\(\int \frac {1}{(d+e x^n)^2 (a+c x^{2 n})^2} \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 347 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^2} \, dx=-\frac {e^2 \left (c d^2-a e^2\right ) x}{a d \left (c d^2+a e^2\right )^2 n \left (d+e x^n\right )}+\frac {c x \left (d-e x^n\right )}{2 a \left (c d^2+a e^2\right ) n \left (d+e x^n\right ) \left (a+c x^{2 n}\right )}+\frac {c \left (a^2 e^4 (1-4 n)-c^2 d^4 (1-2 n)+6 a c d^2 e^2 n\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right )^3 n}-\frac {e^4 \left (c d^2 (1-5 n)+a e^2 (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (c d^2+a e^2\right )^3 n}+\frac {c^2 d e \left (a e^2 (1-5 n)+c d^2 (1-n)\right ) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 \left (c d^2+a e^2\right )^3 n (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (d+e x^n\right )^2 \left (a+c x^{2 n}\right )^2} \, dx=\frac {x \left (\frac {c e^2 \left (3 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a}+4 c e^4 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )-\frac {4 c^2 d e^3 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (1+n)}+\frac {c \left (c d^2-a e^2\right ) \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2}+\frac {e^4 \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}-\frac {2 c^2 d e \left (c d^2+a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 (1+n)}\right )}{\left (c d^2+a e^2\right )^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.18, 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.

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

\(\Big \downarrow \) 1767

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c*x*(c*d^2 - a*e^2 - 2*c*d*e*x^n))/(2*a*(c*d^2 + a*e^2)^2*n*(a + c*x^(2*n 
))) + (c*e^2*(3*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)^3) - (c*(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)^2*n) + (4*c*e^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(- 
1), -((e*x^n)/d)])/(c*d^2 + a*e^2)^3 - (4*c^2*d*e^3*x^(1 + n)*Hypergeometr 
ic2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*(c*d^2 + a*e 
^2)^3*(1 + n)) + (c^2*d*e*(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)^2*n*(1 + n)) 
 + (e^4*x*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^2*(c* 
d^2 + a*e^2)^2)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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