\(\int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [429]

Optimal result

Integrand size = 33, antiderivative size = 248 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(A b-a B) e (b d-a e)^2 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (b d-a e) (a+b x) (d+e x)^2}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) (d+e x)^4}{4 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (b d-a e)^3 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

(A*b-B*a)*e*(-a*e+b*d)^2*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*(-a 
*e+b*d)*(b*x+a)*(e*x+d)^2/b^3/((b*x+a)^2)^(1/2)+1/3*(A*b-B*a)*(b*x+a)*(e*x 
+d)^3/b^2/((b*x+a)^2)^(1/2)+1/4*B*(b*x+a)*(e*x+d)^4/b/e/((b*x+a)^2)^(1/2)+ 
(A*b-B*a)*(-a*e+b*d)^3*(b*x+a)*ln(b*x+a)/b^5/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.75 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (b x \left (-12 a^3 B e^3+6 a^2 b e^2 (6 B d+2 A e+B e x)-2 a b^2 e \left (3 A e (6 d+e x)+B \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+b^3 \left (2 A e \left (18 d^2+9 d e x+2 e^2 x^2\right )+3 B \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right )\right )+12 (A b-a B) (b d-a e)^3 \log (a+b x)\right )}{12 b^5 \sqrt {(a+b x)^2}} \] Input:

Integrate[((A + B*x)*(d + e*x)^3)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

((a + b*x)*(b*x*(-12*a^3*B*e^3 + 6*a^2*b*e^2*(6*B*d + 2*A*e + B*e*x) - 2*a 
*b^2*e*(3*A*e*(6*d + e*x) + B*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + b^3*(2*A*e 
*(18*d^2 + 9*d*e*x + 2*e^2*x^2) + 3*B*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e 
^3*x^3))) + 12*(A*b - a*B)*(b*d - a*e)^3*Log[a + b*x]))/(12*b^5*Sqrt[(a + 
b*x)^2])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.60, 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.

\(\displaystyle \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b (a+b x) \int \frac {(A+B x) (d+e x)^3}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^3}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {(a+b x) \int \left (\frac {(A b-a B) (b d-a e)^3}{b^4 (a+b x)}+\frac {(A b-a B) e (b d-a e)^2}{b^4}+\frac {(A b-a B) e (d+e x) (b d-a e)}{b^3}+\frac {B (d+e x)^3}{b}+\frac {(A b-a B) e (d+e x)^2}{b^2}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) (b d-a e)^3 \log (a+b x)}{b^5}+\frac {e x (A b-a B) (b d-a e)^2}{b^4}+\frac {(d+e x)^2 (A b-a B) (b d-a e)}{2 b^3}+\frac {(d+e x)^3 (A b-a B)}{3 b^2}+\frac {B (d+e x)^4}{4 b e}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

Int[((A + B*x)*(d + e*x)^3)/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

((a + b*x)*(((A*b - a*B)*e*(b*d - a*e)^2*x)/b^4 + ((A*b - a*B)*(b*d - a*e) 
*(d + e*x)^2)/(2*b^3) + ((A*b - a*B)*(d + e*x)^3)/(3*b^2) + (B*(d + e*x)^4 
)/(4*b*e) + ((A*b - a*B)*(b*d - a*e)^3*Log[a + b*x])/b^5))/Sqrt[a^2 + 2*a* 
b*x + b^2*x^2]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 86

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]))))
 

rule 1187

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.31

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{4} b^{3} B \,x^{4} e^{3}+\frac {1}{3} A \,b^{3} e^{3} x^{3}-\frac {1}{3} B a \,b^{2} e^{3} x^{3}+B \,b^{3} d \,e^{2} x^{3}-\frac {1}{2} A a \,b^{2} e^{3} x^{2}+\frac {3}{2} A \,b^{3} d \,e^{2} x^{2}+\frac {1}{2} B \,a^{2} b \,e^{3} x^{2}-\frac {3}{2} B a \,b^{2} d \,e^{2} x^{2}+\frac {3}{2} B \,b^{3} d^{2} e \,x^{2}+A \,a^{2} b \,e^{3} x -3 A a \,b^{2} d \,e^{2} x +3 A \,b^{3} d^{2} e x -B \,e^{3} a^{3} x +3 B \,a^{2} b d \,e^{2} x -3 B a \,b^{2} d^{2} e x +B \,b^{3} d^{3} x \right )}{\left (b x +a \right ) b^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,a^{3} b \,e^{3}-3 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e -A \,d^{3} b^{4}-B \,e^{3} a^{4}+3 B \,a^{3} b d \,e^{2}-3 B \,a^{2} b^{2} d^{2} e +B a \,b^{3} d^{3}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(326\)
default	\(-\frac {\left (b x +a \right ) \left (-3 B \,x^{4} e^{3} b^{4}-4 A \,b^{4} e^{3} x^{3}+4 B \,e^{3} a \,x^{3} b^{3}-12 B \,b^{4} d \,e^{2} x^{3}+6 A \,x^{2} a \,b^{3} e^{3}-18 A \,x^{2} b^{4} d \,e^{2}-6 B \,x^{2} a^{2} b^{2} e^{3}+18 B \,x^{2} a \,b^{3} d \,e^{2}-18 B \,x^{2} b^{4} d^{2} e +12 A \ln \left (b x +a \right ) a^{3} b \,e^{3}-36 A \ln \left (b x +a \right ) a^{2} b^{2} d \,e^{2}+36 A \ln \left (b x +a \right ) a \,b^{3} d^{2} e -12 A \ln \left (b x +a \right ) b^{4} d^{3}-12 A x \,a^{2} b^{2} e^{3}+36 A x a \,b^{3} d \,e^{2}-36 A \,b^{4} d^{2} e x -12 B \ln \left (b x +a \right ) a^{4} e^{3}+36 B \ln \left (b x +a \right ) a^{3} b d \,e^{2}-36 B \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e +12 B \ln \left (b x +a \right ) a \,b^{3} d^{3}+12 B \,a^{3} b \,e^{3} x -36 B x \,a^{2} b^{2} d \,e^{2}+36 B x a \,b^{3} d^{2} e -12 B \,b^{4} d^{3} x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}}\)	\(356\)

Input:

int((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)/b^4*(1/4*b^3*B*x^4*e^3+1/3*A*b^3*e^3*x^3-1/3*B*a 
*b^2*e^3*x^3+B*b^3*d*e^2*x^3-1/2*A*a*b^2*e^3*x^2+3/2*A*b^3*d*e^2*x^2+1/2*B 
*a^2*b*e^3*x^2-3/2*B*a*b^2*d*e^2*x^2+3/2*B*b^3*d^2*e*x^2+A*a^2*b*e^3*x-3*A 
*a*b^2*d*e^2*x+3*A*b^3*d^2*e*x-B*e^3*a^3*x+3*B*a^2*b*d*e^2*x-3*B*a*b^2*d^2 
*e*x+B*b^3*d^3*x)-((b*x+a)^2)^(1/2)/(b*x+a)/b^5*(A*a^3*b*e^3-3*A*a^2*b^2*d 
*e^2+3*A*a*b^3*d^2*e-A*b^4*d^3-B*a^4*e^3+3*B*a^3*b*d*e^2-3*B*a^2*b^2*d^2*e 
+B*a*b^3*d^3)*ln(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, B b^{4} e^{3} x^{4} + 4 \, {\left (3 \, B b^{4} d e^{2} - {\left (B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 6 \, {\left (3 \, B b^{4} d^{2} e - 3 \, {\left (B a b^{3} - A b^{4}\right )} d e^{2} + {\left (B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 12 \, {\left (B b^{4} d^{3} - 3 \, {\left (B a b^{3} - A b^{4}\right )} d^{2} e + 3 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} - {\left (B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x - 12 \, {\left ({\left (B a b^{3} - A b^{4}\right )} d^{3} - 3 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 3 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} d e^{2} - {\left (B a^{4} - A a^{3} b\right )} e^{3}\right )} \log \left (b x + a\right )}{12 \, b^{5}} \] Input:

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

1/12*(3*B*b^4*e^3*x^4 + 4*(3*B*b^4*d*e^2 - (B*a*b^3 - A*b^4)*e^3)*x^3 + 6* 
(3*B*b^4*d^2*e - 3*(B*a*b^3 - A*b^4)*d*e^2 + (B*a^2*b^2 - A*a*b^3)*e^3)*x^ 
2 + 12*(B*b^4*d^3 - 3*(B*a*b^3 - A*b^4)*d^2*e + 3*(B*a^2*b^2 - A*a*b^3)*d* 
e^2 - (B*a^3*b - A*a^2*b^2)*e^3)*x - 12*((B*a*b^3 - A*b^4)*d^3 - 3*(B*a^2* 
b^2 - A*a*b^3)*d^2*e + 3*(B*a^3*b - A*a^2*b^2)*d*e^2 - (B*a^4 - A*a^3*b)*e 
^3)*log(b*x + a))/b^5
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1110 vs. \(2 (184) = 368\).

Time = 2.75 (sec) , antiderivative size = 1110, normalized size of antiderivative = 4.48 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)**3/((b*x+a)**2)**(1/2),x)
 

Output:

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*e**3*x**3/(4*b**2) + x**2*( 
A*e**3 - 7*B*a*e**3/(4*b) + 3*B*d*e**2)/(3*b**2) + x*(3*A*d*e**2 - 3*B*a** 
2*e**3/(4*b**2) + 3*B*d**2*e - 5*a*(A*e**3 - 7*B*a*e**3/(4*b) + 3*B*d*e**2 
)/(3*b))/(2*b**2) + (3*A*d**2*e + B*d**3 - 2*a**2*(A*e**3 - 7*B*a*e**3/(4* 
b) + 3*B*d*e**2)/(3*b**2) - 3*a*(3*A*d*e**2 - 3*B*a**2*e**3/(4*b**2) + 3*B 
*d**2*e - 5*a*(A*e**3 - 7*B*a*e**3/(4*b) + 3*B*d*e**2)/(3*b))/(2*b))/b**2) 
 + (a/b + x)*(A*d**3 - a**2*(3*A*d*e**2 - 3*B*a**2*e**3/(4*b**2) + 3*B*d** 
2*e - 5*a*(A*e**3 - 7*B*a*e**3/(4*b) + 3*B*d*e**2)/(3*b))/(2*b**2) - a*(3* 
A*d**2*e + B*d**3 - 2*a**2*(A*e**3 - 7*B*a*e**3/(4*b) + 3*B*d*e**2)/(3*b** 
2) - 3*a*(3*A*d*e**2 - 3*B*a**2*e**3/(4*b**2) + 3*B*d**2*e - 5*a*(A*e**3 - 
 7*B*a*e**3/(4*b) + 3*B*d*e**2)/(3*b))/(2*b))/b)*log(a/b + x)/sqrt(b**2*(a 
/b + x)**2), Ne(b**2, 0)), ((2*A*d**3*sqrt(a**2 + 2*a*b*x) + 3*A*d**2*e*(- 
a**2*sqrt(a**2 + 2*a*b*x) + (a**2 + 2*a*b*x)**(3/2)/3)/(a*b) + 3*A*d*e**2* 
(a**4*sqrt(a**2 + 2*a*b*x) - 2*a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2* 
a*b*x)**(5/2)/5)/(2*a**2*b**2) + A*e**3*(-a**6*sqrt(a**2 + 2*a*b*x) + a**4 
*(a**2 + 2*a*b*x)**(3/2) - 3*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a* 
b*x)**(7/2)/7)/(4*a**3*b**3) + B*d**3*(-a**2*sqrt(a**2 + 2*a*b*x) + (a**2 
+ 2*a*b*x)**(3/2)/3)/(a*b) + 3*B*d**2*e*(a**4*sqrt(a**2 + 2*a*b*x) - 2*a** 
2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2*a*b*x)**(5/2)/5)/(2*a**2*b**2) + 3 
*B*d*e**2*(-a**6*sqrt(a**2 + 2*a*b*x) + a**4*(a**2 + 2*a*b*x)**(3/2) - ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 438 vs. \(2 (189) = 378\).

Time = 0.05 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.77 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B e^{3} x^{3}}{4 \, b^{2}} + \frac {13 \, B a^{2} e^{3} x^{2}}{12 \, b^{3}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a e^{3} x^{2}}{12 \, b^{3}} - \frac {13 \, B a^{3} e^{3} x}{6 \, b^{4}} + \frac {A d^{3} \log \left (x + \frac {a}{b}\right )}{b} + \frac {B a^{4} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3} e^{3}}{6 \, b^{5}} - \frac {5 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a x^{2}}{6 \, b^{2}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} x^{2}}{2 \, b} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} x^{2}}{3 \, b^{2}} + \frac {5 \, {\left (3 \, B d e^{2} + A e^{3}\right )} a^{2} x}{3 \, b^{3}} - \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} a x}{b^{2}} - \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} a \log \left (x + \frac {a}{b}\right )}{b^{2}} - \frac {2 \, {\left (3 \, B d e^{2} + A e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{3 \, b^{4}} + \frac {{\left (B d^{3} + 3 \, A d^{2} e\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \] Input:

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*e^3*x^3/b^2 + 13/12*B*a^2*e^3*x^2/b^3 
- 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*e^3*x^2/b^3 - 13/6*B*a^3*e^3*x/b^ 
4 + A*d^3*log(x + a/b)/b + B*a^4*e^3*log(x + a/b)/b^5 + 7/6*sqrt(b^2*x^2 + 
 2*a*b*x + a^2)*B*a^3*e^3/b^5 - 5/6*(3*B*d*e^2 + A*e^3)*a*x^2/b^2 + 3/2*(B 
*d^2*e + A*d*e^2)*x^2/b + 1/3*(3*B*d*e^2 + A*e^3)*sqrt(b^2*x^2 + 2*a*b*x + 
 a^2)*x^2/b^2 + 5/3*(3*B*d*e^2 + A*e^3)*a^2*x/b^3 - 3*(B*d^2*e + A*d*e^2)* 
a*x/b^2 - (3*B*d*e^2 + A*e^3)*a^3*log(x + a/b)/b^4 + 3*(B*d^2*e + A*d*e^2) 
*a^2*log(x + a/b)/b^3 - (B*d^3 + 3*A*d^2*e)*a*log(x + a/b)/b^2 - 2/3*(3*B* 
d*e^2 + A*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (189) = 378\).

Time = 0.17 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.78 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, B b^{3} e^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 12 \, B b^{3} d e^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, B a b^{2} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, A b^{3} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, B b^{3} d^{2} e x^{2} \mathrm {sgn}\left (b x + a\right ) - 18 \, B a b^{2} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, A b^{3} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, A a b^{2} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B b^{3} d^{3} x \mathrm {sgn}\left (b x + a\right ) - 36 \, B a b^{2} d^{2} e x \mathrm {sgn}\left (b x + a\right ) + 36 \, A b^{3} d^{2} e x \mathrm {sgn}\left (b x + a\right ) + 36 \, B a^{2} b d e^{2} x \mathrm {sgn}\left (b x + a\right ) - 36 \, A a b^{2} d e^{2} x \mathrm {sgn}\left (b x + a\right ) - 12 \, B a^{3} e^{3} x \mathrm {sgn}\left (b x + a\right ) + 12 \, A a^{2} b e^{3} x \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} - \frac {{\left (B a b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a^{2} b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, A a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{3} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + A a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \] Input:

integrate((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

1/12*(3*B*b^3*e^3*x^4*sgn(b*x + a) + 12*B*b^3*d*e^2*x^3*sgn(b*x + a) - 4*B 
*a*b^2*e^3*x^3*sgn(b*x + a) + 4*A*b^3*e^3*x^3*sgn(b*x + a) + 18*B*b^3*d^2* 
e*x^2*sgn(b*x + a) - 18*B*a*b^2*d*e^2*x^2*sgn(b*x + a) + 18*A*b^3*d*e^2*x^ 
2*sgn(b*x + a) + 6*B*a^2*b*e^3*x^2*sgn(b*x + a) - 6*A*a*b^2*e^3*x^2*sgn(b* 
x + a) + 12*B*b^3*d^3*x*sgn(b*x + a) - 36*B*a*b^2*d^2*e*x*sgn(b*x + a) + 3 
6*A*b^3*d^2*e*x*sgn(b*x + a) + 36*B*a^2*b*d*e^2*x*sgn(b*x + a) - 36*A*a*b^ 
2*d*e^2*x*sgn(b*x + a) - 12*B*a^3*e^3*x*sgn(b*x + a) + 12*A*a^2*b*e^3*x*sg 
n(b*x + a))/b^4 - (B*a*b^3*d^3*sgn(b*x + a) - A*b^4*d^3*sgn(b*x + a) - 3*B 
*a^2*b^2*d^2*e*sgn(b*x + a) + 3*A*a*b^3*d^2*e*sgn(b*x + a) + 3*B*a^3*b*d*e 
^2*sgn(b*x + a) - 3*A*a^2*b^2*d*e^2*sgn(b*x + a) - B*a^4*e^3*sgn(b*x + a) 
+ A*a^3*b*e^3*sgn(b*x + a))*log(abs(b*x + a))/b^5
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \] Input:

int(((A + B*x)*(d + e*x)^3)/((a + b*x)^2)^(1/2),x)
 

Output:

int(((A + B*x)*(d + e*x)^3)/((a + b*x)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.13 \[ \int \frac {(A+B x) (d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right )}{4} \] Input:

int((B*x+A)*(e*x+d)^3/((b*x+a)^2)^(1/2),x)
 

Output:

(x*(4*d**3 + 6*d**2*e*x + 4*d*e**2*x**2 + e**3*x**3))/4