\(\int \frac {x^2}{(a+b x^2) (c+d x^2)^2} \, dx\) [638]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {x}{2 (b c-a d) \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{(b c-a d)^2}+\frac {(b c+a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 \sqrt {c} \sqrt {d} (b c-a d)^2} \] Output:

1/2*x/(-a*d+b*c)/(d*x^2+c)-a^(1/2)*b^(1/2)*arctan(b^(1/2)*x/a^(1/2))/(-a*d 
+b*c)^2+1/2*(a*d+b*c)*arctan(d^(1/2)*x/c^(1/2))/c^(1/2)/d^(1/2)/(-a*d+b*c) 
^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {\frac {(b c-a d) x}{c+d x^2}-2 \sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\frac {(b c+a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}}}{2 (b c-a d)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {373, 397, 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.

\(\displaystyle \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx\)

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

x/(2*(b*c - a*d)*(c + d*x^2)) - ((2*Sqrt[a]*Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqr 
t[a]])/(b*c - a*d) - ((b*c + a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*Sq 
rt[d]*(b*c - a*d)))/(2*(b*c - a*d))
 

Defintions of rubi rules used

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]
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {a b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{2} \sqrt {a b}}+\frac {\frac {\left (-\frac {a d}{2}+\frac {b c}{2}\right ) x}{x^{2} d +c}+\frac {\left (a d +b c \right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{2 \sqrt {c d}}}{\left (a d -b c \right )^{2}}\)	\(85\)
risch	\(-\frac {x}{2 \left (a d -b c \right ) \left (x^{2} d +c \right )}-\frac {\ln \left (c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a d}{4 \sqrt {-c d}\, \left (a d -b c \right )^{2}}-\frac {\ln \left (c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) b c}{4 \sqrt {-c d}\, \left (a d -b c \right )^{2}}+\frac {\ln \left (-c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a d}{4 \sqrt {-c d}\, \left (a d -b c \right )^{2}}+\frac {\ln \left (-c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) b c}{4 \sqrt {-c d}\, \left (a d -b c \right )^{2}}+\frac {\sqrt {-a b}\, \ln \left (\left (-4 \left (-a b \right )^{\frac {3}{2}} a \,d^{2}-4 \left (-a b \right )^{\frac {3}{2}} b c d -5 a^{2} \sqrt {-a b}\, d^{2} b -2 \sqrt {-a b}\, a \,b^{2} c d -b^{3} c^{2} \sqrt {-a b}\right ) x -a^{3} b \,d^{2}+2 a^{2} b^{2} c d -a \,b^{3} c^{2}\right )}{2 \left (a d -b c \right )^{2}}-\frac {\sqrt {-a b}\, \ln \left (\left (4 \left (-a b \right )^{\frac {3}{2}} a \,d^{2}+4 \left (-a b \right )^{\frac {3}{2}} b c d +5 a^{2} \sqrt {-a b}\, d^{2} b +2 \sqrt {-a b}\, a \,b^{2} c d +b^{3} c^{2} \sqrt {-a b}\right ) x -a^{3} b \,d^{2}+2 a^{2} b^{2} c d -a \,b^{3} c^{2}\right )}{2 \left (a d -b c \right )^{2}}\)	\(403\)

Input:

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

Output:

-a*b/(a*d-b*c)^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+1/(a*d-b*c)^2*((-1/2* 
a*d+1/2*b*c)*x/(d*x^2+c)+1/2*(a*d+b*c)/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2)) 
)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 705, normalized size of antiderivative = 6.78 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\left [\frac {2 \, {\left (c d^{2} x^{2} + c^{2} d\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - {\left (b c^{2} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) + 2 \, {\left (b c^{2} d - a c d^{2}\right )} x}{4 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}, \frac {{\left (b c^{2} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) + {\left (c d^{2} x^{2} + c^{2} d\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + {\left (b c^{2} d - a c d^{2}\right )} x}{2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}, -\frac {4 \, {\left (c d^{2} x^{2} + c^{2} d\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (b c^{2} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) - 2 \, {\left (b c^{2} d - a c d^{2}\right )} x}{4 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}, -\frac {2 \, {\left (c d^{2} x^{2} + c^{2} d\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (b c^{2} + a c d + {\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) - {\left (b c^{2} d - a c d^{2}\right )} x}{2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/4*(2*(c*d^2*x^2 + c^2*d)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b 
*x^2 + a)) - (b*c^2 + a*c*d + (b*c*d + a*d^2)*x^2)*sqrt(-c*d)*log((d*x^2 - 
 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) + 2*(b*c^2*d - a*c*d^2)*x)/(b^2*c^4*d - 
2*a*b*c^3*d^2 + a^2*c^2*d^3 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^ 
2), 1/2*((b*c^2 + a*c*d + (b*c*d + a*d^2)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)* 
x/c) + (c*d^2*x^2 + c^2*d)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b* 
x^2 + a)) + (b*c^2*d - a*c*d^2)*x)/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^ 
3 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^2), -1/4*(4*(c*d^2*x^2 + c 
^2*d)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (b*c^2 + a*c*d + (b*c*d + a*d^2)*x 
^2)*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) - 2*(b*c^2*d 
- a*c*d^2)*x)/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3 + (b^2*c^3*d^2 - 2* 
a*b*c^2*d^3 + a^2*c*d^4)*x^2), -1/2*(2*(c*d^2*x^2 + c^2*d)*sqrt(a*b)*arcta 
n(sqrt(a*b)*x/a) - (b*c^2 + a*c*d + (b*c*d + a*d^2)*x^2)*sqrt(c*d)*arctan( 
sqrt(c*d)*x/c) - (b*c^2*d - a*c*d^2)*x)/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c 
^2*d^3 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {a b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b}} + \frac {{\left (b c + a d\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c d}} + \frac {x}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}} \] Input:

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

Output:

-a*b*arctan(b*x/sqrt(a*b))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b)) + 1 
/2*(b*c + a*d)*arctan(d*x/sqrt(c*d))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt 
(c*d)) + 1/2*x/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=-\frac {a b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b}} + \frac {{\left (b c + a d\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c d}} + \frac {x}{2 \, {\left (d x^{2} + c\right )} {\left (b c - a d\right )}} \] Input:

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

Output:

-a*b*arctan(b*x/sqrt(a*b))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b)) + 1 
/2*(b*c + a*d)*arctan(d*x/sqrt(c*d))/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt 
(c*d)) + 1/2*x/((d*x^2 + c)*(b*c - a*d))
 

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 3154, normalized size of antiderivative = 30.33 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(-(((-a*b)^(1/2)*(((-a*b)^(1/2)*((2*a^5*b^2*d^6 + 2*a*b^6*c^4*d^2 - 8 
*a^4*b^3*c*d^5 - 8*a^2*b^5*c^3*d^3 + 12*a^3*b^4*c^2*d^4)/(2*(a^3*d^3 - b^3 
*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)) - (x*(-a*b)^(1/2)*(16*a^5*b^2*d^7 + 
 16*b^7*c^5*d^2 - 48*a*b^6*c^4*d^3 - 48*a^4*b^3*c*d^6 + 32*a^2*b^5*c^3*d^4 
 + 32*a^3*b^4*c^2*d^5))/(8*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2)))/(2*(a^2*d^ 
2 + b^2*c^2 - 2*a*b*c*d)) - (x*(b^5*c^2*d + 5*a^2*b^3*d^3 + 2*a*b^4*c*d^2) 
)/(4*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*1i)/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d) 
 - ((-a*b)^(1/2)*(((-a*b)^(1/2)*((2*a^5*b^2*d^6 + 2*a*b^6*c^4*d^2 - 8*a^4* 
b^3*c*d^5 - 8*a^2*b^5*c^3*d^3 + 12*a^3*b^4*c^2*d^4)/(2*(a^3*d^3 - b^3*c^3 
+ 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)) + (x*(-a*b)^(1/2)*(16*a^5*b^2*d^7 + 16*b 
^7*c^5*d^2 - 48*a*b^6*c^4*d^3 - 48*a^4*b^3*c*d^6 + 32*a^2*b^5*c^3*d^4 + 32 
*a^3*b^4*c^2*d^5))/(8*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2)))/(2*(a^2*d^2 + b 
^2*c^2 - 2*a*b*c*d)) + (x*(b^5*c^2*d + 5*a^2*b^3*d^3 + 2*a*b^4*c*d^2))/(4* 
(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*1i)/(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/((( 
a^2*b^3*d^2)/2 + (a*b^4*c*d)/2)/(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2 
*b*c*d^2) + ((-a*b)^(1/2)*(((-a*b)^(1/2)*((2*a^5*b^2*d^6 + 2*a*b^6*c^4*d^2 
 - 8*a^4*b^3*c*d^5 - 8*a^2*b^5*c^3*d^3 + 12*a^3*b^4*c^2*d^4)/(2*(a^3*d^3 - 
 b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)) - (x*(-a*b)^(1/2)*(16*a^5*b^2*d 
^7 + 16*b^7*c^5*d^2 - 48*a*b^6*c^4*d^3 - 48*a^4*b^3*c*d^6 + 32*a^2*b^5*c^3 
*d^4 + 32*a^3*b^4*c^2*d^5))/(8*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)^2)))/(2*...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.09 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) c^{2} d -2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) c \,d^{2} x^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a c d +\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a \,d^{2} x^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b \,c^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b c d \,x^{2}-a c \,d^{2} x +b \,c^{2} d x}{2 c d \left (a^{2} d^{3} x^{2}-2 a b c \,d^{2} x^{2}+b^{2} c^{2} d \,x^{2}+a^{2} c \,d^{2}-2 a b \,c^{2} d +b^{2} c^{3}\right )} \] Input:

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

Output:

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