\(\int (f x)^m (d+e x^n)^q (a+c x^{2 n})^3 \, dx\) [58]

Optimal result

Integrand size = 26, antiderivative size = 374 \[ \int (f x)^m \left (d+e x^n\right )^q \left (a+c x^{2 n}\right )^3 \, dx=\frac {c^3 x^{5 n} (f x)^{1+m} \left (d+e x^n\right )^{1+q}}{e f (1+m+n (6+q))}+\frac {a^3 (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{n},-q,\frac {1+m+n}{n},-\frac {e x^n}{d}\right )}{f (1+m)}+\frac {3 a^2 c x^{2 n} (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1+m+2 n}{n},-q,\frac {1+m+3 n}{n},-\frac {e x^n}{d}\right )}{f (1+m+2 n)}+\frac {3 a c^2 x^{4 n} (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1+m+4 n}{n},-q,\frac {1+m+5 n}{n},-\frac {e x^n}{d}\right )}{f (1+m+4 n)}-\frac {c^3 d x^{5 n} (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1+m+5 n}{n},-q,\frac {1+m+6 n}{n},-\frac {e x^n}{d}\right )}{e f (1+m+n (6+q))} \] Output:

c^3*x^(5*n)*(f*x)^(1+m)*(d+e*x^n)^(1+q)/e/f/(1+m+n*(6+q))+a^3*(f*x)^(1+m)* 
(d+e*x^n)^q*hypergeom([-q, (1+m)/n],[(1+m+n)/n],-e*x^n/d)/f/(1+m)/((1+e*x^ 
n/d)^q)+3*a^2*c*x^(2*n)*(f*x)^(1+m)*(d+e*x^n)^q*hypergeom([-q, (1+m+2*n)/n 
],[(1+m+3*n)/n],-e*x^n/d)/f/(1+m+2*n)/((1+e*x^n/d)^q)+3*a*c^2*x^(4*n)*(f*x 
)^(1+m)*(d+e*x^n)^q*hypergeom([-q, (1+m+4*n)/n],[(1+m+5*n)/n],-e*x^n/d)/f/ 
(1+m+4*n)/((1+e*x^n/d)^q)-c^3*d*x^(5*n)*(f*x)^(1+m)*(d+e*x^n)^q*hypergeom( 
[-q, (1+m+5*n)/n],[(1+m+6*n)/n],-e*x^n/d)/e/f/(1+m+n*(6+q))/((1+e*x^n/d)^q 
)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.59 \[ \int (f x)^m \left (d+e x^n\right )^q \left (a+c x^{2 n}\right )^3 \, dx=x (f x)^m \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \left (\frac {a^3 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{n},-q,\frac {1+m+n}{n},-\frac {e x^n}{d}\right )}{1+m}+c x^{2 n} \left (\frac {3 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1+m+2 n}{n},-q,\frac {1+m+3 n}{n},-\frac {e x^n}{d}\right )}{1+m+2 n}+c x^{2 n} \left (\frac {3 a \operatorname {Hypergeometric2F1}\left (\frac {1+m+4 n}{n},-q,\frac {1+m+5 n}{n},-\frac {e x^n}{d}\right )}{1+m+4 n}+\frac {c x^{2 n} \operatorname {Hypergeometric2F1}\left (\frac {1+m+6 n}{n},-q,\frac {1+m+7 n}{n},-\frac {e x^n}{d}\right )}{1+m+6 n}\right )\right )\right ) \] Input:

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

Output:

(x*(f*x)^m*(d + e*x^n)^q*((a^3*Hypergeometric2F1[(1 + m)/n, -q, (1 + m + n 
)/n, -((e*x^n)/d)])/(1 + m) + c*x^(2*n)*((3*a^2*Hypergeometric2F1[(1 + m + 
 2*n)/n, -q, (1 + m + 3*n)/n, -((e*x^n)/d)])/(1 + m + 2*n) + c*x^(2*n)*((3 
*a*Hypergeometric2F1[(1 + m + 4*n)/n, -q, (1 + m + 5*n)/n, -((e*x^n)/d)])/ 
(1 + m + 4*n) + (c*x^(2*n)*Hypergeometric2F1[(1 + m + 6*n)/n, -q, (1 + m + 
 7*n)/n, -((e*x^n)/d)])/(1 + m + 6*n)))))/(1 + (e*x^n)/d)^q
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.84, 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.

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

\(\Big \downarrow \) 1885

\(\displaystyle \int \left (a^3 (f x)^m \left (d+e x^n\right )^q+3 a^2 c x^{2 n} (f x)^m \left (d+e x^n\right )^q+3 a c^2 x^{4 n} (f x)^m \left (d+e x^n\right )^q+c^3 x^{6 n} (f x)^m \left (d+e x^n\right )^q\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 (f x)^{m+1} \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{n},-q,\frac {m+n+1}{n},-\frac {e x^n}{d}\right )}{f (m+1)}+\frac {3 a^2 c x^{2 n+1} (f x)^m \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {m+2 n+1}{n},-q,\frac {m+3 n+1}{n},-\frac {e x^n}{d}\right )}{m+2 n+1}+\frac {3 a c^2 x^{4 n+1} (f x)^m \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {m+4 n+1}{n},-q,\frac {m+5 n+1}{n},-\frac {e x^n}{d}\right )}{m+4 n+1}+\frac {c^3 x^{6 n+1} (f x)^m \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {m+6 n+1}{n},-q,\frac {m+7 n+1}{n},-\frac {e x^n}{d}\right )}{m+6 n+1}\)

Input:

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

Output:

(a^3*(f*x)^(1 + m)*(d + e*x^n)^q*Hypergeometric2F1[(1 + m)/n, -q, (1 + m + 
 n)/n, -((e*x^n)/d)])/(f*(1 + m)*(1 + (e*x^n)/d)^q) + (3*a^2*c*x^(1 + 2*n) 
*(f*x)^m*(d + e*x^n)^q*Hypergeometric2F1[(1 + m + 2*n)/n, -q, (1 + m + 3*n 
)/n, -((e*x^n)/d)])/((1 + m + 2*n)*(1 + (e*x^n)/d)^q) + (3*a*c^2*x^(1 + 4* 
n)*(f*x)^m*(d + e*x^n)^q*Hypergeometric2F1[(1 + m + 4*n)/n, -q, (1 + m + 5 
*n)/n, -((e*x^n)/d)])/((1 + m + 4*n)*(1 + (e*x^n)/d)^q) + (c^3*x^(1 + 6*n) 
*(f*x)^m*(d + e*x^n)^q*Hypergeometric2F1[(1 + m + 6*n)/n, -q, (1 + m + 7*n 
)/n, -((e*x^n)/d)])/((1 + m + 6*n)*(1 + (e*x^n)/d)^q)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((c^3*x^(6*n) + 3*a*c^2*x^(4*n) + 3*a^2*c*x^(2*n) + a^3)*(e*x^n + 
d)^q*(f*x)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((f*x)^m*(d+e*x^n)^q*(a+c*x^(2*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,[5,0,12,7,0,5,7,7,3,0]%%%}+%%%{-7,[5,0,12,7,0,5,6,7,3,0 
]%%%}+%%%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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