\(\int \frac {(a+c x^{2 n})^p}{(d+e x^n)^3} \, dx\) [52]

Optimal result

Integrand size = 21, antiderivative size = 357 \[ \int \frac {\left (a+c x^{2 n}\right )^p}{\left (d+e x^n\right )^3} \, dx=\frac {3 e^2 x^{1+2 n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p,3,\frac {1}{2} \left (4+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^5 (1+2 n)}-\frac {e^3 x^{1+3 n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p,3,\frac {1}{2} \left (5+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^6 (1+3 n)}+\frac {x \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2 n},-p,3,\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^3}-\frac {3 e x^{1+n} \left (a+c x^{2 n}\right )^p \left (1+\frac {c x^{2 n}}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+n}{2 n},-p,3,\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^4 (1+n)} \] Output:

3*e^2*x^(1+2*n)*(a+c*x^(2*n))^p*AppellF1(1+1/2/n,3,-p,2+1/2/n,e^2*x^(2*n)/ 
d^2,-c*x^(2*n)/a)/d^5/(1+2*n)/((1+c*x^(2*n)/a)^p)-e^3*x^(1+3*n)*(a+c*x^(2* 
n))^p*AppellF1(3/2+1/2/n,3,-p,5/2+1/2/n,e^2*x^(2*n)/d^2,-c*x^(2*n)/a)/d^6/ 
(1+3*n)/((1+c*x^(2*n)/a)^p)+x*(a+c*x^(2*n))^p*AppellF1(1/2/n,3,-p,1+1/2/n, 
e^2*x^(2*n)/d^2,-c*x^(2*n)/a)/d^3/((1+c*x^(2*n)/a)^p)-3*e*x^(1+n)*(a+c*x^( 
2*n))^p*AppellF1(1/2*(1+n)/n,3,-p,3/2+1/2/n,e^2*x^(2*n)/d^2,-c*x^(2*n)/a)/ 
d^4/(1+n)/((1+c*x^(2*n)/a)^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1768, 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.

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

\(\Big \downarrow \) 1768

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^3 x^{3 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} \left (3+\frac {1}{n}\right ),-p,3,\frac {1}{2} \left (5+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^6 (3 n+1)}+\frac {3 e^2 x^{2 n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} \left (2+\frac {1}{n}\right ),-p,3,\frac {1}{2} \left (4+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^5 (2 n+1)}-\frac {3 e x^{n+1} \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {n+1}{2 n},-p,3,\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^4 (n+1)}+\frac {x \left (a+c x^{2 n}\right )^p \left (\frac {c x^{2 n}}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2 n},-p,3,\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a},\frac {e^2 x^{2 n}}{d^2}\right )}{d^3}\)

Input:

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

Output:

(3*e^2*x^(1 + 2*n)*(a + c*x^(2*n))^p*AppellF1[(2 + n^(-1))/2, -p, 3, (4 + 
n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^5*(1 + 2*n)*(1 + (c*x^ 
(2*n))/a)^p) - (e^3*x^(1 + 3*n)*(a + c*x^(2*n))^p*AppellF1[(3 + n^(-1))/2, 
 -p, 3, (5 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^6*(1 + 3* 
n)*(1 + (c*x^(2*n))/a)^p) + (x*(a + c*x^(2*n))^p*AppellF1[1/(2*n), -p, 3, 
(2 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^3*(1 + (c*x^(2*n) 
)/a)^p) - (3*e*x^(1 + n)*(a + c*x^(2*n))^p*AppellF1[(1 + n)/(2*n), -p, 3, 
(3 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^4*(1 + n)*(1 + (c 
*x^(2*n))/a)^p)
 

Defintions of rubi rules used

rule 1768

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] 
 :> Int[ExpandIntegrand[(a + c*x^(2*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/ 
(d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[n 
2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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