Integrand size = 33, antiderivative size = 172 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{8 (b d-a e) (d+e x)^8}+\frac {3 b \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{56 (b d-a e)^2 (d+e x)^7}+\frac {b^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{56 (b d-a e)^3 (d+e x)^6}+\frac {b^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{280 (b d-a e)^4 (d+e x)^5} \] Output:
1/8*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e+b*d)/(e*x+d)^8+3/56*b*(b^2*x^2+2*a*b *x+a^2)^(5/2)/(-a*e+b*d)^2/(e*x+d)^7+1/56*b^2*(b^2*x^2+2*a*b*x+a^2)^(5/2)/ (-a*e+b*d)^3/(e*x+d)^6+1/280*b^3*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e+b*d)^4/ (e*x+d)^5
Time = 1.07 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {\sqrt {(a+b x)^2} \left (35 a^4 e^4+20 a^3 b e^3 (d+8 e x)+10 a^2 b^2 e^2 \left (d^2+8 d e x+28 e^2 x^2\right )+4 a b^3 e \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )+b^4 \left (d^4+8 d^3 e x+28 d^2 e^2 x^2+56 d e^3 x^3+70 e^4 x^4\right )\right )}{280 e^5 (a+b x) (d+e x)^8} \] Input:
Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^9,x]
Output:
-1/280*(Sqrt[(a + b*x)^2]*(35*a^4*e^4 + 20*a^3*b*e^3*(d + 8*e*x) + 10*a^2* b^2*e^2*(d^2 + 8*d*e*x + 28*e^2*x^2) + 4*a*b^3*e*(d^3 + 8*d^2*e*x + 28*d*e ^2*x^2 + 56*e^3*x^3) + b^4*(d^4 + 8*d^3*e*x + 28*d^2*e^2*x^2 + 56*d*e^3*x^ 3 + 70*e^4*x^4)))/(e^5*(a + b*x)*(d + e*x)^8)
Time = 0.54 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 53, 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.
| \(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^9}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^9}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^4}{e^4 (d+e x)^5}-\frac {4 (b d-a e) b^3}{e^4 (d+e x)^6}+\frac {6 (b d-a e)^2 b^2}{e^4 (d+e x)^7}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)^8}+\frac {(a e-b d)^4}{e^4 (d+e x)^9}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {4 b^3 (b d-a e)}{5 e^5 (d+e x)^5}-\frac {b^2 (b d-a e)^2}{e^5 (d+e x)^6}+\frac {4 b (b d-a e)^3}{7 e^5 (d+e x)^7}-\frac {(b d-a e)^4}{8 e^5 (d+e x)^8}-\frac {b^4}{4 e^5 (d+e x)^4}\right )}{a+b x}\) |
Input:
Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^9,x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/8*(b*d - a*e)^4/(e^5*(d + e*x)^8) + (4* b*(b*d - a*e)^3)/(7*e^5*(d + e*x)^7) - (b^2*(b*d - a*e)^2)/(e^5*(d + e*x)^ 6) + (4*b^3*(b*d - a*e))/(5*e^5*(d + e*x)^5) - b^4/(4*e^5*(d + e*x)^4)))/( a + b*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 2.60 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.09
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{4} x^{4}}{4 e}-\frac {b^{3} \left (4 a e +b d \right ) x^{3}}{5 e^{2}}-\frac {b^{2} \left (10 e^{2} a^{2}+4 a b d e +b^{2} d^{2}\right ) x^{2}}{10 e^{3}}-\frac {b \left (20 e^{3} a^{3}+10 a^{2} b d \,e^{2}+4 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) x}{35 e^{4}}-\frac {35 a^{4} e^{4}+20 a^{3} b d \,e^{3}+10 a^{2} b^{2} d^{2} e^{2}+4 a \,b^{3} d^{3} e +b^{4} d^{4}}{280 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{8}}\) | \(187\) |
| gosper | \(-\frac {\left (70 b^{4} x^{4} e^{4}+224 x^{3} a \,b^{3} e^{4}+56 x^{3} b^{4} d \,e^{3}+280 x^{2} a^{2} b^{2} e^{4}+112 x^{2} a \,b^{3} d \,e^{3}+28 x^{2} b^{4} d^{2} e^{2}+160 x \,a^{3} b \,e^{4}+80 x \,a^{2} b^{2} d \,e^{3}+32 x a \,b^{3} d^{2} e^{2}+8 x \,b^{4} d^{3} e +35 a^{4} e^{4}+20 a^{3} b d \,e^{3}+10 a^{2} b^{2} d^{2} e^{2}+4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{5} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(201\) |
| default | \(-\frac {\left (70 b^{4} x^{4} e^{4}+224 x^{3} a \,b^{3} e^{4}+56 x^{3} b^{4} d \,e^{3}+280 x^{2} a^{2} b^{2} e^{4}+112 x^{2} a \,b^{3} d \,e^{3}+28 x^{2} b^{4} d^{2} e^{2}+160 x \,a^{3} b \,e^{4}+80 x \,a^{2} b^{2} d \,e^{3}+32 x a \,b^{3} d^{2} e^{2}+8 x \,b^{4} d^{3} e +35 a^{4} e^{4}+20 a^{3} b d \,e^{3}+10 a^{2} b^{2} d^{2} e^{2}+4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{5} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(201\) |
| orering | \(-\frac {\left (70 b^{4} x^{4} e^{4}+224 x^{3} a \,b^{3} e^{4}+56 x^{3} b^{4} d \,e^{3}+280 x^{2} a^{2} b^{2} e^{4}+112 x^{2} a \,b^{3} d \,e^{3}+28 x^{2} b^{4} d^{2} e^{2}+160 x \,a^{3} b \,e^{4}+80 x \,a^{2} b^{2} d \,e^{3}+32 x a \,b^{3} d^{2} e^{2}+8 x \,b^{4} d^{3} e +35 a^{4} e^{4}+20 a^{3} b d \,e^{3}+10 a^{2} b^{2} d^{2} e^{2}+4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{280 e^{5} \left (b x +a \right )^{3} \left (e x +d \right )^{8}}\) | \(210\) |
Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^9,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/4/e*b^4*x^4-1/5*b^3/e^2*(4*a*e+b*d)*x^3-1/10 *b^2/e^3*(10*a^2*e^2+4*a*b*d*e+b^2*d^2)*x^2-1/35*b/e^4*(20*a^3*e^3+10*a^2* b*d*e^2+4*a*b^2*d^2*e+b^3*d^3)*x-1/280/e^5*(35*a^4*e^4+20*a^3*b*d*e^3+10*a ^2*b^2*d^2*e^2+4*a*b^3*d^3*e+b^4*d^4))/(e*x+d)^8
Time = 0.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {70 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + 4 \, a b^{3} d^{3} e + 10 \, a^{2} b^{2} d^{2} e^{2} + 20 \, a^{3} b d e^{3} + 35 \, a^{4} e^{4} + 56 \, {\left (b^{4} d e^{3} + 4 \, a b^{3} e^{4}\right )} x^{3} + 28 \, {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} + 10 \, a^{2} b^{2} e^{4}\right )} x^{2} + 8 \, {\left (b^{4} d^{3} e + 4 \, a b^{3} d^{2} e^{2} + 10 \, a^{2} b^{2} d e^{3} + 20 \, a^{3} b e^{4}\right )} x}{280 \, {\left (e^{13} x^{8} + 8 \, d e^{12} x^{7} + 28 \, d^{2} e^{11} x^{6} + 56 \, d^{3} e^{10} x^{5} + 70 \, d^{4} e^{9} x^{4} + 56 \, d^{5} e^{8} x^{3} + 28 \, d^{6} e^{7} x^{2} + 8 \, d^{7} e^{6} x + d^{8} e^{5}\right )}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^9,x, algorithm="fric as")
Output:
-1/280*(70*b^4*e^4*x^4 + b^4*d^4 + 4*a*b^3*d^3*e + 10*a^2*b^2*d^2*e^2 + 20 *a^3*b*d*e^3 + 35*a^4*e^4 + 56*(b^4*d*e^3 + 4*a*b^3*e^4)*x^3 + 28*(b^4*d^2 *e^2 + 4*a*b^3*d*e^3 + 10*a^2*b^2*e^4)*x^2 + 8*(b^4*d^3*e + 4*a*b^3*d^2*e^ 2 + 10*a^2*b^2*d*e^3 + 20*a^3*b*e^4)*x)/(e^13*x^8 + 8*d*e^12*x^7 + 28*d^2* e^11*x^6 + 56*d^3*e^10*x^5 + 70*d^4*e^9*x^4 + 56*d^5*e^8*x^3 + 28*d^6*e^7* x^2 + 8*d^7*e^6*x + d^8*e^5)
Timed out. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**9,x)
Output:
Timed out
Exception generated. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^9,x, algorithm="maxi ma")
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*e-b*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (156) = 312\).
Time = 0.19 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.99 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {b^{8} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (b^{4} d^{4} e^{5} - 4 \, a b^{3} d^{3} e^{6} + 6 \, a^{2} b^{2} d^{2} e^{7} - 4 \, a^{3} b d e^{8} + a^{4} e^{9}\right )}} - \frac {70 \, b^{4} e^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 56 \, b^{4} d e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 224 \, a b^{3} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, b^{4} d^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 112 \, a b^{3} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 280 \, a^{2} b^{2} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, b^{4} d^{3} e x \mathrm {sgn}\left (b x + a\right ) + 32 \, a b^{3} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 80 \, a^{2} b^{2} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 160 \, a^{3} b e^{4} x \mathrm {sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (e x + d\right )}^{8} e^{5}} \] Input:
integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^9,x, algorithm="giac ")
Output:
1/280*b^8*sgn(b*x + a)/(b^4*d^4*e^5 - 4*a*b^3*d^3*e^6 + 6*a^2*b^2*d^2*e^7 - 4*a^3*b*d*e^8 + a^4*e^9) - 1/280*(70*b^4*e^4*x^4*sgn(b*x + a) + 56*b^4*d *e^3*x^3*sgn(b*x + a) + 224*a*b^3*e^4*x^3*sgn(b*x + a) + 28*b^4*d^2*e^2*x^ 2*sgn(b*x + a) + 112*a*b^3*d*e^3*x^2*sgn(b*x + a) + 280*a^2*b^2*e^4*x^2*sg n(b*x + a) + 8*b^4*d^3*e*x*sgn(b*x + a) + 32*a*b^3*d^2*e^2*x*sgn(b*x + a) + 80*a^2*b^2*d*e^3*x*sgn(b*x + a) + 160*a^3*b*e^4*x*sgn(b*x + a) + b^4*d^4 *sgn(b*x + a) + 4*a*b^3*d^3*e*sgn(b*x + a) + 10*a^2*b^2*d^2*e^2*sgn(b*x + a) + 20*a^3*b*d*e^3*sgn(b*x + a) + 35*a^4*e^4*sgn(b*x + a))/((e*x + d)^8*e ^5)
Time = 11.44 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.61 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {\left (\frac {-4\,a^3\,b\,e^3+6\,a^2\,b^2\,d\,e^2-4\,a\,b^3\,d^2\,e+b^4\,d^3}{7\,e^5}+\frac {d\,\left (\frac {d\,\left (\frac {b^4\,d}{7\,e^3}-\frac {b^3\,\left (4\,a\,e-b\,d\right )}{7\,e^3}\right )}{e}+\frac {b^2\,\left (6\,a^2\,e^2-4\,a\,b\,d\,e+b^2\,d^2\right )}{7\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {a^4}{8\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {a\,b^3}{2\,e}-\frac {b^4\,d}{8\,e^2}\right )}{e}-\frac {3\,a^2\,b^2}{4\,e}\right )}{e}+\frac {a^3\,b}{2\,e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}-\frac {\left (\frac {6\,a^2\,b^2\,e^2-8\,a\,b^3\,d\,e+3\,b^4\,d^2}{6\,e^5}+\frac {d\,\left (\frac {b^4\,d}{6\,e^4}-\frac {b^3\,\left (2\,a\,e-b\,d\right )}{3\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}+\frac {\left (\frac {3\,b^4\,d-4\,a\,b^3\,e}{5\,e^5}+\frac {b^4\,d}{5\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \] Input:
int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^9,x)
Output:
(((b^4*d^3 - 4*a^3*b*e^3 + 6*a^2*b^2*d*e^2 - 4*a*b^3*d^2*e)/(7*e^5) + (d*( (d*((b^4*d)/(7*e^3) - (b^3*(4*a*e - b*d))/(7*e^3)))/e + (b^2*(6*a^2*e^2 + b^2*d^2 - 4*a*b*d*e))/(7*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^7) - ((a^4/(8*e) - (d*((d*((d*((a*b^3)/(2*e) - (b^4*d)/(8*e ^2)))/e - (3*a^2*b^2)/(4*e)))/e + (a^3*b)/(2*e)))/e)*(a^2 + b^2*x^2 + 2*a* b*x)^(1/2))/((a + b*x)*(d + e*x)^8) - (((3*b^4*d^2 + 6*a^2*b^2*e^2 - 8*a*b ^3*d*e)/(6*e^5) + (d*((b^4*d)/(6*e^4) - (b^3*(2*a*e - b*d))/(3*e^4)))/e)*( a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) + (((3*b^4*d - 4*a *b^3*e)/(5*e^5) + (b^4*d)/(5*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^5) - (b^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*e^5*(a + b*x) *(d + e*x)^4)
Time = 0.29 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {-70 b^{4} e^{4} x^{4}-224 a \,b^{3} e^{4} x^{3}-56 b^{4} d \,e^{3} x^{3}-280 a^{2} b^{2} e^{4} x^{2}-112 a \,b^{3} d \,e^{3} x^{2}-28 b^{4} d^{2} e^{2} x^{2}-160 a^{3} b \,e^{4} x -80 a^{2} b^{2} d \,e^{3} x -32 a \,b^{3} d^{2} e^{2} x -8 b^{4} d^{3} e x -35 a^{4} e^{4}-20 a^{3} b d \,e^{3}-10 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e -b^{4} d^{4}}{280 e^{5} \left (e^{8} x^{8}+8 d \,e^{7} x^{7}+28 d^{2} e^{6} x^{6}+56 d^{3} e^{5} x^{5}+70 d^{4} e^{4} x^{4}+56 d^{5} e^{3} x^{3}+28 d^{6} e^{2} x^{2}+8 d^{7} e x +d^{8}\right )} \] Input:
int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^9,x)
Output:
( - 35*a**4*e**4 - 20*a**3*b*d*e**3 - 160*a**3*b*e**4*x - 10*a**2*b**2*d** 2*e**2 - 80*a**2*b**2*d*e**3*x - 280*a**2*b**2*e**4*x**2 - 4*a*b**3*d**3*e - 32*a*b**3*d**2*e**2*x - 112*a*b**3*d*e**3*x**2 - 224*a*b**3*e**4*x**3 - b**4*d**4 - 8*b**4*d**3*e*x - 28*b**4*d**2*e**2*x**2 - 56*b**4*d*e**3*x** 3 - 70*b**4*e**4*x**4)/(280*e**5*(d**8 + 8*d**7*e*x + 28*d**6*e**2*x**2 + 56*d**5*e**3*x**3 + 70*d**4*e**4*x**4 + 56*d**3*e**5*x**5 + 28*d**2*e**6*x **6 + 8*d*e**7*x**7 + e**8*x**8))