\(\int (A+B x+C x^2) (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3)^p \, dx\) [66]

Optimal result

Integrand size = 40, antiderivative size = 167 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (A b^2-a (b B-a C)\right ) (a+b x) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p}{b^3 (1+3 p)}+\frac {(b B-2 a C) (a+b x)^2 \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p}{b^3 (2+3 p)}+\frac {C (a+b x)^3 \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p}{3 b^3 (1+p)} \] Output:

(A*b^2-a*(B*b-C*a))*(b*x+a)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p/b^3/(1+3 
*p)+(B*b-2*C*a)*(b*x+a)^2*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p/b^3/(2+3*p 
)+1/3*C*(b*x+a)^3*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p/b^3/(p+1)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.65 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^3\right )^p \left (2 a^2 C+3 A b^2 \left (2+5 p+3 p^2\right )-a b (3 B (1+p)+2 C (1+3 p) x)+b^2 (1+3 p) x (3 B (1+p)+C (2+3 p) x)\right )}{3 b^3 (1+p) (1+3 p) (2+3 p)} \] Input:

Integrate[(A + B*x + C*x^2)*(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^p,x]
 

Output:

((a + b*x)*((a + b*x)^3)^p*(2*a^2*C + 3*A*b^2*(2 + 5*p + 3*p^2) - a*b*(3*B 
*(1 + p) + 2*C*(1 + 3*p)*x) + b^2*(1 + 3*p)*x*(3*B*(1 + p) + C*(2 + 3*p)*x 
)))/(3*b^3*(1 + p)*(1 + 3*p)*(2 + 3*p))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2008, 1140, 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.

\(\displaystyle \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \left (A+B x+C x^2\right ) \, dx\)

\(\Big \downarrow \) 2008

\(\displaystyle (a+b x)^{-3 p} \left ((a+b x)^3\right )^p \int (a+b x)^{3 p} \left (C x^2+B x+A\right )dx\)

\(\Big \downarrow \) 1140

\(\displaystyle (a+b x)^{-3 p} \left ((a+b x)^3\right )^p \int \left (\frac {\left (A b^2-a (b B-a C)\right ) (a+b x)^{3 p}}{b^2}+\frac {(b B-2 a C) (a+b x)^{3 p+1}}{b^2}+\frac {C (a+b x)^{3 p+2}}{b^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle (a+b x)^{-3 p} \left ((a+b x)^3\right )^p \left (\frac {(a+b x)^{3 p+1} \left (A b^2-a (b B-a C)\right )}{b^3 (3 p+1)}+\frac {(b B-2 a C) (a+b x)^{3 p+2}}{b^3 (3 p+2)}+\frac {C (a+b x)^{3 (p+1)}}{3 b^3 (p+1)}\right )\)

Input:

Int[(A + B*x + C*x^2)*(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)^p,x]
 

Output:

(((a + b*x)^3)^p*((C*(a + b*x)^(3*(1 + p)))/(3*b^3*(1 + p)) + ((A*b^2 - a* 
(b*B - a*C))*(a + b*x)^(1 + 3*p))/(b^3*(1 + 3*p)) + ((b*B - 2*a*C)*(a + b* 
x)^(2 + 3*p))/(b^3*(2 + 3*p))))/(a + b*x)^(3*p)
 

Defintions of rubi rules used

rule 1140

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

rule 2008

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp 
on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p)   Int[u*(a + b*x)^(Expon[Px, x]*p), 
 x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /;  !IntegerQ[p] && PolyQ[Px, x 
] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99

method	result	size
gosper	\(\frac {\left (b x +a \right ) \left (9 C \,b^{2} p^{2} x^{2}+9 B \,b^{2} p^{2} x +9 C \,b^{2} p \,x^{2}+9 A \,b^{2} p^{2}+12 B \,b^{2} p x -6 C a b p x +2 C \,b^{2} x^{2}+15 A \,b^{2} p -3 B a b p +3 B \,b^{2} x -2 C a b x +6 A \,b^{2}-3 a b B +2 C \,a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p}}{3 b^{3} \left (9 p^{3}+18 p^{2}+11 p +2\right )}\)	\(166\)
orering	\(\frac {\left (b x +a \right ) \left (9 C \,b^{2} p^{2} x^{2}+9 B \,b^{2} p^{2} x +9 C \,b^{2} p \,x^{2}+9 A \,b^{2} p^{2}+12 B \,b^{2} p x -6 C a b p x +2 C \,b^{2} x^{2}+15 A \,b^{2} p -3 B a b p +3 B \,b^{2} x -2 C a b x +6 A \,b^{2}-3 a b B +2 C \,a^{2}\right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p}}{3 b^{3} \left (9 p^{3}+18 p^{2}+11 p +2\right )}\)	\(166\)
risch	\(\frac {\left (9 C \,b^{3} p^{2} x^{3}+9 B \,b^{3} p^{2} x^{2}+9 C a \,b^{2} p^{2} x^{2}+9 C \,b^{3} p \,x^{3}+9 A \,b^{3} p^{2} x +9 B a \,b^{2} p^{2} x +12 B \,b^{3} p \,x^{2}+3 C a \,b^{2} p \,x^{2}+2 C \,x^{3} b^{3}+9 A a \,b^{2} p^{2}+15 A \,b^{3} p x +9 B a \,b^{2} p x +3 B \,b^{3} x^{2}-6 C \,a^{2} b p x +15 A a \,b^{2} p +6 A \,b^{3} x -3 B \,a^{2} b p +6 A a \,b^{2}-3 B \,a^{2} b +2 C \,a^{3}\right ) \left (\left (b x +a \right )^{3}\right )^{p}}{3 \left (2+3 p \right ) \left (p +1\right ) \left (1+3 p \right ) b^{3}}\)	\(221\)
norman	\(\frac {\left (B b p +C a p +B b \right ) x^{2} {\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )}}{b \left (3 p^{2}+5 p +2\right )}+\frac {\left (3 A \,b^{2} p^{2}+3 B a b \,p^{2}+5 A \,b^{2} p +3 B a b p -2 C \,a^{2} p +2 A \,b^{2}\right ) x \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )}}{b^{2} \left (9 p^{3}+18 p^{2}+11 p +2\right )}+\frac {C \,x^{3} {\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )}}{3 p +3}+\frac {a \left (9 A \,b^{2} p^{2}+15 A \,b^{2} p -3 B a b p +6 A \,b^{2}-3 a b B +2 C \,a^{2}\right ) {\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )}}{3 b^{3} \left (9 p^{3}+18 p^{2}+11 p +2\right )}\)	\(297\)
parallelrisch	\(\frac {9 C \,x^{3} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p^{2}+9 B \,x^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p^{2}+9 C \,x^{3} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p +9 C \,x^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p^{2}+9 A x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p^{2}+12 B \,x^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p +9 B x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p^{2}+2 C \,x^{3} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3}+3 C \,x^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p +15 A x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3} p +9 A \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p^{2}+3 B \,x^{2} \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3}+9 B x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p -6 C x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a^{2} b p +6 A x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} b^{3}+15 A \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2} p -3 B \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a^{2} b p +6 A \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a \,b^{2}-3 B \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a^{2} b +2 C \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}\right )^{p} a^{3}}{3 b^{3} \left (9 p^{3}+18 p^{2}+11 p +2\right )}\)	\(790\)

Input:

int((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p,x,method=_RETURNVE 
RBOSE)
 

Output:

1/3*(b*x+a)*(9*C*b^2*p^2*x^2+9*B*b^2*p^2*x+9*C*b^2*p*x^2+9*A*b^2*p^2+12*B* 
b^2*p*x-6*C*a*b*p*x+2*C*b^2*x^2+15*A*b^2*p-3*B*a*b*p+3*B*b^2*x-2*C*a*b*x+6 
*A*b^2-3*B*a*b+2*C*a^2)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p/b^3/(9*p^3+1 
8*p^2+11*p+2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.38 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (9 \, A a b^{2} p^{2} + 2 \, C a^{3} - 3 \, B a^{2} b + 6 \, A a b^{2} + {\left (9 \, C b^{3} p^{2} + 9 \, C b^{3} p + 2 \, C b^{3}\right )} x^{3} + 3 \, {\left (B b^{3} + 3 \, {\left (C a b^{2} + B b^{3}\right )} p^{2} + {\left (C a b^{2} + 4 \, B b^{3}\right )} p\right )} x^{2} - 3 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} p + 3 \, {\left (2 \, A b^{3} + 3 \, {\left (B a b^{2} + A b^{3}\right )} p^{2} - {\left (2 \, C a^{2} b - 3 \, B a b^{2} - 5 \, A b^{3}\right )} p\right )} x\right )} {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p}}{3 \, {\left (9 \, b^{3} p^{3} + 18 \, b^{3} p^{2} + 11 \, b^{3} p + 2 \, b^{3}\right )}} \] Input:

integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p,x, algorithm 
="fricas")
 

Output:

1/3*(9*A*a*b^2*p^2 + 2*C*a^3 - 3*B*a^2*b + 6*A*a*b^2 + (9*C*b^3*p^2 + 9*C* 
b^3*p + 2*C*b^3)*x^3 + 3*(B*b^3 + 3*(C*a*b^2 + B*b^3)*p^2 + (C*a*b^2 + 4*B 
*b^3)*p)*x^2 - 3*(B*a^2*b - 5*A*a*b^2)*p + 3*(2*A*b^3 + 3*(B*a*b^2 + A*b^3 
)*p^2 - (2*C*a^2*b - 3*B*a*b^2 - 5*A*b^3)*p)*x)*(b^3*x^3 + 3*a*b^2*x^2 + 3 
*a^2*b*x + a^3)^p/(9*b^3*p^3 + 18*b^3*p^2 + 11*b^3*p + 2*b^3)
 

Sympy [F]

\[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\text {Too large to display} \] Input:

integrate((C*x**2+B*x+A)*(b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3)**p,x)
 

Output:

Piecewise(((A*x + B*x**2/2 + C*x**3/3)*(a**3)**p, Eq(b, 0)), (-A*b**2/(2*a 
**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) - B*a*b/(2*a**2*b**3 + 4*a*b**4*x + 2 
*b**5*x**2) - 2*B*b**2*x/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*C*a* 
*2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 3*C*a**2/(2*a** 
2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*C*a*b*x*log(a/b + x)/(2*a**2*b**3 + 
 4*a*b**4*x + 2*b**5*x**2) + 4*C*a*b*x/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5* 
x**2) + 2*C*b**2*x**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2 
), Eq(p, -1)), (Integral((A + B*x + C*x**2)/((a + b*x)**3)**(2/3), x), Eq( 
p, -2/3)), (Integral((A + B*x + C*x**2)/((a + b*x)**3)**(1/3), x), Eq(p, - 
1/3)), (9*A*a*b**2*p**2*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p 
/(27*b**3*p**3 + 54*b**3*p**2 + 33*b**3*p + 6*b**3) + 15*A*a*b**2*p*(a**3 
+ 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(27*b**3*p**3 + 54*b**3*p**2 
+ 33*b**3*p + 6*b**3) + 6*A*a*b**2*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b* 
*3*x**3)**p/(27*b**3*p**3 + 54*b**3*p**2 + 33*b**3*p + 6*b**3) + 9*A*b**3* 
p**2*x*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(27*b**3*p**3 + 
54*b**3*p**2 + 33*b**3*p + 6*b**3) + 15*A*b**3*p*x*(a**3 + 3*a**2*b*x + 3* 
a*b**2*x**2 + b**3*x**3)**p/(27*b**3*p**3 + 54*b**3*p**2 + 33*b**3*p + 6*b 
**3) + 6*A*b**3*x*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(27*b 
**3*p**3 + 54*b**3*p**2 + 33*b**3*p + 6*b**3) - 3*B*a**2*b*p*(a**3 + 3*a** 
2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(27*b**3*p**3 + 54*b**3*p**2 + 33...
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{3 \, p} A}{b {\left (3 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (3 \, p + 1\right )} x^{2} + 3 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{3 \, p} B}{{\left (9 \, p^{2} + 9 \, p + 2\right )} b^{2}} + \frac {{\left ({\left (9 \, p^{2} + 9 \, p + 2\right )} b^{3} x^{3} + 3 \, {\left (3 \, p^{2} + p\right )} a b^{2} x^{2} - 6 \, a^{2} b p x + 2 \, a^{3}\right )} {\left (b x + a\right )}^{3 \, p} C}{3 \, {\left (9 \, p^{3} + 18 \, p^{2} + 11 \, p + 2\right )} b^{3}} \] Input:

integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p,x, algorithm 
="maxima")
 

Output:

(b*x + a)*(b*x + a)^(3*p)*A/(b*(3*p + 1)) + (b^2*(3*p + 1)*x^2 + 3*a*b*p*x 
 - a^2)*(b*x + a)^(3*p)*B/((9*p^2 + 9*p + 2)*b^2) + 1/3*((9*p^2 + 9*p + 2) 
*b^3*x^3 + 3*(3*p^2 + p)*a*b^2*x^2 - 6*a^2*b*p*x + 2*a^3)*(b*x + a)^(3*p)* 
C/((9*p^3 + 18*p^2 + 11*p + 2)*b^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (166) = 332\).

Time = 0.13 (sec) , antiderivative size = 799, normalized size of antiderivative = 4.78 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx =\text {Too large to display} \] Input:

integrate((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p,x, algorithm 
="giac")
 

Output:

1/3*(9*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*C*b^3*p^2*x^3 + 9*(b^3* 
x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*C*a*b^2*p^2*x^2 + 9*(b^3*x^3 + 3*a* 
b^2*x^2 + 3*a^2*b*x + a^3)^p*B*b^3*p^2*x^2 + 9*(b^3*x^3 + 3*a*b^2*x^2 + 3* 
a^2*b*x + a^3)^p*C*b^3*p*x^3 + 9*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3) 
^p*B*a*b^2*p^2*x + 9*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*A*b^3*p^2 
*x + 3*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*C*a*b^2*p*x^2 + 12*(b^3 
*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*B*b^3*p*x^2 + 2*(b^3*x^3 + 3*a*b^2 
*x^2 + 3*a^2*b*x + a^3)^p*C*b^3*x^3 + 9*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x 
 + a^3)^p*A*a*b^2*p^2 - 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*C*a^ 
2*b*p*x + 9*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*B*a*b^2*p*x + 15*( 
b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*A*b^3*p*x + 3*(b^3*x^3 + 3*a*b^ 
2*x^2 + 3*a^2*b*x + a^3)^p*B*b^3*x^2 - 3*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b* 
x + a^3)^p*B*a^2*b*p + 15*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*A*a* 
b^2*p + 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*A*b^3*x + 2*(b^3*x^3 
 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*C*a^3 - 3*(b^3*x^3 + 3*a*b^2*x^2 + 3*a 
^2*b*x + a^3)^p*B*a^2*b + 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*A* 
a*b^2)/(9*b^3*p^3 + 18*b^3*p^2 + 11*b^3*p + 2*b^3)
 

Mupad [B] (verification not implemented)

Time = 12.91 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.43 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx={\left (a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3\right )}^p\,\left (\frac {x\,\left (-6\,C\,a^2\,b\,p+9\,B\,a\,b^2\,p^2+9\,B\,a\,b^2\,p+9\,A\,b^3\,p^2+15\,A\,b^3\,p+6\,A\,b^3\right )}{3\,b^3\,\left (9\,p^3+18\,p^2+11\,p+2\right )}+\frac {C\,x^3\,\left (9\,p^2+9\,p+2\right )}{3\,\left (9\,p^3+18\,p^2+11\,p+2\right )}+\frac {a\,\left (2\,C\,a^2-3\,B\,a\,b\,p-3\,B\,a\,b+9\,A\,b^2\,p^2+15\,A\,b^2\,p+6\,A\,b^2\right )}{3\,b^3\,\left (9\,p^3+18\,p^2+11\,p+2\right )}+\frac {x^2\,\left (3\,p+1\right )\,\left (B\,b+B\,b\,p+C\,a\,p\right )}{b\,\left (9\,p^3+18\,p^2+11\,p+2\right )}\right ) \] Input:

int((A + B*x + C*x^2)*(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^p,x)
 

Output:

(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^p*((x*(6*A*b^3 + 9*A*b^3*p^2 + 1 
5*A*b^3*p + 9*B*a*b^2*p - 6*C*a^2*b*p + 9*B*a*b^2*p^2))/(3*b^3*(11*p + 18* 
p^2 + 9*p^3 + 2)) + (C*x^3*(9*p + 9*p^2 + 2))/(3*(11*p + 18*p^2 + 9*p^3 + 
2)) + (a*(6*A*b^2 + 2*C*a^2 + 9*A*b^2*p^2 - 3*B*a*b + 15*A*b^2*p - 3*B*a*b 
*p))/(3*b^3*(11*p + 18*p^2 + 9*p^3 + 2)) + (x^2*(3*p + 1)*(B*b + B*b*p + C 
*a*p))/(b*(11*p + 18*p^2 + 9*p^3 + 2)))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.22 \[ \int \left (A+B x+C x^2\right ) \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p} \left (9 b^{3} c \,p^{2} x^{3}+9 a \,b^{2} c \,p^{2} x^{2}+9 b^{4} p^{2} x^{2}+9 b^{3} c p \,x^{3}+18 a \,b^{3} p^{2} x +3 a \,b^{2} c p \,x^{2}+12 b^{4} p \,x^{2}+2 b^{3} c \,x^{3}+9 a^{2} b^{2} p^{2}-6 a^{2} b c p x +24 a \,b^{3} p x +3 b^{4} x^{2}+12 a^{2} b^{2} p +6 a \,b^{3} x +2 a^{3} c +3 a^{2} b^{2}\right )}{3 b^{3} \left (9 p^{3}+18 p^{2}+11 p +2\right )} \] Input:

int((C*x^2+B*x+A)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p,x)
 

Output:

((a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p*(2*a**3*c + 9*a**2*b** 
2*p**2 + 12*a**2*b**2*p + 3*a**2*b**2 - 6*a**2*b*c*p*x + 18*a*b**3*p**2*x 
+ 24*a*b**3*p*x + 6*a*b**3*x + 9*a*b**2*c*p**2*x**2 + 3*a*b**2*c*p*x**2 + 
9*b**4*p**2*x**2 + 12*b**4*p*x**2 + 3*b**4*x**2 + 9*b**3*c*p**2*x**3 + 9*b 
**3*c*p*x**3 + 2*b**3*c*x**3))/(3*b**3*(9*p**3 + 18*p**2 + 11*p + 2))