3.17.62 \(\int (A+B x) (d+e x)^2 (a^2+2 a b x+b^2 x^2) \, dx\) [1662]

3.17.62.1 Optimal result
3.17.62.2 Mathematica [A] (verified)
3.17.62.3 Rubi [A] (verified)
3.17.62.4 Maple [A] (verified)
3.17.62.5 Fricas [A] (verification not implemented)
3.17.62.6 Sympy [A] (verification not implemented)
3.17.62.7 Maxima [A] (verification not implemented)
3.17.62.8 Giac [A] (verification not implemented)
3.17.62.9 Mupad [B] (verification not implemented)

3.17.62.1 Optimal result

Integrand size = 29, antiderivative size = 118 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(A b-a B) (b d-a e)^2 (a+b x)^3}{3 b^4}+\frac {(b d-a e) (b B d+2 A b e-3 a B e) (a+b x)^4}{4 b^4}+\frac {e (2 b B d+A b e-3 a B e) (a+b x)^5}{5 b^4}+\frac {B e^2 (a+b x)^6}{6 b^4} \]

output
1/3*(A*b-B*a)*(-a*e+b*d)^2*(b*x+a)^3/b^4+1/4*(-a*e+b*d)*(2*A*b*e-3*B*a*e+B 
*b*d)*(b*x+a)^4/b^4+1/5*e*(A*b*e-3*B*a*e+2*B*b*d)*(b*x+a)^5/b^4+1/6*B*e^2* 
(b*x+a)^6/b^4
 
3.17.62.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.33 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^2 A d^2 x+\frac {1}{2} a d (a B d+2 A (b d+a e)) x^2+\frac {1}{3} \left (2 a B d (b d+a e)+A \left (b^2 d^2+4 a b d e+a^2 e^2\right )\right ) x^3+\frac {1}{4} \left (a^2 B e^2+2 a b e (2 B d+A e)+b^2 d (B d+2 A e)\right ) x^4+\frac {1}{5} b e (2 b B d+A b e+2 a B e) x^5+\frac {1}{6} b^2 B e^2 x^6 \]

input
Integrate[(A + B*x)*(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]
 
output
a^2*A*d^2*x + (a*d*(a*B*d + 2*A*(b*d + a*e))*x^2)/2 + ((2*a*B*d*(b*d + a*e 
) + A*(b^2*d^2 + 4*a*b*d*e + a^2*e^2))*x^3)/3 + ((a^2*B*e^2 + 2*a*b*e*(2*B 
*d + A*e) + b^2*d*(B*d + 2*A*e))*x^4)/4 + (b*e*(2*b*B*d + A*b*e + 2*a*B*e) 
*x^5)/5 + (b^2*B*e^2*x^6)/6
 
3.17.62.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 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.

\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right ) (A+B x) (d+e x)^2 \, dx\)

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

\(\displaystyle \int (a+b x)^2 (A+B x) (d+e x)^2dx\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e (a+b x)^5 (-3 a B e+A b e+2 b B d)}{5 b^4}+\frac {(a+b x)^4 (b d-a e) (-3 a B e+2 A b e+b B d)}{4 b^4}+\frac {(a+b x)^3 (A b-a B) (b d-a e)^2}{3 b^4}+\frac {B e^2 (a+b x)^6}{6 b^4}\)

input
Int[(A + B*x)*(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]
 
output
((A*b - a*B)*(b*d - a*e)^2*(a + b*x)^3)/(3*b^4) + ((b*d - a*e)*(b*B*d + 2* 
A*b*e - 3*a*B*e)*(a + b*x)^4)/(4*b^4) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*(a 
+ b*x)^5)/(5*b^4) + (B*e^2*(a + b*x)^6)/(6*b^4)
 

3.17.62.3.1 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 1184
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^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}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.17.62.4 Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.43

method result size
default \(\frac {B \,b^{2} e^{2} x^{6}}{6}+\frac {\left (\left (A \,e^{2}+2 B d e \right ) b^{2}+2 B a b \,e^{2}\right ) x^{5}}{5}+\frac {\left (\left (2 A d e +B \,d^{2}\right ) b^{2}+2 \left (A \,e^{2}+2 B d e \right ) b a +a^{2} B \,e^{2}\right ) x^{4}}{4}+\frac {\left (A \,b^{2} d^{2}+2 \left (2 A d e +B \,d^{2}\right ) b a +\left (A \,e^{2}+2 B d e \right ) a^{2}\right ) x^{3}}{3}+\frac {\left (2 A a b \,d^{2}+\left (2 A d e +B \,d^{2}\right ) a^{2}\right ) x^{2}}{2}+A \,d^{2} a^{2} x\) \(169\)
norman \(\frac {B \,b^{2} e^{2} x^{6}}{6}+\left (\frac {1}{5} A \,b^{2} e^{2}+\frac {2}{5} B a b \,e^{2}+\frac {2}{5} B \,b^{2} d e \right ) x^{5}+\left (\frac {1}{2} A a b \,e^{2}+\frac {1}{2} A \,b^{2} d e +\frac {1}{4} a^{2} B \,e^{2}+B a b d e +\frac {1}{4} B \,b^{2} d^{2}\right ) x^{4}+\left (\frac {1}{3} A \,a^{2} e^{2}+\frac {4}{3} A a b d e +\frac {1}{3} A \,b^{2} d^{2}+\frac {2}{3} B \,a^{2} d e +\frac {2}{3} B a b \,d^{2}\right ) x^{3}+\left (A \,a^{2} d e +A a b \,d^{2}+\frac {1}{2} B \,a^{2} d^{2}\right ) x^{2}+A \,d^{2} a^{2} x\) \(172\)
gosper \(\frac {x \left (10 B \,b^{2} e^{2} x^{5}+12 x^{4} A \,b^{2} e^{2}+24 x^{4} B a b \,e^{2}+24 x^{4} B \,b^{2} d e +30 x^{3} A a b \,e^{2}+30 x^{3} A \,b^{2} d e +15 x^{3} a^{2} B \,e^{2}+60 x^{3} B a b d e +15 x^{3} B \,b^{2} d^{2}+20 x^{2} A \,a^{2} e^{2}+80 x^{2} A a b d e +20 x^{2} A \,b^{2} d^{2}+40 x^{2} B \,a^{2} d e +40 x^{2} B a b \,d^{2}+60 x A \,a^{2} d e +60 x A a b \,d^{2}+30 x B \,a^{2} d^{2}+60 A \,d^{2} a^{2}\right )}{60}\) \(200\)
risch \(\frac {1}{6} B \,b^{2} e^{2} x^{6}+\frac {1}{5} x^{5} A \,b^{2} e^{2}+\frac {2}{5} x^{5} B a b \,e^{2}+\frac {2}{5} x^{5} B \,b^{2} d e +\frac {1}{2} x^{4} A a b \,e^{2}+\frac {1}{2} x^{4} A \,b^{2} d e +\frac {1}{4} x^{4} a^{2} B \,e^{2}+x^{4} B a b d e +\frac {1}{4} x^{4} B \,b^{2} d^{2}+\frac {1}{3} x^{3} A \,a^{2} e^{2}+\frac {4}{3} x^{3} A a b d e +\frac {1}{3} A \,b^{2} d^{2} x^{3}+\frac {2}{3} x^{3} B \,a^{2} d e +\frac {2}{3} x^{3} B a b \,d^{2}+x^{2} A \,a^{2} d e +x^{2} A a b \,d^{2}+\frac {1}{2} x^{2} B \,a^{2} d^{2}+A \,d^{2} a^{2} x\) \(200\)
parallelrisch \(\frac {1}{6} B \,b^{2} e^{2} x^{6}+\frac {1}{5} x^{5} A \,b^{2} e^{2}+\frac {2}{5} x^{5} B a b \,e^{2}+\frac {2}{5} x^{5} B \,b^{2} d e +\frac {1}{2} x^{4} A a b \,e^{2}+\frac {1}{2} x^{4} A \,b^{2} d e +\frac {1}{4} x^{4} a^{2} B \,e^{2}+x^{4} B a b d e +\frac {1}{4} x^{4} B \,b^{2} d^{2}+\frac {1}{3} x^{3} A \,a^{2} e^{2}+\frac {4}{3} x^{3} A a b d e +\frac {1}{3} A \,b^{2} d^{2} x^{3}+\frac {2}{3} x^{3} B \,a^{2} d e +\frac {2}{3} x^{3} B a b \,d^{2}+x^{2} A \,a^{2} d e +x^{2} A a b \,d^{2}+\frac {1}{2} x^{2} B \,a^{2} d^{2}+A \,d^{2} a^{2} x\) \(200\)

input
int((B*x+A)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
 
output
1/6*B*b^2*e^2*x^6+1/5*((A*e^2+2*B*d*e)*b^2+2*B*a*b*e^2)*x^5+1/4*((2*A*d*e+ 
B*d^2)*b^2+2*(A*e^2+2*B*d*e)*b*a+a^2*B*e^2)*x^4+1/3*(A*b^2*d^2+2*(2*A*d*e+ 
B*d^2)*b*a+(A*e^2+2*B*d*e)*a^2)*x^3+1/2*(2*A*a*b*d^2+(2*A*d*e+B*d^2)*a^2)* 
x^2+A*d^2*a^2*x
 
3.17.62.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.42 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{6} \, B b^{2} e^{2} x^{6} + A a^{2} d^{2} x + \frac {1}{5} \, {\left (2 \, B b^{2} d e + {\left (2 \, B a b + A b^{2}\right )} e^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{2} d^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (A a^{2} e^{2} + {\left (2 \, B a b + A b^{2}\right )} d^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, A a^{2} d e + {\left (B a^{2} + 2 \, A a b\right )} d^{2}\right )} x^{2} \]

input
integrate((B*x+A)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
 
output
1/6*B*b^2*e^2*x^6 + A*a^2*d^2*x + 1/5*(2*B*b^2*d*e + (2*B*a*b + A*b^2)*e^2 
)*x^5 + 1/4*(B*b^2*d^2 + 2*(2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a*b)*e^2)* 
x^4 + 1/3*(A*a^2*e^2 + (2*B*a*b + A*b^2)*d^2 + 2*(B*a^2 + 2*A*a*b)*d*e)*x^ 
3 + 1/2*(2*A*a^2*d*e + (B*a^2 + 2*A*a*b)*d^2)*x^2
 
3.17.62.6 Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.71 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=A a^{2} d^{2} x + \frac {B b^{2} e^{2} x^{6}}{6} + x^{5} \left (\frac {A b^{2} e^{2}}{5} + \frac {2 B a b e^{2}}{5} + \frac {2 B b^{2} d e}{5}\right ) + x^{4} \left (\frac {A a b e^{2}}{2} + \frac {A b^{2} d e}{2} + \frac {B a^{2} e^{2}}{4} + B a b d e + \frac {B b^{2} d^{2}}{4}\right ) + x^{3} \left (\frac {A a^{2} e^{2}}{3} + \frac {4 A a b d e}{3} + \frac {A b^{2} d^{2}}{3} + \frac {2 B a^{2} d e}{3} + \frac {2 B a b d^{2}}{3}\right ) + x^{2} \left (A a^{2} d e + A a b d^{2} + \frac {B a^{2} d^{2}}{2}\right ) \]

input
integrate((B*x+A)*(e*x+d)**2*(b**2*x**2+2*a*b*x+a**2),x)
 
output
A*a**2*d**2*x + B*b**2*e**2*x**6/6 + x**5*(A*b**2*e**2/5 + 2*B*a*b*e**2/5 
+ 2*B*b**2*d*e/5) + x**4*(A*a*b*e**2/2 + A*b**2*d*e/2 + B*a**2*e**2/4 + B* 
a*b*d*e + B*b**2*d**2/4) + x**3*(A*a**2*e**2/3 + 4*A*a*b*d*e/3 + A*b**2*d* 
*2/3 + 2*B*a**2*d*e/3 + 2*B*a*b*d**2/3) + x**2*(A*a**2*d*e + A*a*b*d**2 + 
B*a**2*d**2/2)
 
3.17.62.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.42 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{6} \, B b^{2} e^{2} x^{6} + A a^{2} d^{2} x + \frac {1}{5} \, {\left (2 \, B b^{2} d e + {\left (2 \, B a b + A b^{2}\right )} e^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{2} d^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (A a^{2} e^{2} + {\left (2 \, B a b + A b^{2}\right )} d^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, A a^{2} d e + {\left (B a^{2} + 2 \, A a b\right )} d^{2}\right )} x^{2} \]

input
integrate((B*x+A)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
 
output
1/6*B*b^2*e^2*x^6 + A*a^2*d^2*x + 1/5*(2*B*b^2*d*e + (2*B*a*b + A*b^2)*e^2 
)*x^5 + 1/4*(B*b^2*d^2 + 2*(2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a*b)*e^2)* 
x^4 + 1/3*(A*a^2*e^2 + (2*B*a*b + A*b^2)*d^2 + 2*(B*a^2 + 2*A*a*b)*d*e)*x^ 
3 + 1/2*(2*A*a^2*d*e + (B*a^2 + 2*A*a*b)*d^2)*x^2
 
3.17.62.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.69 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{6} \, B b^{2} e^{2} x^{6} + \frac {2}{5} \, B b^{2} d e x^{5} + \frac {2}{5} \, B a b e^{2} x^{5} + \frac {1}{5} \, A b^{2} e^{2} x^{5} + \frac {1}{4} \, B b^{2} d^{2} x^{4} + B a b d e x^{4} + \frac {1}{2} \, A b^{2} d e x^{4} + \frac {1}{4} \, B a^{2} e^{2} x^{4} + \frac {1}{2} \, A a b e^{2} x^{4} + \frac {2}{3} \, B a b d^{2} x^{3} + \frac {1}{3} \, A b^{2} d^{2} x^{3} + \frac {2}{3} \, B a^{2} d e x^{3} + \frac {4}{3} \, A a b d e x^{3} + \frac {1}{3} \, A a^{2} e^{2} x^{3} + \frac {1}{2} \, B a^{2} d^{2} x^{2} + A a b d^{2} x^{2} + A a^{2} d e x^{2} + A a^{2} d^{2} x \]

input
integrate((B*x+A)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
 
output
1/6*B*b^2*e^2*x^6 + 2/5*B*b^2*d*e*x^5 + 2/5*B*a*b*e^2*x^5 + 1/5*A*b^2*e^2* 
x^5 + 1/4*B*b^2*d^2*x^4 + B*a*b*d*e*x^4 + 1/2*A*b^2*d*e*x^4 + 1/4*B*a^2*e^ 
2*x^4 + 1/2*A*a*b*e^2*x^4 + 2/3*B*a*b*d^2*x^3 + 1/3*A*b^2*d^2*x^3 + 2/3*B* 
a^2*d*e*x^3 + 4/3*A*a*b*d*e*x^3 + 1/3*A*a^2*e^2*x^3 + 1/2*B*a^2*d^2*x^2 + 
A*a*b*d^2*x^2 + A*a^2*d*e*x^2 + A*a^2*d^2*x
 
3.17.62.9 Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.33 \[ \int (A+B x) (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^3\,\left (\frac {2\,B\,a^2\,d\,e}{3}+\frac {A\,a^2\,e^2}{3}+\frac {2\,B\,a\,b\,d^2}{3}+\frac {4\,A\,a\,b\,d\,e}{3}+\frac {A\,b^2\,d^2}{3}\right )+x^4\,\left (\frac {B\,a^2\,e^2}{4}+B\,a\,b\,d\,e+\frac {A\,a\,b\,e^2}{2}+\frac {B\,b^2\,d^2}{4}+\frac {A\,b^2\,d\,e}{2}\right )+\frac {a\,d\,x^2\,\left (2\,A\,a\,e+2\,A\,b\,d+B\,a\,d\right )}{2}+\frac {b\,e\,x^5\,\left (A\,b\,e+2\,B\,a\,e+2\,B\,b\,d\right )}{5}+A\,a^2\,d^2\,x+\frac {B\,b^2\,e^2\,x^6}{6} \]

input
int((A + B*x)*(d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x),x)
 
output
x^3*((A*a^2*e^2)/3 + (A*b^2*d^2)/3 + (2*B*a*b*d^2)/3 + (2*B*a^2*d*e)/3 + ( 
4*A*a*b*d*e)/3) + x^4*((B*a^2*e^2)/4 + (B*b^2*d^2)/4 + (A*a*b*e^2)/2 + (A* 
b^2*d*e)/2 + B*a*b*d*e) + (a*d*x^2*(2*A*a*e + 2*A*b*d + B*a*d))/2 + (b*e*x 
^5*(A*b*e + 2*B*a*e + 2*B*b*d))/5 + A*a^2*d^2*x + (B*b^2*e^2*x^6)/6