\(\int \frac {(f x)^m (d+e x^n)^q}{(a+c x^{2 n})^2} \, dx\) [62]

Optimal result

Integrand size = 26, antiderivative size = 467 \[ \int \frac {(f x)^m \left (d+e x^n\right )^q}{\left (a+c x^{2 n}\right )^2} \, dx=\frac {c (f x)^{1+m} \left (d-e x^n\right ) \left (d+e x^n\right )^{1+q}}{2 a \left (c d^2+a e^2\right ) f n \left (a+c x^{2 n}\right )}-\frac {\left (c d^2 (1+m-2 n)+a e^2 (1+m-n (2-q))-\sqrt {-a} \sqrt {c} d e n q\right ) (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1+m}{n},-q,1,\frac {1+m+n}{n},-\frac {e x^n}{d},-\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{4 a^2 \left (c d^2+a e^2\right ) f (1+m) n}-\frac {\left (c d^2 (1+m-2 n)+a e^2 (1+m-n (2-q))+\sqrt {-a} \sqrt {c} d e n q\right ) (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1+m}{n},-q,1,\frac {1+m+n}{n},-\frac {e x^n}{d},\frac {\sqrt {c} x^n}{\sqrt {-a}}\right )}{4 a^2 \left (c d^2+a e^2\right ) f (1+m) n}+\frac {e^2 (1+m+n q) (f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{n},-q,\frac {1+m+n}{n},-\frac {e x^n}{d}\right )}{2 a \left (c d^2+a e^2\right ) f (1+m) n} \] Output:

1/2*c*(f*x)^(1+m)*(d-e*x^n)*(d+e*x^n)^(1+q)/a/(a*e^2+c*d^2)/f/n/(a+c*x^(2* 
n))-1/4*(c*d^2*(1+m-2*n)+a*e^2*(1+m-n*(2-q))-(-a)^(1/2)*c^(1/2)*d*e*n*q)*( 
f*x)^(1+m)*(d+e*x^n)^q*AppellF1((1+m)/n,1,-q,(1+m+n)/n,-c^(1/2)*x^n/(-a)^( 
1/2),-e*x^n/d)/a^2/(a*e^2+c*d^2)/f/(1+m)/n/((1+e*x^n/d)^q)-1/4*(c*d^2*(1+m 
-2*n)+a*e^2*(1+m-n*(2-q))+(-a)^(1/2)*c^(1/2)*d*e*n*q)*(f*x)^(1+m)*(d+e*x^n 
)^q*AppellF1((1+m)/n,1,-q,(1+m+n)/n,c^(1/2)*x^n/(-a)^(1/2),-e*x^n/d)/a^2/( 
a*e^2+c*d^2)/f/(1+m)/n/((1+e*x^n/d)^q)+1/2*e^2*(n*q+m+1)*(f*x)^(1+m)*(d+e* 
x^n)^q*hypergeom([-q, (1+m)/n],[(1+m+n)/n],-e*x^n/d)/a/(a*e^2+c*d^2)/f/(1+ 
m)/n/((1+e*x^n/d)^q)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 1888

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1888

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )^q}{\left (a+c x^{2 n}\right )^2} \, dx=f^{m} \left (\int \frac {x^{m} \left (x^{n} e +d \right )^{q}}{x^{4 n} c^{2}+2 x^{2 n} a c +a^{2}}d x \right ) \] Input:

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

Output:

f**m*int((x**m*(x**n*e + d)**q)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)