Integrand size = 24, antiderivative size = 85 \[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\frac {(c+d x) \left (a-\frac {c^3}{d}+\frac {(c+d x)^3}{d}\right )^p \left (1-\frac {(c+d x)^3}{c^3-a d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-p,\frac {4}{3},\frac {(c+d x)^3}{c^3-a d}\right )}{d} \] Output:
(d*x+c)*(a-c^3/d+(d*x+c)^3/d)^p*hypergeom([1/3, -p],[4/3],(d*x+c)^3/(c^3-a *d))/d/((1-(d*x+c)^3/(c^3-a*d))^p)
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.15 (sec) , antiderivative size = 459, normalized size of antiderivative = 5.40 \[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\frac {\left (a+x \left (3 c^2+3 c d x+d^2 x^2\right )\right )^p \operatorname {AppellF1}\left (1+p,-p,-p,2+p,\frac {-x+\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]}{\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,2\right ]},\frac {-x+\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]}{\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,3\right ]}\right ) \left (x-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]\right ) \left (\frac {x-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,2\right ]}{\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,2\right ]}\right )^{-p} \left (\frac {x-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,3\right ]}{\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,1\right ]-\text {Root}\left [a+3 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3\&,3\right ]}\right )^{-p}}{1+p} \] Input:
Integrate[(a + 3*c^2*x + 3*c*d*x^2 + d^2*x^3)^p,x]
Output:
((a + x*(3*c^2 + 3*c*d*x + d^2*x^2))^p*AppellF1[1 + p, -p, -p, 2 + p, (-x + Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1])/(Root[a + 3*c^2*#1 + 3 *c*d*#1^2 + d^2*#1^3 & , 1] - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 2]), (-x + Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1])/(Root[a + 3 *c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1] - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 3])]*(x - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1]))/ ((1 + p)*((x - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 2])/(Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1] - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 2]))^p*((x - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 3])/(Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 1] - Root[a + 3*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3 & , 3]))^p)
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2458, 779, 778}
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.
\(\displaystyle \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx\) |
\(\Big \downarrow \) 2458 |
\(\displaystyle \int \left (a-\frac {c^3}{d}+d^2 \left (\frac {c}{d}+x\right )^3\right )^pd\left (\frac {c}{d}+x\right )\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \left (a-\frac {c^3}{d}+d^2 \left (\frac {c}{d}+x\right )^3\right )^p \left (1-\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3-a d}\right )^{-p} \int \left (1-\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3-a d}\right )^pd\left (\frac {c}{d}+x\right )\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \left (\frac {c}{d}+x\right ) \left (a-\frac {c^3}{d}+d^2 \left (\frac {c}{d}+x\right )^3\right )^p \left (1-\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3-a d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-p,\frac {4}{3},\frac {d^3 \left (\frac {c}{d}+x\right )^3}{c^3-a d}\right )\) |
Input:
Int[(a + 3*c^2*x + 3*c*d*x^2 + d^2*x^3)^p,x]
Output:
((c/d + x)*(a - c^3/d + d^2*(c/d + x)^3)^p*Hypergeometric2F1[1/3, -p, 4/3, (d^3*(c/d + x)^3)/(c^3 - a*d)])/(1 - (d^3*(c/d + x)^3)/(c^3 - a*d))^p
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x ^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p, x], x ] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Si mplify[1/n + p], 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
\[\int \left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p}d x\]
Input:
int((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x)
Output:
int((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x)
\[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\int { {\left (d^{2} x^{3} + 3 \, c d x^{2} + 3 \, c^{2} x + a\right )}^{p} \,d x } \] Input:
integrate((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x, algorithm="fricas")
Output:
integral((d^2*x^3 + 3*c*d*x^2 + 3*c^2*x + a)^p, x)
Timed out. \[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\text {Timed out} \] Input:
integrate((d**2*x**3+3*c*d*x**2+3*c**2*x+a)**p,x)
Output:
Timed out
\[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\int { {\left (d^{2} x^{3} + 3 \, c d x^{2} + 3 \, c^{2} x + a\right )}^{p} \,d x } \] Input:
integrate((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x, algorithm="maxima")
Output:
integrate((d^2*x^3 + 3*c*d*x^2 + 3*c^2*x + a)^p, x)
\[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\int { {\left (d^{2} x^{3} + 3 \, c d x^{2} + 3 \, c^{2} x + a\right )}^{p} \,d x } \] Input:
integrate((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x, algorithm="giac")
Output:
integrate((d^2*x^3 + 3*c*d*x^2 + 3*c^2*x + a)^p, x)
Timed out. \[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\int {\left (3\,c^2\,x+3\,c\,d\,x^2+d^2\,x^3+a\right )}^p \,d x \] Input:
int((a + 3*c^2*x + d^2*x^3 + 3*c*d*x^2)^p,x)
Output:
int((a + 3*c^2*x + d^2*x^3 + 3*c*d*x^2)^p, x)
\[ \int \left (a+3 c^2 x+3 c d x^2+d^2 x^3\right )^p \, dx=\frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} a +\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} c^{2} x -9 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x^{2}}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) a \,d^{2} p^{2}-3 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x^{2}}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) a \,d^{2} p +9 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x^{2}}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) c^{3} d \,p^{2}+3 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x^{2}}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) c^{3} d p -18 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) a c d \,p^{2}-6 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) a c d p +18 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) c^{4} p^{2}+6 \left (\int \frac {\left (d^{2} x^{3}+3 c d \,x^{2}+3 c^{2} x +a \right )^{p} x}{3 d^{2} p \,x^{3}+9 c d p \,x^{2}+d^{2} x^{3}+9 c^{2} p x +3 c d \,x^{2}+3 c^{2} x +3 a p +a}d x \right ) c^{4} p}{c^{2} \left (3 p +1\right )} \] Input:
int((d^2*x^3+3*c*d*x^2+3*c^2*x+a)^p,x)
Output:
((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*a + (a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*c**2*x - 9*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)** p*x**2)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3 *d**2*p*x**3 + d**2*x**3),x)*a*d**2*p**2 - 3*int(((a + 3*c**2*x + 3*c*d*x* *2 + d**2*x**3)**p*x**2)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3*d**2*p*x**3 + d**2*x**3),x)*a*d**2*p + 9*int(((a + 3*c** 2*x + 3*c*d*x**2 + d**2*x**3)**p*x**2)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3*d**2*p*x**3 + d**2*x**3),x)*c**3*d*p**2 + 3*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*x**2)/(3*a*p + a + 9*c** 2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3*d**2*p*x**3 + d**2*x**3), x)*c**3*d*p - 18*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*x)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3*d**2*p*x**3 + d**2*x**3),x)*a*c*d*p**2 - 6*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3) **p*x)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c*d*x**2 + 3* d**2*p*x**3 + d**2*x**3),x)*a*c*d*p + 18*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*x)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p*x**2 + 3*c* d*x**2 + 3*d**2*p*x**3 + d**2*x**3),x)*c**4*p**2 + 6*int(((a + 3*c**2*x + 3*c*d*x**2 + d**2*x**3)**p*x)/(3*a*p + a + 9*c**2*p*x + 3*c**2*x + 9*c*d*p *x**2 + 3*c*d*x**2 + 3*d**2*p*x**3 + d**2*x**3),x)*c**4*p)/(c**2*(3*p + 1) )