\(\int x^8 (a+b x^3)^p (c+d x^3) \, dx\) [395]

Optimal result

Integrand size = 20, antiderivative size = 123 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {a^2 (b c-a d) \left (a+b x^3\right )^{1+p}}{3 b^4 (1+p)}-\frac {a (2 b c-3 a d) \left (a+b x^3\right )^{2+p}}{3 b^4 (2+p)}+\frac {(b c-3 a d) \left (a+b x^3\right )^{3+p}}{3 b^4 (3+p)}+\frac {d \left (a+b x^3\right )^{4+p}}{3 b^4 (4+p)} \] Output:

1/3*a^2*(-a*d+b*c)*(b*x^3+a)^(p+1)/b^4/(p+1)-1/3*a*(-3*a*d+2*b*c)*(b*x^3+a 
)^(2+p)/b^4/(2+p)+1/3*(-3*a*d+b*c)*(b*x^3+a)^(3+p)/b^4/(3+p)+1/3*d*(b*x^3+ 
a)^(4+p)/b^4/(4+p)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {\left (a+b x^3\right )^{1+p} \left (\frac {a^2 (b c-a d)}{1+p}+\frac {a (-2 b c+3 a d) \left (a+b x^3\right )}{2+p}+\frac {(b c-3 a d) \left (a+b x^3\right )^2}{3+p}+\frac {d \left (a+b x^3\right )^3}{4+p}\right )}{3 b^4} \] Input:

Integrate[x^8*(a + b*x^3)^p*(c + d*x^3),x]
 

Output:

((a + b*x^3)^(1 + p)*((a^2*(b*c - a*d))/(1 + p) + (a*(-2*b*c + 3*a*d)*(a + 
 b*x^3))/(2 + p) + ((b*c - 3*a*d)*(a + b*x^3)^2)/(3 + p) + (d*(a + b*x^3)^ 
3)/(4 + p)))/(3*b^4)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {948, 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.

\(\displaystyle \int x^8 \left (c+d x^3\right ) \left (a+b x^3\right )^p \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int x^6 \left (b x^3+a\right )^p \left (d x^3+c\right )dx^3\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {a^2 (b c-a d) \left (a+b x^3\right )^{p+1}}{b^4 (p+1)}-\frac {a (2 b c-3 a d) \left (a+b x^3\right )^{p+2}}{b^4 (p+2)}+\frac {(b c-3 a d) \left (a+b x^3\right )^{p+3}}{b^4 (p+3)}+\frac {d \left (a+b x^3\right )^{p+4}}{b^4 (p+4)}\right )\)

Input:

Int[x^8*(a + b*x^3)^p*(c + d*x^3),x]
 

Output:

((a^2*(b*c - a*d)*(a + b*x^3)^(1 + p))/(b^4*(1 + p)) - (a*(2*b*c - 3*a*d)* 
(a + b*x^3)^(2 + p))/(b^4*(2 + p)) + ((b*c - 3*a*d)*(a + b*x^3)^(3 + p))/( 
b^4*(3 + p)) + (d*(a + b*x^3)^(4 + p))/(b^4*(4 + p)))/3
 

Defintions of rubi rules used

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 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(115)=230\).

Time = 1.02 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.90

method	result	size
gosper	\(-\frac {\left (b \,x^{3}+a \right )^{p +1} \left (-b^{3} d \,p^{3} x^{9}-6 b^{3} d \,p^{2} x^{9}-11 b^{3} d p \,x^{9}-b^{3} c \,p^{3} x^{6}-6 b^{3} d \,x^{9}+3 a \,b^{2} d \,p^{2} x^{6}-7 b^{3} c \,p^{2} x^{6}+9 a \,b^{2} d p \,x^{6}-14 b^{3} c p \,x^{6}+6 a \,b^{2} d \,x^{6}-8 b^{3} c \,x^{6}+2 a \,b^{2} c \,p^{2} x^{3}-6 a^{2} b d p \,x^{3}+10 a \,b^{2} c p \,x^{3}-6 a^{2} b d \,x^{3}+8 a \,b^{2} c \,x^{3}-2 a^{2} b c p +6 a^{3} d -8 a^{2} b c \right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\)	\(234\)
orering	\(-\frac {\left (b \,x^{3}+a \right )^{p} \left (-b^{3} d \,p^{3} x^{9}-6 b^{3} d \,p^{2} x^{9}-11 b^{3} d p \,x^{9}-b^{3} c \,p^{3} x^{6}-6 b^{3} d \,x^{9}+3 a \,b^{2} d \,p^{2} x^{6}-7 b^{3} c \,p^{2} x^{6}+9 a \,b^{2} d p \,x^{6}-14 b^{3} c p \,x^{6}+6 a \,b^{2} d \,x^{6}-8 b^{3} c \,x^{6}+2 a \,b^{2} c \,p^{2} x^{3}-6 a^{2} b d p \,x^{3}+10 a \,b^{2} c p \,x^{3}-6 a^{2} b d \,x^{3}+8 a \,b^{2} c \,x^{3}-2 a^{2} b c p +6 a^{3} d -8 a^{2} b c \right ) \left (b \,x^{3}+a \right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )}\)	\(239\)
risch	\(-\frac {\left (-b^{4} d \,p^{3} x^{12}-6 b^{4} d \,p^{2} x^{12}-11 b^{4} d p \,x^{12}-a \,b^{3} d \,p^{3} x^{9}-b^{4} c \,p^{3} x^{9}-6 d \,x^{12} b^{4}-3 a \,b^{3} d \,p^{2} x^{9}-7 b^{4} c \,p^{2} x^{9}-2 a \,b^{3} d p \,x^{9}-14 b^{4} c p \,x^{9}-a \,b^{3} c \,p^{3} x^{6}-8 b^{4} c \,x^{9}+3 a^{2} b^{2} d \,p^{2} x^{6}-5 a \,b^{3} c \,p^{2} x^{6}+3 a^{2} b^{2} d p \,x^{6}-4 a \,b^{3} c p \,x^{6}+2 a^{2} b^{2} c \,p^{2} x^{3}-6 a^{3} b d p \,x^{3}+8 a^{2} b^{2} c p \,x^{3}-2 a^{3} b c p +6 a^{4} d -8 a^{3} b c \right ) \left (b \,x^{3}+a \right )^{p}}{3 \left (3+p \right ) \left (4+p \right ) \left (2+p \right ) \left (p +1\right ) b^{4}}\)	\(284\)
parallelrisch	\(\frac {x^{9} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} d \,p^{3}+x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{4} c \,p^{3}+3 x^{9} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} d \,p^{2}+7 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{4} c \,p^{2}+2 x^{9} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} d p +14 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{4} c p +6 x^{12} \left (b \,x^{3}+a \right )^{p} a \,b^{4} d +8 x^{9} \left (b \,x^{3}+a \right )^{p} a \,b^{4} c +2 \left (b \,x^{3}+a \right )^{p} a^{4} b c p +6 x^{3} \left (b \,x^{3}+a \right )^{p} a^{4} b d p -8 x^{3} \left (b \,x^{3}+a \right )^{p} a^{3} b^{2} c p +x^{12} \left (b \,x^{3}+a \right )^{p} a \,b^{4} d \,p^{3}+6 x^{12} \left (b \,x^{3}+a \right )^{p} a \,b^{4} d \,p^{2}+11 x^{12} \left (b \,x^{3}+a \right )^{p} a \,b^{4} d p -2 x^{3} \left (b \,x^{3}+a \right )^{p} a^{3} b^{2} c \,p^{2}+x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} c \,p^{3}-3 x^{6} \left (b \,x^{3}+a \right )^{p} a^{3} b^{2} d \,p^{2}+5 x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} c \,p^{2}-3 x^{6} \left (b \,x^{3}+a \right )^{p} a^{3} b^{2} d p +4 x^{6} \left (b \,x^{3}+a \right )^{p} a^{2} b^{3} c p +8 \left (b \,x^{3}+a \right )^{p} a^{4} b c -6 \left (b \,x^{3}+a \right )^{p} a^{5} d}{3 \left (p^{3}+9 p^{2}+26 p +24\right ) \left (p +1\right ) a \,b^{4}}\)	\(492\)

Input:

int(x^8*(b*x^3+a)^p*(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-1/3/b^4*(b*x^3+a)^(p+1)/(p^4+10*p^3+35*p^2+50*p+24)*(-b^3*d*p^3*x^9-6*b^3 
*d*p^2*x^9-11*b^3*d*p*x^9-b^3*c*p^3*x^6-6*b^3*d*x^9+3*a*b^2*d*p^2*x^6-7*b^ 
3*c*p^2*x^6+9*a*b^2*d*p*x^6-14*b^3*c*p*x^6+6*a*b^2*d*x^6-8*b^3*c*x^6+2*a*b 
^2*c*p^2*x^3-6*a^2*b*d*p*x^3+10*a*b^2*c*p*x^3-6*a^2*b*d*x^3+8*a*b^2*c*x^3- 
2*a^2*b*c*p+6*a^3*d-8*a^2*b*c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (115) = 230\).

Time = 0.12 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.08 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left ({\left (b^{4} d p^{3} + 6 \, b^{4} d p^{2} + 11 \, b^{4} d p + 6 \, b^{4} d\right )} x^{12} + {\left (8 \, b^{4} c + {\left (b^{4} c + a b^{3} d\right )} p^{3} + {\left (7 \, b^{4} c + 3 \, a b^{3} d\right )} p^{2} + 2 \, {\left (7 \, b^{4} c + a b^{3} d\right )} p\right )} x^{9} + {\left (a b^{3} c p^{3} + {\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d\right )} p^{2} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d\right )} p\right )} x^{6} + 2 \, a^{3} b c p + 8 \, a^{3} b c - 6 \, a^{4} d - 2 \, {\left (a^{2} b^{2} c p^{2} + {\left (4 \, a^{2} b^{2} c - 3 \, a^{3} b d\right )} p\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{p}}{3 \, {\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )}} \] Input:

integrate(x^8*(b*x^3+a)^p*(d*x^3+c),x, algorithm="fricas")
 

Output:

1/3*((b^4*d*p^3 + 6*b^4*d*p^2 + 11*b^4*d*p + 6*b^4*d)*x^12 + (8*b^4*c + (b 
^4*c + a*b^3*d)*p^3 + (7*b^4*c + 3*a*b^3*d)*p^2 + 2*(7*b^4*c + a*b^3*d)*p) 
*x^9 + (a*b^3*c*p^3 + (5*a*b^3*c - 3*a^2*b^2*d)*p^2 + (4*a*b^3*c - 3*a^2*b 
^2*d)*p)*x^6 + 2*a^3*b*c*p + 8*a^3*b*c - 6*a^4*d - 2*(a^2*b^2*c*p^2 + (4*a 
^2*b^2*c - 3*a^3*b*d)*p)*x^3)*(b*x^3 + a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4 
*p^2 + 50*b^4*p + 24*b^4)
 

Sympy [F(-1)]

Timed out. \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\text {Timed out} \] Input:

integrate(x**8*(b*x**3+a)**p*(d*x**3+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.48 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{9} + {\left (p^{2} + p\right )} a b^{2} x^{6} - 2 \, a^{2} b p x^{3} + 2 \, a^{3}\right )} {\left (b x^{3} + a\right )}^{p} c}{3 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} + \frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{12} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{9} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} x^{6} + 6 \, a^{3} b p x^{3} - 6 \, a^{4}\right )} {\left (b x^{3} + a\right )}^{p} d}{3 \, {\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \] Input:

integrate(x^8*(b*x^3+a)^p*(d*x^3+c),x, algorithm="maxima")
 

Output:

1/3*((p^2 + 3*p + 2)*b^3*x^9 + (p^2 + p)*a*b^2*x^6 - 2*a^2*b*p*x^3 + 2*a^3 
)*(b*x^3 + a)^p*c/((p^3 + 6*p^2 + 11*p + 6)*b^3) + 1/3*((p^3 + 6*p^2 + 11* 
p + 6)*b^4*x^12 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^9 - 3*(p^2 + p)*a^2*b^2*x^6 
+ 6*a^3*b*p*x^3 - 6*a^4)*(b*x^3 + a)^p*d/((p^4 + 10*p^3 + 35*p^2 + 50*p + 
24)*b^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (115) = 230\).

Time = 0.14 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.55 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {{\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} b c p^{2} - 2 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a b c p^{2} + {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} d p^{2} - 3 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a d p^{2} + 3 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} d p^{2} + 6 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} b c p - 14 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a b c p + 5 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} d p - 18 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a d p + 21 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} d p + 8 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} b c - 24 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a b c + 6 \, {\left (b x^{3} + a\right )}^{4} {\left (b x^{3} + a\right )}^{p} d - 24 \, {\left (b x^{3} + a\right )}^{3} {\left (b x^{3} + a\right )}^{p} a d + 36 \, {\left (b x^{3} + a\right )}^{2} {\left (b x^{3} + a\right )}^{p} a^{2} d}{3 \, {\left (b^{4} p^{3} + 9 \, b^{4} p^{2} + 26 \, b^{4} p + 24 \, b^{4}\right )}} + \frac {\frac {{\left (b x^{3} + a\right )}^{p + 1} a^{2} b c}{p + 1} - \frac {{\left (b x^{3} + a\right )}^{p + 1} a^{3} d}{p + 1}}{3 \, b^{4}} \] Input:

integrate(x^8*(b*x^3+a)^p*(d*x^3+c),x, algorithm="giac")
 

Output:

1/3*((b*x^3 + a)^3*(b*x^3 + a)^p*b*c*p^2 - 2*(b*x^3 + a)^2*(b*x^3 + a)^p*a 
*b*c*p^2 + (b*x^3 + a)^4*(b*x^3 + a)^p*d*p^2 - 3*(b*x^3 + a)^3*(b*x^3 + a) 
^p*a*d*p^2 + 3*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2*d*p^2 + 6*(b*x^3 + a)^3*(b* 
x^3 + a)^p*b*c*p - 14*(b*x^3 + a)^2*(b*x^3 + a)^p*a*b*c*p + 5*(b*x^3 + a)^ 
4*(b*x^3 + a)^p*d*p - 18*(b*x^3 + a)^3*(b*x^3 + a)^p*a*d*p + 21*(b*x^3 + a 
)^2*(b*x^3 + a)^p*a^2*d*p + 8*(b*x^3 + a)^3*(b*x^3 + a)^p*b*c - 24*(b*x^3 
+ a)^2*(b*x^3 + a)^p*a*b*c + 6*(b*x^3 + a)^4*(b*x^3 + a)^p*d - 24*(b*x^3 + 
 a)^3*(b*x^3 + a)^p*a*d + 36*(b*x^3 + a)^2*(b*x^3 + a)^p*a^2*d)/(b^4*p^3 + 
 9*b^4*p^2 + 26*b^4*p + 24*b^4) + 1/3*((b*x^3 + a)^(p + 1)*a^2*b*c/(p + 1) 
 - (b*x^3 + a)^(p + 1)*a^3*d/(p + 1))/b^4
 

Mupad [B] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.89 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx={\left (b\,x^3+a\right )}^p\,\left (\frac {d\,x^{12}\,\left (p^3+6\,p^2+11\,p+6\right )}{3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {2\,a^3\,\left (4\,b\,c-3\,a\,d+b\,c\,p\right )}{3\,b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {x^9\,\left (4\,b\,c+a\,d\,p+b\,c\,p\right )\,\left (p^2+3\,p+2\right )}{3\,b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {2\,a^2\,p\,x^3\,\left (4\,b\,c-3\,a\,d+b\,c\,p\right )}{3\,b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^6\,\left (p+1\right )\,\left (4\,b\,c-3\,a\,d+b\,c\,p\right )}{3\,b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right ) \] Input:

int(x^8*(a + b*x^3)^p*(c + d*x^3),x)
 

Output:

(a + b*x^3)^p*((d*x^12*(11*p + 6*p^2 + p^3 + 6))/(3*(50*p + 35*p^2 + 10*p^ 
3 + p^4 + 24)) + (2*a^3*(4*b*c - 3*a*d + b*c*p))/(3*b^4*(50*p + 35*p^2 + 1 
0*p^3 + p^4 + 24)) + (x^9*(4*b*c + a*d*p + b*c*p)*(3*p + p^2 + 2))/(3*b*(5 
0*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (2*a^2*p*x^3*(4*b*c - 3*a*d + b*c*p)) 
/(3*b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^6*(p + 1)*(4*b*c - 3 
*a*d + b*c*p))/(3*b^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.27 \[ \int x^8 \left (a+b x^3\right )^p \left (c+d x^3\right ) \, dx=\frac {\left (b \,x^{3}+a \right )^{p} \left (b^{4} d \,p^{3} x^{12}+6 b^{4} d \,p^{2} x^{12}+11 b^{4} d p \,x^{12}+a \,b^{3} d \,p^{3} x^{9}+b^{4} c \,p^{3} x^{9}+6 b^{4} d \,x^{12}+3 a \,b^{3} d \,p^{2} x^{9}+7 b^{4} c \,p^{2} x^{9}+2 a \,b^{3} d p \,x^{9}+14 b^{4} c p \,x^{9}+a \,b^{3} c \,p^{3} x^{6}+8 b^{4} c \,x^{9}-3 a^{2} b^{2} d \,p^{2} x^{6}+5 a \,b^{3} c \,p^{2} x^{6}-3 a^{2} b^{2} d p \,x^{6}+4 a \,b^{3} c p \,x^{6}-2 a^{2} b^{2} c \,p^{2} x^{3}+6 a^{3} b d p \,x^{3}-8 a^{2} b^{2} c p \,x^{3}+2 a^{3} b c p -6 a^{4} d +8 a^{3} b c \right )}{3 b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:

int(x^8*(b*x^3+a)^p*(d*x^3+c),x)
 

Output:

((a + b*x**3)**p*( - 6*a**4*d + 2*a**3*b*c*p + 8*a**3*b*c + 6*a**3*b*d*p*x 
**3 - 2*a**2*b**2*c*p**2*x**3 - 8*a**2*b**2*c*p*x**3 - 3*a**2*b**2*d*p**2* 
x**6 - 3*a**2*b**2*d*p*x**6 + a*b**3*c*p**3*x**6 + 5*a*b**3*c*p**2*x**6 + 
4*a*b**3*c*p*x**6 + a*b**3*d*p**3*x**9 + 3*a*b**3*d*p**2*x**9 + 2*a*b**3*d 
*p*x**9 + b**4*c*p**3*x**9 + 7*b**4*c*p**2*x**9 + 14*b**4*c*p*x**9 + 8*b** 
4*c*x**9 + b**4*d*p**3*x**12 + 6*b**4*d*p**2*x**12 + 11*b**4*d*p*x**12 + 6 
*b**4*d*x**12))/(3*b**4*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))