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

Optimal. Leaf size=402 \[ \frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}-\frac {x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (a^3 d^3 \left (2 n^2+3 n+1\right )-3 a^2 b c d^2 (n+1) (n (p+3)+1)+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b x^n}{a}\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}-\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b c (n (p+5)+1))}{b^2 (n (p+2)+1) (n (p+3)+1)}+\frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n p+3 n+1)} \]

[Out]

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

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Rubi [A]  time = 0.58, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {416, 528, 388, 246, 245} \[ -\frac {x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (-3 a^2 b c d^2 (n+1) (n (p+3)+1)+a^3 d^3 \left (2 n^2+3 n+1\right )+3 a b^2 c^2 d \left (n^2 \left (p^2+5 p+6\right )+n (2 p+5)+1\right )-b^3 c^3 \left (n^3 \left (p^3+6 p^2+11 p+6\right )+n^2 \left (3 p^2+12 p+11\right )+3 n (p+2)+1\right )\right ) \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b x^n}{a}\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}+\frac {d x \left (a+b x^n\right )^{p+1} \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (n^2 (p+7)+n (2 p+9)+2\right )+b^2 c^2 \left (n^2 \left (p^2+6 p+11\right )+2 n (p+3)+1\right )\right )}{b^3 (n p+n+1) (n (p+2)+1) (n (p+3)+1)}-\frac {d x \left (c+d x^n\right ) \left (a+b x^n\right )^{p+1} (a d (2 n+1)-b (c n (p+5)+c))}{b^2 (n (p+2)+1) (n (p+3)+1)}+\frac {d x \left (c+d x^n\right )^2 \left (a+b x^n\right )^{p+1}}{b (n (p+3)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^p*(c + d*x^n)^3,x]

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 388

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

Rule 416

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] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[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] /; F
reeQ[{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] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 528

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] + Dist[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]

Rubi steps

\begin {align*} \int \left (a+b x^n\right )^p \left (c+d x^n\right )^3 \, dx &=\frac {d x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )^2}{b (1+n (3+p))}+\frac {\int \left (a+b x^n\right )^p \left (c+d x^n\right ) \left (-c (a d-b (c+c n (3+p)))-d (a d (1+2 n)-b (c+c n (5+p))) x^n\right ) \, dx}{b (1+n (3+p))}\\ &=-\frac {d (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{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 {\int \left (a+b x^n\right )^p \left (c \left (a^2 d^2 (1+2 n)-a b c d (2+n (7+2 p))+b^2 c^2 \left (1+n (5+2 p)+n^2 \left (6+5 p+p^2\right )\right )\right )+d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+6 p+p^2\right )\right )\right ) x^n\right ) \, dx}{b^2 (1+n (2+p)) (1+n (3+p))}\\ &=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+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 (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{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 ) \int \left (a+b x^n\right )^p \, dx}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}\\ &=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+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 (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{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 (\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 ) \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^n}{a}\right )^p \, dx}{b^3 (1+n+n p) (1+n (2+p)) (1+n (3+p))}\\ &=\frac {d \left (a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+n^2 (7+p)+n (9+2 p)\right )+b^2 c^2 \left (1+2 n (3+p)+n^2 \left (11+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 (a d (1+2 n)-b (c+c n (5+p))) x \left (a+b x^n\right )^{1+p} \left (c+d x^n\right )}{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} \, _2F_1\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))}\\ \end {align*}

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Mathematica [A]  time = 5.32, size = 168, normalized size = 0.42 \[ x \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \left (c^3 \, _2F_1\left (\frac {1}{n},-p;1+\frac {1}{n};-\frac {b x^n}{a}\right )+\frac {3 c^2 d x^n \, _2F_1\left (1+\frac {1}{n},-p;2+\frac {1}{n};-\frac {b x^n}{a}\right )}{n+1}+\frac {3 c d^2 x^{2 n} \, _2F_1\left (2+\frac {1}{n},-p;3+\frac {1}{n};-\frac {b x^n}{a}\right )}{2 n+1}+\frac {d^3 x^{3 n} \, _2F_1\left (3+\frac {1}{n},-p;4+\frac {1}{n};-\frac {b x^n}{a}\right )}{3 n+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)*Hypergeometric2
F1[3 + n^(-1), -p, 4 + n^(-1), -((b*x^n)/a)])/(1 + 3*n) + c^3*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*x
^n)/a)]))/(1 + (b*x^n)/a)^p

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{3} x^{3 \, n} + 3 \, c d^{2} x^{2 \, n} + 3 \, c^{2} d x^{n} + c^{3}\right )} {\left (b x^{n} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{1,[2,0,6,4,2,4,4,3,0]%%%}+%%%{4,[2,0,6,4,2,3,4,3,0]%%%}+%%%{6,[
2,0,6,4,2,2,4,3,0]%%%}+%%%{4,[2,0,6,4,2,1,4,3,0]%%%}+%%%{1,[2,0,6,4,2,0,4,3,0]%%%}+%%%{3,[1,0,6,4,2,4,4,2,1]%%
%}+%%%{12,[1,0,6,4,2,3,4,2,1]%%%}+%%%{18,[1,0,6,4,2,2,4,2,1]%%%}+%%%{12,[1,0,6,4,2,1,4,2,1]%%%}+%%%{3,[1,0,6,4
,2,0,4,2,1]%%%}+%%%{-1,[1,0,6,4,1,4,5,3,0]%%%}+%%%{-4,[1,0,6,4,1,3,5,3,0]%%%}+%%%{-6,[1,0,6,4,1,2,5,3,0]%%%}+%
%%{-4,[1,0,6,4,1,1,5,3,0]%%%}+%%%{-1,[1,0,6,4,1,0,5,3,0]%%%}+%%%{1,[0,0,6,4,3,4,3,0,3]%%%}+%%%{4,[0,0,6,4,3,3,
3,0,3]%%%}+%%%{6,[0,0,6,4,3,2,3,0,3]%%%}+%%%{4,[0,0,6,4,3,1,3,0,3]%%%}+%%%{1,[0,0,6,4,3,0,3,0,3]%%%}+%%%{1,[0,
0,6,3,3,3,3,0,3]%%%}+%%%{3,[0,0,6,3,3,2,3,0,3]%%%}+%%%{3,[0,0,6,3,3,1,3,0,3]%%%}+%%%{1,[0,0,6,3,3,0,3,0,3]%%%}
+%%%{-3,[0,0,6,3,2,3,4,1,2]%%%}+%%%{-9,[0,0,6,3,2,2,4,1,2]%%%}+%%%{-9,[0,0,6,3,2,1,4,1,2]%%%}+%%%{-3,[0,0,6,3,
2,0,4,1,2]%%%}+%%%{3,[0,0,6,3,1,3,5,2,1]%%%}+%%%{9,[0,0,6,3,1,2,5,2,1]%%%}+%%%{9,[0,0,6,3,1,1,5,2,1]%%%}+%%%{3
,[0,0,6,3,1,0,5,2,1]%%%}+%%%{-1,[0,0,6,3,0,3,6,3,0]%%%}+%%%{-3,[0,0,6,3,0,2,6,3,0]%%%}+%%%{-3,[0,0,6,3,0,1,6,3
,0]%%%}+%%%{-1,[0,0,6,3,0,0,6,3,0]%%%} / %%%{1,[0,0,7,4,3,4,4,0,0]%%%}+%%%{4,[0,0,7,4,3,3,4,0,0]%%%}+%%%{6,[0,
0,7,4,3,2,4,0,0]%%%}+%%%{4,[0,0,7,4,3,1,4,0,0]%%%}+%%%{1,[0,0,7,4,3,0,4,0,0]%%%} Error: Bad Argument Value

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \left (d \,x^{n}+c \right )^{3} \left (b \,x^{n}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{n} + c\right )}^{3} {\left (b x^{n} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,x^n\right )}^p\,{\left (c+d\,x^n\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^p*(c + d*x^n)^3,x)

[Out]

int((a + b*x^n)^p*(c + d*x^n)^3, x)

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sympy [C]  time = 93.23, size = 199, normalized size = 0.50 \[ \frac {a^{p} c^{3} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{n}, - p \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {3 a^{p} c^{2} d x x^{n} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} + \frac {3 a^{p} c d^{2} x x^{2 n} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {a^{p} d^{3} x x^{3 n} \Gamma \left (3 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 3 + \frac {1}{n} \\ 4 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (4 + \frac {1}{n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*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**p*c**2
*d*x*x**n*gamma(1 + 1/n)*hyper((-p, 1 + 1/n), (2 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(2 + 1/n)) + 3*a**
p*c*d**2*x*x**(2*n)*gamma(2 + 1/n)*hyper((-p, 2 + 1/n), (3 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(3 + 1/n
)) + a**p*d**3*x*x**(3*n)*gamma(3 + 1/n)*hyper((-p, 3 + 1/n), (4 + 1/n,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(4
+ 1/n))

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