\(\int (A+B x+C x^2) (3 c^2 x+3 c d x^2+d^2 x^3)^p \, dx\) [81]

Optimal result

Integrand size = 34, antiderivative size = 295 \[ \int \left (A+B x+C x^2\right ) \left (3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\frac {C \left (3 c^2 x+3 c d x^2+d^2 x^3\right )^{1+p}}{3 d^2 (1+p)}-\frac {(2 c C-B d) x^2 \left (1+\frac {2 d x}{3 c-\sqrt {3} \sqrt {-c^2}}\right )^{-p} \left (1+\frac {2 d x}{3 c+\sqrt {3} \sqrt {-c^2}}\right )^{-p} \left (3 c^2 x+3 c d x^2+d^2 x^3\right )^p \operatorname {AppellF1}\left (2+p,-p,-p,3+p,-\frac {2 d x}{3 c-\sqrt {3} \sqrt {-c^2}},-\frac {2 d x}{3 c+\sqrt {3} \sqrt {-c^2}}\right )}{d (2+p)}-\frac {\left (c^2 C-A d^2\right ) (c+d x) \left (3 c^2 x+3 c d x^2+d^2 x^3\right )^p \left (1-\frac {(c+d x)^3}{c^3}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-p,\frac {4}{3},\frac {(c+d x)^3}{c^3}\right )}{d^3} \] Output:

1/3*C*(d^2*x^3+3*c*d*x^2+3*c^2*x)^(p+1)/d^2/(p+1)-(-B*d+2*C*c)*x^2*(d^2*x^ 
3+3*c*d*x^2+3*c^2*x)^p*AppellF1(2+p,-p,-p,3+p,-2*d*x/(3*c-3^(1/2)*(-c^2)^( 
1/2)),-2*d*x/(3*c+3^(1/2)*(-c^2)^(1/2)))/d/(2+p)/((1+2*d*x/(3*c-3^(1/2)*(- 
c^2)^(1/2)))^p)/((1+2*d*x/(3*c+3^(1/2)*(-c^2)^(1/2)))^p)-(-A*d^2+C*c^2)*(d 
*x+c)*(d^2*x^3+3*c*d*x^2+3*c^2*x)^p*hypergeom([1/3, -p],[4/3],(d*x+c)^3/c^ 
3)/d^3/((1-(d*x+c)^3/c^3)^p)
 

Mathematica [A] (warning: unable to verify)

Time = 0.96 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.86 \[ \int \left (A+B x+C x^2\right ) \left (3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\frac {x \left (\frac {3 c d-\sqrt {3} \sqrt {-c^2 d^2}}{2 d^2}+x\right )^{-p} \left (\frac {3 c}{d}+\frac {\sqrt {3} c^2}{\sqrt {-c^2 d^2}}+2 x\right )^p \left (\frac {3 c}{d}+\frac {\sqrt {3} \sqrt {-c^2 d^2}}{d^2}+2 x\right )^p \left (\frac {6 c}{d}+\frac {2 \sqrt {3} \sqrt {-c^2 d^2}}{d^2}+4 x\right )^{-p} \left (\frac {-3 c d+\sqrt {3} \sqrt {-c^2 d^2}-2 d^2 x}{-3 c d+\sqrt {3} \sqrt {-c^2 d^2}}\right )^{-p} \left (\frac {3 c d+\sqrt {3} \sqrt {-c^2 d^2}+2 d^2 x}{3 c d+\sqrt {3} \sqrt {-c^2 d^2}}\right )^{-p} \left (x \left (3 c^2+3 c d x+d^2 x^2\right )\right )^p \left (A \left (6+5 p+p^2\right ) \operatorname {AppellF1}\left (1+p,-p,-p,2+p,-\frac {2 d^2 x}{3 c d+\sqrt {3} \sqrt {-c^2 d^2}},\frac {2 d^2 x}{-3 c d+\sqrt {3} \sqrt {-c^2 d^2}}\right )+(1+p) x \left (B (3+p) \operatorname {AppellF1}\left (2+p,-p,-p,3+p,-\frac {2 d^2 x}{3 c d+\sqrt {3} \sqrt {-c^2 d^2}},\frac {2 d^2 x}{-3 c d+\sqrt {3} \sqrt {-c^2 d^2}}\right )+C (2+p) x \operatorname {AppellF1}\left (3+p,-p,-p,4+p,-\frac {2 d^2 x}{3 c d+\sqrt {3} \sqrt {-c^2 d^2}},\frac {2 d^2 x}{-3 c d+\sqrt {3} \sqrt {-c^2 d^2}}\right )\right )\right )}{(1+p) (2+p) (3+p)} \] Input:

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

Output:

(x*((3*c)/d + (Sqrt[3]*c^2)/Sqrt[-(c^2*d^2)] + 2*x)^p*((3*c)/d + (Sqrt[3]* 
Sqrt[-(c^2*d^2)])/d^2 + 2*x)^p*(x*(3*c^2 + 3*c*d*x + d^2*x^2))^p*(A*(6 + 5 
*p + p^2)*AppellF1[1 + p, -p, -p, 2 + p, (-2*d^2*x)/(3*c*d + Sqrt[3]*Sqrt[ 
-(c^2*d^2)]), (2*d^2*x)/(-3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)])] + (1 + p)*x*( 
B*(3 + p)*AppellF1[2 + p, -p, -p, 3 + p, (-2*d^2*x)/(3*c*d + Sqrt[3]*Sqrt[ 
-(c^2*d^2)]), (2*d^2*x)/(-3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)])] + C*(2 + p)*x 
*AppellF1[3 + p, -p, -p, 4 + p, (-2*d^2*x)/(3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2 
)]), (2*d^2*x)/(-3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)])])))/((1 + p)*(2 + p)*(3 
 + p)*((3*c*d - Sqrt[3]*Sqrt[-(c^2*d^2)])/(2*d^2) + x)^p*((6*c)/d + (2*Sqr 
t[3]*Sqrt[-(c^2*d^2)])/d^2 + 4*x)^p*((-3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)] - 
2*d^2*x)/(-3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)]))^p*((3*c*d + Sqrt[3]*Sqrt[-(c 
^2*d^2)] + 2*d^2*x)/(3*c*d + Sqrt[3]*Sqrt[-(c^2*d^2)]))^p)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2459, 2425, 793, 2432, 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.

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

\(\Big \downarrow \) 2459

\(\displaystyle \int \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^p \left (A+\frac {c (c C-B d)}{d^2}+\left (\frac {c}{d}+x\right ) \left (B-\frac {2 c C}{d}\right )+C \left (\frac {c}{d}+x\right )^2\right )d\left (\frac {c}{d}+x\right )\)

\(\Big \downarrow \) 2425

\(\displaystyle \int \left (A+\frac {c (c C-B d)}{d^2}+\left (B-\frac {2 c C}{d}\right ) \left (\frac {c}{d}+x\right )\right ) \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^pd\left (\frac {c}{d}+x\right )+C \int \left (\frac {c}{d}+x\right )^2 \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^pd\left (\frac {c}{d}+x\right )\)

\(\Big \downarrow \) 793

\(\displaystyle \int \left (A+\frac {c (c C-B d)}{d^2}+\left (B-\frac {2 c C}{d}\right ) \left (\frac {c}{d}+x\right )\right ) \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^pd\left (\frac {c}{d}+x\right )+\frac {C \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^{p+1}}{3 d^2 (p+1)}\)

\(\Big \downarrow \) 2432

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (\frac {c}{d}+x\right ) \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^p \left (1-\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3}\right )^{-p} \left (A d^2-B c d+c^2 C\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-p,\frac {4}{3},\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3}\right )}{d^2}-\frac {\left (\frac {c}{d}+x\right )^2 (2 c C-B d) \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^p \left (1-\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},-p,\frac {5}{3},\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3}\right )}{2 d}+\frac {C \left (d^2 \left (\frac {c}{d}+x\right )^3-\frac {c^3}{d}\right )^{p+1}}{3 d^2 (p+1)}\)

Input:

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

Output:

(C*(-(c^3/d) + d^2*(c/d + x)^3)^(1 + p))/(3*d^2*(1 + p)) + ((c^2*C - B*c*d 
 + A*d^2)*(c/d + x)*(-(c^3/d) + d^2*(c/d + x)^3)^p*Hypergeometric2F1[1/3, 
-p, 4/3, (d^3*(c/d + x)^3)/c^3])/(d^2*(1 - (d^3*(c/d + x)^3)/c^3)^p) - ((2 
*c*C - B*d)*(c/d + x)^2*(-(c^3/d) + d^2*(c/d + x)^3)^p*Hypergeometric2F1[2 
/3, -p, 5/3, (d^3*(c/d + x)^3)/c^3])/(2*d*(1 - (d^3*(c/d + x)^3)/c^3)^p)
 

Defintions of rubi rules used

rule 793

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

rule 2009

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

rule 2425

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Pq, x, n - 
 1]   Int[x^(n - 1)*(a + b*x^n)^p, x], x] + Int[ExpandToSum[Pq - Coeff[Pq, 
x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && PolyQ[P 
q, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

rule 2459

Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 
]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x 
 -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial 
Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - 
> x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ 
[Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] &&  !(MonomialQ[Qx, x] 
&& IGtQ[p, 0])
 

Maple [F]

\[\int \left (C \,x^{2}+B x +A \right ) \left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x \right )^{p}d x\]

Input:

int((C*x^2+B*x+A)*(d^2*x^3+3*c*d*x^2+3*c^2*x)^p,x)
 

Output:

int((C*x^2+B*x+A)*(d^2*x^3+3*c*d*x^2+3*c^2*x)^p,x)
 

Fricas [F]

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

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

Output:

integral((C*x^2 + B*x + A)*(d^2*x^3 + 3*c*d*x^2 + 3*c^2*x)^p, x)
 

Sympy [F]

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

integrate((C*x**2+B*x+A)*(d**2*x**3+3*c*d*x**2+3*c**2*x)**p,x)
 

Output:

Integral((x*(3*c**2 + 3*c*d*x + d**2*x**2))**p*(A + B*x + C*x**2), x)
 

Maxima [F]

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

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

Output:

integrate((C*x^2 + B*x + A)*(d^2*x^3 + 3*c*d*x^2 + 3*c^2*x)^p, x)
 

Giac [F]

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

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

Output:

integrate((C*x^2 + B*x + A)*(d^2*x^3 + 3*c*d*x^2 + 3*c^2*x)^p, x)
 

Mupad [F(-1)]

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

int((A + B*x + C*x^2)*(3*c^2*x + d^2*x^3 + 3*c*d*x^2)^p,x)
 

Output:

int((A + B*x + C*x^2)*(3*c^2*x + d^2*x^3 + 3*c*d*x^2)^p, x)
 

Reduce [F]

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

int((C*x^2+B*x+A)*(d^2*x^3+3*c*d*x^2+3*c^2*x)^p,x)
 

Output:

(18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*a*c*d**2*p**2 + 30*(3*c**2*x + 
3*c*d*x**2 + d**2*x**3)**p*a*c*d**2*p + 12*(3*c**2*x + 3*c*d*x**2 + d**2*x 
**3)**p*a*c*d**2 + 18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*a*d**3*p**2*x 
 + 30*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*a*d**3*p*x + 12*(3*c**2*x + 3 
*c*d*x**2 + d**2*x**3)**p*a*d**3*x - 9*(3*c**2*x + 3*c*d*x**2 + d**2*x**3) 
**p*b*c**2*d*p**2 - 18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*b*c**2*d*p - 
 9*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*b*c**2*d + 18*(3*c**2*x + 3*c*d* 
x**2 + d**2*x**3)**p*b*c*d**2*p**2*x + 18*(3*c**2*x + 3*c*d*x**2 + d**2*x* 
*3)**p*b*c*d**2*p*x + 18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*b*d**3*p** 
2*x**2 + 24*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*b*d**3*p*x**2 + 6*(3*c* 
*2*x + 3*c*d*x**2 + d**2*x**3)**p*b*d**3*x**2 + 6*(3*c**2*x + 3*c*d*x**2 + 
 d**2*x**3)**p*c**4*p + 6*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c**4 - 12 
*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c**3*d*p*x + 18*(3*c**2*x + 3*c*d* 
x**2 + d**2*x**3)**p*c**2*d**2*p**2*x**2 + 6*(3*c**2*x + 3*c*d*x**2 + d**2 
*x**3)**p*c**2*d**2*p*x**2 + 18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c*d 
**3*p**2*x**3 + 18*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c*d**3*p*x**3 + 
4*(3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c*d**3*x**3 - 486*int((3*c**2*x + 
 3*c*d*x**2 + d**2*x**3)**p/(27*c**2*p**2*x + 27*c**2*p*x + 6*c**2*x + 27* 
c*d*p**2*x**2 + 27*c*d*p*x**2 + 6*c*d*x**2 + 9*d**2*p**2*x**3 + 9*d**2*p*x 
**3 + 2*d**2*x**3),x)*a*c**3*d**2*p**5 - 1296*int((3*c**2*x + 3*c*d*x**...