Integrand size = 33, antiderivative size = 298 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=-\frac {(b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^5 (a+b x) (d+e x)^9}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{8 e^5 (a+b x) (d+e x)^8}-\frac {3 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}+\frac {b^2 (4 b B d-A b e-3 a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{6 e^5 (a+b x) (d+e x)^6}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^5} \]
-1/9*(-a*e+b*d)^3*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^9+1/8*( -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) ^8-3/7*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)^7+1/6*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)^6-1/5*b^3*B*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^5
Time = 1.08 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (35 a^3 e^3 (8 A e+B (d+9 e x))+15 a^2 b e^2 \left (7 A e (d+9 e x)+2 B \left (d^2+9 d e x+36 e^2 x^2\right )\right )+15 a b^2 e \left (2 A e \left (d^2+9 d e x+36 e^2 x^2\right )+B \left (d^3+9 d^2 e x+36 d e^2 x^2+84 e^3 x^3\right )\right )+b^3 \left (5 A e \left (d^3+9 d^2 e x+36 d e^2 x^2+84 e^3 x^3\right )+4 B \left (d^4+9 d^3 e x+36 d^2 e^2 x^2+84 d e^3 x^3+126 e^4 x^4\right )\right )\right )}{2520 e^5 (a+b x) (d+e x)^9} \]
-1/2520*(Sqrt[(a + b*x)^2]*(35*a^3*e^3*(8*A*e + B*(d + 9*e*x)) + 15*a^2*b* e^2*(7*A*e*(d + 9*e*x) + 2*B*(d^2 + 9*d*e*x + 36*e^2*x^2)) + 15*a*b^2*e*(2 *A*e*(d^2 + 9*d*e*x + 36*e^2*x^2) + B*(d^3 + 9*d^2*e*x + 36*d*e^2*x^2 + 84 *e^3*x^3)) + b^3*(5*A*e*(d^3 + 9*d^2*e*x + 36*d*e^2*x^2 + 84*e^3*x^3) + 4* B*(d^4 + 9*d^3*e*x + 36*d^2*e^2*x^2 + 84*d*e^3*x^3 + 126*e^4*x^4))))/(e^5* (a + b*x)*(d + e*x)^9)
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^{10}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^{10}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{10}}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B b^3}{e^4 (d+e x)^6}+\frac {(-4 b B d+A b e+3 a B e) b^2}{e^4 (d+e x)^7}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 (d+e x)^8}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^9}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^{10}}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b^2 (-3 a B e-A b e+4 b B d)}{6 e^5 (d+e x)^6}-\frac {3 b (b d-a e) (-a B e-A b e+2 b B d)}{7 e^5 (d+e x)^7}+\frac {(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{8 e^5 (d+e x)^8}-\frac {(b d-a e)^3 (B d-A e)}{9 e^5 (d+e x)^9}-\frac {b^3 B}{5 e^5 (d+e x)^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/9*((b*d - a*e)^3*(B*d - A*e))/(e^5*(d + e*x)^9) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(8*e^5*(d + e*x)^8) - (3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e))/(7*e^5*(d + e*x)^7) + (b^2* (4*b*B*d - A*b*e - 3*a*B*e))/(6*e^5*(d + e*x)^6) - (b^3*B)/(5*e^5*(d + e*x )^5)))/(a + b*x)
3.18.35.3.1 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 = 2.75 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{5 e}-\frac {b^{2} \left (5 A b e +15 B a e +4 B b d \right ) x^{3}}{30 e^{2}}-\frac {b \left (30 A a b \,e^{2}+5 A \,b^{2} d e +30 a^{2} B \,e^{2}+15 B a b d e +4 B \,b^{2} d^{2}\right ) x^{2}}{70 e^{3}}-\frac {\left (105 A \,a^{2} b \,e^{3}+30 A a \,b^{2} d \,e^{2}+5 A \,b^{3} d^{2} e +35 B \,e^{3} a^{3}+30 B \,a^{2} b d \,e^{2}+15 B a \,b^{2} d^{2} e +4 B \,b^{3} d^{3}\right ) x}{280 e^{4}}-\frac {280 A \,a^{3} e^{4}+105 A \,a^{2} b d \,e^{3}+30 A a \,b^{2} d^{2} e^{2}+5 A \,b^{3} d^{3} e +35 B \,a^{3} d \,e^{3}+30 B \,a^{2} b \,d^{2} e^{2}+15 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}}{2520 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{9}}\) | \(286\) |
gosper | \(-\frac {\left (504 B \,b^{3} e^{4} x^{4}+420 A \,b^{3} e^{4} x^{3}+1260 B a \,b^{2} e^{4} x^{3}+336 B \,b^{3} d \,e^{3} x^{3}+1080 A a \,b^{2} e^{4} x^{2}+180 A \,b^{3} d \,e^{3} x^{2}+1080 B \,a^{2} b \,e^{4} x^{2}+540 B a \,b^{2} d \,e^{3} x^{2}+144 B \,b^{3} d^{2} e^{2} x^{2}+945 A \,a^{2} b \,e^{4} x +270 A a \,b^{2} d \,e^{3} x +45 A \,b^{3} d^{2} e^{2} x +315 B \,a^{3} e^{4} x +270 B \,a^{2} b d \,e^{3} x +135 B a \,b^{2} d^{2} e^{2} x +36 B \,b^{3} d^{3} e x +280 A \,a^{3} e^{4}+105 A \,a^{2} b d \,e^{3}+30 A a \,b^{2} d^{2} e^{2}+5 A \,b^{3} d^{3} e +35 B \,a^{3} d \,e^{3}+30 B \,a^{2} b \,d^{2} e^{2}+15 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 e^{5} \left (e x +d \right )^{9} \left (b x +a \right )^{3}}\) | \(317\) |
default | \(-\frac {\left (504 B \,b^{3} e^{4} x^{4}+420 A \,b^{3} e^{4} x^{3}+1260 B a \,b^{2} e^{4} x^{3}+336 B \,b^{3} d \,e^{3} x^{3}+1080 A a \,b^{2} e^{4} x^{2}+180 A \,b^{3} d \,e^{3} x^{2}+1080 B \,a^{2} b \,e^{4} x^{2}+540 B a \,b^{2} d \,e^{3} x^{2}+144 B \,b^{3} d^{2} e^{2} x^{2}+945 A \,a^{2} b \,e^{4} x +270 A a \,b^{2} d \,e^{3} x +45 A \,b^{3} d^{2} e^{2} x +315 B \,a^{3} e^{4} x +270 B \,a^{2} b d \,e^{3} x +135 B a \,b^{2} d^{2} e^{2} x +36 B \,b^{3} d^{3} e x +280 A \,a^{3} e^{4}+105 A \,a^{2} b d \,e^{3}+30 A a \,b^{2} d^{2} e^{2}+5 A \,b^{3} d^{3} e +35 B \,a^{3} d \,e^{3}+30 B \,a^{2} b \,d^{2} e^{2}+15 B a \,b^{2} d^{3} e +4 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 e^{5} \left (e x +d \right )^{9} \left (b x +a \right )^{3}}\) | \(317\) |
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/5/e*B*b^3*x^4-1/30*b^2/e^2*(5*A*b*e+15*B*a*e +4*B*b*d)*x^3-1/70*b/e^3*(30*A*a*b*e^2+5*A*b^2*d*e+30*B*a^2*e^2+15*B*a*b*d *e+4*B*b^2*d^2)*x^2-1/280/e^4*(105*A*a^2*b*e^3+30*A*a*b^2*d*e^2+5*A*b^3*d^ 2*e+35*B*a^3*e^3+30*B*a^2*b*d*e^2+15*B*a*b^2*d^2*e+4*B*b^3*d^3)*x-1/2520/e ^5*(280*A*a^3*e^4+105*A*a^2*b*d*e^3+30*A*a*b^2*d^2*e^2+5*A*b^3*d^3*e+35*B* a^3*d*e^3+30*B*a^2*b*d^2*e^2+15*B*a*b^2*d^3*e+4*B*b^3*d^4))/(e*x+d)^9
Time = 0.33 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=-\frac {504 \, B b^{3} e^{4} x^{4} + 4 \, B b^{3} d^{4} + 280 \, A a^{3} e^{4} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 30 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 35 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 84 \, {\left (4 \, B b^{3} d e^{3} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 36 \, {\left (4 \, B b^{3} d^{2} e^{2} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 30 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 9 \, {\left (4 \, B b^{3} d^{3} e + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 30 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 35 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{2520 \, {\left (e^{14} x^{9} + 9 \, d e^{13} x^{8} + 36 \, d^{2} e^{12} x^{7} + 84 \, d^{3} e^{11} x^{6} + 126 \, d^{4} e^{10} x^{5} + 126 \, d^{5} e^{9} x^{4} + 84 \, d^{6} e^{8} x^{3} + 36 \, d^{7} e^{7} x^{2} + 9 \, d^{8} e^{6} x + d^{9} e^{5}\right )}} \]
-1/2520*(504*B*b^3*e^4*x^4 + 4*B*b^3*d^4 + 280*A*a^3*e^4 + 5*(3*B*a*b^2 + A*b^3)*d^3*e + 30*(B*a^2*b + A*a*b^2)*d^2*e^2 + 35*(B*a^3 + 3*A*a^2*b)*d*e ^3 + 84*(4*B*b^3*d*e^3 + 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 36*(4*B*b^3*d^2* e^2 + 5*(3*B*a*b^2 + A*b^3)*d*e^3 + 30*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 9*(4 *B*b^3*d^3*e + 5*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 30*(B*a^2*b + A*a*b^2)*d*e^ 3 + 35*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^14*x^9 + 9*d*e^13*x^8 + 36*d^2*e^12* x^7 + 84*d^3*e^11*x^6 + 126*d^4*e^10*x^5 + 126*d^5*e^9*x^4 + 84*d^6*e^8*x^ 3 + 36*d^7*e^7*x^2 + 9*d^8*e^6*x + d^9*e^5)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=\text {Exception raised: ValueError} \]
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. 560 vs. \(2 (233) = 466\).
Time = 0.29 (sec) , antiderivative size = 560, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=\frac {{\left (4 \, B b^{9} d - 9 \, B a b^{8} e + 5 \, A b^{9} e\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, {\left (b^{6} d^{6} e^{5} - 6 \, a b^{5} d^{5} e^{6} + 15 \, a^{2} b^{4} d^{4} e^{7} - 20 \, a^{3} b^{3} d^{3} e^{8} + 15 \, a^{4} b^{2} d^{2} e^{9} - 6 \, a^{5} b d e^{10} + a^{6} e^{11}\right )}} - \frac {504 \, B b^{3} e^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 336 \, B b^{3} d e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1260 \, B a b^{2} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 420 \, A b^{3} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 144 \, B b^{3} d^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 540 \, B a b^{2} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 180 \, A b^{3} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 1080 \, B a^{2} b e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 1080 \, A a b^{2} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 36 \, B b^{3} d^{3} e x \mathrm {sgn}\left (b x + a\right ) + 135 \, B a b^{2} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 45 \, A b^{3} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 270 \, B a^{2} b d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 270 \, A a b^{2} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 315 \, B a^{3} e^{4} x \mathrm {sgn}\left (b x + a\right ) + 945 \, A a^{2} b e^{4} x \mathrm {sgn}\left (b x + a\right ) + 4 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 5 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 30 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 30 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 280 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )}{2520 \, {\left (e x + d\right )}^{9} e^{5}} \]
1/2520*(4*B*b^9*d - 9*B*a*b^8*e + 5*A*b^9*e)*sgn(b*x + a)/(b^6*d^6*e^5 - 6 *a*b^5*d^5*e^6 + 15*a^2*b^4*d^4*e^7 - 20*a^3*b^3*d^3*e^8 + 15*a^4*b^2*d^2* e^9 - 6*a^5*b*d*e^10 + a^6*e^11) - 1/2520*(504*B*b^3*e^4*x^4*sgn(b*x + a) + 336*B*b^3*d*e^3*x^3*sgn(b*x + a) + 1260*B*a*b^2*e^4*x^3*sgn(b*x + a) + 4 20*A*b^3*e^4*x^3*sgn(b*x + a) + 144*B*b^3*d^2*e^2*x^2*sgn(b*x + a) + 540*B *a*b^2*d*e^3*x^2*sgn(b*x + a) + 180*A*b^3*d*e^3*x^2*sgn(b*x + a) + 1080*B* a^2*b*e^4*x^2*sgn(b*x + a) + 1080*A*a*b^2*e^4*x^2*sgn(b*x + a) + 36*B*b^3* d^3*e*x*sgn(b*x + a) + 135*B*a*b^2*d^2*e^2*x*sgn(b*x + a) + 45*A*b^3*d^2*e ^2*x*sgn(b*x + a) + 270*B*a^2*b*d*e^3*x*sgn(b*x + a) + 270*A*a*b^2*d*e^3*x *sgn(b*x + a) + 315*B*a^3*e^4*x*sgn(b*x + a) + 945*A*a^2*b*e^4*x*sgn(b*x + a) + 4*B*b^3*d^4*sgn(b*x + a) + 15*B*a*b^2*d^3*e*sgn(b*x + a) + 5*A*b^3*d ^3*e*sgn(b*x + a) + 30*B*a^2*b*d^2*e^2*sgn(b*x + a) + 30*A*a*b^2*d^2*e^2*s gn(b*x + a) + 35*B*a^3*d*e^3*sgn(b*x + a) + 105*A*a^2*b*d*e^3*sgn(b*x + a) + 280*A*a^3*e^4*sgn(b*x + a))/((e*x + d)^9*e^5)
Time = 10.94 (sec) , antiderivative size = 577, normalized size of antiderivative = 1.94 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{10}} \, dx=-\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{6\,e^5}-\frac {B\,b^3\,d}{6\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {A\,a^3}{9\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{9\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{9\,e}-\frac {B\,b^3\,d}{9\,e^2}\right )}{e}-\frac {a\,b\,\left (A\,b+B\,a\right )}{3\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^9}-\frac {\left (\frac {B\,a^3\,e^3-3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e-3\,A\,a\,b^2\,d\,e^2-B\,b^3\,d^3+A\,b^3\,d^2\,e}{8\,e^5}-\frac {d\,\left (\frac {3\,B\,a^2\,b\,e^3-3\,B\,a\,b^2\,d\,e^2+3\,A\,a\,b^2\,e^3+B\,b^3\,d^2\,e-A\,b^3\,d\,e^2}{8\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{8\,e^3}-\frac {B\,b^3\,d}{8\,e^3}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}-\frac {\left (\frac {3\,B\,a^2\,b\,e^2-6\,B\,a\,b^2\,d\,e+3\,A\,a\,b^2\,e^2+3\,B\,b^3\,d^2-2\,A\,b^3\,d\,e}{7\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{7\,e^4}-\frac {B\,b^3\,d}{7\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \]
- (((A*b^3*e - 3*B*b^3*d + 3*B*a*b^2*e)/(6*e^5) - (B*b^3*d)/(6*e^5))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) - (((A*a^3)/(9*e) - (d *((B*a^3 + 3*A*a^2*b)/(9*e) + (d*((d*((A*b^3 + 3*B*a*b^2)/(9*e) - (B*b^3*d )/(9*e^2)))/e - (a*b*(A*b + B*a))/(3*e)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x) ^(1/2))/((a + b*x)*(d + e*x)^9) - (((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)/(8*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)/(8*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - B*b*d))/(8*e^3) - (B*b ^3*d)/(8*e^3)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e* x)^8) - (((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)/(7*e^5) - (d*((b^2*(A*b*e + 3*B*a*e - 2*B*b*d))/(7*e^4) - (B*b ^3*d)/(7*e^4)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^7 ) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*e^5*(a + b*x)*(d + e*x)^5)