\(\int \frac {1}{(d+e x^n)^2 (a+c x^{2 n})^3} \, dx\) [41]

Optimal result

Integrand size = 21, antiderivative size = 660 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\frac {c x \left (d-e x^n\right )}{4 a \left (c d^2+a e^2\right ) n \left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2}+\frac {c x \left (a^2 e^4 (1-8 n)-c^2 d^4 (1-4 n)+8 a c d^2 e^2 n+2 c d e \left (a e^2 (1-9 n)+c d^2 (1-3 n)\right ) x^n\right )}{8 a^2 \left (c d^2+a e^2\right )^3 n^2 \left (a+c x^{2 n}\right )}-\frac {c (1-2 n) \left (a^2 e^4 (1-8 n)-c^2 d^4 (1-4 n)+8 a c d^2 e^2 n\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 \left (c d^2+a e^2\right )^3 n^2}-\frac {c e^2 \left (a c d^2 e^2 (1-12 n)+c^2 d^4 (1-2 n)+2 a^2 e^4 n\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right )^4 n}-\frac {c e^4 \left (a e^2 (1-13 n)+c d^2 (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{2 a \left (c d^2+a e^2\right )^4 n}-\frac {c^2 d e \left (a e^2 (1-9 n)+c d^2 (1-3 n)\right ) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 \left (c d^2+a e^2\right )^3 n^2 (1+n)}+\frac {c^2 d e^3 \left (a e^2 (1-13 n)+c d^2 (1-n)\right ) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right )^4 n (1+n)}-\frac {e^4 \left (c d^2-2 a e^2\right ) x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{2 a d^2 \left (c d^2+a e^2\right )^3} \] Output:

1/4*c*x*(d-e*x^n)/a/(a*e^2+c*d^2)/n/(d+e*x^n)/(a+c*x^(2*n))^2+1/8*c*x*(a^2 
*e^4*(1-8*n)-c^2*d^4*(1-4*n)+8*a*c*d^2*e^2*n+2*c*d*e*(a*e^2*(1-9*n)+c*d^2* 
(1-3*n))*x^n)/a^2/(a*e^2+c*d^2)^3/n^2/(a+c*x^(2*n))-1/8*c*(1-2*n)*(a^2*e^4 
*(1-8*n)-c^2*d^4*(1-4*n)+8*a*c*d^2*e^2*n)*x*hypergeom([1, 1/2/n],[1+1/2/n] 
,-c*x^(2*n)/a)/a^3/(a*e^2+c*d^2)^3/n^2-1/2*c*e^2*(a*c*d^2*e^2*(1-12*n)+c^2 
*d^4*(1-2*n)+2*a^2*e^4*n)*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a 
^2/(a*e^2+c*d^2)^4/n-1/2*c*e^4*(a*e^2*(1-13*n)+c*d^2*(1-n))*x*hypergeom([1 
, 1/n],[1+1/n],-e*x^n/d)/a/(a*e^2+c*d^2)^4/n-1/4*c^2*d*e*(a*e^2*(1-9*n)+c* 
d^2*(1-3*n))*(1-n)*x^(1+n)*hypergeom([1, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2* 
n)/a)/a^3/(a*e^2+c*d^2)^3/n^2/(1+n)+1/2*c^2*d*e^3*(a*e^2*(1-13*n)+c*d^2*(1 
-n))*x^(1+n)*hypergeom([1, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/a^2/(a*e 
^2+c*d^2)^4/n/(1+n)-1/2*e^4*(-2*a*e^2+c*d^2)*x*hypergeom([2, 1/n],[1+1/n], 
-e*x^n/d)/a/d^2/(a*e^2+c*d^2)^3
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 426, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (\frac {c e^4 \left (5 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a}+6 c e^6 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )-\frac {6 c^2 d e^5 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (1+n)}+\frac {c e^2 \left (3 c d^2-a e^2\right ) \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2}+\frac {e^6 \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}-\frac {4 c^2 d e^3 \left (c d^2+a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 (1+n)}+\frac {c \left (c d^2-a e^2\right ) \left (c d^2+a e^2\right )^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3}-\frac {2 c^2 d e \left (c d^2+a e^2\right )^2 x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3 (1+n)}\right )}{\left (c d^2+a e^2\right )^4} \] Input:

Integrate[1/((d + e*x^n)^2*(a + c*x^(2*n))^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 701, normalized size of antiderivative = 1.06, 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.

\(\displaystyle \int \frac {1}{\left (a+c x^{2 n}\right )^3 \left (d+e x^n\right )^2} \, dx\)

\(\Big \downarrow \) 1767

\(\displaystyle \int \left (-\frac {c e^2 \left (a e^2-3 c d^2+4 c d e x^n\right )}{\left (a e^2+c d^2\right )^3 \left (a+c x^{2 n}\right )^2}-\frac {c \left (a e^2-c d^2+2 c d e x^n\right )}{\left (a e^2+c d^2\right )^2 \left (a+c x^{2 n}\right )^3}+\frac {6 c d e^6}{\left (a e^2+c d^2\right )^4 \left (d+e x^n\right )}+\frac {e^6}{\left (a e^2+c d^2\right )^3 \left (d+e x^n\right )^2}-\frac {c e^4 \left (a e^2-5 c d^2+6 c d e x^n\right )}{\left (a e^2+c d^2\right )^4 \left (a+c x^{2 n}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {c^2 d e (1-3 n) (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{4 a^3 n^2 (n+1) \left (a e^2+c d^2\right )^2}+\frac {c (1-4 n) (1-2 n) x \left (c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 n^2 \left (a e^2+c d^2\right )^2}+\frac {2 c^2 d e^3 (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 n (n+1) \left (a e^2+c d^2\right )^3}-\frac {c e^2 (1-2 n) x \left (3 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 n \left (a e^2+c d^2\right )^3}-\frac {c x \left ((1-4 n) \left (c d^2-a e^2\right )-2 c d e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a e^2+c d^2\right )^2 \left (a+c x^{2 n}\right )}-\frac {6 c^2 d e^5 x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )^4}+\frac {c e^2 x \left (-a e^2+3 c d^2-4 c d e x^n\right )}{2 a n \left (a e^2+c d^2\right )^3 \left (a+c x^{2 n}\right )}+\frac {c x \left (-a e^2+c d^2-2 c d e x^n\right )}{4 a n \left (a e^2+c d^2\right )^2 \left (a+c x^{2 n}\right )^2}+\frac {6 c e^6 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{\left (a e^2+c d^2\right )^4}+\frac {e^6 x \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (a e^2+c d^2\right )^3}+\frac {c e^4 x \left (5 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )^4}\)

Input:

Int[1/((d + e*x^n)^2*(a + c*x^(2*n))^3),x]
 

Output:

(c*x*(c*d^2 - a*e^2 - 2*c*d*e*x^n))/(4*a*(c*d^2 + a*e^2)^2*n*(a + c*x^(2*n 
))^2) + (c*e^2*x*(3*c*d^2 - a*e^2 - 4*c*d*e*x^n))/(2*a*(c*d^2 + a*e^2)^3*n 
*(a + c*x^(2*n))) - (c*x*((c*d^2 - a*e^2)*(1 - 4*n) - 2*c*d*e*(1 - 3*n)*x^ 
n))/(8*a^2*(c*d^2 + a*e^2)^2*n^2*(a + c*x^(2*n))) + (c*e^4*(5*c*d^2 - a*e^ 
2)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*( 
c*d^2 + a*e^2)^4) + (c*(c*d^2 - a*e^2)*(1 - 4*n)*(1 - 2*n)*x*Hypergeometri 
c2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*(c*d^2 + a*e^2) 
^2*n^2) - (c*e^2*(3*c*d^2 - a*e^2)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n 
), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(2*a^2*(c*d^2 + a*e^2)^3*n) + (6*c*e 
^6*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(c*d^2 + a*e^ 
2)^4 - (6*c^2*d*e^5*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^( 
-1))/2, -((c*x^(2*n))/a)])/(a*(c*d^2 + a*e^2)^4*(1 + n)) - (c^2*d*e*(1 - 3 
*n)*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, 
-((c*x^(2*n))/a)])/(4*a^3*(c*d^2 + a*e^2)^2*n^2*(1 + n)) + (2*c^2*d*e^3*(1 
 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x 
^(2*n))/a)])/(a^2*(c*d^2 + a*e^2)^3*n*(1 + n)) + (e^6*x*Hypergeometric2F1[ 
2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^2*(c*d^2 + a*e^2)^3)
 

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]

\[\int \frac {1}{\left (d +e \,x^{n}\right )^{2} \left (a +c \,x^{2 n}\right )^{3}}d x\]

Input:

int(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x)
 

Output:

int(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x)
 

Fricas [F]

\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:

integrate(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="fricas")
 

Output:

integral(1/(a^3*e^2*x^(2*n) + 2*a^3*d*e*x^n + a^3*d^2 + (c^3*e^2*x^(2*n) + 
 2*c^3*d*e*x^n + c^3*d^2)*x^(6*n) + 3*(a*c^2*e^2*x^(2*n) + 2*a*c^2*d*e*x^n 
 + a*c^2*d^2)*x^(4*n) + 3*(a^2*c*e^2*x^(2*n) + 2*a^2*c*d*e*x^n + a^2*c*d^2 
)*x^(2*n)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(d+e*x**n)**2/(a+c*x**(2*n))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:

integrate(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="maxima")
 

Output:

(c*d^2*e^6*(7*n - 1) + a*e^8*(n - 1))*integrate(1/(c^4*d^10*n + 4*a*c^3*d^ 
8*e^2*n + 6*a^2*c^2*d^6*e^4*n + 4*a^3*c*d^4*e^6*n + a^4*d^2*e^8*n + (c^4*d 
^9*e*n + 4*a*c^3*d^7*e^3*n + 6*a^2*c^2*d^5*e^5*n + 4*a^3*c*d^3*e^7*n + a^4 
*d*e^9*n)*x^n), x) - 1/8*(2*(a*c^3*d^2*e^4*(11*n - 1) + c^4*d^4*e^2*(3*n - 
 1) - 4*a^2*c^2*e^6*n)*x*x^(4*n) + (a^2*c^2*d*e^5*(8*n - 1) + 2*a*c^3*d^3* 
e^3*(5*n - 1) + c^4*d^5*e*(2*n - 1))*x*x^(3*n) + (a^2*c^2*d^2*e^4*(34*n - 
3) - c^4*d^6*(4*n - 1) - 2*a*c^3*d^4*e^2*(n + 1) - 16*a^3*c*e^6*n)*x*x^(2* 
n) + (a^3*c*d*e^5*(10*n - 1) + 2*a^2*c^2*d^3*e^3*(7*n - 1) + a*c^3*d^5*e*( 
4*n - 1))*x*x^n + (a^3*c*d^2*e^4*(10*n - 1) - a*c^3*d^6*(6*n - 1) - 12*a^2 
*c^2*d^4*e^2*n - 8*a^4*e^6*n)*x)/(a^4*c^3*d^8*n^2 + 3*a^5*c^2*d^6*e^2*n^2 
+ 3*a^6*c*d^4*e^4*n^2 + a^7*d^2*e^6*n^2 + (a^2*c^5*d^7*e*n^2 + 3*a^3*c^4*d 
^5*e^3*n^2 + 3*a^4*c^3*d^3*e^5*n^2 + a^5*c^2*d*e^7*n^2)*x^(5*n) + (a^2*c^5 
*d^8*n^2 + 3*a^3*c^4*d^6*e^2*n^2 + 3*a^4*c^3*d^4*e^4*n^2 + a^5*c^2*d^2*e^6 
*n^2)*x^(4*n) + 2*(a^3*c^4*d^7*e*n^2 + 3*a^4*c^3*d^5*e^3*n^2 + 3*a^5*c^2*d 
^3*e^5*n^2 + a^6*c*d*e^7*n^2)*x^(3*n) + 2*(a^3*c^4*d^8*n^2 + 3*a^4*c^3*d^6 
*e^2*n^2 + 3*a^5*c^2*d^4*e^4*n^2 + a^6*c*d^2*e^6*n^2)*x^(2*n) + (a^4*c^3*d 
^7*e*n^2 + 3*a^5*c^2*d^5*e^3*n^2 + 3*a^6*c*d^3*e^5*n^2 + a^7*d*e^7*n^2)*x^ 
n) - integrate(-1/8*((8*n^2 - 6*n + 1)*c^4*d^6 + (32*n^2 - 18*n + 1)*a*c^3 
*d^4*e^2 + (48*n^2 - 2*n - 1)*a^2*c^2*d^2*e^4 - (24*n^2 - 10*n + 1)*a^3*c* 
e^6 - 2*((3*n^2 - 4*n + 1)*c^4*d^5*e + 2*(7*n^2 - 8*n + 1)*a*c^3*d^3*e^...
 

Giac [F]

\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{3} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:

integrate(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x, algorithm="giac")
 

Output:

integrate(1/((c*x^(2*n) + a)^3*(e*x^n + d)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int \frac {1}{{\left (a+c\,x^{2\,n}\right )}^3\,{\left (d+e\,x^n\right )}^2} \,d x \] Input:

int(1/((a + c*x^(2*n))^3*(d + e*x^n)^2),x)
 

Output:

int(1/((a + c*x^(2*n))^3*(d + e*x^n)^2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^3} \, dx=\int \frac {1}{x^{8 n} c^{3} e^{2}+2 x^{7 n} c^{3} d e +3 x^{6 n} a \,c^{2} e^{2}+x^{6 n} c^{3} d^{2}+6 x^{5 n} a \,c^{2} d e +3 x^{4 n} a^{2} c \,e^{2}+3 x^{4 n} a \,c^{2} d^{2}+6 x^{3 n} a^{2} c d e +x^{2 n} a^{3} e^{2}+3 x^{2 n} a^{2} c \,d^{2}+2 x^{n} a^{3} d e +a^{3} d^{2}}d x \] Input:

int(1/(d+e*x^n)^2/(a+c*x^(2*n))^3,x)
 

Output:

int(1/(x**(8*n)*c**3*e**2 + 2*x**(7*n)*c**3*d*e + 3*x**(6*n)*a*c**2*e**2 + 
 x**(6*n)*c**3*d**2 + 6*x**(5*n)*a*c**2*d*e + 3*x**(4*n)*a**2*c*e**2 + 3*x 
**(4*n)*a*c**2*d**2 + 6*x**(3*n)*a**2*c*d*e + x**(2*n)*a**3*e**2 + 3*x**(2 
*n)*a**2*c*d**2 + 2*x**n*a**3*d*e + a**3*d**2),x)