\(\int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx\) [98]

Optimal result

Integrand size = 32, antiderivative size = 210 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^3 (b c-a d) (c+d x)^{3/2}}+\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (c^2 C d-A d^3-2 c^3 D\right )\right )}{d^3 (b c-a d)^2 \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x}}{b d^3}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}} \] Output:

2/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^3/(-a*d+b*c)/(d*x+c)^(3/2)+2*(a*d*(-B* 
d^2+2*C*c*d-3*D*c^2)-b*(-A*d^3+C*c^2*d-2*D*c^3))/d^3/(-a*d+b*c)^2/(d*x+c)^ 
(1/2)+2*D*(d*x+c)^(1/2)/b/d^3-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*arctanh(b^(1 
/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(3/2)/(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=\frac {6 a^2 d^2 D (c+d x)^2-2 a b d \left (14 c^3 D+d^3 (A+3 B x)+c^2 (-5 C d+21 d D x)+2 c d^2 \left (B-3 C x+3 D x^2\right )\right )+2 b^2 \left (4 A c d^3+8 c^4 D+3 A d^4 x-2 c^3 d (C-6 D x)-c^2 d^2 (B+3 x (C-D x))\right )}{3 b d^3 (b c-a d)^2 (c+d x)^{3/2}}+\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2} (-b c+a d)^{5/2}} \] Input:

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

Output:

(6*a^2*d^2*D*(c + d*x)^2 - 2*a*b*d*(14*c^3*D + d^3*(A + 3*B*x) + c^2*(-5*C 
*d + 21*d*D*x) + 2*c*d^2*(B - 3*C*x + 3*D*x^2)) + 2*b^2*(4*A*c*d^3 + 8*c^4 
*D + 3*A*d^4*x - 2*c^3*d*(C - 6*D*x) - c^2*d^2*(B + 3*x*(C - D*x))))/(3*b* 
d^3*(b*c - a*d)^2*(c + d*x)^(3/2)) + (2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D) 
)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(b^(3/2)*(-(b*c) + a 
*d)^(5/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2122, 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.

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

\(\Big \downarrow \) 2122

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{5/2}}+\frac {2 \left (a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (-A d^3-2 c^3 D+c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d^3 (c+d x)^{3/2} (b c-a d)}+\frac {2 D \sqrt {c+d x}}{b d^3}\)

Input:

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

Output:

(2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(3*d^3*(b*c - a*d)*(c + d*x)^(3/2) 
) + (2*(a*d*(2*c*C*d - B*d^2 - 3*c^2*D) - b*(c^2*C*d - A*d^3 - 2*c^3*D)))/ 
(d^3*(b*c - a*d)^2*Sqrt[c + d*x]) + (2*D*Sqrt[c + d*x])/(b*d^3) - (2*(A*b^ 
3 - a*(b^2*B - a*b*C + a^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - 
a*d]])/(b^(3/2)*(b*c - a*d)^(5/2))
 

Defintions of rubi rules used

rule 2009

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

rule 2122

Int[((Px_)*((c_.) + (d_.)*(x_))^(n_))/((a_.) + (b_.)*(x_)), x_Symbol] :> In 
t[ExpandIntegrand[1/Sqrt[c + d*x], Px*((c + d*x)^(n + 1/2)/(a + b*x)), x], 
x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0]
 

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 D \sqrt {x d +c}}{b}+\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) d^{3} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}-\frac {2 \left (-A b \,d^{3}+B a \,d^{3}-2 C a c \,d^{2}+C b \,c^{2} d +3 D a \,c^{2} d -2 D b \,c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {x d +c}}}{d^{3}}\)	\(202\)
default	\(\frac {\frac {2 D \sqrt {x d +c}}{b}+\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) d^{3} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}-\frac {2 \left (-A b \,d^{3}+B a \,d^{3}-2 C a c \,d^{2}+C b \,c^{2} d +3 D a \,c^{2} d -2 D b \,c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {x d +c}}}{d^{3}}\)	\(202\)
pseudoelliptic	\(\frac {\frac {2 D \sqrt {x d +c}}{b}+\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right ) d^{3} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}+\frac {2 \left (A b \,d^{3}-B a \,d^{3}+2 C a c \,d^{2}-C b \,c^{2} d -3 D a \,c^{2} d +2 D b \,c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {x d +c}}}{d^{3}}\)	\(202\)

Input:

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

Output:

2/d^3*(1/b*D*(d*x+c)^(1/2)-1/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/(a*d-b*c)/(d* 
x+c)^(3/2)-1/(a*d-b*c)^2*(-A*b*d^3+B*a*d^3-2*C*a*c*d^2+C*b*c^2*d+3*D*a*c^2 
*d-2*D*b*c^3)/(d*x+c)^(1/2)+1/b*(A*b^3-B*a*b^2+C*a^2*b-D*a^3)/(a*d-b*c)^2* 
d^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 607 vs. \(2 (191) = 382\).

Time = 0.12 (sec) , antiderivative size = 1227, normalized size of antiderivative = 5.84 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/3*(3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*d^5*x^2 + 2*(D*a^3 - C*a^2*b 
 + B*a*b^2 - A*b^3)*c*d^4*x + (D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*c^2*d^3) 
*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt 
(d*x + c))/(b*x + a)) - 2*(8*D*b^4*c^5 + A*a^2*b^2*d^5 - 2*(11*D*a*b^3 + C 
*b^4)*c^4*d + (17*D*a^2*b^2 + 7*C*a*b^3 - B*b^4)*c^3*d^2 - (3*D*a^3*b + 5* 
C*a^2*b^2 + B*a*b^3 - 4*A*b^4)*c^2*d^3 + (2*B*a^2*b^2 - 5*A*a*b^3)*c*d^4 + 
 3*(D*b^4*c^3*d^2 - 3*D*a*b^3*c^2*d^3 + 3*D*a^2*b^2*c*d^4 - D*a^3*b*d^5)*x 
^2 + 3*(4*D*b^4*c^4*d - (11*D*a*b^3 + C*b^4)*c^3*d^2 + 3*(3*D*a^2*b^2 + C* 
a*b^3)*c^2*d^3 - (2*D*a^3*b + 2*C*a^2*b^2 + B*a*b^3 - A*b^4)*c*d^4 + (B*a^ 
2*b^2 - A*a*b^3)*d^5)*x)*sqrt(d*x + c))/(b^5*c^5*d^3 - 3*a*b^4*c^4*d^4 + 3 
*a^2*b^3*c^3*d^5 - a^3*b^2*c^2*d^6 + (b^5*c^3*d^5 - 3*a*b^4*c^2*d^6 + 3*a^ 
2*b^3*c*d^7 - a^3*b^2*d^8)*x^2 + 2*(b^5*c^4*d^4 - 3*a*b^4*c^3*d^5 + 3*a^2* 
b^3*c^2*d^6 - a^3*b^2*c*d^7)*x), -2/3*(3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b 
^3)*d^5*x^2 + 2*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*c*d^4*x + (D*a^3 - C*a 
^2*b + B*a*b^2 - A*b^3)*c^2*d^3)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + 
 a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (8*D*b^4*c^5 + A*a^2*b^2*d^5 - 2*(1 
1*D*a*b^3 + C*b^4)*c^4*d + (17*D*a^2*b^2 + 7*C*a*b^3 - B*b^4)*c^3*d^2 - (3 
*D*a^3*b + 5*C*a^2*b^2 + B*a*b^3 - 4*A*b^4)*c^2*d^3 + (2*B*a^2*b^2 - 5*A*a 
*b^3)*c*d^4 + 3*(D*b^4*c^3*d^2 - 3*D*a*b^3*c^2*d^3 + 3*D*a^2*b^2*c*d^4 - D 
*a^3*b*d^5)*x^2 + 3*(4*D*b^4*c^4*d - (11*D*a*b^3 + C*b^4)*c^3*d^2 + 3*(...
 

Sympy [A] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.46 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D \sqrt {c + d x}}{b d^{2}} - \frac {- A b d^{3} + B a d^{3} - 2 C a c d^{2} + C b c^{2} d + 3 D a c^{2} d - 2 D b c^{3}}{d^{2} \sqrt {c + d x} \left (a d - b c\right )^{2}} + \frac {- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}}{3 d^{2} \left (c + d x\right )^{\frac {3}{2}} \left (a d - b c\right )} - \frac {d \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{2} \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )^{2}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\frac {D x^{3}}{3 b} + \frac {x^{2} \left (C b - D a\right )}{2 b^{2}} + \frac {x \left (B b^{2} - C a b + D a^{2}\right )}{b^{3}} - \frac {\left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3}}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)/(d*x+c)**(5/2),x)
 

Output:

Piecewise((2*(D*sqrt(c + d*x)/(b*d**2) - (-A*b*d**3 + B*a*d**3 - 2*C*a*c*d 
**2 + C*b*c**2*d + 3*D*a*c**2*d - 2*D*b*c**3)/(d**2*sqrt(c + d*x)*(a*d - b 
*c)**2) + (-A*d**3 + B*c*d**2 - C*c**2*d + D*c**3)/(3*d**2*(c + d*x)**(3/2 
)*(a*d - b*c)) - d*(-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*atan(sqrt(c + 
d*x)/sqrt((a*d - b*c)/b))/(b**2*sqrt((a*d - b*c)/b)*(a*d - b*c)**2))/d, Ne 
(d, 0)), ((D*x**3/(3*b) + x**2*(C*b - D*a)/(2*b**2) + x*(B*b**2 - C*a*b + 
D*a**2)/b**3 - (-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*Piecewise((x/a, Eq 
(b, 0)), (log(a + b*x)/b, True))/b**3)/c**(5/2), True))
 

Maxima [F(-2)]

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

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (6 \, {\left (d x + c\right )} D b c^{3} - D b c^{4} - 9 \, {\left (d x + c\right )} D a c^{2} d - 3 \, {\left (d x + c\right )} C b c^{2} d + D a c^{3} d + C b c^{3} d + 6 \, {\left (d x + c\right )} C a c d^{2} - C a c^{2} d^{2} - B b c^{2} d^{2} - 3 \, {\left (d x + c\right )} B a d^{3} + 3 \, {\left (d x + c\right )} A b d^{3} + B a c d^{3} + A b c d^{3} - A a d^{4}\right )}}{3 \, {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} + \frac {2 \, \sqrt {d x + c} D}{b d^{3}} \] Input:

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

Output:

-2*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c 
+ a*b*d))/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*sqrt(-b^2*c + a*b*d)) + 2/3 
*(6*(d*x + c)*D*b*c^3 - D*b*c^4 - 9*(d*x + c)*D*a*c^2*d - 3*(d*x + c)*C*b* 
c^2*d + D*a*c^3*d + C*b*c^3*d + 6*(d*x + c)*C*a*c*d^2 - C*a*c^2*d^2 - B*b* 
c^2*d^2 - 3*(d*x + c)*B*a*d^3 + 3*(d*x + c)*A*b*d^3 + B*a*c*d^3 + A*b*c*d^ 
3 - A*a*d^4)/((b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)*(d*x + c)^(3/2)) + 2*s 
qrt(d*x + c)*D/(b*d^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.44 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{5/2}} \, dx=\frac {-2 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} c \,d^{2}-2 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{3} x -\frac {2 a^{2} b^{2} d^{3}}{3}+2 a^{2} b \,c^{2} d^{2}+4 a^{2} b c \,d^{3} x +2 a^{2} b \,d^{4} x^{2}+\frac {4 a \,b^{3} c \,d^{2}}{3}-6 a \,b^{2} c^{3} d -10 a \,b^{2} c^{2} d^{2} x -4 a \,b^{2} c \,d^{3} x^{2}-\frac {2 b^{4} c^{2} d}{3}+4 b^{3} c^{4}+6 b^{3} c^{3} d x +2 b^{3} c^{2} d^{2} x^{2}}{\sqrt {d x +c}\, b^{2} d^{2} \left (a^{2} d^{3} x -2 a b c \,d^{2} x +b^{2} c^{2} d x +a^{2} c \,d^{2}-2 a b \,c^{2} d +b^{2} c^{3}\right )} \] Input:

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

Output:

(2*( - 3*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqr 
t(b)*sqrt(a*d - b*c)))*a**2*c*d**2 - 3*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b* 
c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*d**3*x - a**2*b* 
*2*d**3 + 3*a**2*b*c**2*d**2 + 6*a**2*b*c*d**3*x + 3*a**2*b*d**4*x**2 + 2* 
a*b**3*c*d**2 - 9*a*b**2*c**3*d - 15*a*b**2*c**2*d**2*x - 6*a*b**2*c*d**3* 
x**2 - b**4*c**2*d + 6*b**3*c**4 + 9*b**3*c**3*d*x + 3*b**3*c**2*d**2*x**2 
))/(3*sqrt(c + d*x)*b**2*d**2*(a**2*c*d**2 + a**2*d**3*x - 2*a*b*c**2*d - 
2*a*b*c*d**2*x + b**2*c**3 + b**2*c**2*d*x))