\(\int \frac {(f x)^m (d+e x^n)^q}{(a+b x^n+c x^{2 n})^2} \, dx\) [183]

Optimal result

Integrand size = 31, antiderivative size = 907 \[ \int \frac {(f x)^m \left (d+e x^n\right )^q}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {(f x)^{1+m} \left (d+e x^n\right )^{1+q} \left (c \left (b^2-2 a c\right ) d-b \left (b^2-3 a c\right ) e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) f n \left (a+b x^n+c x^{2 n}\right )}+\frac {c \left (b^3 d e (1+m-n)-b^2 \left (c d^2 (1+m-n)-e \left (\sqrt {b^2-4 a c} d (1+m-n)-a e (1+m-n (1-q))\right )\right )-b \left (c d \left (4 a e (1+m-2 n)+\sqrt {b^2-4 a c} d (1+m-n)\right )+a \sqrt {b^2-4 a c} e^2 (1+m-n (1-q))\right )+2 a c \left (2 c d^2 (1+m-2 n)+e \left (2 a e (1+m-n (2-q))+\sqrt {b^2-4 a c} d n q\right )\right )\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 {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) f (1+m) n}-\frac {c \left (b^3 d e (1+m-n)-b^2 \left (c d^2 (1+m-n)+e \left (\sqrt {b^2-4 a c} d (1+m-n)+a e (1+m-n (1-q))\right )\right )-b \left (c d \left (4 a e (1+m-2 n)-\sqrt {b^2-4 a c} d (1+m-n)\right )-a \sqrt {b^2-4 a c} e^2 (1+m-n (1-q))\right )+2 a c \left (2 c d^2 (1+m-2 n)+e \left (2 a e (1+m-n (2-q))-\sqrt {b^2-4 a c} d n q\right )\right )\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 {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) f (1+m) n}-\frac {e \left (b c d-b^2 e+2 a c e\right ) (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 )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) f (1+m) n} \] Output:

(f*x)^(1+m)*(d+e*x^n)^(1+q)*(c*(-2*a*c+b^2)*d-b*(-3*a*c+b^2)*e+c*(2*a*c*e- 
b^2*e+b*c*d)*x^n)/a/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)/f/n/(a+b*x^n+c*x^(2*n 
))+c*(b^3*d*e*(1+m-n)-b^2*(c*d^2*(1+m-n)-e*((-4*a*c+b^2)^(1/2)*d*(1+m-n)-a 
*e*(1+m-n*(1-q))))-b*(c*d*(4*a*e*(1+m-2*n)+(-4*a*c+b^2)^(1/2)*d*(1+m-n))+a 
*(-4*a*c+b^2)^(1/2)*e^2*(1+m-n*(1-q)))+2*a*c*(2*c*d^2*(1+m-2*n)+e*(2*a*e*( 
1+m-n*(2-q))+(-4*a*c+b^2)^(1/2)*d*n*q)))*(f*x)^(1+m)*(d+e*x^n)^q*AppellF1( 
(1+m)/n,1,-q,(1+m+n)/n,-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)),-e*x^n/d)/a/(-4*a*c 
+b^2)^(3/2)/(b-(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/f/(1+m)/n/((1+e*x^n 
/d)^q)-c*(b^3*d*e*(1+m-n)-b^2*(c*d^2*(1+m-n)+e*((-4*a*c+b^2)^(1/2)*d*(1+m- 
n)+a*e*(1+m-n*(1-q))))-b*(c*d*(4*a*e*(1+m-2*n)-(-4*a*c+b^2)^(1/2)*d*(1+m-n 
))-a*(-4*a*c+b^2)^(1/2)*e^2*(1+m-n*(1-q)))+2*a*c*(2*c*d^2*(1+m-2*n)+e*(2*a 
*e*(1+m-n*(2-q))-(-4*a*c+b^2)^(1/2)*d*n*q)))*(f*x)^(1+m)*(d+e*x^n)^q*Appel 
lF1((1+m)/n,1,-q,(1+m+n)/n,-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)),-e*x^n/d)/a/(-4 
*a*c+b^2)^(3/2)/(b+(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/f/(1+m)/n/((1+e 
*x^n/d)^q)-e*(2*a*c*e-b^2*e+b*c*d)*(n*q+m+1)*(f*x)^(1+m)*(d+e*x^n)^q*hyper 
geom([-q, (1+m)/n],[(1+m+n)/n],-e*x^n/d)/a/(-4*a*c+b^2)/(a*e^2-b*d*e+c*d^2 
)/f/(1+m)/n/((1+e*x^n/d)^q)
 

Mathematica [F]

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

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

Output:

Integrate[((f*x)^m*(d + e*x^n)^q)/(a + b*x^n + 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.

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

\(\Big \downarrow \) 1887

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1887

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(f x)^m \left (d+e x^n\right )^q}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=f^{m} \left (\int \frac {x^{m} \left (x^{n} e +d \right )^{q}}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) \] Input:

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

Output:

f**m*int((x**m*(x**n*e + d)**q)/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2* 
n)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)