Integrand size = 35, antiderivative size = 304 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt {d+e x}}+\frac {6 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac {2 b^2 (4 b B d-A b e-3 a B e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}+\frac {2 b^3 B (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)} \]
-2/3*(-a*e+b*d)^3*(-A*e+B*d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)/(e*x+d)^(3/2)-2 /3*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a )+2/5*b^3*B*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)+2*(-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)^(1/2)+6*b*(-a* e+b*d)*(-A*b*e-B*a*e+2*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 = 243, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (5 a^3 e^3 (2 B d+A e+3 B e x)-15 a^2 b e^2 \left (-A e (2 d+3 e x)+B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )-15 a b^2 e \left (A e \left (8 d^2+12 d e x+3 e^2 x^2\right )+B \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (5 A e \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )-B \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]
(-2*Sqrt[(a + b*x)^2]*(5*a^3*e^3*(2*B*d + A*e + 3*B*e*x) - 15*a^2*b*e^2*(- (A*e*(2*d + 3*e*x)) + B*(8*d^2 + 12*d*e*x + 3*e^2*x^2)) - 15*a*b^2*e*(A*e* (8*d^2 + 12*d*e*x + 3*e^2*x^2) + B*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e ^3*x^3)) + b^3*(5*A*e*(16*d^3 + 24*d^2*e*x + 6*d*e^2*x^2 - e^3*x^3) - B*(1 28*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4))))/(15*e^ 5*(a + b*x)*(d + e*x)^(3/2))
Time = 0.34 (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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^{5/2}}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B (d+e x)^{3/2} b^3}{e^4}+\frac {(-4 b B d+A b e+3 a B e) \sqrt {d+e x} b^2}{e^4}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 \sqrt {d+e x}}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^{3/2}}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^{5/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac {6 b \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac {2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^3*(B*d - A*e))/(3*e^5*(d + e*x)^(3/2)) + (2*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(e^5*Sqrt[d + e*x]) + (6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*Sqrt[d + e*x])/e^5 - ( 2*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^(3/2))/(3*e^5) + (2*b^3*B*(d + e*x)^(5/2))/(5*e^5)))/(a + b*x)
3.19.49.3.1 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 = 211, normalized size of antiderivative = 0.69
| method | result | size |
| risch | \(\frac {2 b \left (3 B \,b^{2} e^{2} x^{2}+5 A \,b^{2} e^{2} x +15 B a b \,e^{2} x -14 B \,b^{2} d e x +45 A a b \,e^{2}-40 A \,b^{2} d e +45 a^{2} B \,e^{2}-120 B a b d e +73 B \,b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 e^{5} \left (b x +a \right )}-\frac {2 \left (9 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+8 A b d e +2 B a d e -11 B b \,d^{2}\right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\) | \(211\) |
| gosper | \(-\frac {2 \left (-3 B \,b^{3} e^{4} x^{4}-5 A \,b^{3} e^{4} x^{3}-15 B a \,b^{2} e^{4} x^{3}+8 B \,b^{3} d \,e^{3} x^{3}-45 A a \,b^{2} e^{4} x^{2}+30 A \,b^{3} d \,e^{3} x^{2}-45 B \,a^{2} b \,e^{4} x^{2}+90 B a \,b^{2} d \,e^{3} x^{2}-48 B \,b^{3} d^{2} e^{2} x^{2}+45 A \,a^{2} b \,e^{4} x -180 A a \,b^{2} d \,e^{3} x +120 A \,b^{3} d^{2} e^{2} x +15 B \,a^{3} e^{4} x -180 B \,a^{2} b d \,e^{3} x +360 B a \,b^{2} d^{2} e^{2} x -192 B \,b^{3} d^{3} e x +5 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(317\) |
| default | \(-\frac {2 \left (-3 B \,b^{3} e^{4} x^{4}-5 A \,b^{3} e^{4} x^{3}-15 B a \,b^{2} e^{4} x^{3}+8 B \,b^{3} d \,e^{3} x^{3}-45 A a \,b^{2} e^{4} x^{2}+30 A \,b^{3} d \,e^{3} x^{2}-45 B \,a^{2} b \,e^{4} x^{2}+90 B a \,b^{2} d \,e^{3} x^{2}-48 B \,b^{3} d^{2} e^{2} x^{2}+45 A \,a^{2} b \,e^{4} x -180 A a \,b^{2} d \,e^{3} x +120 A \,b^{3} d^{2} e^{2} x +15 B \,a^{3} e^{4} x -180 B \,a^{2} b d \,e^{3} x +360 B a \,b^{2} d^{2} e^{2} x -192 B \,b^{3} d^{3} e x +5 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5} \left (b x +a \right )^{3}}\) | \(317\) |
2/15*b*(3*B*b^2*e^2*x^2+5*A*b^2*e^2*x+15*B*a*b*e^2*x-14*B*b^2*d*e*x+45*A*a *b*e^2-40*A*b^2*d*e+45*B*a^2*e^2-120*B*a*b*d*e+73*B*b^2*d^2)*(e*x+d)^(1/2) /e^5*((b*x+a)^2)^(1/2)/(b*x+a)-2/3*(9*A*b*e^2*x+3*B*a*e^2*x-12*B*b*d*e*x+A *a*e^2+8*A*b*d*e+2*B*a*d*e-11*B*b*d^2)*(a^2*e^2-2*a*b*d*e+b^2*d^2)/e^5/(e* x+d)^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.30 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (8 \, B b^{3} d e^{3} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (16 \, B b^{3} d^{2} e^{2} - 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \, {\left (64 \, B b^{3} d^{3} e - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]
2/15*(3*B*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^ 3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2 - 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b ^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^ 5)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} A}{3 \, {\left (e^{5} x + d e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 240 \, a b^{2} d^{3} e + 120 \, a^{2} b d^{2} e^{2} - 10 \, a^{3} d e^{3} - {\left (8 \, b^{3} d e^{3} - 15 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{3} d^{2} e^{2} - 30 \, a b^{2} d e^{3} + 15 \, a^{2} b e^{4}\right )} x^{2} + 3 \, {\left (64 \, b^{3} d^{3} e - 120 \, a b^{2} d^{2} e^{2} + 60 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x\right )} B}{15 \, {\left (e^{6} x + d e^{5}\right )} \sqrt {e x + d}} \]
2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 - 3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a ^2*b*e^3)*x)*A/((e^5*x + d*e^4)*sqrt(e*x + d)) + 2/15*(3*b^3*e^4*x^4 + 128 *b^3*d^4 - 240*a*b^2*d^3*e + 120*a^2*b*d^2*e^2 - 10*a^3*d*e^3 - (8*b^3*d*e ^3 - 15*a*b^2*e^4)*x^3 + 3*(16*b^3*d^2*e^2 - 30*a*b^2*d*e^3 + 15*a^2*b*e^4 )*x^2 + 3*(64*b^3*d^3*e - 120*a*b^2*d^2*e^2 + 60*a^2*b*d*e^3 - 5*a^3*e^4)* x)*B/((e^6*x + d*e^5)*sqrt(e*x + d))
Leaf count of result is larger than twice the leaf count of optimal. 508 vs. \(2 (233) = 466\).
Time = 0.30 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} B b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 27 \, {\left (e x + d\right )} B a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 9 \, {\left (e x + d\right )} A b^{3} d^{2} e \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 ) + 18 \, {\left (e x + d\right )} B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\left (e x + d\right )} A a b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} B a^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, {\left (e x + d\right )} A a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} e^{20} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d e^{20} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {e x + d} B b^{3} d^{2} e^{20} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} e^{21} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} e^{21} \mathrm {sgn}\left (b x + a\right ) - 135 \, \sqrt {e x + d} B a b^{2} d e^{21} \mathrm {sgn}\left (b x + a\right ) - 45 \, \sqrt {e x + d} A b^{3} d e^{21} \mathrm {sgn}\left (b x + a\right ) + 45 \, \sqrt {e x + d} B a^{2} b e^{22} \mathrm {sgn}\left (b x + a\right ) + 45 \, \sqrt {e x + d} A a b^{2} e^{22} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, e^{25}} \]
2/3*(12*(e*x + d)*B*b^3*d^3*sgn(b*x + a) - B*b^3*d^4*sgn(b*x + a) - 27*(e* x + d)*B*a*b^2*d^2*e*sgn(b*x + a) - 9*(e*x + d)*A*b^3*d^2*e*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) + 18*(e*x + d)*B* a^2*b*d*e^2*sgn(b*x + a) + 18*(e*x + d)*A*a*b^2*d*e^2*sgn(b*x + a) - 3*B*a ^2*b*d^2*e^2*sgn(b*x + a) - 3*A*a*b^2*d^2*e^2*sgn(b*x + a) - 3*(e*x + d)*B *a^3*e^3*sgn(b*x + a) - 9*(e*x + d)*A*a^2*b*e^3*sgn(b*x + a) + B*a^3*d*e^3 *sgn(b*x + a) + 3*A*a^2*b*d*e^3*sgn(b*x + a) - A*a^3*e^4*sgn(b*x + a))/((e *x + d)^(3/2)*e^5) + 2/15*(3*(e*x + d)^(5/2)*B*b^3*e^20*sgn(b*x + a) - 20* (e*x + d)^(3/2)*B*b^3*d*e^20*sgn(b*x + a) + 90*sqrt(e*x + d)*B*b^3*d^2*e^2 0*sgn(b*x + a) + 15*(e*x + d)^(3/2)*B*a*b^2*e^21*sgn(b*x + a) + 5*(e*x + d )^(3/2)*A*b^3*e^21*sgn(b*x + a) - 135*sqrt(e*x + d)*B*a*b^2*d*e^21*sgn(b*x + a) - 45*sqrt(e*x + d)*A*b^3*d*e^21*sgn(b*x + a) + 45*sqrt(e*x + d)*B*a^ 2*b*e^22*sgn(b*x + a) + 45*sqrt(e*x + d)*A*a*b^2*e^22*sgn(b*x + a))/e^25
Time = 11.84 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {x^2\,\left (6\,B\,a^2\,b\,e^4-12\,B\,a\,b^2\,d\,e^3+6\,A\,a\,b^2\,e^4+\frac {32\,B\,b^3\,d^2\,e^2}{5}-4\,A\,b^3\,d\,e^3\right )}{b\,e^6}-\frac {x\,\left (30\,B\,a^3\,e^4-360\,B\,a^2\,b\,d\,e^3+90\,A\,a^2\,b\,e^4+720\,B\,a\,b^2\,d^2\,e^2-360\,A\,a\,b^2\,d\,e^3-384\,B\,b^3\,d^3\,e+240\,A\,b^3\,d^2\,e^2\right )}{15\,b\,e^6}-\frac {\frac {4\,B\,a^3\,d\,e^3}{3}+\frac {2\,A\,a^3\,e^4}{3}-16\,B\,a^2\,b\,d^2\,e^2+4\,A\,a^2\,b\,d\,e^3+32\,B\,a\,b^2\,d^3\,e-16\,A\,a\,b^2\,d^2\,e^2-\frac {256\,B\,b^3\,d^4}{15}+\frac {32\,A\,b^3\,d^3\,e}{3}}{b\,e^6}+\frac {2\,b\,x^3\,\left (5\,A\,b\,e+15\,B\,a\,e-8\,B\,b\,d\right )}{15\,e^3}+\frac {2\,B\,b^2\,x^4}{5\,e^2}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (15\,a\,e^6+15\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{15\,b\,e^6}} \]
((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((x^2*(6*A*a*b^2*e^4 + 6*B*a^2*b*e^4 - 4* A*b^3*d*e^3 + (32*B*b^3*d^2*e^2)/5 - 12*B*a*b^2*d*e^3))/(b*e^6) - (x*(30*B *a^3*e^4 + 90*A*a^2*b*e^4 - 384*B*b^3*d^3*e + 240*A*b^3*d^2*e^2 + 720*B*a* b^2*d^2*e^2 - 360*A*a*b^2*d*e^3 - 360*B*a^2*b*d*e^3))/(15*b*e^6) - ((2*A*a ^3*e^4)/3 - (256*B*b^3*d^4)/15 + (32*A*b^3*d^3*e)/3 + (4*B*a^3*d*e^3)/3 - 16*A*a*b^2*d^2*e^2 - 16*B*a^2*b*d^2*e^2 + 4*A*a^2*b*d*e^3 + 32*B*a*b^2*d^3 *e)/(b*e^6) + (2*b*x^3*(5*A*b*e + 15*B*a*e - 8*B*b*d))/(15*e^3) + (2*B*b^2 *x^4)/(5*e^2)))/(x^2*(d + e*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(1 5*a*e^6 + 15*b*d*e^5)*(d + e*x)^(1/2))/(15*b*e^6))