\(\int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx\) [148]

Optimal result

Integrand size = 46, antiderivative size = 103 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {b^4 x}{d^4}-\frac {(b c-a d)^4}{3 d^5 (c+d x)^3}+\frac {2 b (b c-a d)^3}{d^5 (c+d x)^2}-\frac {6 b^2 (b c-a d)^2}{d^5 (c+d x)}-\frac {4 b^3 (b c-a d) \log (c+d x)}{d^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.58 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=-\frac {a^4 d^4+2 a^3 b d^3 (c+3 d x)+6 a^2 b^2 d^2 \left (c^2+3 c d x+3 d^2 x^2\right )-2 a b^3 c d \left (11 c^2+27 c d x+18 d^2 x^2\right )+b^4 \left (13 c^4+27 c^3 d x+9 c^2 d^2 x^2-9 c d^3 x^3-3 d^4 x^4\right )+12 b^3 (b c-a d) (c+d x)^3 \log (c+d x)}{3 d^5 (c+d x)^3} \] Input:

Integrate[(a^4 + 4*a^3*b*x + 6*a^2*b^2*x^2 + 4*a*b^3*x^3 + b^4*x^4)/(c + d 
*x)^4,x]
 

Output:

-1/3*(a^4*d^4 + 2*a^3*b*d^3*(c + 3*d*x) + 6*a^2*b^2*d^2*(c^2 + 3*c*d*x + 3 
*d^2*x^2) - 2*a*b^3*c*d*(11*c^2 + 27*c*d*x + 18*d^2*x^2) + b^4*(13*c^4 + 2 
7*c^3*d*x + 9*c^2*d^2*x^2 - 9*c*d^3*x^3 - 3*d^4*x^4) + 12*b^3*(b*c - a*d)* 
(c + d*x)^3*Log[c + d*x])/(d^5*(c + d*x)^3)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2006, 49, 2009}

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.

\(\displaystyle \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx\)

\(\Big \downarrow \) 2006

\(\displaystyle \int \frac {(a+b x)^4}{(c+d x)^4}dx\)

\(\Big \downarrow \) 49

\(\displaystyle \int \left (-\frac {4 b^3 (b c-a d)}{d^4 (c+d x)}+\frac {6 b^2 (b c-a d)^2}{d^4 (c+d x)^2}-\frac {4 b (b c-a d)^3}{d^4 (c+d x)^3}+\frac {(a d-b c)^4}{d^4 (c+d x)^4}+\frac {b^4}{d^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 b^3 (b c-a d) \log (c+d x)}{d^5}-\frac {6 b^2 (b c-a d)^2}{d^5 (c+d x)}+\frac {2 b (b c-a d)^3}{d^5 (c+d x)^2}-\frac {(b c-a d)^4}{3 d^5 (c+d x)^3}+\frac {b^4 x}{d^4}\)

Input:

Int[(a^4 + 4*a^3*b*x + 6*a^2*b^2*x^2 + 4*a*b^3*x^3 + b^4*x^4)/(c + d*x)^4, 
x]
 

Output:

(b^4*x)/d^4 - (b*c - a*d)^4/(3*d^5*(c + d*x)^3) + (2*b*(b*c - a*d)^3)/(d^5 
*(c + d*x)^2) - (6*b^2*(b*c - a*d)^2)/(d^5*(c + d*x)) - (4*b^3*(b*c - a*d) 
*Log[c + d*x])/d^5
 

Defintions of rubi rules used

rule 49

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

rule 2006

Int[(u_.)*(Px_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], 
b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^Expon[Px 
, x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; PolyQ[Px, x] && GtQ[Expon[P 
x, x], 1] && NeQ[Coeff[Px, x, 0], 0] &&  !MatchQ[Px, (a_.)*(v_)^Expon[Px, x 
] /; FreeQ[a, x] && LinearQ[v, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {b^{4} x}{d^{4}}-\frac {6 b^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{d^{5} \left (x d +c \right )}+\frac {4 b^{3} \left (a d -b c \right ) \ln \left (x d +c \right )}{d^{5}}-\frac {2 b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{d^{5} \left (x d +c \right )^{2}}-\frac {d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}}{3 d^{5} \left (x d +c \right )^{3}}\)	\(178\)
norman	\(\frac {\frac {b^{4} x^{4}}{d}-\frac {d^{4} a^{4}+2 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-22 a \,b^{3} c^{3} d +22 c^{4} b^{4}}{3 d^{5}}-\frac {3 \left (2 a^{2} b^{2} d^{2}-4 a \,b^{3} c d +4 b^{4} c^{2}\right ) x^{2}}{d^{3}}-\frac {\left (2 a^{3} b \,d^{3}+6 a^{2} b^{2} c \,d^{2}-18 a \,b^{3} c^{2} d +18 b^{4} c^{3}\right ) x}{d^{4}}}{\left (x d +c \right )^{3}}+\frac {4 b^{3} \left (a d -b c \right ) \ln \left (x d +c \right )}{d^{5}}\)	\(180\)
risch	\(\frac {b^{4} x}{d^{4}}+\frac {\left (-6 a^{2} b^{2} d^{3}+12 a \,b^{3} c \,d^{2}-6 b^{4} c^{2} d \right ) x^{2}-2 b \left (a^{3} d^{3}+3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right ) x -\frac {d^{4} a^{4}+2 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-22 a \,b^{3} c^{3} d +13 c^{4} b^{4}}{3 d}}{d^{4} \left (x d +c \right )^{3}}+\frac {4 b^{3} \ln \left (x d +c \right ) a}{d^{4}}-\frac {4 b^{4} \ln \left (x d +c \right ) c}{d^{5}}\)	\(182\)
parallelrisch	\(\frac {-22 c^{4} b^{4}-d^{4} a^{4}+3 d^{4} x^{4} b^{4}+36 a \,b^{3} c \,d^{3} x^{2}-18 a^{2} b^{2} c \,d^{3} x +54 a \,b^{3} c^{2} d^{2} x -36 \ln \left (x d +c \right ) x \,b^{4} c^{3} d -12 \ln \left (x d +c \right ) b^{4} c^{4}+12 \ln \left (x d +c \right ) a \,b^{3} c^{3} d -2 a^{3} b c \,d^{3}-6 a^{2} b^{2} c^{2} d^{2}+22 a \,b^{3} c^{3} d -18 a^{2} b^{2} d^{4} x^{2}-36 b^{4} c^{2} d^{2} x^{2}-6 a^{3} b \,d^{4} x -54 b^{4} c^{3} d x +36 \ln \left (x d +c \right ) x a \,b^{3} c^{2} d^{2}+36 \ln \left (x d +c \right ) x^{2} a \,b^{3} c \,d^{3}-36 \ln \left (x d +c \right ) x^{2} b^{4} c^{2} d^{2}+12 \ln \left (x d +c \right ) x^{3} a \,b^{3} d^{4}-12 \ln \left (x d +c \right ) x^{3} b^{4} c \,d^{3}}{3 d^{5} \left (x d +c \right )^{3}}\)	\(302\)

Input:

int((b^4*x^4+4*a*b^3*x^3+6*a^2*b^2*x^2+4*a^3*b*x+a^4)/(d*x+c)^4,x,method=_ 
RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (101) = 202\).

Time = 0.07 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.83 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {3 \, b^{4} d^{4} x^{4} + 9 \, b^{4} c d^{3} x^{3} - 13 \, b^{4} c^{4} + 22 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} - a^{4} d^{4} - 9 \, {\left (b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} + 2 \, a^{2} b^{2} d^{4}\right )} x^{2} - 3 \, {\left (9 \, b^{4} c^{3} d - 18 \, a b^{3} c^{2} d^{2} + 6 \, a^{2} b^{2} c d^{3} + 2 \, a^{3} b d^{4}\right )} x - 12 \, {\left (b^{4} c^{4} - a b^{3} c^{3} d + {\left (b^{4} c d^{3} - a b^{3} d^{4}\right )} x^{3} + 3 \, {\left (b^{4} c^{2} d^{2} - a b^{3} c d^{3}\right )} x^{2} + 3 \, {\left (b^{4} c^{3} d - a b^{3} c^{2} d^{2}\right )} x\right )} \log \left (d x + c\right )}{3 \, {\left (d^{8} x^{3} + 3 \, c d^{7} x^{2} + 3 \, c^{2} d^{6} x + c^{3} d^{5}\right )}} \] Input:

integrate((b^4*x^4+4*a*b^3*x^3+6*a^2*b^2*x^2+4*a^3*b*x+a^4)/(d*x+c)^4,x, a 
lgorithm="fricas")
 

Output:

1/3*(3*b^4*d^4*x^4 + 9*b^4*c*d^3*x^3 - 13*b^4*c^4 + 22*a*b^3*c^3*d - 6*a^2 
*b^2*c^2*d^2 - 2*a^3*b*c*d^3 - a^4*d^4 - 9*(b^4*c^2*d^2 - 4*a*b^3*c*d^3 + 
2*a^2*b^2*d^4)*x^2 - 3*(9*b^4*c^3*d - 18*a*b^3*c^2*d^2 + 6*a^2*b^2*c*d^3 + 
 2*a^3*b*d^4)*x - 12*(b^4*c^4 - a*b^3*c^3*d + (b^4*c*d^3 - a*b^3*d^4)*x^3 
+ 3*(b^4*c^2*d^2 - a*b^3*c*d^3)*x^2 + 3*(b^4*c^3*d - a*b^3*c^2*d^2)*x)*log 
(d*x + c))/(d^8*x^3 + 3*c*d^7*x^2 + 3*c^2*d^6*x + c^3*d^5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (94) = 188\).

Time = 0.88 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.03 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {b^{4} x}{d^{4}} + \frac {4 b^{3} \left (a d - b c\right ) \log {\left (c + d x \right )}}{d^{5}} + \frac {- a^{4} d^{4} - 2 a^{3} b c d^{3} - 6 a^{2} b^{2} c^{2} d^{2} + 22 a b^{3} c^{3} d - 13 b^{4} c^{4} + x^{2} \left (- 18 a^{2} b^{2} d^{4} + 36 a b^{3} c d^{3} - 18 b^{4} c^{2} d^{2}\right ) + x \left (- 6 a^{3} b d^{4} - 18 a^{2} b^{2} c d^{3} + 54 a b^{3} c^{2} d^{2} - 30 b^{4} c^{3} d\right )}{3 c^{3} d^{5} + 9 c^{2} d^{6} x + 9 c d^{7} x^{2} + 3 d^{8} x^{3}} \] Input:

integrate((b**4*x**4+4*a*b**3*x**3+6*a**2*b**2*x**2+4*a**3*b*x+a**4)/(d*x+ 
c)**4,x)
 

Output:

b**4*x/d**4 + 4*b**3*(a*d - b*c)*log(c + d*x)/d**5 + (-a**4*d**4 - 2*a**3* 
b*c*d**3 - 6*a**2*b**2*c**2*d**2 + 22*a*b**3*c**3*d - 13*b**4*c**4 + x**2* 
(-18*a**2*b**2*d**4 + 36*a*b**3*c*d**3 - 18*b**4*c**2*d**2) + x*(-6*a**3*b 
*d**4 - 18*a**2*b**2*c*d**3 + 54*a*b**3*c**2*d**2 - 30*b**4*c**3*d))/(3*c* 
*3*d**5 + 9*c**2*d**6*x + 9*c*d**7*x**2 + 3*d**8*x**3)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.95 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {b^{4} x}{d^{4}} - \frac {13 \, b^{4} c^{4} - 22 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 2 \, a^{3} b c d^{3} + a^{4} d^{4} + 18 \, {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x^{2} + 6 \, {\left (5 \, b^{4} c^{3} d - 9 \, a b^{3} c^{2} d^{2} + 3 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4}\right )} x}{3 \, {\left (d^{8} x^{3} + 3 \, c d^{7} x^{2} + 3 \, c^{2} d^{6} x + c^{3} d^{5}\right )}} - \frac {4 \, {\left (b^{4} c - a b^{3} d\right )} \log \left (d x + c\right )}{d^{5}} \] Input:

integrate((b^4*x^4+4*a*b^3*x^3+6*a^2*b^2*x^2+4*a^3*b*x+a^4)/(d*x+c)^4,x, a 
lgorithm="maxima")
 

Output:

b^4*x/d^4 - 1/3*(13*b^4*c^4 - 22*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 2*a^3*b 
*c*d^3 + a^4*d^4 + 18*(b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 6* 
(5*b^4*c^3*d - 9*a*b^3*c^2*d^2 + 3*a^2*b^2*c*d^3 + a^3*b*d^4)*x)/(d^8*x^3 
+ 3*c*d^7*x^2 + 3*c^2*d^6*x + c^3*d^5) - 4*(b^4*c - a*b^3*d)*log(d*x + c)/ 
d^5
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.72 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {b^{4} x}{d^{4}} - \frac {4 \, {\left (b^{4} c - a b^{3} d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} - \frac {13 \, b^{4} c^{4} - 22 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} + 2 \, a^{3} b c d^{3} + a^{4} d^{4} + 18 \, {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x^{2} + 6 \, {\left (5 \, b^{4} c^{3} d - 9 \, a b^{3} c^{2} d^{2} + 3 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4}\right )} x}{3 \, {\left (d x + c\right )}^{3} d^{5}} \] Input:

integrate((b^4*x^4+4*a*b^3*x^3+6*a^2*b^2*x^2+4*a^3*b*x+a^4)/(d*x+c)^4,x, a 
lgorithm="giac")
 

Output:

b^4*x/d^4 - 4*(b^4*c - a*b^3*d)*log(abs(d*x + c))/d^5 - 1/3*(13*b^4*c^4 - 
22*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 2*a^3*b*c*d^3 + a^4*d^4 + 18*(b^4*c^2 
*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 6*(5*b^4*c^3*d - 9*a*b^3*c^2*d^2 
 + 3*a^2*b^2*c*d^3 + a^3*b*d^4)*x)/((d*x + c)^3*d^5)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.98 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {b^4\,x}{d^4}-\frac {\ln \left (c+d\,x\right )\,\left (4\,b^4\,c-4\,a\,b^3\,d\right )}{d^5}-\frac {\frac {a^4\,d^4+2\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-22\,a\,b^3\,c^3\,d+13\,b^4\,c^4}{3\,d}+x\,\left (2\,a^3\,b\,d^3+6\,a^2\,b^2\,c\,d^2-18\,a\,b^3\,c^2\,d+10\,b^4\,c^3\right )+x^2\,\left (6\,a^2\,b^2\,d^3-12\,a\,b^3\,c\,d^2+6\,b^4\,c^2\,d\right )}{c^3\,d^4+3\,c^2\,d^5\,x+3\,c\,d^6\,x^2+d^7\,x^3} \] Input:

int((a^4 + b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x)/(c + d*x)^4, 
x)
 

Output:

(b^4*x)/d^4 - (log(c + d*x)*(4*b^4*c - 4*a*b^3*d))/d^5 - ((a^4*d^4 + 13*b^ 
4*c^4 + 6*a^2*b^2*c^2*d^2 - 22*a*b^3*c^3*d + 2*a^3*b*c*d^3)/(3*d) + x*(10* 
b^4*c^3 + 2*a^3*b*d^3 + 6*a^2*b^2*c*d^2 - 18*a*b^3*c^2*d) + x^2*(6*b^4*c^2 
*d + 6*a^2*b^2*d^3 - 12*a*b^3*c*d^2))/(c^3*d^4 + d^7*x^3 + 3*c^2*d^5*x + 3 
*c*d^6*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.00 \[ \int \frac {a^4+4 a^3 b x+6 a^2 b^2 x^2+4 a b^3 x^3+b^4 x^4}{(c+d x)^4} \, dx=\frac {12 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{4} d +36 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{3} d^{2} x +36 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{2} d^{3} x^{2}+12 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c \,d^{4} x^{3}-12 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{5}-36 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{4} d x -36 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{3} d^{2} x^{2}-12 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{2} d^{3} x^{3}-a^{4} c \,d^{4}-2 a^{3} b \,c^{2} d^{3}-6 a^{3} b c \,d^{4} x +6 a^{2} b^{2} d^{5} x^{3}+10 a \,b^{3} c^{4} d +18 a \,b^{3} c^{3} d^{2} x -12 a \,b^{3} c \,d^{4} x^{3}-10 b^{4} c^{5}-18 b^{4} c^{4} d x +12 b^{4} c^{2} d^{3} x^{3}+3 b^{4} c \,d^{4} x^{4}}{3 c \,d^{5} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:

int((b^4*x^4+4*a*b^3*x^3+6*a^2*b^2*x^2+4*a^3*b*x+a^4)/(d*x+c)^4,x)
 

Output:

(12*log(c + d*x)*a*b**3*c**4*d + 36*log(c + d*x)*a*b**3*c**3*d**2*x + 36*l 
og(c + d*x)*a*b**3*c**2*d**3*x**2 + 12*log(c + d*x)*a*b**3*c*d**4*x**3 - 1 
2*log(c + d*x)*b**4*c**5 - 36*log(c + d*x)*b**4*c**4*d*x - 36*log(c + d*x) 
*b**4*c**3*d**2*x**2 - 12*log(c + d*x)*b**4*c**2*d**3*x**3 - a**4*c*d**4 - 
 2*a**3*b*c**2*d**3 - 6*a**3*b*c*d**4*x + 6*a**2*b**2*d**5*x**3 + 10*a*b** 
3*c**4*d + 18*a*b**3*c**3*d**2*x - 12*a*b**3*c*d**4*x**3 - 10*b**4*c**5 - 
18*b**4*c**4*d*x + 12*b**4*c**2*d**3*x**3 + 3*b**4*c*d**4*x**4)/(3*c*d**5* 
(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))