2)/(d*x+c)^(5/2),x)
 

Output:

(2*(5*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**4*c**2*d**3 + 10*sqrt(d)*sqrt(b)*sq 
rt(a + b*x)*a**4*c*d**4*x + 5*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**4*d**5*x**2 
 + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b*c**3*d**2 + 11*sqrt(d)*sqrt(b)*s 
qrt(a + b*x)*a**3*b*c**2*d**3*x + 13*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b* 
c*d**4*x**2 + 5*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b*d**5*x**3 + 3*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*a**2*b**2*c**4*d + 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)* 
a**2*b**2*c**3*d**2*x + 9*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b**2*c**2*d** 
3*x**2 + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b**2*c*d**4*x**3 + 5*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*a*b**3*c**5 + 13*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b* 
*3*c**4*d*x + 11*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**3*c**3*d**2*x**2 + 3*s 
qrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**3*c**2*d**3*x**3 + 5*sqrt(d)*sqrt(b)*sqr 
t(a + b*x)*b**4*c**5*x + 10*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**4*c**4*d*x**2 
 + 5*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**4*c**3*d**2*x**3 - 16*sqrt(c + d*x)* 
a**3*b**2*c**3*d**2 - 24*sqrt(c + d*x)*a**3*b**2*c**2*d**3*x - 6*sqrt(c + 
d*x)*a**3*b**2*c*d**4*x**2 + sqrt(c + d*x)*a**3*b**2*d**5*x**3 - 24*sqrt(c 
 + d*x)*a**2*b**3*c**3*d**2*x - 36*sqrt(c + d*x)*a**2*b**3*c**2*d**3*x**2 
- 9*sqrt(c + d*x)*a**2*b**3*c*d**4*x**3 - 6*sqrt(c + d*x)*a*b**4*c**3*d**2 
*x**2 - 9*sqrt(c + d*x)*a*b**4*c**2*d**3*x**3 + sqrt(c + d*x)*b**5*c**3*d* 
*2*x**3))/(3*sqrt(a + b*x)*b**2*d**2*(a**5*c**2*d**4 + 2*a**5*c*d**5*x + a 
**5*d**6*x**2 - 4*a**4*b*c**3*d**3 - 7*a**4*b*c**2*d**4*x - 2*a**4*b*c*...