\(\int \frac {(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [447]

Optimal result

Integrand size = 33, antiderivative size = 227 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {3 B e^2 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {B \left (25 a^4 e^3+a^3 b e^2 (-9 d+88 e x)-3 a^2 b^2 e \left (d^2+12 d e x-36 e^2 x^2\right )-2 b^4 d x \left (2 d^2+9 d e x+18 e^2 x^2\right )-a b^3 \left (d^3+12 d^2 e x+54 d e^2 x^2-48 e^3 x^3\right )\right )-3 A b \left (a^3 e^3+a^2 b e^2 (d+4 e x)+a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+12 B e^3 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

Integrate[((A + B*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(B*(25*a^4*e^3 + a^3*b*e^2*(-9*d + 88*e*x) - 3*a^2*b^2*e*(d^2 + 12*d*e*x - 
 36*e^2*x^2) - 2*b^4*d*x*(2*d^2 + 9*d*e*x + 18*e^2*x^2) - a*b^3*(d^3 + 12* 
d^2*e*x + 54*d*e^2*x^2 - 48*e^3*x^3)) - 3*A*b*(a^3*e^3 + a^2*b*e^2*(d + 4* 
e*x) + a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + b^3*(d^3 + 4*d^2*e*x + 6*d*e^ 
2*x^2 + 4*e^3*x^3)) + 12*B*e^3*(a + b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x) 
^3*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.69, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1187, 27, 87, 49, 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.

\(\displaystyle \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {b^5 (a+b x) \int \frac {(A+B x) (d+e x)^3}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b x) \int \frac {(A+B x) (d+e x)^3}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

Int[((A + B*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

((a + b*x)*(-1/4*((A*b - a*B)*(d + e*x)^4)/(b*(b*d - a*e)*(a + b*x)^4) + ( 
B*(-1/3*(b*d - a*e)^3/(b^4*(a + b*x)^3) - (3*e*(b*d - a*e)^2)/(2*b^4*(a + 
b*x)^2) - (3*e^2*(b*d - a*e))/(b^4*(a + b*x)) + (e^3*Log[a + b*x])/b^4))/b 
))/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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 = 1.55 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.29

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {e^{2} \left (A b e -4 B a e +3 B b d \right ) x^{3}}{b^{2}}-\frac {3 e \left (A a b \,e^{2}+A \,b^{2} d e -6 B \,e^{2} a^{2}+3 B a b d e +B \,b^{2} d^{2}\right ) x^{2}}{2 b^{3}}-\frac {\left (3 A \,a^{2} b \,e^{3}+3 A a \,b^{2} d \,e^{2}+3 A \,b^{3} d^{2} e -22 B \,e^{3} a^{3}+9 B \,a^{2} b d \,e^{2}+3 B a \,b^{2} d^{2} e +B \,b^{3} d^{3}\right ) x}{3 b^{4}}-\frac {3 A \,a^{3} b \,e^{3}+3 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e +3 A \,d^{3} b^{4}-25 B \,e^{3} a^{4}+9 B \,a^{3} b d \,e^{2}+3 B \,a^{2} b^{2} d^{2} e +B a \,b^{3} d^{3}}{12 b^{5}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B \,e^{3} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(292\)
default	\(-\frac {\left (12 A \,b^{4} d^{2} e x -88 B \,a^{3} b \,e^{3} x -12 B \ln \left (b x +a \right ) b^{4} e^{3} x^{4}+3 B \,a^{2} b^{2} d^{2} e +9 B \,a^{3} b d \,e^{2}+3 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e -12 B \ln \left (b x +a \right ) a^{4} e^{3}+4 B \,b^{4} d^{3} x +36 B \,b^{4} d \,e^{2} x^{3}+12 A \,b^{4} e^{3} x^{3}+12 A x \,a^{2} b^{2} e^{3}+B a \,b^{3} d^{3}+3 A \,a^{3} b \,e^{3}+18 A \,x^{2} a \,b^{3} e^{3}+18 A \,x^{2} b^{4} d \,e^{2}-108 B \,x^{2} a^{2} b^{2} e^{3}+18 B \,x^{2} b^{4} d^{2} e +54 B \,x^{2} a \,b^{3} d \,e^{2}-48 B \,e^{3} a \,x^{3} b^{3}+12 A x a \,b^{3} d \,e^{2}+36 B x \,a^{2} b^{2} d \,e^{2}+12 B x a \,b^{3} d^{2} e -48 B \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{3}-25 B \,e^{3} a^{4}-72 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{3}-48 B \ln \left (b x +a \right ) x \,a^{3} b \,e^{3}+3 A \,d^{3} b^{4}\right ) \left (b x +a \right )}{12 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(385\)

Input:

int((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x+a)^2)^(1/2)/(b*x+a)^5*(-e^2*(A*b*e-4*B*a*e+3*B*b*d)/b^2*x^3-3/2*e*(A 
*a*b*e^2+A*b^2*d*e-6*B*a^2*e^2+3*B*a*b*d*e+B*b^2*d^2)/b^3*x^2-1/3*(3*A*a^2 
*b*e^3+3*A*a*b^2*d*e^2+3*A*b^3*d^2*e-22*B*a^3*e^3+9*B*a^2*b*d*e^2+3*B*a*b^ 
2*d^2*e+B*b^3*d^3)/b^4*x-1/12*(3*A*a^3*b*e^3+3*A*a^2*b^2*d*e^2+3*A*a*b^3*d 
^2*e+3*A*b^4*d^3-25*B*a^4*e^3+9*B*a^3*b*d*e^2+3*B*a^2*b^2*d^2*e+B*a*b^3*d^ 
3)/b^5)+((b*x+a)^2)^(1/2)/(b*x+a)*B*e^3/b^5*ln(b*x+a)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (166) = 332\).

Time = 0.09 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.58 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {{\left (B a b^{3} + 3 \, A b^{4}\right )} d^{3} + 3 \, {\left (B a^{2} b^{2} + A a b^{3}\right )} d^{2} e + 3 \, {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} d e^{2} - {\left (25 \, B a^{4} - 3 \, A a^{3} b\right )} e^{3} + 12 \, {\left (3 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 18 \, {\left (B b^{4} d^{2} e + {\left (3 \, B a b^{3} + A b^{4}\right )} d e^{2} - {\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 4 \, {\left (B b^{4} d^{3} + 3 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} e + 3 \, {\left (3 \, B a^{2} b^{2} + A a b^{3}\right )} d e^{2} - {\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} e^{3}\right )} x - 12 \, {\left (B b^{4} e^{3} x^{4} + 4 \, B a b^{3} e^{3} x^{3} + 6 \, B a^{2} b^{2} e^{3} x^{2} + 4 \, B a^{3} b e^{3} x + B a^{4} e^{3}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \] Input:

integrate((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fric 
as")
 

Output:

-1/12*((B*a*b^3 + 3*A*b^4)*d^3 + 3*(B*a^2*b^2 + A*a*b^3)*d^2*e + 3*(3*B*a^ 
3*b + A*a^2*b^2)*d*e^2 - (25*B*a^4 - 3*A*a^3*b)*e^3 + 12*(3*B*b^4*d*e^2 - 
(4*B*a*b^3 - A*b^4)*e^3)*x^3 + 18*(B*b^4*d^2*e + (3*B*a*b^3 + A*b^4)*d*e^2 
 - (6*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 4*(B*b^4*d^3 + 3*(B*a*b^3 + A*b^4)*d 
^2*e + 3*(3*B*a^2*b^2 + A*a*b^3)*d*e^2 - (22*B*a^3*b - 3*A*a^2*b^2)*e^3)*x 
 - 12*(B*b^4*e^3*x^4 + 4*B*a*b^3*e^3*x^3 + 6*B*a^2*b^2*e^3*x^2 + 4*B*a^3*b 
*e^3*x + B*a^4*e^3)*log(b*x + a))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 
 4*a^3*b^6*x + a^4*b^5)
 

Sympy [F]

\[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((B*x+A)*(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Integral((A + B*x)*(d + e*x)**3/((a + b*x)**2)**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (166) = 332\).

Time = 0.06 (sec) , antiderivative size = 533, normalized size of antiderivative = 2.35 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, B e^{3} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{4} \, B d e^{2} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, A e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, B d^{3} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, A d^{2} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, B d^{2} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, A d e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {A d^{3}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:

integrate((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxi 
ma")
 

Output:

1/12*B*e^3*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^ 
4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a) 
/b^5) - 1/4*B*d*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/ 
((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7 
*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) - 1/12*A*e^3*(12*x^2/((b^2*x^2 + 
2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 
6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) 
 - 1/12*B*d^3*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b 
)^4)) - 1/4*A*d^2*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x 
 + a/b)^4)) - 1/4*B*d^2*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3 
*a^2/(b^7*(x + a/b)^4)) - 1/4*A*d*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + 
 a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*A*d^3/(b^5*(x + a/b)^4)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {B e^{3} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {12 \, {\left (3 \, B b^{3} d e^{2} - 4 \, B a b^{2} e^{3} + A b^{3} e^{3}\right )} x^{3} + 18 \, {\left (B b^{3} d^{2} e + 3 \, B a b^{2} d e^{2} + A b^{3} d e^{2} - 6 \, B a^{2} b e^{3} + A a b^{2} e^{3}\right )} x^{2} + 4 \, {\left (B b^{3} d^{3} + 3 \, B a b^{2} d^{2} e + 3 \, A b^{3} d^{2} e + 9 \, B a^{2} b d e^{2} + 3 \, A a b^{2} d e^{2} - 22 \, B a^{3} e^{3} + 3 \, A a^{2} b e^{3}\right )} x + \frac {B a b^{3} d^{3} + 3 \, A b^{4} d^{3} + 3 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 9 \, B a^{3} b d e^{2} + 3 \, A a^{2} b^{2} d e^{2} - 25 \, B a^{4} e^{3} + 3 \, A a^{3} b e^{3}}{b}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac 
")
 

Output:

B*e^3*log(abs(b*x + a))/(b^5*sgn(b*x + a)) - 1/12*(12*(3*B*b^3*d*e^2 - 4*B 
*a*b^2*e^3 + A*b^3*e^3)*x^3 + 18*(B*b^3*d^2*e + 3*B*a*b^2*d*e^2 + A*b^3*d* 
e^2 - 6*B*a^2*b*e^3 + A*a*b^2*e^3)*x^2 + 4*(B*b^3*d^3 + 3*B*a*b^2*d^2*e + 
3*A*b^3*d^2*e + 9*B*a^2*b*d*e^2 + 3*A*a*b^2*d*e^2 - 22*B*a^3*e^3 + 3*A*a^2 
*b*e^3)*x + (B*a*b^3*d^3 + 3*A*b^4*d^3 + 3*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2 
*e + 9*B*a^3*b*d*e^2 + 3*A*a^2*b^2*d*e^2 - 25*B*a^4*e^3 + 3*A*a^3*b*e^3)/b 
)/((b*x + a)^4*b^4*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((A + B*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int(((A + B*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {6 \,\mathrm {log}\left (b x +a \right ) a^{4} e^{3}+18 \,\mathrm {log}\left (b x +a \right ) a^{3} b \,e^{3} x +18 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} e^{3} x^{2}+6 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} e^{3} x^{3}+5 a^{4} e^{3}+9 a^{3} b \,e^{3} x -3 a^{2} b^{2} d^{2} e -2 a \,b^{3} d^{3}-9 a \,b^{3} d^{2} e x -6 a \,b^{3} e^{3} x^{3}+6 b^{4} d \,e^{2} x^{3}}{6 a \,b^{4} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

int((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

(6*log(a + b*x)*a**4*e**3 + 18*log(a + b*x)*a**3*b*e**3*x + 18*log(a + b*x 
)*a**2*b**2*e**3*x**2 + 6*log(a + b*x)*a*b**3*e**3*x**3 + 5*a**4*e**3 + 9* 
a**3*b*e**3*x - 3*a**2*b**2*d**2*e - 2*a*b**3*d**3 - 9*a*b**3*d**2*e*x - 6 
*a*b**3*e**3*x**3 + 6*b**4*d*e**2*x**3)/(6*a*b**4*(a**3 + 3*a**2*b*x + 3*a 
*b**2*x**2 + b**3*x**3))