\(\int x^{3/2} (A+B x) (a^2+2 a b x+b^2 x^2)^3 \, dx\) [378]

Optimal result

Integrand size = 29, antiderivative size = 159 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{5} a^6 A x^{5/2}+\frac {2}{7} a^5 (6 A b+a B) x^{7/2}+\frac {2}{3} a^4 b (5 A b+2 a B) x^{9/2}+\frac {10}{11} a^3 b^2 (4 A b+3 a B) x^{11/2}+\frac {10}{13} a^2 b^3 (3 A b+4 a B) x^{13/2}+\frac {2}{5} a b^4 (2 A b+5 a B) x^{15/2}+\frac {2}{17} b^5 (A b+6 a B) x^{17/2}+\frac {2}{19} b^6 B x^{19/2} \] Output:

2/5*a^6*A*x^(5/2)+2/7*a^5*(6*A*b+B*a)*x^(7/2)+2/3*a^4*b*(5*A*b+2*B*a)*x^(9 
/2)+10/11*a^3*b^2*(4*A*b+3*B*a)*x^(11/2)+10/13*a^2*b^3*(3*A*b+4*B*a)*x^(13 
/2)+2/5*a*b^4*(2*A*b+5*B*a)*x^(15/2)+2/17*b^5*(A*b+6*B*a)*x^(17/2)+2/19*b^ 
6*B*x^(19/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.96 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{35} a^6 x^{5/2} (7 A+5 B x)+\frac {4}{21} a^5 b x^{7/2} (9 A+7 B x)+\frac {10}{33} a^4 b^2 x^{9/2} (11 A+9 B x)+\frac {40}{143} a^3 b^3 x^{11/2} (13 A+11 B x)+\frac {2}{13} a^2 b^4 x^{13/2} (15 A+13 B x)+\frac {4}{85} a b^5 x^{15/2} (17 A+15 B x)+\frac {2}{323} b^6 x^{17/2} (19 A+17 B x) \] Input:

Integrate[x^(3/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

(2*a^6*x^(5/2)*(7*A + 5*B*x))/35 + (4*a^5*b*x^(7/2)*(9*A + 7*B*x))/21 + (1 
0*a^4*b^2*x^(9/2)*(11*A + 9*B*x))/33 + (40*a^3*b^3*x^(11/2)*(13*A + 11*B*x 
))/143 + (2*a^2*b^4*x^(13/2)*(15*A + 13*B*x))/13 + (4*a*b^5*x^(15/2)*(17*A 
 + 15*B*x))/85 + (2*b^6*x^(17/2)*(19*A + 17*B*x))/323
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 85, 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.

\(\displaystyle \int x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x) \, dx\)

\(\Big \downarrow \) 1184

\(\displaystyle \frac {\int b^6 x^{3/2} (a+b x)^6 (A+B x)dx}{b^6}\)

\(\Big \downarrow \) 27

\(\displaystyle \int x^{3/2} (a+b x)^6 (A+B x)dx\)

\(\Big \downarrow \) 85

\(\displaystyle \int \left (a^6 A x^{3/2}+a^5 x^{5/2} (a B+6 A b)+3 a^4 b x^{7/2} (2 a B+5 A b)+5 a^3 b^2 x^{9/2} (3 a B+4 A b)+5 a^2 b^3 x^{11/2} (4 a B+3 A b)+b^5 x^{15/2} (6 a B+A b)+3 a b^4 x^{13/2} (5 a B+2 A b)+b^6 B x^{17/2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} a^6 A x^{5/2}+\frac {2}{7} a^5 x^{7/2} (a B+6 A b)+\frac {2}{3} a^4 b x^{9/2} (2 a B+5 A b)+\frac {10}{11} a^3 b^2 x^{11/2} (3 a B+4 A b)+\frac {10}{13} a^2 b^3 x^{13/2} (4 a B+3 A b)+\frac {2}{17} b^5 x^{17/2} (6 a B+A b)+\frac {2}{5} a b^4 x^{15/2} (5 a B+2 A b)+\frac {2}{19} b^6 B x^{19/2}\)

Input:

Int[x^(3/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

(2*a^6*A*x^(5/2))/5 + (2*a^5*(6*A*b + a*B)*x^(7/2))/7 + (2*a^4*b*(5*A*b + 
2*a*B)*x^(9/2))/3 + (10*a^3*b^2*(4*A*b + 3*a*B)*x^(11/2))/11 + (10*a^2*b^3 
*(3*A*b + 4*a*B)*x^(13/2))/13 + (2*a*b^4*(2*A*b + 5*a*B)*x^(15/2))/5 + (2* 
b^5*(A*b + 6*a*B)*x^(17/2))/17 + (2*b^6*B*x^(19/2))/19
 

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 85

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* 
f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n 
 + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 
1])
 

rule 1184

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

rule 2009

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

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.93

method	result	size
gosper	\(\frac {2 x^{\frac {5}{2}} \left (255255 b^{6} B \,x^{7}+285285 A \,b^{6} x^{6}+1711710 B a \,b^{5} x^{6}+1939938 A a \,b^{5} x^{5}+4849845 B \,a^{2} b^{4} x^{5}+5595975 A \,a^{2} b^{4} x^{4}+7461300 B \,a^{3} b^{3} x^{4}+8817900 A \,a^{3} b^{3} x^{3}+6613425 B \,a^{4} b^{2} x^{3}+8083075 A \,a^{4} b^{2} x^{2}+3233230 B \,a^{5} b \,x^{2}+4157010 A \,a^{5} b x +692835 B \,a^{6} x +969969 A \,a^{6}\right )}{4849845}\)	\(148\)
derivativedivides	\(\frac {2 b^{6} B \,x^{\frac {19}{2}}}{19}+\frac {2 \left (A \,b^{6}+6 B a \,b^{5}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (6 A a \,b^{5}+15 B \,a^{2} b^{4}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (6 A \,a^{5} b +B \,a^{6}\right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{6} A \,x^{\frac {5}{2}}}{5}\)	\(148\)
default	\(\frac {2 b^{6} B \,x^{\frac {19}{2}}}{19}+\frac {2 \left (A \,b^{6}+6 B a \,b^{5}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (6 A a \,b^{5}+15 B \,a^{2} b^{4}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (6 A \,a^{5} b +B \,a^{6}\right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{6} A \,x^{\frac {5}{2}}}{5}\)	\(148\)
trager	\(\frac {2 x^{\frac {5}{2}} \left (255255 b^{6} B \,x^{7}+285285 A \,b^{6} x^{6}+1711710 B a \,b^{5} x^{6}+1939938 A a \,b^{5} x^{5}+4849845 B \,a^{2} b^{4} x^{5}+5595975 A \,a^{2} b^{4} x^{4}+7461300 B \,a^{3} b^{3} x^{4}+8817900 A \,a^{3} b^{3} x^{3}+6613425 B \,a^{4} b^{2} x^{3}+8083075 A \,a^{4} b^{2} x^{2}+3233230 B \,a^{5} b \,x^{2}+4157010 A \,a^{5} b x +692835 B \,a^{6} x +969969 A \,a^{6}\right )}{4849845}\)	\(148\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (255255 b^{6} B \,x^{7}+285285 A \,b^{6} x^{6}+1711710 B a \,b^{5} x^{6}+1939938 A a \,b^{5} x^{5}+4849845 B \,a^{2} b^{4} x^{5}+5595975 A \,a^{2} b^{4} x^{4}+7461300 B \,a^{3} b^{3} x^{4}+8817900 A \,a^{3} b^{3} x^{3}+6613425 B \,a^{4} b^{2} x^{3}+8083075 A \,a^{4} b^{2} x^{2}+3233230 B \,a^{5} b \,x^{2}+4157010 A \,a^{5} b x +692835 B \,a^{6} x +969969 A \,a^{6}\right )}{4849845}\)	\(148\)
orering	\(\frac {2 \left (255255 b^{6} B \,x^{7}+285285 A \,b^{6} x^{6}+1711710 B a \,b^{5} x^{6}+1939938 A a \,b^{5} x^{5}+4849845 B \,a^{2} b^{4} x^{5}+5595975 A \,a^{2} b^{4} x^{4}+7461300 B \,a^{3} b^{3} x^{4}+8817900 A \,a^{3} b^{3} x^{3}+6613425 B \,a^{4} b^{2} x^{3}+8083075 A \,a^{4} b^{2} x^{2}+3233230 B \,a^{5} b \,x^{2}+4157010 A \,a^{5} b x +692835 B \,a^{6} x +969969 A \,a^{6}\right ) x^{\frac {5}{2}} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{3}}{4849845 \left (b x +a \right )^{6}}\)	\(173\)

Input:

int(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2/4849845*x^(5/2)*(255255*B*b^6*x^7+285285*A*b^6*x^6+1711710*B*a*b^5*x^6+1 
939938*A*a*b^5*x^5+4849845*B*a^2*b^4*x^5+5595975*A*a^2*b^4*x^4+7461300*B*a 
^3*b^3*x^4+8817900*A*a^3*b^3*x^3+6613425*B*a^4*b^2*x^3+8083075*A*a^4*b^2*x 
^2+3233230*B*a^5*b*x^2+4157010*A*a^5*b*x+692835*B*a^6*x+969969*A*a^6)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.96 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{4849845} \, {\left (255255 \, B b^{6} x^{9} + 969969 \, A a^{6} x^{2} + 285285 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{8} + 969969 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{7} + 1865325 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{6} + 2204475 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{5} + 1616615 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{4} + 692835 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{3}\right )} \sqrt {x} \] Input:

integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
 

Output:

2/4849845*(255255*B*b^6*x^9 + 969969*A*a^6*x^2 + 285285*(6*B*a*b^5 + A*b^6 
)*x^8 + 969969*(5*B*a^2*b^4 + 2*A*a*b^5)*x^7 + 1865325*(4*B*a^3*b^3 + 3*A* 
a^2*b^4)*x^6 + 2204475*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^5 + 1616615*(2*B*a^5* 
b + 5*A*a^4*b^2)*x^4 + 692835*(B*a^6 + 6*A*a^5*b)*x^3)*sqrt(x)
 

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.35 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2 A a^{6} x^{\frac {5}{2}}}{5} + \frac {12 A a^{5} b x^{\frac {7}{2}}}{7} + \frac {10 A a^{4} b^{2} x^{\frac {9}{2}}}{3} + \frac {40 A a^{3} b^{3} x^{\frac {11}{2}}}{11} + \frac {30 A a^{2} b^{4} x^{\frac {13}{2}}}{13} + \frac {4 A a b^{5} x^{\frac {15}{2}}}{5} + \frac {2 A b^{6} x^{\frac {17}{2}}}{17} + \frac {2 B a^{6} x^{\frac {7}{2}}}{7} + \frac {4 B a^{5} b x^{\frac {9}{2}}}{3} + \frac {30 B a^{4} b^{2} x^{\frac {11}{2}}}{11} + \frac {40 B a^{3} b^{3} x^{\frac {13}{2}}}{13} + 2 B a^{2} b^{4} x^{\frac {15}{2}} + \frac {12 B a b^{5} x^{\frac {17}{2}}}{17} + \frac {2 B b^{6} x^{\frac {19}{2}}}{19} \] Input:

integrate(x**(3/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3,x)
 

Output:

2*A*a**6*x**(5/2)/5 + 12*A*a**5*b*x**(7/2)/7 + 10*A*a**4*b**2*x**(9/2)/3 + 
 40*A*a**3*b**3*x**(11/2)/11 + 30*A*a**2*b**4*x**(13/2)/13 + 4*A*a*b**5*x* 
*(15/2)/5 + 2*A*b**6*x**(17/2)/17 + 2*B*a**6*x**(7/2)/7 + 4*B*a**5*b*x**(9 
/2)/3 + 30*B*a**4*b**2*x**(11/2)/11 + 40*B*a**3*b**3*x**(13/2)/13 + 2*B*a* 
*2*b**4*x**(15/2) + 12*B*a*b**5*x**(17/2)/17 + 2*B*b**6*x**(19/2)/19
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{19} \, B b^{6} x^{\frac {19}{2}} + \frac {2}{5} \, A a^{6} x^{\frac {5}{2}} + \frac {2}{17} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {17}{2}} + \frac {2}{5} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {15}{2}} + \frac {10}{13} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {13}{2}} + \frac {10}{11} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac {11}{2}} + \frac {2}{3} \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{\frac {7}{2}} \] Input:

integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
 

Output:

2/19*B*b^6*x^(19/2) + 2/5*A*a^6*x^(5/2) + 2/17*(6*B*a*b^5 + A*b^6)*x^(17/2 
) + 2/5*(5*B*a^2*b^4 + 2*A*a*b^5)*x^(15/2) + 10/13*(4*B*a^3*b^3 + 3*A*a^2* 
b^4)*x^(13/2) + 10/11*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^(11/2) + 2/3*(2*B*a^5* 
b + 5*A*a^4*b^2)*x^(9/2) + 2/7*(B*a^6 + 6*A*a^5*b)*x^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.94 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{19} \, B b^{6} x^{\frac {19}{2}} + \frac {12}{17} \, B a b^{5} x^{\frac {17}{2}} + \frac {2}{17} \, A b^{6} x^{\frac {17}{2}} + 2 \, B a^{2} b^{4} x^{\frac {15}{2}} + \frac {4}{5} \, A a b^{5} x^{\frac {15}{2}} + \frac {40}{13} \, B a^{3} b^{3} x^{\frac {13}{2}} + \frac {30}{13} \, A a^{2} b^{4} x^{\frac {13}{2}} + \frac {30}{11} \, B a^{4} b^{2} x^{\frac {11}{2}} + \frac {40}{11} \, A a^{3} b^{3} x^{\frac {11}{2}} + \frac {4}{3} \, B a^{5} b x^{\frac {9}{2}} + \frac {10}{3} \, A a^{4} b^{2} x^{\frac {9}{2}} + \frac {2}{7} \, B a^{6} x^{\frac {7}{2}} + \frac {12}{7} \, A a^{5} b x^{\frac {7}{2}} + \frac {2}{5} \, A a^{6} x^{\frac {5}{2}} \] Input:

integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
 

Output:

2/19*B*b^6*x^(19/2) + 12/17*B*a*b^5*x^(17/2) + 2/17*A*b^6*x^(17/2) + 2*B*a 
^2*b^4*x^(15/2) + 4/5*A*a*b^5*x^(15/2) + 40/13*B*a^3*b^3*x^(13/2) + 30/13* 
A*a^2*b^4*x^(13/2) + 30/11*B*a^4*b^2*x^(11/2) + 40/11*A*a^3*b^3*x^(11/2) + 
 4/3*B*a^5*b*x^(9/2) + 10/3*A*a^4*b^2*x^(9/2) + 2/7*B*a^6*x^(7/2) + 12/7*A 
*a^5*b*x^(7/2) + 2/5*A*a^6*x^(5/2)
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.82 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=x^{7/2}\,\left (\frac {2\,B\,a^6}{7}+\frac {12\,A\,b\,a^5}{7}\right )+x^{17/2}\,\left (\frac {2\,A\,b^6}{17}+\frac {12\,B\,a\,b^5}{17}\right )+\frac {2\,A\,a^6\,x^{5/2}}{5}+\frac {2\,B\,b^6\,x^{19/2}}{19}+\frac {10\,a^3\,b^2\,x^{11/2}\,\left (4\,A\,b+3\,B\,a\right )}{11}+\frac {10\,a^2\,b^3\,x^{13/2}\,\left (3\,A\,b+4\,B\,a\right )}{13}+\frac {2\,a^4\,b\,x^{9/2}\,\left (5\,A\,b+2\,B\,a\right )}{3}+\frac {2\,a\,b^4\,x^{15/2}\,\left (2\,A\,b+5\,B\,a\right )}{5} \] Input:

int(x^(3/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
 

Output:

x^(7/2)*((2*B*a^6)/7 + (12*A*a^5*b)/7) + x^(17/2)*((2*A*b^6)/17 + (12*B*a* 
b^5)/17) + (2*A*a^6*x^(5/2))/5 + (2*B*b^6*x^(19/2))/19 + (10*a^3*b^2*x^(11 
/2)*(4*A*b + 3*B*a))/11 + (10*a^2*b^3*x^(13/2)*(3*A*b + 4*B*a))/13 + (2*a^ 
4*b*x^(9/2)*(5*A*b + 2*B*a))/3 + (2*a*b^4*x^(15/2)*(2*A*b + 5*B*a))/5
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.51 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2 \sqrt {x}\, x^{2} \left (36465 b^{7} x^{7}+285285 a \,b^{6} x^{6}+969969 a^{2} b^{5} x^{5}+1865325 a^{3} b^{4} x^{4}+2204475 a^{4} b^{3} x^{3}+1616615 a^{5} b^{2} x^{2}+692835 a^{6} b x +138567 a^{7}\right )}{692835} \] Input:

int(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x)
 

Output:

(2*sqrt(x)*x**2*(138567*a**7 + 692835*a**6*b*x + 1616615*a**5*b**2*x**2 + 
2204475*a**4*b**3*x**3 + 1865325*a**3*b**4*x**4 + 969969*a**2*b**5*x**5 + 
285285*a*b**6*x**6 + 36465*b**7*x**7))/692835