Integrand size = 33, antiderivative size = 298 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {(b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{8 e^5 (a+b x) (d+e x)^8}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}-\frac {b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^6}+\frac {b^2 (4 b B d-A b e-3 a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^5}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4} \]
-1/8*(-a*e+b*d)^3*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^8+1/7*( -a*e+b*d)^2*(-3*A*b*e-B*a*e+4*B*b*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d) ^7-1/2*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/( e*x+d)^6+1/5*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e *x+d)^5-1/4*b^3*B*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^4
Time = 1.07 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {\sqrt {(a+b x)^2} \left (5 a^3 e^3 (7 A e+B (d+8 e x))+5 a^2 b e^2 \left (3 A e (d+8 e x)+B \left (d^2+8 d e x+28 e^2 x^2\right )\right )+a b^2 e \left (5 A e \left (d^2+8 d e x+28 e^2 x^2\right )+3 B \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )\right )+b^3 \left (A e \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )+B \left (d^4+8 d^3 e x+28 d^2 e^2 x^2+56 d e^3 x^3+70 e^4 x^4\right )\right )\right )}{280 e^5 (a+b x) (d+e x)^8} \]
-1/280*(Sqrt[(a + b*x)^2]*(5*a^3*e^3*(7*A*e + B*(d + 8*e*x)) + 5*a^2*b*e^2 *(3*A*e*(d + 8*e*x) + B*(d^2 + 8*d*e*x + 28*e^2*x^2)) + a*b^2*e*(5*A*e*(d^ 2 + 8*d*e*x + 28*e^2*x^2) + 3*B*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x ^3)) + b^3*(A*e*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3) + B*(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 = 0.38 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 86, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^9} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^9}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^9}dx}{a+b x}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B b^3}{e^4 (d+e x)^5}+\frac {(-4 b B d+A b e+3 a B e) b^2}{e^4 (d+e x)^6}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 (d+e x)^7}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^8}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^9}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b^2 (-3 a B e-A b e+4 b B d)}{5 e^5 (d+e x)^5}-\frac {b (b d-a e) (-a B e-A b e+2 b B d)}{2 e^5 (d+e x)^6}+\frac {(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (d+e x)^7}-\frac {(b d-a e)^3 (B d-A e)}{8 e^5 (d+e x)^8}-\frac {b^3 B}{4 e^5 (d+e x)^4}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/8*((b*d - a*e)^3*(B*d - A*e))/(e^5*(d + e*x)^8) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(7*e^5*(d + e*x)^7) - (b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e))/(2*e^5*(d + e*x)^6) + (b^2*(4 *b*B*d - A*b*e - 3*a*B*e))/(5*e^5*(d + e*x)^5) - (b^3*B)/(4*e^5*(d + e*x)^ 4)))/(a + b*x)
3.18.34.3.1 Defintions of rubi rules used
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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 2.11 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.93
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{4 e}-\frac {b^{2} \left (A b e +3 B a e +B b d \right ) x^{3}}{5 e^{2}}-\frac {b \left (5 A a b \,e^{2}+A \,b^{2} d e +5 a^{2} B \,e^{2}+3 B a b d e +B \,b^{2} d^{2}\right ) x^{2}}{10 e^{3}}-\frac {\left (15 A \,a^{2} b \,e^{3}+5 A a \,b^{2} d \,e^{2}+A \,b^{3} d^{2} e +5 B \,e^{3} a^{3}+5 B \,a^{2} b d \,e^{2}+3 B a \,b^{2} d^{2} e +B \,b^{3} d^{3}\right ) x}{35 e^{4}}-\frac {35 A \,a^{3} e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +B \,b^{3} d^{4}}{280 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{8}}\) | \(278\) |
| gosper | \(-\frac {\left (70 B \,b^{3} e^{4} x^{4}+56 A \,b^{3} e^{4} x^{3}+168 B a \,b^{2} e^{4} x^{3}+56 B \,b^{3} d \,e^{3} x^{3}+140 A a \,b^{2} e^{4} x^{2}+28 A \,b^{3} d \,e^{3} x^{2}+140 B \,a^{2} b \,e^{4} x^{2}+84 B a \,b^{2} d \,e^{3} x^{2}+28 B \,b^{3} d^{2} e^{2} x^{2}+120 A \,a^{2} b \,e^{4} x +40 A a \,b^{2} d \,e^{3} x +8 A \,b^{3} d^{2} e^{2} x +40 B \,a^{3} e^{4} x +40 B \,a^{2} b d \,e^{3} x +24 B a \,b^{2} d^{2} e^{2} x +8 B \,b^{3} d^{3} e x +35 A \,a^{3} e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{5} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(315\) |
| default | \(-\frac {\left (70 B \,b^{3} e^{4} x^{4}+56 A \,b^{3} e^{4} x^{3}+168 B a \,b^{2} e^{4} x^{3}+56 B \,b^{3} d \,e^{3} x^{3}+140 A a \,b^{2} e^{4} x^{2}+28 A \,b^{3} d \,e^{3} x^{2}+140 B \,a^{2} b \,e^{4} x^{2}+84 B a \,b^{2} d \,e^{3} x^{2}+28 B \,b^{3} d^{2} e^{2} x^{2}+120 A \,a^{2} b \,e^{4} x +40 A a \,b^{2} d \,e^{3} x +8 A \,b^{3} d^{2} e^{2} x +40 B \,a^{3} e^{4} x +40 B \,a^{2} b d \,e^{3} x +24 B a \,b^{2} d^{2} e^{2} x +8 B \,b^{3} d^{3} e x +35 A \,a^{3} e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 e^{5} \left (e x +d \right )^{8} \left (b x +a \right )^{3}}\) | \(315\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/4/e*B*b^3*x^4-1/5*b^2/e^2*(A*b*e+3*B*a*e+B*b *d)*x^3-1/10*b/e^3*(5*A*a*b*e^2+A*b^2*d*e+5*B*a^2*e^2+3*B*a*b*d*e+B*b^2*d^ 2)*x^2-1/35/e^4*(15*A*a^2*b*e^3+5*A*a*b^2*d*e^2+A*b^3*d^2*e+5*B*a^3*e^3+5* B*a^2*b*d*e^2+3*B*a*b^2*d^2*e+B*b^3*d^3)*x-1/280/e^5*(35*A*a^3*e^4+15*A*a^ 2*b*d*e^3+5*A*a*b^2*d^2*e^2+A*b^3*d^3*e+5*B*a^3*d*e^3+5*B*a^2*b*d^2*e^2+3* B*a*b^2*d^3*e+B*b^3*d^4))/(e*x+d)^8
Time = 0.36 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {70 \, B b^{3} e^{4} x^{4} + B b^{3} d^{4} + 35 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 56 \, {\left (B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 28 \, {\left (B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 8 \, {\left (B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{280 \, {\left (e^{13} x^{8} + 8 \, d e^{12} x^{7} + 28 \, d^{2} e^{11} x^{6} + 56 \, d^{3} e^{10} x^{5} + 70 \, d^{4} e^{9} x^{4} + 56 \, d^{5} e^{8} x^{3} + 28 \, d^{6} e^{7} x^{2} + 8 \, d^{7} e^{6} x + d^{8} e^{5}\right )}} \]
-1/280*(70*B*b^3*e^4*x^4 + B*b^3*d^4 + 35*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)* d^3*e + 5*(B*a^2*b + A*a*b^2)*d^2*e^2 + 5*(B*a^3 + 3*A*a^2*b)*d*e^3 + 56*( B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 28*(B*b^3*d^2*e^2 + (3*B*a*b^ 2 + A*b^3)*d*e^3 + 5*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 8*(B*b^3*d^3*e + (3*B* a*b^2 + A*b^3)*d^2*e^2 + 5*(B*a^2*b + A*a*b^2)*d*e^3 + 5*(B*a^3 + 3*A*a^2* 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. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\text {Exception raised: ValueError} \]
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. 543 vs. \(2 (233) = 466\).
Time = 0.28 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=\frac {{\left (B b^{8} d - 2 \, B a b^{7} e + A b^{8} e\right )} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (b^{5} d^{5} e^{5} - 5 \, a b^{4} d^{4} e^{6} + 10 \, a^{2} b^{3} d^{3} e^{7} - 10 \, a^{3} b^{2} d^{2} e^{8} + 5 \, a^{4} b d e^{9} - a^{5} e^{10}\right )}} - \frac {70 \, B b^{3} e^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 56 \, B b^{3} d e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 168 \, B a b^{2} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 56 \, A b^{3} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, B b^{3} d^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 84 \, B a b^{2} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 28 \, A b^{3} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 140 \, B a^{2} b e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 140 \, A a b^{2} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, B b^{3} d^{3} e x \mathrm {sgn}\left (b x + a\right ) + 24 \, B a b^{2} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 8 \, A b^{3} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{2} b d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 40 \, A a b^{2} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{3} e^{4} x \mathrm {sgn}\left (b x + a\right ) + 120 \, A a^{2} b e^{4} x \mathrm {sgn}\left (b x + a\right ) + B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) + A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )}{280 \, {\left (e x + d\right )}^{8} e^{5}} \]
1/280*(B*b^8*d - 2*B*a*b^7*e + A*b^8*e)*sgn(b*x + a)/(b^5*d^5*e^5 - 5*a*b^ 4*d^4*e^6 + 10*a^2*b^3*d^3*e^7 - 10*a^3*b^2*d^2*e^8 + 5*a^4*b*d*e^9 - a^5* e^10) - 1/280*(70*B*b^3*e^4*x^4*sgn(b*x + a) + 56*B*b^3*d*e^3*x^3*sgn(b*x + a) + 168*B*a*b^2*e^4*x^3*sgn(b*x + a) + 56*A*b^3*e^4*x^3*sgn(b*x + a) + 28*B*b^3*d^2*e^2*x^2*sgn(b*x + a) + 84*B*a*b^2*d*e^3*x^2*sgn(b*x + a) + 28 *A*b^3*d*e^3*x^2*sgn(b*x + a) + 140*B*a^2*b*e^4*x^2*sgn(b*x + a) + 140*A*a *b^2*e^4*x^2*sgn(b*x + a) + 8*B*b^3*d^3*e*x*sgn(b*x + a) + 24*B*a*b^2*d^2* e^2*x*sgn(b*x + a) + 8*A*b^3*d^2*e^2*x*sgn(b*x + a) + 40*B*a^2*b*d*e^3*x*s gn(b*x + a) + 40*A*a*b^2*d*e^3*x*sgn(b*x + a) + 40*B*a^3*e^4*x*sgn(b*x + a ) + 120*A*a^2*b*e^4*x*sgn(b*x + a) + B*b^3*d^4*sgn(b*x + a) + 3*B*a*b^2*d^ 3*e*sgn(b*x + a) + A*b^3*d^3*e*sgn(b*x + a) + 5*B*a^2*b*d^2*e^2*sgn(b*x + a) + 5*A*a*b^2*d^2*e^2*sgn(b*x + a) + 5*B*a^3*d*e^3*sgn(b*x + a) + 15*A*a^ 2*b*d*e^3*sgn(b*x + a) + 35*A*a^3*e^4*sgn(b*x + a))/((e*x + d)^8*e^5)
Time = 10.81 (sec) , antiderivative size = 577, normalized size of antiderivative = 1.94 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^9} \, dx=-\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{5\,e^5}-\frac {B\,b^3\,d}{5\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {A\,a^3}{8\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{8\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{8\,e}-\frac {B\,b^3\,d}{8\,e^2}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{8\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}-\frac {\left (\frac {B\,a^3\,e^3-3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e-3\,A\,a\,b^2\,d\,e^2-B\,b^3\,d^3+A\,b^3\,d^2\,e}{7\,e^5}-\frac {d\,\left (\frac {3\,B\,a^2\,b\,e^3-3\,B\,a\,b^2\,d\,e^2+3\,A\,a\,b^2\,e^3+B\,b^3\,d^2\,e-A\,b^3\,d\,e^2}{7\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{7\,e^3}-\frac {B\,b^3\,d}{7\,e^3}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {3\,B\,a^2\,b\,e^2-6\,B\,a\,b^2\,d\,e+3\,A\,a\,b^2\,e^2+3\,B\,b^3\,d^2-2\,A\,b^3\,d\,e}{6\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{6\,e^4}-\frac {B\,b^3\,d}{6\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \]
- (((A*b^3*e - 3*B*b^3*d + 3*B*a*b^2*e)/(5*e^5) - (B*b^3*d)/(5*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^5) - (((A*a^3)/(8*e) - (d *((B*a^3 + 3*A*a^2*b)/(8*e) + (d*((d*((A*b^3 + 3*B*a*b^2)/(8*e) - (B*b^3*d )/(8*e^2)))/e - (3*a*b*(A*b + B*a))/(8*e)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b* x)^(1/2))/((a + b*x)*(d + e*x)^8) - (((B*a^3*e^3 - B*b^3*d^3 + 3*A*a^2*b*e ^3 + A*b^3*d^2*e - 3*A*a*b^2*d*e^2 + 3*B*a*b^2*d^2*e - 3*B*a^2*b*d*e^2)/(7 *e^5) - (d*((3*A*a*b^2*e^3 + 3*B*a^2*b*e^3 - A*b^3*d*e^2 + B*b^3*d^2*e - 3 *B*a*b^2*d*e^2)/(7*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - B*b*d))/(7*e^3) - (B *b^3*d)/(7*e^3)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^7) - (((3*B*b^3*d^2 - 2*A*b^3*d*e + 3*A*a*b^2*e^2 + 3*B*a^2*b*e^2 - 6 *B*a*b^2*d*e)/(6*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - 2*B*b*d))/(6*e^4) - (B *b^3*d)/(6*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x) ^6) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*e^5*(a + b*x)*(d + e*x)^4 )