\(\int \frac {1}{(d+e x^n) (a+c x^{2 n})^3} \, dx\) [40]

Optimal result

Integrand size = 21, antiderivative size = 462 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^3} \, dx=\frac {c x \left (d-e x^n\right )}{4 a \left (c d^2+a e^2\right ) n \left (a+c x^{2 n}\right )^2}-\frac {c x \left (d \left (a e^2 (1-8 n)+c d^2 (1-4 n)\right )-e \left (a e^2 (1-7 n)+c d^2 (1-3 n)\right ) x^n\right )}{8 a^2 \left (c d^2+a e^2\right )^2 n^2 \left (a+c x^{2 n}\right )}+\frac {c d e^4 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^3}+\frac {c d \left (a e^2 (1-8 n)+c d^2 (1-4 n)\right ) (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 \left (c d^2+a e^2\right )^2 n^2}+\frac {e^6 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )^3}-\frac {c e^5 x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^3 (1+n)}-\frac {c e \left (a e^2 (1-7 n)+c d^2 (1-3 n)\right ) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 \left (c d^2+a e^2\right )^2 n^2 (1+n)} \] Output:

1/4*c*x*(d-e*x^n)/a/(a*e^2+c*d^2)/n/(a+c*x^(2*n))^2-1/8*c*x*(d*(a*e^2*(1-8 
*n)+c*d^2*(1-4*n))-e*(a*e^2*(1-7*n)+c*d^2*(1-3*n))*x^n)/a^2/(a*e^2+c*d^2)^ 
2/n^2/(a+c*x^(2*n))+c*d*e^4*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a) 
/a/(a*e^2+c*d^2)^3+1/8*c*d*(a*e^2*(1-8*n)+c*d^2*(1-4*n))*(1-2*n)*x*hyperge 
om([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a^3/(a*e^2+c*d^2)^2/n^2+e^6*x*hyperg 
eom([1, 1/n],[1+1/n],-e*x^n/d)/d/(a*e^2+c*d^2)^3-c*e^5*x^(1+n)*hypergeom([ 
1, 1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/a/(a*e^2+c*d^2)^3/(1+n)-1/8*c*e* 
(a*e^2*(1-7*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)^2/n^2/(1+n)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 346, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (\frac {c d e^4 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a}+\frac {e^6 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}-\frac {c e^5 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (1+n)}+\frac {c d e^2 \left (c d^2+a e^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2}-\frac {c e^3 \left (c d^2+a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 (1+n)}+\frac {c d \left (c d^2+a e^2\right )^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3}-\frac {c e \left (c d^2+a e^2\right )^2 x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3 (1+n)}\right )}{\left (c d^2+a e^2\right )^3} \] Input:

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

Output:

(x*((c*d*e^4*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a 
)])/a + (e^6*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/d - ( 
c*e^5*x^n*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n) 
)/a)])/(a*(1 + n)) + (c*d*e^2*(c*d^2 + a*e^2)*Hypergeometric2F1[2, 1/(2*n) 
, (2 + n^(-1))/2, -((c*x^(2*n))/a)])/a^2 - (c*e^3*(c*d^2 + a*e^2)*x^n*Hype 
rgeometric2F1[2, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(a^2*(1 
 + n)) + (c*d*(c*d^2 + a*e^2)^2*Hypergeometric2F1[3, 1/(2*n), (2 + n^(-1)) 
/2, -((c*x^(2*n))/a)])/a^3 - (c*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)^3
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 582, normalized size of antiderivative = 1.26, 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.

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

\(\Big \downarrow \) 1767

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

e^6*integrate(1/(c^3*d^7 + 3*a*c^2*d^5*e^2 + 3*a^2*c*d^3*e^4 + a^3*d*e^6 + 
 (c^3*d^6*e + 3*a*c^2*d^4*e^3 + 3*a^2*c*d^2*e^5 + a^3*e^7)*x^n), x) - 1/8* 
((a*c^2*e^3*(7*n - 1) + c^3*d^2*e*(3*n - 1))*x*x^(3*n) - (a*c^2*d*e^2*(8*n 
 - 1) + c^3*d^3*(4*n - 1))*x*x^(2*n) + (a^2*c*e^3*(9*n - 1) + a*c^2*d^2*e* 
(5*n - 1))*x*x^n - (a^2*c*d*e^2*(10*n - 1) + a*c^2*d^3*(6*n - 1))*x)/(a^4* 
c^2*d^4*n^2 + 2*a^5*c*d^2*e^2*n^2 + a^6*e^4*n^2 + (a^2*c^4*d^4*n^2 + 2*a^3 
*c^3*d^2*e^2*n^2 + a^4*c^2*e^4*n^2)*x^(4*n) + 2*(a^3*c^3*d^4*n^2 + 2*a^4*c 
^2*d^2*e^2*n^2 + a^5*c*e^4*n^2)*x^(2*n)) - integrate(-1/8*((8*n^2 - 6*n + 
1)*c^3*d^5 + 2*(12*n^2 - 8*n + 1)*a*c^2*d^3*e^2 + (24*n^2 - 10*n + 1)*a^2* 
c*d*e^4 - ((3*n^2 - 4*n + 1)*c^3*d^4*e + 2*(5*n^2 - 6*n + 1)*a*c^2*d^2*e^3 
 + (15*n^2 - 8*n + 1)*a^2*c*e^5)*x^n)/(a^3*c^3*d^6*n^2 + 3*a^4*c^2*d^4*e^2 
*n^2 + 3*a^5*c*d^2*e^4*n^2 + a^6*e^6*n^2 + (a^2*c^4*d^6*n^2 + 3*a^3*c^3*d^ 
4*e^2*n^2 + 3*a^4*c^2*d^2*e^4*n^2 + a^5*c*e^6*n^2)*x^(2*n)), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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