Integrand size = 22, antiderivative size = 205 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 a^4 (b c-a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^6 x^5}-\frac {2 a^3 (4 b c-5 a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^6 x^7}+\frac {4 a^2 (3 b c-5 a d) \left (a x^2+b x^3\right )^{9/2}}{9 b^6 x^9}-\frac {4 a (2 b c-5 a d) \left (a x^2+b x^3\right )^{11/2}}{11 b^6 x^{11}}+\frac {2 (b c-5 a d) \left (a x^2+b x^3\right )^{13/2}}{13 b^6 x^{13}}+\frac {2 d \left (a x^2+b x^3\right )^{15/2}}{15 b^6 x^{15}} \] Output:
2/5*a^4*(-a*d+b*c)*(b*x^3+a*x^2)^(5/2)/b^6/x^5-2/7*a^3*(-5*a*d+4*b*c)*(b*x ^3+a*x^2)^(7/2)/b^6/x^7+4/9*a^2*(-5*a*d+3*b*c)*(b*x^3+a*x^2)^(9/2)/b^6/x^9 -4/11*a*(-5*a*d+2*b*c)*(b*x^3+a*x^2)^(11/2)/b^6/x^11+2/13*(-5*a*d+b*c)*(b* x^3+a*x^2)^(13/2)/b^6/x^13+2/15*d*(b*x^3+a*x^2)^(15/2)/b^6/x^15
Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.56 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 x (a+b x)^3 \left (-256 a^5 d+1680 a^2 b^3 x^2 (c+d x)+128 a^4 b (3 c+5 d x)-160 a^3 b^2 x (6 c+7 d x)-210 a b^4 x^3 (12 c+11 d x)+231 b^5 x^4 (15 c+13 d x)\right )}{45045 b^6 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[x*(c + d*x)*(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*x*(a + b*x)^3*(-256*a^5*d + 1680*a^2*b^3*x^2*(c + d*x) + 128*a^4*b*(3*c + 5*d*x) - 160*a^3*b^2*x*(6*c + 7*d*x) - 210*a*b^4*x^3*(12*c + 11*d*x) + 231*b^5*x^4*(15*c + 13*d*x)))/(45045*b^6*Sqrt[x^2*(a + b*x)])
Time = 0.70 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1945, 1922, 1908, 1922, 1922, 1920}
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.
\(\displaystyle \int x \left (a x^2+b x^3\right )^{3/2} (c+d x) \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle \frac {(3 b c-2 a d) \int x \left (b x^3+a x^2\right )^{3/2}dx}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \int \left (b x^3+a x^2\right )^{3/2}dx}{13 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
\(\Big \downarrow \) 1908 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x}dx}{11 b}\right )}{13 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^2}dx}{9 b}\right )}{11 b}\right )}{13 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {2 a \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^3}dx}{7 b}\right )}{9 b}\right )}{11 b}\right )}{13 b}\right )}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {\left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac {8 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {6 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}-\frac {4 a \left (\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}\right )}{9 b}\right )}{11 b}\right )}{13 b}\right ) (3 b c-2 a d)}{3 b}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{15 b}\) |
Input:
Int[x*(c + d*x)*(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*d*(a*x^2 + b*x^3)^(5/2))/(15*b) + ((3*b*c - 2*a*d)*((2*(a*x^2 + b*x^3)^ (5/2))/(13*b*x) - (8*a*((2*(a*x^2 + b*x^3)^(5/2))/(11*b*x^2) - (6*a*((2*(a *x^2 + b*x^3)^(5/2))/(9*b*x^3) - (4*a*((-4*a*(a*x^2 + b*x^3)^(5/2))/(35*b^ 2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)))/(9*b)))/(11*b)))/(13*b)))/( 3*b)
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j - 1)), x] - Simp[b*((n*p + n - j + 1)/(a*( j*p + 1))) Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.60 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.20
method | result | size |
pseudoelliptic | \(\frac {16 \left (b x +a \right )^{\frac {5}{2}} \left (\frac {45 x \left (c +\frac {7 d x}{9}\right ) b^{2}}{8}-\frac {9 \left (\frac {10 d x}{9}+c \right ) a b}{4}+a^{2} d \right )}{315 b^{3}}\) | \(41\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-3003 d \,x^{5} b^{5}+2310 a \,b^{4} d \,x^{4}-3465 b^{5} c \,x^{4}-1680 a^{2} b^{3} d \,x^{3}+2520 a \,b^{4} c \,x^{3}+1120 a^{3} b^{2} d \,x^{2}-1680 a^{2} b^{3} c \,x^{2}-640 a^{4} b d x +960 a^{3} b^{2} c x +256 a^{5} d -384 a^{4} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}}\) | \(133\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-3003 d \,x^{5} b^{5}+2310 a \,b^{4} d \,x^{4}-3465 b^{5} c \,x^{4}-1680 a^{2} b^{3} d \,x^{3}+2520 a \,b^{4} c \,x^{3}+1120 a^{3} b^{2} d \,x^{2}-1680 a^{2} b^{3} c \,x^{2}-640 a^{4} b d x +960 a^{3} b^{2} c x +256 a^{5} d -384 a^{4} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}}\) | \(133\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-3003 d \,x^{5} b^{5}+2310 a \,b^{4} d \,x^{4}-3465 b^{5} c \,x^{4}-1680 a^{2} b^{3} d \,x^{3}+2520 a \,b^{4} c \,x^{3}+1120 a^{3} b^{2} d \,x^{2}-1680 a^{2} b^{3} c \,x^{2}-640 a^{4} b d x +960 a^{3} b^{2} c x +256 a^{5} d -384 a^{4} b c \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}}\) | \(133\) |
risch | \(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-3003 b^{7} d \,x^{7}-3696 a \,b^{6} d \,x^{6}-3465 b^{7} c \,x^{6}-63 a^{2} b^{5} d \,x^{5}-4410 a \,b^{6} c \,x^{5}+70 a^{3} b^{4} d \,x^{4}-105 a^{2} b^{5} c \,x^{4}-80 a^{4} b^{3} d \,x^{3}+120 a^{3} b^{4} c \,x^{3}+96 a^{5} b^{2} d \,x^{2}-144 a^{4} b^{3} c \,x^{2}-128 a^{6} b d x +192 a^{5} b^{2} c x +256 a^{7} d -384 a^{6} b c \right )}{45045 x \,b^{6}}\) | \(174\) |
trager | \(-\frac {2 \left (-3003 b^{7} d \,x^{7}-3696 a \,b^{6} d \,x^{6}-3465 b^{7} c \,x^{6}-63 a^{2} b^{5} d \,x^{5}-4410 a \,b^{6} c \,x^{5}+70 a^{3} b^{4} d \,x^{4}-105 a^{2} b^{5} c \,x^{4}-80 a^{4} b^{3} d \,x^{3}+120 a^{3} b^{4} c \,x^{3}+96 a^{5} b^{2} d \,x^{2}-144 a^{4} b^{3} c \,x^{2}-128 a^{6} b d x +192 a^{5} b^{2} c x +256 a^{7} d -384 a^{6} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{45045 b^{6} x}\) | \(176\) |
Input:
int(x*(d*x+c)*(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
16/315*(b*x+a)^(5/2)*(45/8*x*(c+7/9*d*x)*b^2-9/4*(10/9*d*x+c)*a*b+a^2*d)/b ^3
Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.86 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (3003 \, b^{7} d x^{7} + 384 \, a^{6} b c - 256 \, a^{7} d + 231 \, {\left (15 \, b^{7} c + 16 \, a b^{6} d\right )} x^{6} + 63 \, {\left (70 \, a b^{6} c + a^{2} b^{5} d\right )} x^{5} + 35 \, {\left (3 \, a^{2} b^{5} c - 2 \, a^{3} b^{4} d\right )} x^{4} - 40 \, {\left (3 \, a^{3} b^{4} c - 2 \, a^{4} b^{3} d\right )} x^{3} + 48 \, {\left (3 \, a^{4} b^{3} c - 2 \, a^{5} b^{2} d\right )} x^{2} - 64 \, {\left (3 \, a^{5} b^{2} c - 2 \, a^{6} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{45045 \, b^{6} x} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/45045*(3003*b^7*d*x^7 + 384*a^6*b*c - 256*a^7*d + 231*(15*b^7*c + 16*a*b ^6*d)*x^6 + 63*(70*a*b^6*c + a^2*b^5*d)*x^5 + 35*(3*a^2*b^5*c - 2*a^3*b^4* d)*x^4 - 40*(3*a^3*b^4*c - 2*a^4*b^3*d)*x^3 + 48*(3*a^4*b^3*c - 2*a^5*b^2* d)*x^2 - 64*(3*a^5*b^2*c - 2*a^6*b*d)*x)*sqrt(b*x^3 + a*x^2)/(b^6*x)
\[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\int x \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )\, dx \] Input:
integrate(x*(d*x+c)*(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x*(x**2*(a + b*x))**(3/2)*(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.80 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt {b x + a} c}{15015 \, b^{5}} + \frac {2 \, {\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt {b x + a} d}{45045 \, b^{6}} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a^6)*sqrt(b*x + a)*c/b^5 + 2/45045*(300 3*b^7*x^7 + 3696*a*b^6*x^6 + 63*a^2*b^5*x^5 - 70*a^3*b^4*x^4 + 80*a^4*b^3* x^3 - 96*a^5*b^2*x^2 + 128*a^6*b*x - 256*a^7)*sqrt(b*x + a)*d/b^6
Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (181) = 362\).
Time = 0.19 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.58 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx =\text {Too large to display} \] Input:
integrate(x*(d*x+c)*(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
2/45045*(143*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^( 5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a^2*c*sgn(x)/b ^4 + 130*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(b*x + a)^(7/2 )*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a*c*sgn(x)/b^4 + 65*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)* a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3 /2)*a^4 - 693*sqrt(b*x + a)*a^5)*a^2*d*sgn(x)/b^5 + 15*(231*(b*x + a)^(13/ 2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^( 7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt (b*x + a)*a^6)*c*sgn(x)/b^4 + 30*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(1 1/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)*a*d*sgn (x)/b^5 + 7*(429*(b*x + a)^(15/2) - 3465*(b*x + a)^(13/2)*a + 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/2)*a^3 + 32175*(b*x + a)^(7/2)*a^4 - 2 7027*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2)*a^6 - 6435*sqrt(b*x + a)* a^7)*d*sgn(x)/b^5)/b - 256/45045*(3*a^(13/2)*b*c - 2*a^(15/2)*d)*sgn(x)/b^ 6
Time = 9.21 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.74 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (x^6\,\left (\frac {32\,a\,d}{195}+\frac {2\,b\,c}{13}\right )-\frac {512\,a^7\,d-768\,a^6\,b\,c}{45045\,b^6}+\frac {2\,b\,d\,x^7}{15}+\frac {128\,a^5\,x\,\left (2\,a\,d-3\,b\,c\right )}{45045\,b^5}+\frac {2\,a\,x^5\,\left (a\,d+70\,b\,c\right )}{715\,b}-\frac {2\,a^2\,x^4\,\left (2\,a\,d-3\,b\,c\right )}{1287\,b^2}+\frac {16\,a^3\,x^3\,\left (2\,a\,d-3\,b\,c\right )}{9009\,b^3}-\frac {32\,a^4\,x^2\,\left (2\,a\,d-3\,b\,c\right )}{15015\,b^4}\right )}{x} \] Input:
int(x*(a*x^2 + b*x^3)^(3/2)*(c + d*x),x)
Output:
((a*x^2 + b*x^3)^(1/2)*(x^6*((32*a*d)/195 + (2*b*c)/13) - (512*a^7*d - 768 *a^6*b*c)/(45045*b^6) + (2*b*d*x^7)/15 + (128*a^5*x*(2*a*d - 3*b*c))/(4504 5*b^5) + (2*a*x^5*(a*d + 70*b*c))/(715*b) - (2*a^2*x^4*(2*a*d - 3*b*c))/(1 287*b^2) + (16*a^3*x^3*(2*a*d - 3*b*c))/(9009*b^3) - (32*a^4*x^2*(2*a*d - 3*b*c))/(15015*b^4)))/x
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.80 \[ \int x (c+d x) \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \sqrt {b x +a}\, \left (3003 b^{7} d \,x^{7}+3696 a \,b^{6} d \,x^{6}+3465 b^{7} c \,x^{6}+63 a^{2} b^{5} d \,x^{5}+4410 a \,b^{6} c \,x^{5}-70 a^{3} b^{4} d \,x^{4}+105 a^{2} b^{5} c \,x^{4}+80 a^{4} b^{3} d \,x^{3}-120 a^{3} b^{4} c \,x^{3}-96 a^{5} b^{2} d \,x^{2}+144 a^{4} b^{3} c \,x^{2}+128 a^{6} b d x -192 a^{5} b^{2} c x -256 a^{7} d +384 a^{6} b c \right )}{45045 b^{6}} \] Input:
int(x*(d*x+c)*(b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(a + b*x)*( - 256*a**7*d + 384*a**6*b*c + 128*a**6*b*d*x - 192*a**5 *b**2*c*x - 96*a**5*b**2*d*x**2 + 144*a**4*b**3*c*x**2 + 80*a**4*b**3*d*x* *3 - 120*a**3*b**4*c*x**3 - 70*a**3*b**4*d*x**4 + 105*a**2*b**5*c*x**4 + 6 3*a**2*b**5*d*x**5 + 4410*a*b**6*c*x**5 + 3696*a*b**6*d*x**6 + 3465*b**7*c *x**6 + 3003*b**7*d*x**7))/(45045*b**6)