\(\int (a+b x^n)^p (c+d x^n)^3 \, dx\) [119]

Optimal result

Integrand size = 19, antiderivative size = 395 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-3 a b c d (1+n) (1+n (3+p))+2 b^2 c^2 \left (1+2 n (3+p)+n^2 \left (8+6 p+p^2\right )\right )\right ) x \left (a+b x^n\right )^{1+p}}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}-\frac {d^2 (a d (1+2 n)-b (c+c n (5+p))) x^{1+n} \left (a+b x^n\right )^{1+p}}{b^2 (1+n (2+p)) (1+n (3+p))}+\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}-\frac {\left (a^3 d^3 \left (1+3 n+2 n^2\right )-3 a^2 b c d^2 (1+n) (1+n (3+p))+3 a b^2 c^2 d \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )-b^3 c^3 \left (1+3 n (2+p)+n^2 \left (11+12 p+3 p^2\right )+n^3 \left (6+11 p+6 p^2+p^3\right )\right )\right ) x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))} \] Output:

d*(a^2*d^2*(2*n^2+3*n+1)-3*a*b*c*d*(1+n)*(1+n*(3+p))+2*b^2*c^2*(1+2*n*(3+p 
)+n^2*(p^2+6*p+8)))*x*(a+b*x^n)^(p+1)/b^3/(n*p+n+1)/(1+n*(2+p))/(1+n*(3+p) 
)-d^2*(a*d*(1+2*n)-b*(c+c*n*(5+p)))*x^(1+n)*(a+b*x^n)^(p+1)/b^2/(1+n*(2+p) 
)/(1+n*(3+p))+d*x*(a+b*x^n)^(p+1)*(c+d*x^n)^2/b/(1+n*(3+p))-(a^3*d^3*(2*n^ 
2+3*n+1)-3*a^2*b*c*d^2*(1+n)*(1+n*(3+p))+3*a*b^2*c^2*d*(1+n*(5+2*p)+n^2*(p 
^2+5*p+6))-b^3*c^3*(1+3*n*(2+p)+n^2*(3*p^2+12*p+11)+n^3*(p^3+6*p^2+11*p+6) 
))*x*(a+b*x^n)^p*hypergeom([-p, 1/n],[1+1/n],-b*x^n/a)/b^3/(n*p+n+1)/(1+n* 
(2+p))/(1+n*(3+p))/((1+b*x^n/a)^p)
 

Mathematica [A] (verified)

Time = 5.32 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.43 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=x \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (\frac {3 c^2 d x^n \operatorname {Hypergeometric2F1}\left (1+\frac {1}{n},-p,2+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+n}+\frac {3 c d^2 x^{2 n} \operatorname {Hypergeometric2F1}\left (2+\frac {1}{n},-p,3+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+2 n}+\frac {d^3 x^{3 n} \operatorname {Hypergeometric2F1}\left (3+\frac {1}{n},-p,4+\frac {1}{n},-\frac {b x^n}{a}\right )}{1+3 n}+c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )\right ) \] Input:

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

Output:

(x*(a + b*x^n)^p*((3*c^2*d*x^n*Hypergeometric2F1[1 + n^(-1), -p, 2 + n^(-1 
), -((b*x^n)/a)])/(1 + n) + (3*c*d^2*x^(2*n)*Hypergeometric2F1[2 + n^(-1), 
 -p, 3 + n^(-1), -((b*x^n)/a)])/(1 + 2*n) + (d^3*x^(3*n)*Hypergeometric2F1 
[3 + n^(-1), -p, 4 + n^(-1), -((b*x^n)/a)])/(1 + 3*n) + c^3*Hypergeometric 
2F1[n^(-1), -p, 1 + n^(-1), -((b*x^n)/a)]))/(1 + (b*x^n)/a)^p
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 385, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {933, 25, 1025, 25, 913, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^n+a\right )^p \left (d x^n+c\right ) \left (d (a d (2 n+1)-b (n (p+5) c+c)) x^n+c (a d-b (n (p+3) c+c))\right )dx}{b (n (p+3)+1)}+\frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\int \left (b x^n+a\right )^p \left (d x^n+c\right ) \left (d (a d (2 n+1)-b (n (p+5) c+c)) x^n+c (a d-b (n (p+3) c+c))\right )dx}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\frac {\int -\left (b x^n+a\right )^p \left (d \left (b^2 \left (\left (p^2+6 p+11\right ) n^2+2 (p+3) n+1\right ) c^2-a b d \left ((p+7) n^2+(2 p+9) n+2\right ) c+a^2 d^2 \left (2 n^2+3 n+1\right )\right ) x^n+c \left (b^2 \left (\left (p^2+5 p+6\right ) n^2+(2 p+5) n+1\right ) c^2-a b d (n (2 p+7)+2) c+a^2 d^2 (2 n+1)\right )\right )dx}{b (n (p+2)+1)}+\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b (n (p+2)+1)}}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b (n (p+2)+1)}-\frac {\int \left (b x^n+a\right )^p \left (d \left (b^2 \left (\left (p^2+6 p+11\right ) n^2+2 (p+3) n+1\right ) c^2-a b d \left ((p+7) n^2+(2 p+9) n+2\right ) c+a^2 d^2 \left (2 n^2+3 n+1\right )\right ) x^n+c \left (b^2 \left (\left (p^2+5 p+6\right ) n^2+(2 p+5) n+1\right ) c^2-a b d (n (2 p+7)+2) c+a^2 d^2 (2 n+1)\right )\right )dx}{b (n (p+2)+1)}}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b (n (p+2)+1)}-\frac {\frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b (n p+n+1)}-\frac {\left (a^3 d^3 \left (2 n^2+3 n+1\right )-3 a^2 b c d^2 (n+1) (n (p+3)+1)+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \int \left (b x^n+a\right )^pdx}{b (n p+n+1)}}{b (n (p+2)+1)}}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b (n (p+2)+1)}-\frac {\frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b (n p+n+1)}-\frac {\left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (a^3 d^3 \left (2 n^2+3 n+1\right )-3 a^2 b c d^2 (n+1) (n (p+3)+1)+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \int \left (\frac {b x^n}{a}+1\right )^pdx}{b (n p+n+1)}}{b (n (p+2)+1)}}{b (n (p+3)+1)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)}-\frac {\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b (n (p+2)+1)}-\frac {\frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b (n p+n+1)}-\frac {x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (a^3 d^3 \left (2 n^2+3 n+1\right )-3 a^2 b c d^2 (n+1) (n (p+3)+1)+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b (n p+n+1)}}{b (n (p+2)+1)}}{b (n (p+3)+1)}\)

Input:

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

Output:

(d*x*(a + b*x^n)^(1 + p)*(c + d*x^n)^2)/(b*(1 + n*(3 + p))) - ((d*(a*d*(1 
+ 2*n) - b*(c + c*n*(5 + p)))*x*(a + b*x^n)^(1 + p)*(c + d*x^n))/(b*(1 + n 
*(2 + p))) - ((d*(a^2*d^2*(1 + 3*n + 2*n^2) - a*b*c*d*(2 + n^2*(7 + p) + n 
*(9 + 2*p)) + b^2*c^2*(1 + 2*n*(3 + p) + n^2*(11 + 6*p + p^2)))*x*(a + b*x 
^n)^(1 + p))/(b*(1 + n + n*p)) - ((a^3*d^3*(1 + 3*n + 2*n^2) - 3*a^2*b*c*d 
^2*(1 + n)*(1 + n*(3 + p)) + 3*a*b^2*c^2*d*(1 + n*(5 + 2*p) + n^2*(6 + 5*p 
 + p^2)) - b^3*c^3*(1 + 3*n*(2 + p) + n^2*(11 + 12*p + 3*p^2) + n^3*(6 + 1 
1*p + 6*p^2 + p^3)))*x*(a + b*x^n)^p*Hypergeometric2F1[n^(-1), -p, 1 + n^( 
-1), -((b*x^n)/a)])/(b*(1 + n + n*p)*(1 + (b*x^n)/a)^p))/(b*(1 + n*(2 + p) 
)))/(b*(1 + n*(3 + p)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 778

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])
 

rule 779

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])
 

rule 913

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

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

rule 1025

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

Maple [F]

\[\int \left (a +b \,x^{n}\right )^{p} \left (c +d \,x^{n}\right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3)*(b*x^n + a)^p 
, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 91.68 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.62 \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\frac {a^{\frac {1}{n}} a^{p - \frac {1}{n}} c^{3} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{n}, - p \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 a^{1 + \frac {1}{n}} a^{p - 1 - \frac {1}{n}} c^{2} d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} + \frac {3 a^{2 + \frac {1}{n}} a^{p - 2 - \frac {1}{n}} c d^{2} x^{2 n + 1} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {a^{3 + \frac {1}{n}} a^{p - 3 - \frac {1}{n}} d^{3} x^{3 n + 1} \Gamma \left (3 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 3 + \frac {1}{n} \\ 4 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (4 + \frac {1}{n}\right )} \] Input:

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

Output:

a**(1/n)*a**(p - 1/n)*c**3*x*gamma(1/n)*hyper((1/n, -p), (1 + 1/n,), b*x** 
n*exp_polar(I*pi)/a)/(n*gamma(1 + 1/n)) + 3*a**(1 + 1/n)*a**(p - 1 - 1/n)* 
c**2*d*x**(n + 1)*gamma(1 + 1/n)*hyper((-p, 1 + 1/n), (2 + 1/n,), b*x**n*e 
xp_polar(I*pi)/a)/(n*gamma(2 + 1/n)) + 3*a**(2 + 1/n)*a**(p - 2 - 1/n)*c*d 
**2*x**(2*n + 1)*gamma(2 + 1/n)*hyper((-p, 2 + 1/n), (3 + 1/n,), b*x**n*ex 
p_polar(I*pi)/a)/(n*gamma(3 + 1/n)) + a**(3 + 1/n)*a**(p - 3 - 1/n)*d**3*x 
**(3*n + 1)*gamma(3 + 1/n)*hyper((-p, 3 + 1/n), (4 + 1/n,), b*x**n*exp_pol 
ar(I*pi)/a)/(n*gamma(4 + 1/n))
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[2,0,6,4,2,4,4,3,0]%%%}+%%%{4,[2,0,6,4,2,3,4,3,0]%%%}+%% 
%{6,[2,0,
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx=\text {too large to display} \] Input:

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

Output:

(x**(3*n)*(x**n*b + a)**p*b**3*d**3*n**3*p**3*x + 3*x**(3*n)*(x**n*b + a)* 
*p*b**3*d**3*n**3*p**2*x + 2*x**(3*n)*(x**n*b + a)**p*b**3*d**3*n**3*p*x + 
 3*x**(3*n)*(x**n*b + a)**p*b**3*d**3*n**2*p**2*x + 6*x**(3*n)*(x**n*b + a 
)**p*b**3*d**3*n**2*p*x + 2*x**(3*n)*(x**n*b + a)**p*b**3*d**3*n**2*x + 3* 
x**(3*n)*(x**n*b + a)**p*b**3*d**3*n*p*x + 3*x**(3*n)*(x**n*b + a)**p*b**3 
*d**3*n*x + x**(3*n)*(x**n*b + a)**p*b**3*d**3*x + x**(2*n)*(x**n*b + a)** 
p*a*b**2*d**3*n**3*p**3*x + x**(2*n)*(x**n*b + a)**p*a*b**2*d**3*n**3*p**2 
*x + 2*x**(2*n)*(x**n*b + a)**p*a*b**2*d**3*n**2*p**2*x + x**(2*n)*(x**n*b 
 + a)**p*a*b**2*d**3*n**2*p*x + x**(2*n)*(x**n*b + a)**p*a*b**2*d**3*n*p*x 
 + 3*x**(2*n)*(x**n*b + a)**p*b**3*c*d**2*n**3*p**3*x + 12*x**(2*n)*(x**n* 
b + a)**p*b**3*c*d**2*n**3*p**2*x + 9*x**(2*n)*(x**n*b + a)**p*b**3*c*d**2 
*n**3*p*x + 9*x**(2*n)*(x**n*b + a)**p*b**3*c*d**2*n**2*p**2*x + 24*x**(2* 
n)*(x**n*b + a)**p*b**3*c*d**2*n**2*p*x + 9*x**(2*n)*(x**n*b + a)**p*b**3* 
c*d**2*n**2*x + 9*x**(2*n)*(x**n*b + a)**p*b**3*c*d**2*n*p*x + 12*x**(2*n) 
*(x**n*b + a)**p*b**3*c*d**2*n*x + 3*x**(2*n)*(x**n*b + a)**p*b**3*c*d**2* 
x - 2*x**n*(x**n*b + a)**p*a**2*b*d**3*n**3*p**2*x - x**n*(x**n*b + a)**p* 
a**2*b*d**3*n**2*p**2*x - 2*x**n*(x**n*b + a)**p*a**2*b*d**3*n**2*p*x - x* 
*n*(x**n*b + a)**p*a**2*b*d**3*n*p*x + 3*x**n*(x**n*b + a)**p*a*b**2*c*d** 
2*n**3*p**3*x + 9*x**n*(x**n*b + a)**p*a*b**2*c*d**2*n**3*p**2*x + 6*x**n* 
(x**n*b + a)**p*a*b**2*c*d**2*n**2*p**2*x + 9*x**n*(x**n*b + a)**p*a*b*...