Integrand size = 33, antiderivative size = 249 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(b d-a e)^2 (b B d+3 A b e-4 a B e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (3 b B d+A b e-3 a B e) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (b d-a e) (b B d+A b e-2 a B e) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-(-a*e+b*d)^2*(3*A*b*e-4*B*a*e+B*b*d)/b^5/((b*x+a)^2)^(1/2)-1/2*(A*b-B*a)* (-a*e+b*d)^3/b^5/(b*x+a)/((b*x+a)^2)^(1/2)+e^2*(A*b*e-3*B*a*e+3*B*b*d)*x*( b*x+a)/b^4/((b*x+a)^2)^(1/2)+1/2*B*e^3*x^2*(b*x+a)/b^3/((b*x+a)^2)^(1/2)+3 *e*(-a*e+b*d)*(A*b*e-2*B*a*e+B*b*d)*(b*x+a)*ln(b*x+a)/b^5/((b*x+a)^2)^(1/2 )
Time = 1.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-A b \left (5 a^3 e^3+a^2 b e^2 (-9 d+4 e x)+a b^2 e \left (3 d^2-12 d e x-4 e^2 x^2\right )+b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )+B \left (7 a^4 e^3+a^3 b e^2 (-15 d+2 e x)+a^2 b^2 e \left (9 d^2-12 d e x-11 e^2 x^2\right )+b^4 x \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )-a b^3 \left (d^3-12 d^2 e x-12 d e^2 x^2+4 e^3 x^3\right )\right )+6 e (b d-a e) (b B d+A b e-2 a B e) (a+b x)^2 \log (a+b x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \]
(-(A*b*(5*a^3*e^3 + a^2*b*e^2*(-9*d + 4*e*x) + a*b^2*e*(3*d^2 - 12*d*e*x - 4*e^2*x^2) + b^3*(d^3 + 6*d^2*e*x - 2*e^3*x^3))) + B*(7*a^4*e^3 + a^3*b*e ^2*(-15*d + 2*e*x) + a^2*b^2*e*(9*d^2 - 12*d*e*x - 11*e^2*x^2) + b^4*x*(-2 *d^3 + 6*d*e^2*x^2 + e^3*x^3) - a*b^3*(d^3 - 12*d^2*e*x - 12*d*e^2*x^2 + 4 *e^3*x^3)) + 6*e*(b*d - a*e)*(b*B*d + A*b*e - 2*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])
Time = 0.40 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.67, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(A+B x) (d+e x)^3}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^3}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {B x e^3}{b^3}+\frac {(3 b B d+A b e-3 a B e) e^2}{b^4}+\frac {3 (b d-a e) (b B d+A b e-2 a B e) e}{b^4 (a+b x)}+\frac {(b d-a e)^2 (b B d+3 A b e-4 a B e)}{b^4 (a+b x)^2}+\frac {(A b-a B) (b d-a e)^3}{b^4 (a+b x)^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 (a+b x)}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x)^2}+\frac {3 e (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5}+\frac {e^2 x (-3 a B e+A b e+3 b B d)}{b^4}+\frac {B e^3 x^2}{2 b^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((e^2*(3*b*B*d + A*b*e - 3*a*B*e)*x)/b^4 + (B*e^3*x^2)/(2*b^3) - ((A*b - a*B)*(b*d - a*e)^3)/(2*b^5*(a + b*x)^2) - ((b*d - a*e)^2*(b*B*d + 3*A*b*e - 4*a*B*e))/(b^5*(a + b*x)) + (3*e*(b*d - a*e)*(b*B*d + A*b*e - 2*a*B*e)*Log[a + b*x])/b^5))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.18.68.3.1 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 = 0.61 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.22
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (\frac {1}{2} B b e \,x^{2}+A b e x -3 B a e x +3 B b d x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 A \,a^{2} b \,e^{3}+6 A a \,b^{2} d \,e^{2}-3 A \,b^{3} d^{2} e +4 B \,e^{3} a^{3}-9 B \,a^{2} b d \,e^{2}+6 B a \,b^{2} d^{2} e -B \,b^{3} d^{3}\right ) x -\frac {5 A \,a^{3} b \,e^{3}-9 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e +A \,d^{3} b^{4}-7 B \,e^{3} a^{4}+15 B \,a^{3} b d \,e^{2}-9 B \,a^{2} b^{2} d^{2} e +B a \,b^{3} d^{3}}{2 b}\right )}{\left (b x +a \right )^{3} b^{4}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, e \left (A a b \,e^{2}-A \,b^{2} d e -2 a^{2} B \,e^{2}+3 B a b d e -B \,b^{2} d^{2}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(305\) |
| default | \(-\frac {\left (-7 B \,e^{3} a^{4}+A \,d^{3} b^{4}-12 A x a \,b^{3} d \,e^{2}-12 B x a \,b^{3} d^{2} e +12 B x \,a^{2} b^{2} d \,e^{2}-6 B \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e -12 B \,x^{2} a \,b^{3} d \,e^{2}+6 A \ln \left (b x +a \right ) x^{2} a \,b^{3} e^{3}-12 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{3}+18 B \ln \left (b x +a \right ) a^{3} b d \,e^{2}-24 B \ln \left (b x +a \right ) x \,a^{3} b \,e^{3}-6 A \ln \left (b x +a \right ) a^{2} b^{2} d \,e^{2}+12 A \ln \left (b x +a \right ) x \,a^{2} b^{2} e^{3}-6 A \ln \left (b x +a \right ) b^{4} d \,e^{2} x^{2}-6 B \ln \left (b x +a \right ) b^{4} d^{2} e \,x^{2}-12 B \ln \left (b x +a \right ) a^{4} e^{3}-2 A \,x^{3} b^{4} e^{3}-B \,x^{4} e^{3} b^{4}+B a \,b^{3} d^{3}+5 A \,a^{3} b \,e^{3}+2 B \,b^{4} d^{3} x +6 A \ln \left (b x +a \right ) a^{3} b \,e^{3}+4 B \,x^{3} a \,b^{3} e^{3}-6 B \,x^{3} b^{4} d \,e^{2}-4 A \,x^{2} a \,b^{3} e^{3}+11 B \,x^{2} a^{2} b^{2} e^{3}+15 B \,a^{3} b d \,e^{2}-9 B \,a^{2} b^{2} d^{2} e -12 A \ln \left (b x +a \right ) x a \,b^{3} d \,e^{2}+36 B \ln \left (b x +a \right ) x \,a^{2} b^{2} d \,e^{2}-12 B \ln \left (b x +a \right ) x a \,b^{3} d^{2} e +6 A \,b^{4} d^{2} e x -2 B \,a^{3} b \,e^{3} x +3 A a \,b^{3} d^{2} e -9 A \,a^{2} b^{2} d \,e^{2}+18 B \ln \left (b x +a \right ) x^{2} a \,b^{3} d \,e^{2}+4 A x \,a^{2} b^{2} e^{3}\right ) \left (b x +a \right )}{2 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(556\) |
((b*x+a)^2)^(1/2)/(b*x+a)*e^2/b^4*(1/2*B*b*e*x^2+A*b*e*x-3*B*a*e*x+3*B*b*d *x)+((b*x+a)^2)^(1/2)/(b*x+a)^3*((-3*A*a^2*b*e^3+6*A*a*b^2*d*e^2-3*A*b^3*d ^2*e+4*B*a^3*e^3-9*B*a^2*b*d*e^2+6*B*a*b^2*d^2*e-B*b^3*d^3)*x-1/2*(5*A*a^3 *b*e^3-9*A*a^2*b^2*d*e^2+3*A*a*b^3*d^2*e+A*b^4*d^3-7*B*a^4*e^3+15*B*a^3*b* d*e^2-9*B*a^2*b^2*d^2*e+B*a*b^3*d^3)/b)/b^4-3*((b*x+a)^2)^(1/2)/(b*x+a)/b^ 5*e*(A*a*b*e^2-A*b^2*d*e-2*B*a^2*e^2+3*B*a*b*d*e-B*b^2*d^2)*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (190) = 380\).
Time = 0.32 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.78 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B b^{4} e^{3} x^{4} - {\left (B a b^{3} + A b^{4}\right )} d^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e - 3 \, {\left (5 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} d e^{2} + {\left (7 \, B a^{4} - 5 \, A a^{3} b\right )} e^{3} + 2 \, {\left (3 \, B b^{4} d e^{2} - {\left (2 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + {\left (12 \, B a b^{3} d e^{2} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} e^{3}\right )} x^{2} - 2 \, {\left (B b^{4} d^{3} - 3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} d^{2} e + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} - {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (B a^{2} b^{2} d^{2} e - {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} d e^{2} + {\left (2 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 2 \, {\left (B a b^{3} d^{2} e - {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d e^{2} + {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]
1/2*(B*b^4*e^3*x^4 - (B*a*b^3 + A*b^4)*d^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*d^2 *e - 3*(5*B*a^3*b - 3*A*a^2*b^2)*d*e^2 + (7*B*a^4 - 5*A*a^3*b)*e^3 + 2*(3* B*b^4*d*e^2 - (2*B*a*b^3 - A*b^4)*e^3)*x^3 + (12*B*a*b^3*d*e^2 - (11*B*a^2 *b^2 - 4*A*a*b^3)*e^3)*x^2 - 2*(B*b^4*d^3 - 3*(2*B*a*b^3 - A*b^4)*d^2*e + 6*(B*a^2*b^2 - A*a*b^3)*d*e^2 - (B*a^3*b - 2*A*a^2*b^2)*e^3)*x + 6*(B*a^2* b^2*d^2*e - (3*B*a^3*b - A*a^2*b^2)*d*e^2 + (2*B*a^4 - A*a^3*b)*e^3 + (B*b ^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 2*(B*a*b^3*d^2*e - (3*B*a^2*b^2 - A*a*b^3)*d*e^2 + (2*B*a^3*b - A*a^2*b^2) *e^3)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)
\[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (190) = 380\).
Time = 0.20 (sec) , antiderivative size = 487, normalized size of antiderivative = 1.96 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B e^{3} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, B a e^{3} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {6 \, B a^{2} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {5 \, B a^{3} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, B a^{3} e^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, B a^{4} e^{3}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {B d^{3} + 3 \, A d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {6 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (B d^{2} e + A d e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, {\left (B d^{2} e + A d e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \]
1/2*B*e^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*e^3*x^2/(sqrt( b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*B*a^2*e^3*log(x + a/b)/b^5 - 5*B*a^3*e^3 /(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + (3*B*d*e^2 + A*e^3)*x^2/(sqrt(b^2*x ^2 + 2*a*b*x + a^2)*b^2) + 12*B*a^3*e^3*x/(b^6*(x + a/b)^2) - 1/2*A*d^3/(b ^3*(x + a/b)^2) + 23/2*B*a^4*e^3/(b^7*(x + a/b)^2) - 3*(3*B*d*e^2 + A*e^3) *a*log(x + a/b)/b^4 + 3*(B*d^2*e + A*d*e^2)*log(x + a/b)/b^3 + 2*(3*B*d*e^ 2 + A*e^3)*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (B*d^3 + 3*A*d^2*e)/( sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 6*(3*B*d*e^2 + A*e^3)*a^2*x/(b^5*(x + a/b)^2) + 6*(B*d^2*e + A*d*e^2)*a*x/(b^4*(x + a/b)^2) - 11/2*(3*B*d*e^2 + A*e^3)*a^3/(b^6*(x + a/b)^2) + 9/2*(B*d^2*e + A*d*e^2)*a^2/(b^5*(x + a/b) ^2) + 1/2*(B*d^3 + 3*A*d^2*e)*a/(b^4*(x + a/b)^2)
Time = 0.29 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, {\left (B b^{2} d^{2} e - 3 \, B a b d e^{2} + A b^{2} d e^{2} + 2 \, B a^{2} e^{3} - A a b e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {B b^{3} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B b^{3} d e^{2} x \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b^{2} e^{3} x \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} e^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{6}} - \frac {B a b^{3} d^{3} + A b^{4} d^{3} - 9 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 15 \, B a^{3} b d e^{2} - 9 \, A a^{2} b^{2} d e^{2} - 7 \, B a^{4} e^{3} + 5 \, A a^{3} b e^{3} + 2 \, {\left (B b^{4} d^{3} - 6 \, B a b^{3} d^{2} e + 3 \, A b^{4} d^{2} e + 9 \, B a^{2} b^{2} d e^{2} - 6 \, A a b^{3} d e^{2} - 4 \, B a^{3} b e^{3} + 3 \, A a^{2} b^{2} e^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} \]
3*(B*b^2*d^2*e - 3*B*a*b*d*e^2 + A*b^2*d*e^2 + 2*B*a^2*e^3 - A*a*b*e^3)*lo g(abs(b*x + a))/(b^5*sgn(b*x + a)) + 1/2*(B*b^3*e^3*x^2*sgn(b*x + a) + 6*B *b^3*d*e^2*x*sgn(b*x + a) - 6*B*a*b^2*e^3*x*sgn(b*x + a) + 2*A*b^3*e^3*x*s gn(b*x + a))/b^6 - 1/2*(B*a*b^3*d^3 + A*b^4*d^3 - 9*B*a^2*b^2*d^2*e + 3*A* a*b^3*d^2*e + 15*B*a^3*b*d*e^2 - 9*A*a^2*b^2*d*e^2 - 7*B*a^4*e^3 + 5*A*a^3 *b*e^3 + 2*(B*b^4*d^3 - 6*B*a*b^3*d^2*e + 3*A*b^4*d^2*e + 9*B*a^2*b^2*d*e^ 2 - 6*A*a*b^3*d*e^2 - 4*B*a^3*b*e^3 + 3*A*a^2*b^2*e^3)*x)/((b*x + a)^2*b^5 *sgn(b*x + a))
Timed out. \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]