Integrand size = 35, antiderivative size = 304 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 (b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{5/2}}+\frac {2 (b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}-\frac {6 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}-\frac {2 b^2 (4 b B d-A b e-3 a B e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {2 b^3 B (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)} \]
-2/5*(-a*e+b*d)^3*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(5/2)+2 /3*(-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)^(3/2)+2/3*b^3*B*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)-6*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)^(1/2)-2 *b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (a^3 e^3 (2 B d+3 A e+5 B e x)+3 a^2 b e^2 \left (A e (2 d+5 e x)+B \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )-3 a b^2 e \left (-A e \left (8 d^2+20 d e x+15 e^2 x^2\right )+3 B \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )+b^3 \left (-3 A e \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )+B \left (128 d^4+320 d^3 e x+240 d^2 e^2 x^2+40 d e^3 x^3-5 e^4 x^4\right )\right )\right )}{15 e^5 (a+b x) (d+e x)^{5/2}} \]
(-2*Sqrt[(a + b*x)^2]*(a^3*e^3*(2*B*d + 3*A*e + 5*B*e*x) + 3*a^2*b*e^2*(A* e*(2*d + 5*e*x) + B*(8*d^2 + 20*d*e*x + 15*e^2*x^2)) - 3*a*b^2*e*(-(A*e*(8 *d^2 + 20*d*e*x + 15*e^2*x^2)) + 3*B*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3)) + b^3*(-3*A*e*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3 ) + B*(128*d^4 + 320*d^3*e*x + 240*d^2*e^2*x^2 + 40*d*e^3*x^3 - 5*e^4*x^4) )))/(15*e^5*(a + b*x)*(d + e*x)^(5/2))
Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^{7/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{7/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B \sqrt {d+e x} b^3}{e^4}+\frac {(-4 b B d+A b e+3 a B e) b^2}{e^4 \sqrt {d+e x}}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 (d+e x)^{3/2}}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^{5/2}}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^{7/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 b^2 \sqrt {d+e x} (-3 a B e-A b e+4 b B d)}{e^5}-\frac {6 b (b d-a e) (-a B e-A b e+2 b B d)}{e^5 \sqrt {d+e x}}+\frac {2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{3 e^5 (d+e x)^{3/2}}-\frac {2 (b d-a e)^3 (B d-A e)}{5 e^5 (d+e x)^{5/2}}+\frac {2 b^3 B (d+e x)^{3/2}}{3 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^3*(B*d - A*e))/(5*e^5*(d + e*x)^(5/2)) + (2*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(3*e^5*(d + e *x)^(3/2)) - (6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e))/(e^5*Sqrt[d + e*x ]) - (2*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*Sqrt[d + e*x])/e^5 + (2*b^3*B*(d + e*x)^(3/2))/(3*e^5)))/(a + b*x)
3.19.50.3.1 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 = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.83
| method | result | size |
| risch | \(\frac {2 b^{2} \left (B b e x +3 A b e +9 B a e -11 B b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 e^{5} \left (b x +a \right )}-\frac {2 \left (45 A \,b^{2} e^{3} x^{2}+45 B \,x^{2} a b \,e^{3}-90 B \,x^{2} b^{2} d \,e^{2}+15 A x a b \,e^{3}+75 A x \,b^{2} d \,e^{2}+5 B x \,a^{2} e^{3}+65 B x a b d \,e^{2}-160 B x \,b^{2} d^{2} e +3 A \,a^{2} e^{3}+9 A a b d \,e^{2}+33 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+26 B a b \,d^{2} e -73 B \,b^{2} d^{3}\right ) \left (a e -b d \right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{5} \sqrt {e x +d}\, \left (e^{2} x^{2}+2 d e x +d^{2}\right ) \left (b x +a \right )}\) | \(252\) |
| gosper | \(-\frac {2 \left (-5 B \,b^{3} e^{4} x^{4}-15 A \,b^{3} e^{4} x^{3}-45 B a \,b^{2} e^{4} x^{3}+40 B \,b^{3} d \,e^{3} x^{3}+45 A a \,b^{2} e^{4} x^{2}-90 A \,b^{3} d \,e^{3} x^{2}+45 B \,a^{2} b \,e^{4} x^{2}-270 B a \,b^{2} d \,e^{3} x^{2}+240 B \,b^{3} d^{2} e^{2} x^{2}+15 A \,a^{2} b \,e^{4} x +60 A a \,b^{2} d \,e^{3} x -120 A \,b^{3} d^{2} e^{2} x +5 B \,a^{3} e^{4} x +60 B \,a^{2} b d \,e^{3} x -360 B a \,b^{2} d^{2} e^{2} x +320 B \,b^{3} d^{3} e x +3 A \,a^{3} e^{4}+6 A \,a^{2} b d \,e^{3}+24 A a \,b^{2} d^{2} e^{2}-48 A \,b^{3} d^{3} e +2 B \,a^{3} d \,e^{3}+24 B \,a^{2} b \,d^{2} e^{2}-144 B a \,b^{2} d^{3} e +128 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(317\) |
| default | \(-\frac {2 \left (-5 B \,b^{3} e^{4} x^{4}-15 A \,b^{3} e^{4} x^{3}-45 B a \,b^{2} e^{4} x^{3}+40 B \,b^{3} d \,e^{3} x^{3}+45 A a \,b^{2} e^{4} x^{2}-90 A \,b^{3} d \,e^{3} x^{2}+45 B \,a^{2} b \,e^{4} x^{2}-270 B a \,b^{2} d \,e^{3} x^{2}+240 B \,b^{3} d^{2} e^{2} x^{2}+15 A \,a^{2} b \,e^{4} x +60 A a \,b^{2} d \,e^{3} x -120 A \,b^{3} d^{2} e^{2} x +5 B \,a^{3} e^{4} x +60 B \,a^{2} b d \,e^{3} x -360 B a \,b^{2} d^{2} e^{2} x +320 B \,b^{3} d^{3} e x +3 A \,a^{3} e^{4}+6 A \,a^{2} b d \,e^{3}+24 A a \,b^{2} d^{2} e^{2}-48 A \,b^{3} d^{3} e +2 B \,a^{3} d \,e^{3}+24 B \,a^{2} b \,d^{2} e^{2}-144 B a \,b^{2} d^{3} e +128 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(317\) |
2/3*b^2*(B*b*e*x+3*A*b*e+9*B*a*e-11*B*b*d)*(e*x+d)^(1/2)/e^5*((b*x+a)^2)^( 1/2)/(b*x+a)-2/15*(45*A*b^2*e^3*x^2+45*B*a*b*e^3*x^2-90*B*b^2*d*e^2*x^2+15 *A*a*b*e^3*x+75*A*b^2*d*e^2*x+5*B*a^2*e^3*x+65*B*a*b*d*e^2*x-160*B*b^2*d^2 *e*x+3*A*a^2*e^3+9*A*a*b*d*e^2+33*A*b^2*d^2*e+2*B*a^2*d*e^2+26*B*a*b*d^2*e -73*B*b^2*d^3)*(a*e-b*d)/e^5/(e*x+d)^(1/2)/(e^2*x^2+2*d*e*x+d^2)*((b*x+a)^ 2)^(1/2)/(b*x+a)
Time = 0.32 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, B b^{3} e^{4} x^{4} - 128 \, B b^{3} d^{4} - 3 \, A a^{3} e^{4} + 48 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 24 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 5 \, {\left (8 \, B b^{3} d e^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} - 15 \, {\left (16 \, B b^{3} d^{2} e^{2} - 6 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} - 5 \, {\left (64 \, B b^{3} d^{3} e - 24 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 12 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]
2/15*(5*B*b^3*e^4*x^4 - 128*B*b^3*d^4 - 3*A*a^3*e^4 + 48*(3*B*a*b^2 + A*b^ 3)*d^3*e - 24*(B*a^2*b + A*a*b^2)*d^2*e^2 - 2*(B*a^3 + 3*A*a^2*b)*d*e^3 - 5*(8*B*b^3*d*e^3 - 3*(3*B*a*b^2 + A*b^3)*e^4)*x^3 - 15*(16*B*b^3*d^2*e^2 - 6*(3*B*a*b^2 + A*b^3)*d*e^3 + 3*(B*a^2*b + A*a*b^2)*e^4)*x^2 - 5*(64*B*b^ 3*d^3*e - 24*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 12*(B*a^2*b + A*a*b^2)*d*e^3 + (B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e ^6*x + d^3*e^5)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} A}{5 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 144 \, a b^{2} d^{3} e - 24 \, a^{2} b d^{2} e^{2} - 2 \, a^{3} d e^{3} - 5 \, {\left (8 \, b^{3} d e^{3} - 9 \, a b^{2} e^{4}\right )} x^{3} - 15 \, {\left (16 \, b^{3} d^{2} e^{2} - 18 \, a b^{2} d e^{3} + 3 \, a^{2} b e^{4}\right )} x^{2} - 5 \, {\left (64 \, b^{3} d^{3} e - 72 \, a b^{2} d^{2} e^{2} + 12 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x\right )} B}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )} \sqrt {e x + d}} \]
2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 + 5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2* b*e^3)*x)*A/((e^6*x^2 + 2*d*e^5*x + d^2*e^4)*sqrt(e*x + d)) + 2/15*(5*b^3* e^4*x^4 - 128*b^3*d^4 + 144*a*b^2*d^3*e - 24*a^2*b*d^2*e^2 - 2*a^3*d*e^3 - 5*(8*b^3*d*e^3 - 9*a*b^2*e^4)*x^3 - 15*(16*b^3*d^2*e^2 - 18*a*b^2*d*e^3 + 3*a^2*b*e^4)*x^2 - 5*(64*b^3*d^3*e - 72*a*b^2*d^2*e^2 + 12*a^2*b*d*e^3 + a^3*e^4)*x)*B/((e^7*x^2 + 2*d*e^6*x + d^2*e^5)*sqrt(e*x + d))
Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (233) = 466\).
Time = 0.29 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (90 \, {\left (e x + d\right )}^{2} B b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (e x + d\right )} B b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 135 \, {\left (e x + d\right )}^{2} B a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 45 \, {\left (e x + d\right )}^{2} A b^{3} d e \mathrm {sgn}\left (b x + a\right ) + 45 \, {\left (e x + d\right )} B a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (e x + d\right )} A b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 45 \, {\left (e x + d\right )}^{2} B a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) + 45 \, {\left (e x + d\right )}^{2} A a b^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 30 \, {\left (e x + d\right )} B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 30 \, {\left (e x + d\right )} A a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )} B a^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (e x + d\right )} A a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{3} e^{10} \mathrm {sgn}\left (b x + a\right ) - 12 \, \sqrt {e x + d} B b^{3} d e^{10} \mathrm {sgn}\left (b x + a\right ) + 9 \, \sqrt {e x + d} B a b^{2} e^{11} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {e x + d} A b^{3} e^{11} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, e^{15}} \]
-2/15*(90*(e*x + d)^2*B*b^3*d^2*sgn(b*x + a) - 20*(e*x + d)*B*b^3*d^3*sgn( b*x + a) + 3*B*b^3*d^4*sgn(b*x + a) - 135*(e*x + d)^2*B*a*b^2*d*e*sgn(b*x + a) - 45*(e*x + d)^2*A*b^3*d*e*sgn(b*x + a) + 45*(e*x + d)*B*a*b^2*d^2*e* sgn(b*x + a) + 15*(e*x + d)*A*b^3*d^2*e*sgn(b*x + a) - 9*B*a*b^2*d^3*e*sgn (b*x + a) - 3*A*b^3*d^3*e*sgn(b*x + a) + 45*(e*x + d)^2*B*a^2*b*e^2*sgn(b* x + a) + 45*(e*x + d)^2*A*a*b^2*e^2*sgn(b*x + a) - 30*(e*x + d)*B*a^2*b*d* e^2*sgn(b*x + a) - 30*(e*x + d)*A*a*b^2*d*e^2*sgn(b*x + a) + 9*B*a^2*b*d^2 *e^2*sgn(b*x + a) + 9*A*a*b^2*d^2*e^2*sgn(b*x + a) + 5*(e*x + d)*B*a^3*e^3 *sgn(b*x + a) + 15*(e*x + d)*A*a^2*b*e^3*sgn(b*x + a) - 3*B*a^3*d*e^3*sgn( b*x + a) - 9*A*a^2*b*d*e^3*sgn(b*x + a) + 3*A*a^3*e^4*sgn(b*x + a))/((e*x + d)^(5/2)*e^5) + 2/3*((e*x + d)^(3/2)*B*b^3*e^10*sgn(b*x + a) - 12*sqrt(e *x + d)*B*b^3*d*e^10*sgn(b*x + a) + 9*sqrt(e*x + d)*B*a*b^2*e^11*sgn(b*x + a) + 3*sqrt(e*x + d)*A*b^3*e^11*sgn(b*x + a))/e^15
Time = 11.87 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {4\,B\,a^3\,d\,e^3+6\,A\,a^3\,e^4+48\,B\,a^2\,b\,d^2\,e^2+12\,A\,a^2\,b\,d\,e^3-288\,B\,a\,b^2\,d^3\,e+48\,A\,a\,b^2\,d^2\,e^2+256\,B\,b^3\,d^4-96\,A\,b^3\,d^3\,e}{15\,b\,e^7}+\frac {2\,x^2\,\left (3\,B\,a^2\,e^2-18\,B\,a\,b\,d\,e+3\,A\,a\,b\,e^2+16\,B\,b^2\,d^2-6\,A\,b^2\,d\,e\right )}{e^5}+\frac {x\,\left (10\,B\,a^3\,e^4+120\,B\,a^2\,b\,d\,e^3+30\,A\,a^2\,b\,e^4-720\,B\,a\,b^2\,d^2\,e^2+120\,A\,a\,b^2\,d\,e^3+640\,B\,b^3\,d^3\,e-240\,A\,b^3\,d^2\,e^2\right )}{15\,b\,e^7}-\frac {2\,b\,x^3\,\left (3\,A\,b\,e+9\,B\,a\,e-8\,B\,b\,d\right )}{3\,e^4}-\frac {2\,B\,b^2\,x^4}{3\,e^3}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^7+30\,b\,d\,e^6\right )\,\sqrt {d+e\,x}}{15\,b\,e^7}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((6*A*a^3*e^4 + 256*B*b^3*d^4 - 96*A*b^3 *d^3*e + 4*B*a^3*d*e^3 + 48*A*a*b^2*d^2*e^2 + 48*B*a^2*b*d^2*e^2 + 12*A*a^ 2*b*d*e^3 - 288*B*a*b^2*d^3*e)/(15*b*e^7) + (2*x^2*(3*B*a^2*e^2 + 16*B*b^2 *d^2 + 3*A*a*b*e^2 - 6*A*b^2*d*e - 18*B*a*b*d*e))/e^5 + (x*(10*B*a^3*e^4 + 30*A*a^2*b*e^4 + 640*B*b^3*d^3*e - 240*A*b^3*d^2*e^2 - 720*B*a*b^2*d^2*e^ 2 + 120*A*a*b^2*d*e^3 + 120*B*a^2*b*d*e^3))/(15*b*e^7) - (2*b*x^3*(3*A*b*e + 9*B*a*e - 8*B*b*d))/(3*e^4) - (2*B*b^2*x^4)/(3*e^3)))/(x^3*(d + e*x)^(1 /2) + (a*d^2*(d + e*x)^(1/2))/(b*e^2) + (x^2*(15*a*e^7 + 30*b*d*e^6)*(d + e*x)^(1/2))/(15*b*e^7) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))