Integrand size = 29, antiderivative size = 174 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=\frac {c^3 \left (c^3+4 a d^2\right )^3 x}{d^6}-\frac {2 c^4 \left (c^3+4 a d^2\right )^2 (c+d x)^3}{d^7}+\frac {3 c^2 \left (c^3+4 a d^2\right ) \left (5 c^3+4 a d^2\right ) (c+d x)^5}{5 d^7}-\frac {4 c^3 \left (5 c^3+12 a d^2\right ) (c+d x)^7}{7 d^7}+\frac {c \left (5 c^3+4 a d^2\right ) (c+d x)^9}{3 d^7}-\frac {6 c^2 (c+d x)^{11}}{11 d^7}+\frac {(c+d x)^{13}}{13 d^7} \] Output:
c^3*(4*a*d^2+c^3)^3*x/d^6-2*c^4*(4*a*d^2+c^3)^2*(d*x+c)^3/d^7+3/5*c^2*(4*a *d^2+c^3)*(4*a*d^2+5*c^3)*(d*x+c)^5/d^7-4/7*c^3*(12*a*d^2+5*c^3)*(d*x+c)^7 /d^7+1/3*c*(4*a*d^2+5*c^3)*(d*x+c)^9/d^7-6/11*c^2*(d*x+c)^11/d^7+1/13*(d*x +c)^13/d^7
Time = 0.02 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.98 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=64 a^3 c^3 x+64 a^2 c^4 x^3+48 a^2 c^3 d x^4+\frac {48}{5} a c^2 \left (4 c^3+a d^2\right ) x^5+64 a c^4 d x^6+\frac {32}{7} c^3 \left (2 c^3+9 a d^2\right ) x^7+12 c^2 d \left (2 c^3+a d^2\right ) x^8+\frac {4}{3} c d^2 \left (20 c^3+a d^2\right ) x^9+16 c^3 d^3 x^{10}+\frac {60}{11} c^2 d^4 x^{11}+c d^5 x^{12}+\frac {d^6 x^{13}}{13} \] Input:
Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^3,x]
Output:
64*a^3*c^3*x + 64*a^2*c^4*x^3 + 48*a^2*c^3*d*x^4 + (48*a*c^2*(4*c^3 + a*d^ 2)*x^5)/5 + 64*a*c^4*d*x^6 + (32*c^3*(2*c^3 + 9*a*d^2)*x^7)/7 + 12*c^2*d*( 2*c^3 + a*d^2)*x^8 + (4*c*d^2*(20*c^3 + a*d^2)*x^9)/3 + 16*c^3*d^3*x^10 + (60*c^2*d^4*x^11)/11 + c*d^5*x^12 + (d^6*x^13)/13
Time = 0.43 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2458, 1403, 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.
| \(\displaystyle \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx\) |
| \(\Big \downarrow \) 2458 |
| \(\displaystyle \int \left (c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4\right )^3d\left (\frac {c}{d}+x\right )\) |
| \(\Big \downarrow \) 1403 |
| \(\displaystyle \int \left (\frac {\left (4 a c d^2+c^4\right )^3}{d^6}-8 c^6 \left (\frac {6 a d^2}{c^3}+\frac {5}{2}\right ) \left (\frac {c}{d}+x\right )^6+\frac {12 c^5 \left (4 a d^2+c^3\right ) \left (\frac {a d^2}{c^3}+\frac {5}{4}\right ) \left (\frac {c}{d}+x\right )^4}{d^2}+12 c^4 d^2 \left (\frac {a d^2}{c^3}+\frac {5}{4}\right ) \left (\frac {c}{d}+x\right )^8-\frac {6 c^4 \left (4 a d^2+c^3\right )^2 \left (\frac {c}{d}+x\right )^2}{d^4}-6 c^2 d^4 \left (\frac {c}{d}+x\right )^{10}+d^6 \left (\frac {c}{d}+x\right )^{12}\right )d\left (\frac {c}{d}+x\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} c d^2 \left (4 a d^2+5 c^3\right ) \left (\frac {c}{d}+x\right )^9-\frac {4}{7} c^3 \left (12 a d^2+5 c^3\right ) \left (\frac {c}{d}+x\right )^7+\frac {c^3 \left (4 a d^2+c^3\right )^3 \left (\frac {c}{d}+x\right )}{d^6}-\frac {2 c^4 \left (4 a d^2+c^3\right )^2 \left (\frac {c}{d}+x\right )^3}{d^4}+\frac {3 c^2 \left (4 a d^2+c^3\right ) \left (4 a d^2+5 c^3\right ) \left (\frac {c}{d}+x\right )^5}{5 d^2}-\frac {6}{11} c^2 d^4 \left (\frac {c}{d}+x\right )^{11}+\frac {1}{13} d^6 \left (\frac {c}{d}+x\right )^{13}\) |
Input:
Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^3,x]
Output:
(c^3*(c^3 + 4*a*d^2)^3*(c/d + x))/d^6 - (2*c^4*(c^3 + 4*a*d^2)^2*(c/d + x) ^3)/d^4 + (3*c^2*(c^3 + 4*a*d^2)*(5*c^3 + 4*a*d^2)*(c/d + x)^5)/(5*d^2) - (4*c^3*(5*c^3 + 12*a*d^2)*(c/d + x)^7)/7 + (c*d^2*(5*c^3 + 4*a*d^2)*(c/d + x)^9)/3 - (6*c^2*d^4*(c/d + x)^11)/11 + (d^6*(c/d + x)^13)/13
Int[((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandInte grand[(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a *c, 0] && IGtQ[p, 0]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Time = 0.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94
| method | result | size |
| norman | \(\frac {d^{6} x^{13}}{13}+c \,d^{5} x^{12}+\frac {60 c^{2} d^{4} x^{11}}{11}+16 c^{3} d^{3} x^{10}+\left (\frac {4}{3} d^{4} c a +\frac {80}{3} c^{4} d^{2}\right ) x^{9}+\left (12 a \,c^{2} d^{3}+24 c^{5} d \right ) x^{8}+\left (\frac {288}{7} a \,c^{3} d^{2}+\frac {64}{7} c^{6}\right ) x^{7}+64 a \,c^{4} d \,x^{6}+\left (\frac {48}{5} a^{2} c^{2} d^{2}+\frac {192}{5} a \,c^{5}\right ) x^{5}+48 a^{2} c^{3} d \,x^{4}+64 a^{2} c^{4} x^{3}+64 a^{3} c^{3} x\) | \(163\) |
| gosper | \(\frac {1}{13} d^{6} x^{13}+c \,d^{5} x^{12}+\frac {60}{11} c^{2} d^{4} x^{11}+16 c^{3} d^{3} x^{10}+\frac {4}{3} x^{9} d^{4} c a +\frac {80}{3} x^{9} c^{4} d^{2}+12 a \,c^{2} d^{3} x^{8}+24 c^{5} d \,x^{8}+\frac {288}{7} x^{7} a \,c^{3} d^{2}+\frac {64}{7} x^{7} c^{6}+64 a \,c^{4} d \,x^{6}+\frac {48}{5} x^{5} a^{2} c^{2} d^{2}+\frac {192}{5} x^{5} a \,c^{5}+48 a^{2} c^{3} d \,x^{4}+64 a^{2} c^{4} x^{3}+64 a^{3} c^{3} x\) | \(167\) |
| risch | \(\frac {1}{13} d^{6} x^{13}+c \,d^{5} x^{12}+\frac {60}{11} c^{2} d^{4} x^{11}+16 c^{3} d^{3} x^{10}+\frac {4}{3} x^{9} d^{4} c a +\frac {80}{3} x^{9} c^{4} d^{2}+12 a \,c^{2} d^{3} x^{8}+24 c^{5} d \,x^{8}+\frac {288}{7} x^{7} a \,c^{3} d^{2}+\frac {64}{7} x^{7} c^{6}+64 a \,c^{4} d \,x^{6}+\frac {48}{5} x^{5} a^{2} c^{2} d^{2}+\frac {192}{5} x^{5} a \,c^{5}+48 a^{2} c^{3} d \,x^{4}+64 a^{2} c^{4} x^{3}+64 a^{3} c^{3} x\) | \(167\) |
| parallelrisch | \(\frac {1}{13} d^{6} x^{13}+c \,d^{5} x^{12}+\frac {60}{11} c^{2} d^{4} x^{11}+16 c^{3} d^{3} x^{10}+\frac {4}{3} x^{9} d^{4} c a +\frac {80}{3} x^{9} c^{4} d^{2}+12 a \,c^{2} d^{3} x^{8}+24 c^{5} d \,x^{8}+\frac {288}{7} x^{7} a \,c^{3} d^{2}+\frac {64}{7} x^{7} c^{6}+64 a \,c^{4} d \,x^{6}+\frac {48}{5} x^{5} a^{2} c^{2} d^{2}+\frac {192}{5} x^{5} a \,c^{5}+48 a^{2} c^{3} d \,x^{4}+64 a^{2} c^{4} x^{3}+64 a^{3} c^{3} x\) | \(167\) |
| orering | \(\frac {x \left (1155 d^{6} x^{12}+15015 c \,d^{5} x^{11}+81900 c^{2} d^{4} x^{10}+240240 c^{3} d^{3} x^{9}+20020 a c \,d^{4} x^{8}+400400 c^{4} d^{2} x^{8}+180180 a \,c^{2} d^{3} x^{7}+360360 c^{5} d \,x^{7}+617760 a \,c^{3} d^{2} x^{6}+137280 c^{6} x^{6}+960960 a \,c^{4} d \,x^{5}+144144 a^{2} c^{2} d^{2} x^{4}+576576 a \,c^{5} x^{4}+720720 a^{2} c^{3} d \,x^{3}+960960 a^{2} c^{4} x^{2}+960960 a^{3} c^{3}\right )}{15015}\) | \(170\) |
| default | \(\frac {d^{6} x^{13}}{13}+c \,d^{5} x^{12}+\frac {60 c^{2} d^{4} x^{11}}{11}+16 c^{3} d^{3} x^{10}+\frac {\left (4 d^{4} c a +224 c^{4} d^{2}+d^{2} \left (8 a c \,d^{2}+16 c^{4}\right )\right ) x^{9}}{9}+\frac {\left (64 a \,c^{2} d^{3}+128 c^{5} d +4 c d \left (8 a c \,d^{2}+16 c^{4}\right )\right ) x^{8}}{8}+\frac {\left (256 a \,c^{3} d^{2}+4 c^{2} \left (8 a c \,d^{2}+16 c^{4}\right )\right ) x^{7}}{7}+64 a \,c^{4} d \,x^{6}+\frac {\left (4 a c \left (8 a c \,d^{2}+16 c^{4}\right )+128 a \,c^{5}+16 a^{2} c^{2} d^{2}\right ) x^{5}}{5}+48 a^{2} c^{3} d \,x^{4}+64 a^{2} c^{4} x^{3}+64 a^{3} c^{3} x\) | \(231\) |
Input:
int((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^3,x,method=_RETURNVERBOSE)
Output:
1/13*d^6*x^13+c*d^5*x^12+60/11*c^2*d^4*x^11+16*c^3*d^3*x^10+(4/3*d^4*c*a+8 0/3*c^4*d^2)*x^9+(12*a*c^2*d^3+24*c^5*d)*x^8+(288/7*a*c^3*d^2+64/7*c^6)*x^ 7+64*a*c^4*d*x^6+(48/5*a^2*c^2*d^2+192/5*a*c^5)*x^5+48*a^2*c^3*d*x^4+64*a^ 2*c^4*x^3+64*a^3*c^3*x
Time = 0.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=\frac {1}{13} \, d^{6} x^{13} + c d^{5} x^{12} + \frac {60}{11} \, c^{2} d^{4} x^{11} + 16 \, c^{3} d^{3} x^{10} + 64 \, a c^{4} d x^{6} + 48 \, a^{2} c^{3} d x^{4} + \frac {4}{3} \, {\left (20 \, c^{4} d^{2} + a c d^{4}\right )} x^{9} + 64 \, a^{2} c^{4} x^{3} + 12 \, {\left (2 \, c^{5} d + a c^{2} d^{3}\right )} x^{8} + \frac {32}{7} \, {\left (2 \, c^{6} + 9 \, a c^{3} d^{2}\right )} x^{7} + 64 \, a^{3} c^{3} x + \frac {48}{5} \, {\left (4 \, a c^{5} + a^{2} c^{2} d^{2}\right )} x^{5} \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^3,x, algorithm="fricas")
Output:
1/13*d^6*x^13 + c*d^5*x^12 + 60/11*c^2*d^4*x^11 + 16*c^3*d^3*x^10 + 64*a*c ^4*d*x^6 + 48*a^2*c^3*d*x^4 + 4/3*(20*c^4*d^2 + a*c*d^4)*x^9 + 64*a^2*c^4* x^3 + 12*(2*c^5*d + a*c^2*d^3)*x^8 + 32/7*(2*c^6 + 9*a*c^3*d^2)*x^7 + 64*a ^3*c^3*x + 48/5*(4*a*c^5 + a^2*c^2*d^2)*x^5
Time = 0.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.03 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=64 a^{3} c^{3} x + 64 a^{2} c^{4} x^{3} + 48 a^{2} c^{3} d x^{4} + 64 a c^{4} d x^{6} + 16 c^{3} d^{3} x^{10} + \frac {60 c^{2} d^{4} x^{11}}{11} + c d^{5} x^{12} + \frac {d^{6} x^{13}}{13} + x^{9} \cdot \left (\frac {4 a c d^{4}}{3} + \frac {80 c^{4} d^{2}}{3}\right ) + x^{8} \cdot \left (12 a c^{2} d^{3} + 24 c^{5} d\right ) + x^{7} \cdot \left (\frac {288 a c^{3} d^{2}}{7} + \frac {64 c^{6}}{7}\right ) + x^{5} \cdot \left (\frac {48 a^{2} c^{2} d^{2}}{5} + \frac {192 a c^{5}}{5}\right ) \] Input:
integrate((d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**3,x)
Output:
64*a**3*c**3*x + 64*a**2*c**4*x**3 + 48*a**2*c**3*d*x**4 + 64*a*c**4*d*x** 6 + 16*c**3*d**3*x**10 + 60*c**2*d**4*x**11/11 + c*d**5*x**12 + d**6*x**13 /13 + x**9*(4*a*c*d**4/3 + 80*c**4*d**2/3) + x**8*(12*a*c**2*d**3 + 24*c** 5*d) + x**7*(288*a*c**3*d**2/7 + 64*c**6/7) + x**5*(48*a**2*c**2*d**2/5 + 192*a*c**5/5)
Time = 0.04 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=\frac {1}{13} \, d^{6} x^{13} + c d^{5} x^{12} + \frac {48}{11} \, c^{2} d^{4} x^{11} + \frac {32}{5} \, c^{3} d^{3} x^{10} + \frac {64}{7} \, c^{6} x^{7} + 64 \, a^{3} c^{3} x + \frac {16}{5} \, {\left (3 \, d^{2} x^{5} + 15 \, c d x^{4} + 20 \, c^{2} x^{3}\right )} a^{2} c^{2} + \frac {8}{3} \, {\left (2 \, d^{2} x^{9} + 9 \, c d x^{8}\right )} c^{4} + \frac {4}{105} \, {\left (35 \, d^{4} x^{9} + 315 \, c d^{3} x^{8} + 720 \, c^{2} d^{2} x^{7} + 1008 \, c^{4} x^{5} + 120 \, {\left (3 \, d^{2} x^{7} + 14 \, c d x^{6}\right )} c^{2}\right )} a c + \frac {4}{165} \, {\left (45 \, d^{4} x^{11} + 396 \, c d^{3} x^{10} + 880 \, c^{2} d^{2} x^{9}\right )} c^{2} \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^3,x, algorithm="maxima")
Output:
1/13*d^6*x^13 + c*d^5*x^12 + 48/11*c^2*d^4*x^11 + 32/5*c^3*d^3*x^10 + 64/7 *c^6*x^7 + 64*a^3*c^3*x + 16/5*(3*d^2*x^5 + 15*c*d*x^4 + 20*c^2*x^3)*a^2*c ^2 + 8/3*(2*d^2*x^9 + 9*c*d*x^8)*c^4 + 4/105*(35*d^4*x^9 + 315*c*d^3*x^8 + 720*c^2*d^2*x^7 + 1008*c^4*x^5 + 120*(3*d^2*x^7 + 14*c*d*x^6)*c^2)*a*c + 4/165*(45*d^4*x^11 + 396*c*d^3*x^10 + 880*c^2*d^2*x^9)*c^2
Time = 0.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.95 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=\frac {1}{13} \, d^{6} x^{13} + c d^{5} x^{12} + \frac {60}{11} \, c^{2} d^{4} x^{11} + 16 \, c^{3} d^{3} x^{10} + \frac {80}{3} \, c^{4} d^{2} x^{9} + \frac {4}{3} \, a c d^{4} x^{9} + 24 \, c^{5} d x^{8} + 12 \, a c^{2} d^{3} x^{8} + \frac {64}{7} \, c^{6} x^{7} + \frac {288}{7} \, a c^{3} d^{2} x^{7} + 64 \, a c^{4} d x^{6} + \frac {192}{5} \, a c^{5} x^{5} + \frac {48}{5} \, a^{2} c^{2} d^{2} x^{5} + 48 \, a^{2} c^{3} d x^{4} + 64 \, a^{2} c^{4} x^{3} + 64 \, a^{3} c^{3} x \] Input:
integrate((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^3,x, algorithm="giac")
Output:
1/13*d^6*x^13 + c*d^5*x^12 + 60/11*c^2*d^4*x^11 + 16*c^3*d^3*x^10 + 80/3*c ^4*d^2*x^9 + 4/3*a*c*d^4*x^9 + 24*c^5*d*x^8 + 12*a*c^2*d^3*x^8 + 64/7*c^6* x^7 + 288/7*a*c^3*d^2*x^7 + 64*a*c^4*d*x^6 + 192/5*a*c^5*x^5 + 48/5*a^2*c^ 2*d^2*x^5 + 48*a^2*c^3*d*x^4 + 64*a^2*c^4*x^3 + 64*a^3*c^3*x
Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.92 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=x^8\,\left (24\,c^5\,d+12\,a\,c^2\,d^3\right )+x^9\,\left (\frac {80\,c^4\,d^2}{3}+\frac {4\,a\,c\,d^4}{3}\right )+\frac {d^6\,x^{13}}{13}+x^7\,\left (\frac {64\,c^6}{7}+\frac {288\,a\,c^3\,d^2}{7}\right )+64\,a^3\,c^3\,x+c\,d^5\,x^{12}+64\,a^2\,c^4\,x^3+16\,c^3\,d^3\,x^{10}+\frac {60\,c^2\,d^4\,x^{11}}{11}+48\,a^2\,c^3\,d\,x^4+\frac {48\,a\,c^2\,x^5\,\left (4\,c^3+a\,d^2\right )}{5}+64\,a\,c^4\,d\,x^6 \] Input:
int((4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^3,x)
Output:
x^8*(24*c^5*d + 12*a*c^2*d^3) + x^9*((80*c^4*d^2)/3 + (4*a*c*d^4)/3) + (d^ 6*x^13)/13 + x^7*((64*c^6)/7 + (288*a*c^3*d^2)/7) + 64*a^3*c^3*x + c*d^5*x ^12 + 64*a^2*c^4*x^3 + 16*c^3*d^3*x^10 + (60*c^2*d^4*x^11)/11 + 48*a^2*c^3 *d*x^4 + (48*a*c^2*x^5*(a*d^2 + 4*c^3))/5 + 64*a*c^4*d*x^6
Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.97 \[ \int \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )^3 \, dx=\frac {x \left (1155 d^{6} x^{12}+15015 c \,d^{5} x^{11}+81900 c^{2} d^{4} x^{10}+240240 c^{3} d^{3} x^{9}+20020 a c \,d^{4} x^{8}+400400 c^{4} d^{2} x^{8}+180180 a \,c^{2} d^{3} x^{7}+360360 c^{5} d \,x^{7}+617760 a \,c^{3} d^{2} x^{6}+137280 c^{6} x^{6}+960960 a \,c^{4} d \,x^{5}+144144 a^{2} c^{2} d^{2} x^{4}+576576 a \,c^{5} x^{4}+720720 a^{2} c^{3} d \,x^{3}+960960 a^{2} c^{4} x^{2}+960960 a^{3} c^{3}\right )}{15015} \] Input:
int((d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^3,x)
Output:
(x*(960960*a**3*c**3 + 960960*a**2*c**4*x**2 + 720720*a**2*c**3*d*x**3 + 1 44144*a**2*c**2*d**2*x**4 + 576576*a*c**5*x**4 + 960960*a*c**4*d*x**5 + 61 7760*a*c**3*d**2*x**6 + 180180*a*c**2*d**3*x**7 + 20020*a*c*d**4*x**8 + 13 7280*c**6*x**6 + 360360*c**5*d*x**7 + 400400*c**4*d**2*x**8 + 240240*c**3* d**3*x**9 + 81900*c**2*d**4*x**10 + 15015*c*d**5*x**11 + 1155*d**6*x**12)) /15015