\(\int (f x)^m (d+e x^n)^3 (a+c x^{2 n})^p \, dx\) [42]

Optimal result

Integrand size = 26, antiderivative size = 367 \[ \int (f x)^m \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=\frac {d^3 (f x)^{1+m} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2 n},-p,1+\frac {1+m}{2 n},-\frac {c x^{2 n}}{a}\right )}{f (1+m)}+\frac {3 d^2 e x^n (f x)^{1+m} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m+n}{2 n},-p,\frac {1+m+3 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{f (1+m+n)}+\frac {3 d e^2 x^{2 n} (f x)^{1+m} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m+2 n}{2 n},-p,\frac {1+m+4 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{f (1+m+2 n)}+\frac {e^3 x^{3 n} (f x)^{1+m} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m+3 n}{2 n},-p,\frac {1+m+5 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{f (1+m+3 n)} \] Output:

d^3*(f*x)^(1+m)*(a+c*x^(2*n))^p*hypergeom([-p, 1/2*(1+m)/n],[1+1/2*(1+m)/n 
],-c*x^(2*n)/a)/f/(1+m)/((1+c*x^(2*n)/a)^p)+3*d^2*e*x^n*(f*x)^(1+m)*(a+c*x 
^(2*n))^p*hypergeom([-p, 1/2*(1+m+n)/n],[1/2*(1+m+3*n)/n],-c*x^(2*n)/a)/f/ 
(1+m+n)/((1+c*x^(2*n)/a)^p)+3*d*e^2*x^(2*n)*(f*x)^(1+m)*(a+c*x^(2*n))^p*hy 
pergeom([-p, 1/2*(1+m+2*n)/n],[1/2*(1+m+4*n)/n],-c*x^(2*n)/a)/f/(1+m+2*n)/ 
((1+c*x^(2*n)/a)^p)+e^3*x^(3*n)*(f*x)^(1+m)*(a+c*x^(2*n))^p*hypergeom([-p, 
 1/2*(1+m+3*n)/n],[1/2*(1+m+5*n)/n],-c*x^(2*n)/a)/f/(1+m+3*n)/((1+c*x^(2*n 
)/a)^p)
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.68 \[ \int (f x)^m \left (d+e x^n\right )^3 \left (a+c x^{2 n}\right )^p \, dx=x (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \left (\frac {d^3 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2 n},-p,1+\frac {1+m}{2 n},-\frac {c x^{2 n}}{a}\right )}{1+m}+e x^n \left (\frac {3 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m+n}{2 n},-p,\frac {1+m+3 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{1+m+n}+e x^n \left (\frac {3 d \operatorname {Hypergeometric2F1}\left (\frac {1+m+2 n}{2 n},-p,\frac {1+m+4 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{1+m+2 n}+\frac {e x^n \operatorname {Hypergeometric2F1}\left (\frac {1+m+3 n}{2 n},-p,\frac {1+m+5 n}{2 n},-\frac {c x^{2 n}}{a}\right )}{1+m+3 n}\right )\right )\right ) \] Input:

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

Output:

(x*(f*x)^m*(a + c*x^(2*n))^p*((d^3*Hypergeometric2F1[(1 + m)/(2*n), -p, 1 
+ (1 + m)/(2*n), -((c*x^(2*n))/a)])/(1 + m) + e*x^n*((3*d^2*Hypergeometric 
2F1[(1 + m + n)/(2*n), -p, (1 + m + 3*n)/(2*n), -((c*x^(2*n))/a)])/(1 + m 
+ n) + e*x^n*((3*d*Hypergeometric2F1[(1 + m + 2*n)/(2*n), -p, (1 + m + 4*n 
)/(2*n), -((c*x^(2*n))/a)])/(1 + m + 2*n) + (e*x^n*Hypergeometric2F1[(1 + 
m + 3*n)/(2*n), -p, (1 + m + 5*n)/(2*n), -((c*x^(2*n))/a)])/(1 + m + 3*n)) 
)))/(1 + (c*x^(2*n))/a)^p
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1885, 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.

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

\(\Big \downarrow \) 1885

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1885

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

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((f*x)^m*(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*(f*x)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((f*x)^m*(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*(f*x)^m, x)
 

Giac [F(-2)]

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

integrate((f*x)^m*(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,[1,0,6,4,0,3,5,4,1,2]%%%}+%%%{480,[1,0,6,4,0,3,4,4,1,2] 
%%%}+%%%{
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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