\(\int (A+B x+C x^2) (27 a^3+27 a^2 b x-4 b^3 x^3)^p \, dx\) [14]

Optimal result

Integrand size = 34, antiderivative size = 259 \[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=-\frac {(2 b B (1+p)+3 a C (2+3 p)) (3 a-b x) (3 a+2 b x) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p}{4 b^3 (1+p) (2+3 p)}+\frac {C (3 a-b x)^2 (3 a+2 b x) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p}{6 b^3 (1+p)}-\frac {3^{2 p} \left (4 A b^2 (2+3 p)+3 a (3 a C (2+3 p)+2 b (B+3 B p))\right ) (3 a-b x) \left (1+\frac {2 b x}{3 a}\right )^{-2 p} \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \operatorname {Hypergeometric2F1}\left (-2 p,1+p,2+p,\frac {2 (3 a-b x)}{9 a}\right )}{4 b^3 (1+p) (2+3 p)} \] Output:

-1/4*(2*b*B*(p+1)+3*a*C*(2+3*p))*(-b*x+3*a)*(2*b*x+3*a)*(-4*b^3*x^3+27*a^2 
*b*x+27*a^3)^p/b^3/(p+1)/(2+3*p)+1/6*C*(-b*x+3*a)^2*(2*b*x+3*a)*(-4*b^3*x^ 
3+27*a^2*b*x+27*a^3)^p/b^3/(p+1)-1/4*3^(2*p)*(4*A*b^2*(2+3*p)+3*a*(3*a*C*( 
2+3*p)+2*b*(3*B*p+B)))*(-b*x+3*a)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p*hyperge 
om([-2*p, p+1],[2+p],2/9*(-b*x+3*a)/a)/b^3/(p+1)/(2+3*p)/((1+2/3*b*x/a)^(2 
*p))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.64 \[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\frac {81^p (-3 a+b x) \left ((3 a-b x) (3 a+2 b x)^2\right )^p \left (3+\frac {2 b x}{a}\right )^{-2 p} \left (18 a (b B-3 a C) \operatorname {Hypergeometric2F1}\left (-1-2 p,1+p,2+p,\frac {2}{3}-\frac {2 b x}{9 a}\right )+\left (4 A b^2-6 a b B+9 a^2 C\right ) \operatorname {Hypergeometric2F1}\left (-2 p,1+p,2+p,\frac {2}{3}-\frac {2 b x}{9 a}\right )+81 a^2 C \operatorname {Hypergeometric2F1}\left (-2 (1+p),1+p,2+p,\frac {2}{3}-\frac {2 b x}{9 a}\right )\right )}{4 b^3 (1+p)} \] Input:

Integrate[(A + B*x + C*x^2)*(27*a^3 + 27*a^2*b*x - 4*b^3*x^3)^p,x]
 

Output:

(81^p*(-3*a + b*x)*((3*a - b*x)*(3*a + 2*b*x)^2)^p*(18*a*(b*B - 3*a*C)*Hyp 
ergeometric2F1[-1 - 2*p, 1 + p, 2 + p, 2/3 - (2*b*x)/(9*a)] + (4*A*b^2 - 6 
*a*b*B + 9*a^2*C)*Hypergeometric2F1[-2*p, 1 + p, 2 + p, 2/3 - (2*b*x)/(9*a 
)] + 81*a^2*C*Hypergeometric2F1[-2*(1 + p), 1 + p, 2 + p, 2/3 - (2*b*x)/(9 
*a)]))/(4*b^3*(1 + p)*(3 + (2*b*x)/a)^(2*p))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2526, 27, 2483, 90, 80, 79}

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.

\(\displaystyle \int \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \left (A+B x+C x^2\right ) \, dx\)

\(\Big \downarrow \) 2526

\(\displaystyle -\frac {\int -3 b \left (9 C a^2+4 A b^2+4 b^2 B x\right ) \left (27 a^3+27 b x a^2-4 b^3 x^3\right )^pdx}{12 b^3}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \left (9 C a^2+4 A b^2+4 b^2 B x\right ) \left (27 a^3+27 b x a^2-4 b^3 x^3\right )^pdx}{4 b^2}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

\(\Big \downarrow \) 2483

\(\displaystyle \frac {\left (81 a^3-27 a^2 b x\right )^{-p} \left (81 a^3+54 a^2 b x\right )^{-2 p} \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \int \left (81 a^3-27 a^2 b x\right )^p \left (81 a^3+54 b x a^2\right )^{2 p} \left (9 C a^2+4 A b^2+4 b^2 B x\right )dx}{4 b^2}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\left (81 a^3-27 a^2 b x\right )^{-p} \left (81 a^3+54 a^2 b x\right )^{-2 p} \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \left (\left (9 a^2 C+\frac {6 a b B (3 p+1)}{3 p+2}+4 A b^2\right ) \int \left (81 a^3-27 a^2 b x\right )^p \left (81 a^3+54 b x a^2\right )^{2 p}dx-\frac {2 B \left (81 a^3-27 a^2 b x\right )^{p+1} \left (81 a^3+54 a^2 b x\right )^{2 p+1}}{729 a^4 (3 p+2)}\right )}{4 b^2}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\left (81 a^3-27 a^2 b x\right )^{-p} \left (81 a^3+54 a^2 b x\right )^{-2 p} \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \left (81^p \left (\frac {3 a+2 b x}{a}\right )^{-2 p} \left (81 a^3+54 a^2 b x\right )^{2 p} \left (9 a^2 C+\frac {6 a b B (3 p+1)}{3 p+2}+4 A b^2\right ) \int \left (\frac {2 b x}{9 a}+\frac {1}{3}\right )^{2 p} \left (81 a^3-27 a^2 b x\right )^pdx-\frac {2 B \left (81 a^3-27 a^2 b x\right )^{p+1} \left (81 a^3+54 a^2 b x\right )^{2 p+1}}{729 a^4 (3 p+2)}\right )}{4 b^2}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\left (81 a^3-27 a^2 b x\right )^{-p} \left (81 a^3+54 a^2 b x\right )^{-2 p} \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \left (-\frac {3^{4 p-3} \left (81 a^3-27 a^2 b x\right )^{p+1} \left (81 a^3+54 a^2 b x\right )^{2 p} \left (\frac {3 a+2 b x}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-2 p,p+1,p+2,\frac {2 (3 a-b x)}{9 a}\right ) \left (9 a^2 C+\frac {6 a b B (3 p+1)}{3 p+2}+4 A b^2\right )}{a^2 b (p+1)}-\frac {2 B \left (81 a^3-27 a^2 b x\right )^{p+1} \left (81 a^3+54 a^2 b x\right )^{2 p+1}}{729 a^4 (3 p+2)}\right )}{4 b^2}-\frac {C \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{p+1}}{12 b^3 (p+1)}\)

Input:

Int[(A + B*x + C*x^2)*(27*a^3 + 27*a^2*b*x - 4*b^3*x^3)^p,x]
 

Output:

-1/12*(C*(27*a^3 + 27*a^2*b*x - 4*b^3*x^3)^(1 + p))/(b^3*(1 + p)) + ((27*a 
^3 + 27*a^2*b*x - 4*b^3*x^3)^p*((-2*B*(81*a^3 - 27*a^2*b*x)^(1 + p)*(81*a^ 
3 + 54*a^2*b*x)^(1 + 2*p))/(729*a^4*(2 + 3*p)) - (3^(-3 + 4*p)*(4*A*b^2 + 
9*a^2*C + (6*a*b*B*(1 + 3*p))/(2 + 3*p))*(81*a^3 - 27*a^2*b*x)^(1 + p)*(81 
*a^3 + 54*a^2*b*x)^(2*p)*Hypergeometric2F1[-2*p, 1 + p, 2 + p, (2*(3*a - b 
*x))/(9*a)])/(a^2*b*(1 + p)*((3*a + 2*b*x)/a)^(2*p))))/(4*b^2*(81*a^3 - 27 
*a^2*b*x)^p*(81*a^3 + 54*a^2*b*x)^(2*p))
 

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 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 90

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

rule 2483

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_S 
ymbol] :> Simp[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*x)^(2*p))   In 
t[(e + f*x)^m*(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, 
e, f, m, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0] &&  !IntegerQ[p]
 

rule 2526

Int[(Pm_)*(Qn_)^(p_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x] 
}, Simp[Coeff[Pm, x, m]*(Qn^(p + 1)/(n*(p + 1)*Coeff[Qn, x, n])), x] + Simp 
[1/(n*Coeff[Qn, x, n])   Int[ExpandToSum[n*Coeff[Qn, x, n]*Pm - Coeff[Pm, x 
, m]*D[Qn, x], x]*Qn^p, x], x] /; EqQ[m, n - 1]] /; FreeQ[p, x] && PolyQ[Pm 
, x] && PolyQ[Qn, x] && NeQ[p, -1]
 

Maple [F]

\[\int \left (C \,x^{2}+B x +A \right ) \left (-4 b^{3} x^{3}+27 b \,a^{2} x +27 a^{3}\right )^{p}d x\]

Input:

int((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x)
 

Output:

int((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x)
 

Fricas [F]

\[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}\right )}^{p} \,d x } \] Input:

integrate((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x, algorithm="fri 
cas")
 

Output:

integral((C*x^2 + B*x + A)*(-4*b^3*x^3 + 27*a^2*b*x + 27*a^3)^p, x)
 

Sympy [F]

\[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\int \left (- \left (- 3 a + b x\right ) \left (3 a + 2 b x\right )^{2}\right )^{p} \left (A + B x + C x^{2}\right )\, dx \] Input:

integrate((C*x**2+B*x+A)*(-4*b**3*x**3+27*a**2*b*x+27*a**3)**p,x)
 

Output:

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

Maxima [F]

\[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}\right )}^{p} \,d x } \] Input:

integrate((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x, algorithm="max 
ima")
 

Output:

integrate((C*x^2 + B*x + A)*(-4*b^3*x^3 + 27*a^2*b*x + 27*a^3)^p, x)
 

Giac [F]

\[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\int { {\left (C x^{2} + B x + A\right )} {\left (-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}\right )}^{p} \,d x } \] Input:

integrate((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x, algorithm="gia 
c")
 

Output:

integrate((C*x^2 + B*x + A)*(-4*b^3*x^3 + 27*a^2*b*x + 27*a^3)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx=\int \left (C\,x^2+B\,x+A\right )\,{\left (27\,a^3+27\,a^2\,b\,x-4\,b^3\,x^3\right )}^p \,d x \] Input:

int((A + B*x + C*x^2)*(27*a^3 - 4*b^3*x^3 + 27*a^2*b*x)^p,x)
 

Output:

int((A + B*x + C*x^2)*(27*a^3 - 4*b^3*x^3 + 27*a^2*b*x)^p, x)
 

Reduce [F]

\[ \int \left (A+B x+C x^2\right ) \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^p \, dx =\text {Too large to display} \] Input:

int((C*x^2+B*x+A)*(-4*b^3*x^3+27*a^2*b*x+27*a^3)^p,x)
                                                                                    
                                                                                    
 

Output:

(81*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*a**3*c*p + 54*(27*a**3 + 27*a 
**2*b*x - 4*b**3*x**3)**p*a**3*c + 54*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3 
)**p*a**2*b**2*p**2 + 90*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*a**2*b** 
2*p + 36*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*a**2*b**2 - 81*(27*a**3 
+ 27*a**2*b*x - 4*b**3*x**3)**p*a**2*b*c*p**2*x - 54*(27*a**3 + 27*a**2*b* 
x - 4*b**3*x**3)**p*a**2*b*c*p*x + 18*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3 
)**p*a*b**3*p**2*x + 30*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*a*b**3*p* 
x + 12*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*a*b**3*x + 18*(27*a**3 + 2 
7*a**2*b*x - 4*b**3*x**3)**p*b**4*p**2*x**2 + 24*(27*a**3 + 27*a**2*b*x - 
4*b**3*x**3)**p*b**4*p*x**2 + 6*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*b 
**4*x**2 + 18*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*b**3*c*p**2*x**3 + 
18*(27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*b**3*c*p*x**3 + 4*(27*a**3 + 2 
7*a**2*b*x - 4*b**3*x**3)**p*b**3*c*x**3 + 6561*int(((27*a**3 + 27*a**2*b* 
x - 4*b**3*x**3)**p*x)/(81*a**2*p**2 + 81*a**2*p + 18*a**2 + 27*a*b*p**2*x 
 + 27*a*b*p*x + 6*a*b*x - 18*b**2*p**2*x**2 - 18*b**2*p*x**2 - 4*b**2*x**2 
),x)*a**3*b**2*c*p**5 + 17496*int(((27*a**3 + 27*a**2*b*x - 4*b**3*x**3)** 
p*x)/(81*a**2*p**2 + 81*a**2*p + 18*a**2 + 27*a*b*p**2*x + 27*a*b*p*x + 6* 
a*b*x - 18*b**2*p**2*x**2 - 18*b**2*p*x**2 - 4*b**2*x**2),x)*a**3*b**2*c*p 
**4 + 16767*int(((27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**p*x)/(81*a**2*p**2 
 + 81*a**2*p + 18*a**2 + 27*a*b*p**2*x + 27*a*b*p*x + 6*a*b*x - 18*b**2...