\(\int \frac {(A+B x) (d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [448]

Optimal result

Integrand size = 33, antiderivative size = 104 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(A b-a B) (a+b x) (d+e x)^4}{4 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}-\frac {(B d-A e) (d+e x)^3}{3 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-A b \left (a^2 e^2+2 a b e (d+2 e x)+b^2 \left (3 d^2+8 d e x+6 e^2 x^2\right )\right )-B \left (3 a^3 e^2+2 a^2 b e (d+6 e x)+4 b^3 x \left (d^2+3 d e x+3 e^2 x^2\right )+a b^2 \left (d^2+8 d e x+18 e^2 x^2\right )\right )}{12 b^4 (a+b x)^3 \sqrt {(a+b x)^2}} \] Input:

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

Output:

(-(A*b*(a^2*e^2 + 2*a*b*e*(d + 2*e*x) + b^2*(3*d^2 + 8*d*e*x + 6*e^2*x^2)) 
) - B*(3*a^3*e^2 + 2*a^2*b*e*(d + 6*e*x) + 4*b^3*x*(d^2 + 3*d*e*x + 3*e^2* 
x^2) + a*b^2*(d^2 + 8*d*e*x + 18*e^2*x^2)))/(12*b^4*(a + b*x)^3*Sqrt[(a + 
b*x)^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1186, 1102, 27, 48}

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.

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

\(\Big \downarrow \) 1186

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 48

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

Input:

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

Output:

-1/3*((B*d - A*e)*(d + e*x)^3)/((b*d - a*e)^2*(a^2 + 2*a*b*x + b^2*x^2)^(3 
/2)) - ((A*b - a*B)*(d + e*x)^4)/(4*(b*d - a*e)^2*(a + b*x)^3*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 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 1102

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F 
racPart[p]))   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, 
 d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
 

rule 1186

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*((a 
+ b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e)^2)), x] + Simp[(2*c*f - b*g)/ 
(2*c*d - b*e)   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 
0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.55

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,e^{2} x^{3}}{b}-\frac {e \left (A b e +3 B a e +2 B b d \right ) x^{2}}{2 b^{2}}-\frac {\left (A a b \,e^{2}+2 A \,b^{2} d e +3 B \,e^{2} a^{2}+2 B a b d e +B \,b^{2} d^{2}\right ) x}{3 b^{3}}-\frac {A b \,e^{2} a^{2}+2 A a \,b^{2} d e +3 A \,b^{3} d^{2}+3 B \,e^{2} a^{3}+2 B b d e \,a^{2}+B a \,b^{2} d^{2}}{12 b^{4}}\right )}{\left (b x +a \right )^{5}}\)	\(161\)
gosper	\(-\frac {\left (b x +a \right ) \left (12 B \,b^{3} e^{2} x^{3}+6 A \,b^{3} e^{2} x^{2}+18 B a \,b^{2} e^{2} x^{2}+12 B \,b^{3} d e \,x^{2}+4 A a \,b^{2} e^{2} x +8 A \,b^{3} d e x +12 B \,a^{2} b \,e^{2} x +8 B a \,b^{2} d e x +4 B \,b^{3} d^{2} x +A b \,e^{2} a^{2}+2 A a \,b^{2} d e +3 A \,b^{3} d^{2}+3 B \,e^{2} a^{3}+2 B b d e \,a^{2}+B a \,b^{2} d^{2}\right )}{12 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(174\)
default	\(-\frac {\left (b x +a \right ) \left (12 B \,b^{3} e^{2} x^{3}+6 A \,b^{3} e^{2} x^{2}+18 B a \,b^{2} e^{2} x^{2}+12 B \,b^{3} d e \,x^{2}+4 A a \,b^{2} e^{2} x +8 A \,b^{3} d e x +12 B \,a^{2} b \,e^{2} x +8 B a \,b^{2} d e x +4 B \,b^{3} d^{2} x +A b \,e^{2} a^{2}+2 A a \,b^{2} d e +3 A \,b^{3} d^{2}+3 B \,e^{2} a^{3}+2 B b d e \,a^{2}+B a \,b^{2} d^{2}\right )}{12 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(174\)
orering	\(-\frac {\left (12 B \,b^{3} e^{2} x^{3}+6 A \,b^{3} e^{2} x^{2}+18 B a \,b^{2} e^{2} x^{2}+12 B \,b^{3} d e \,x^{2}+4 A a \,b^{2} e^{2} x +8 A \,b^{3} d e x +12 B \,a^{2} b \,e^{2} x +8 B a \,b^{2} d e x +4 B \,b^{3} d^{2} x +A b \,e^{2} a^{2}+2 A a \,b^{2} d e +3 A \,b^{3} d^{2}+3 B \,e^{2} a^{3}+2 B b d e \,a^{2}+B a \,b^{2} d^{2}\right ) \left (b x +a \right )}{12 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}\)	\(183\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.82 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {12 \, B b^{3} e^{2} x^{3} + {\left (B a b^{2} + 3 \, A b^{3}\right )} d^{2} + 2 \, {\left (B a^{2} b + A a b^{2}\right )} d e + {\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + 6 \, {\left (2 \, B b^{3} d e + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 4 \, {\left (B b^{3} d^{2} + 2 \, {\left (B a b^{2} + A b^{3}\right )} d e + {\left (3 \, B a^{2} b + A a b^{2}\right )} e^{2}\right )} x}{12 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((A + B*x)*(d + e*x)**2/((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. 279 vs. \(2 (96) = 192\).

Time = 0.04 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.68 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {B e^{2} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, B a^{2} e^{2}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {B d^{2} + 2 \, A d e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {B a e^{2}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a^{2} e^{2}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {A d^{2}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {B a^{3} e^{2}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {2 \, B d e + A e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, {\left (2 \, B d e + A e^{2}\right )} a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {{\left (2 \, B d e + A e^{2}\right )} a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \] Input:

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

Output:

-B*e^2*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*B*a^2*e^2/((b^2*x^2 
 + 2*a*b*x + a^2)^(3/2)*b^4) - 1/3*(B*d^2 + 2*A*d*e)/((b^2*x^2 + 2*a*b*x + 
 a^2)^(3/2)*b^2) - 1/2*B*a*e^2/(b^6*(x + a/b)^2) + 2/3*B*a^2*e^2/(b^7*(x + 
 a/b)^3) - 1/4*A*d^2/(b^5*(x + a/b)^4) + 1/4*B*a^3*e^2/(b^8*(x + a/b)^4) - 
 1/2*(2*B*d*e + A*e^2)/(b^5*(x + a/b)^2) + 2/3*(2*B*d*e + A*e^2)*a/(b^6*(x 
 + a/b)^3) - 1/4*(2*B*d*e + A*e^2)*a^2/(b^7*(x + a/b)^4) + 1/4*(B*d^2 + 2* 
A*d*e)*a/(b^6*(x + a/b)^4)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {12 \, B b^{3} e^{2} x^{3} + 12 \, B b^{3} d e x^{2} + 18 \, B a b^{2} e^{2} x^{2} + 6 \, A b^{3} e^{2} x^{2} + 4 \, B b^{3} d^{2} x + 8 \, B a b^{2} d e x + 8 \, A b^{3} d e x + 12 \, B a^{2} b e^{2} x + 4 \, A a b^{2} e^{2} x + B a b^{2} d^{2} + 3 \, A b^{3} d^{2} + 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + 3 \, B a^{3} e^{2} + A a^{2} b e^{2}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:

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

Output:

-1/12*(12*B*b^3*e^2*x^3 + 12*B*b^3*d*e*x^2 + 18*B*a*b^2*e^2*x^2 + 6*A*b^3* 
e^2*x^2 + 4*B*b^3*d^2*x + 8*B*a*b^2*d*e*x + 8*A*b^3*d*e*x + 12*B*a^2*b*e^2 
*x + 4*A*a*b^2*e^2*x + B*a*b^2*d^2 + 3*A*b^3*d^2 + 2*B*a^2*b*d*e + 2*A*a*b 
^2*d*e + 3*B*a^3*e^2 + A*a^2*b*e^2)/((b*x + a)^4*b^4*sgn(b*x + a))
 

Mupad [B] (verification not implemented)

Time = 11.62 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.90 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\left (\frac {A\,d^2}{4\,b}-\frac {a\,\left (\frac {B\,d^2+2\,A\,e\,d}{4\,b}-\frac {a\,\left (\frac {A\,e^2+2\,B\,d\,e}{4\,b}-\frac {B\,a\,e^2}{4\,b^2}\right )}{b}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^5}-\frac {\left (\frac {A\,b\,e^2-2\,B\,a\,e^2+2\,B\,b\,d\,e}{2\,b^4}-\frac {B\,a\,e^2}{2\,b^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^3}-\frac {\left (\frac {B\,a^2\,e^2-2\,B\,a\,b\,d\,e-A\,a\,b\,e^2+B\,b^2\,d^2+2\,A\,b^2\,d\,e}{3\,b^4}-\frac {a\,\left (\frac {e\,\left (A\,b\,e-B\,a\,e+2\,B\,b\,d\right )}{3\,b^3}-\frac {B\,a\,e^2}{3\,b^3}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^4}-\frac {B\,e^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.66 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {b^{2} e^{2} x^{3}-3 a b d e x -a^{2} d e -a b \,d^{2}}{3 a \,b^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )} \] Input:

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

Output:

( - a**2*d*e - a*b*d**2 - 3*a*b*d*e*x + b**2*e**2*x**3)/(3*a*b**2*(a**3 + 
3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))