\(\int (d+e x^n)^3 (a+c x^{2 n})^p \, dx\) [47]

Optimal result

Integrand size = 21, antiderivative size = 263 \[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\frac {3 d e^2 x \left (a+c x^{2 n}\right )^{1+p}}{c (1+2 n (1+p))}+\frac {e^3 x^{1+n} \left (a+c x^{2 n}\right )^{1+p}}{c (1+n (3+2 p))}-\frac {d \left (3 a e^2-c d^2 (1+2 n (1+p))\right ) x \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2 n},-p,\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{c (1+2 n (1+p))}+e \left (\frac {3 d^2}{1+n}-\frac {a e^2}{c+c n (3+2 p)}\right ) x^{1+n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2 n},-p,\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right ) \] Output:

3*d*e^2*x*(a+c*x^(2*n))^(p+1)/c/(1+2*n*(p+1))+e^3*x^(1+n)*(a+c*x^(2*n))^(p 
+1)/c/(1+n*(3+2*p))-d*(3*a*e^2-c*d^2*(1+2*n*(p+1)))*x*(a+c*x^(2*n))^p*hype 
rgeom([-p, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/c/(1+2*n*(p+1))/((1+c*x^(2*n)/a) 
^p)+e*(3*d^2/(1+n)-a*e^2/(c+c*n*(3+2*p)))*x^(1+n)*(a+c*x^(2*n))^p*hypergeo 
m([-p, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/((1+c*x^(2*n)/a)^p)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.81 \[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=x \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \left (\frac {3 d e^2 x^{2 n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p,\frac {1}{2} \left (4+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+2 n}+\frac {e^3 x^{3 n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p,\frac {1}{2} \left (5+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+3 n}+d^2 \left (d \operatorname {Hypergeometric2F1}\left (\frac {1}{2 n},-p,\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\frac {3 e x^n \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2 n},-p,\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+n}\right )\right ) \] Input:

Integrate[(d + e*x^n)^3*(a + c*x^(2*n))^p,x]
 

Output:

(x*(a + c*x^(2*n))^p*((3*d*e^2*x^(2*n)*Hypergeometric2F1[(2 + n^(-1))/2, - 
p, (4 + n^(-1))/2, -((c*x^(2*n))/a)])/(1 + 2*n) + (e^3*x^(3*n)*Hypergeomet 
ric2F1[(3 + n^(-1))/2, -p, (5 + n^(-1))/2, -((c*x^(2*n))/a)])/(1 + 3*n) + 
d^2*(d*Hypergeometric2F1[1/(2*n), -p, (2 + n^(-1))/2, -((c*x^(2*n))/a)] + 
(3*e*x^n*Hypergeometric2F1[(1 + n)/(2*n), -p, (3 + n^(-1))/2, -((c*x^(2*n) 
)/a)])/(1 + n))))/(1 + (c*x^(2*n))/a)^p
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1767, 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.

\(\displaystyle \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx\)

\(\Big \downarrow \) 1767

\(\displaystyle \int \left (d^3 \left (a+c x^{2 n}\right )^p+3 d^2 e x^n \left (a+c x^{2 n}\right )^p+3 d e^2 x^{2 n} \left (a+c x^{2 n}\right )^p+e^3 x^{3 n} \left (a+c x^{2 n}\right )^p\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(d + e*x^n)^3*(a + c*x^(2*n))^p,x]
 

Output:

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

Defintions of rubi rules used

rule 1767

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] 
 :> Int[ExpandIntegrand[(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a 
, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && ((Integ 
ersQ[p, q] &&  !IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] &&  !IntegerQ[n]) 
)
 

rule 2009

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

Maple [F]

\[\int \left (d +e \,x^{n}\right )^{3} \left (a +c \,x^{2 n}\right )^{p}d x\]

Input:

int((d+e*x^n)^3*(a+c*x^(2*n))^p,x)
 

Output:

int((d+e*x^n)^3*(a+c*x^(2*n))^p,x)
 

Fricas [F]

\[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{3} {\left (c x^{2 \, n} + a\right )}^{p} \,d x } \] Input:

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\text {Timed out} \] Input:

integrate((d+e*x**n)**3*(a+c*x**(2*n))**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\int { {\left (e x^{n} + d\right )}^{3} {\left (c x^{2 \, n} + a\right )}^{p} \,d x } \] Input:

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,x, algorithm="maxima")
 

Output:

integrate((e*x^n + d)^3*(c*x^(2*n) + a)^p, x)
 

Giac [F(-2)]

Exception generated. \[ \int \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d+e*x^n)^3*(a+c*x^(2*n))^p,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%%%{96,[1,0,6,4,3,5,4,1,2]%%%}+%%%{480,[1,0,6,4,3,4,4,1,2]%%%} 
+%%%{960,
 

Mupad [F(-1)]

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

int((a + c*x^(2*n))^p*(d + e*x^n)^3,x)
 

Output:

int((a + c*x^(2*n))^p*(d + e*x^n)^3, x)
 

Reduce [F]

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

int((d+e*x^n)^3*(a+c*x^(2*n))^p,x)
 

Output:

(8*x**(3*n)*(x**(2*n)*c + a)**p*c*e**3*n**3*p**3*x + 12*x**(3*n)*(x**(2*n) 
*c + a)**p*c*e**3*n**3*p**2*x + 4*x**(3*n)*(x**(2*n)*c + a)**p*c*e**3*n**3 
*p*x + 12*x**(3*n)*(x**(2*n)*c + a)**p*c*e**3*n**2*p**2*x + 12*x**(3*n)*(x 
**(2*n)*c + a)**p*c*e**3*n**2*p*x + 2*x**(3*n)*(x**(2*n)*c + a)**p*c*e**3* 
n**2*x + 6*x**(3*n)*(x**(2*n)*c + a)**p*c*e**3*n*p*x + 3*x**(3*n)*(x**(2*n 
)*c + a)**p*c*e**3*n*x + x**(3*n)*(x**(2*n)*c + a)**p*c*e**3*x + 24*x**(2* 
n)*(x**(2*n)*c + a)**p*c*d*e**2*n**3*p**3*x + 48*x**(2*n)*(x**(2*n)*c + a) 
**p*c*d*e**2*n**3*p**2*x + 18*x**(2*n)*(x**(2*n)*c + a)**p*c*d*e**2*n**3*p 
*x + 36*x**(2*n)*(x**(2*n)*c + a)**p*c*d*e**2*n**2*p**2*x + 48*x**(2*n)*(x 
**(2*n)*c + a)**p*c*d*e**2*n**2*p*x + 9*x**(2*n)*(x**(2*n)*c + a)**p*c*d*e 
**2*n**2*x + 18*x**(2*n)*(x**(2*n)*c + a)**p*c*d*e**2*n*p*x + 12*x**(2*n)* 
(x**(2*n)*c + a)**p*c*d*e**2*n*x + 3*x**(2*n)*(x**(2*n)*c + a)**p*c*d*e**2 
*x + 8*x**n*(x**(2*n)*c + a)**p*a*e**3*n**3*p**3*x + 8*x**n*(x**(2*n)*c + 
a)**p*a*e**3*n**3*p**2*x + 8*x**n*(x**(2*n)*c + a)**p*a*e**3*n**2*p**2*x + 
 4*x**n*(x**(2*n)*c + a)**p*a*e**3*n**2*p*x + 2*x**n*(x**(2*n)*c + a)**p*a 
*e**3*n*p*x + 24*x**n*(x**(2*n)*c + a)**p*c*d**2*e*n**3*p**3*x + 60*x**n*( 
x**(2*n)*c + a)**p*c*d**2*e*n**3*p**2*x + 36*x**n*(x**(2*n)*c + a)**p*c*d* 
*2*e*n**3*p*x + 36*x**n*(x**(2*n)*c + a)**p*c*d**2*e*n**2*p**2*x + 60*x**n 
*(x**(2*n)*c + a)**p*c*d**2*e*n**2*p*x + 18*x**n*(x**(2*n)*c + a)**p*c*d** 
2*e*n**2*x + 18*x**n*(x**(2*n)*c + a)**p*c*d**2*e*n*p*x + 15*x**n*(x**(...