\(\int \frac {(d+e x^n)^2}{(a+c x^{2 n})^2} \, dx\) [33]

Optimal result

Integrand size = 21, antiderivative size = 166 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^2} \, dx=-\frac {e^2 x}{2 a c n}+\frac {x \left (d+e x^n\right )^2}{2 a n \left (a+c x^{2 n}\right )}+\frac {\left (a e^2-c d^2 (1-2 n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}-\frac {d e (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^2 n (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.82 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^2} \, dx=\frac {x \left (a e^2 (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\left (c d^2-a e^2\right ) (1+n) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+2 c d e x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )}{a^2 c (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.22, 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.

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

\(\Big \downarrow \) 1767

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(x*(c*d^2 - a*e^2 + 2*c*d*e*x^n))/(2*a*c*n*(a + c*x^(2*n))) + (e^2*x*Hyper 
geometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*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*n) - (d*e*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 
 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(a^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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*(2*c*d*e*x*x^n + (c*d^2 - a*e^2)*x)/(a*c^2*n*x^(2*n) + a^2*c*n) + inte 
grate(1/2*(2*c*d*e*(n - 1)*x^n + c*d^2*(2*n - 1) + a*e^2)/(a*c^2*n*x^(2*n) 
 + a^2*c*n), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*a*c*e**2 
 + 2*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*c**2 
*d**2*n - x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x) 
*c**2*d**2 + 2*x**(2*n)*int(x**n/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x 
)*a*c*d*e + int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*a**2*e 
**2 + 2*int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*a*c*d**2*n 
 - int(x**(2*n)/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*a*c*d**2 + 2*in 
t(x**n/(x**(4*n)*c**2 + 2*x**(2*n)*a*c + a**2),x)*a**2*d*e + d**2*x)/(a*(x 
**(2*n)*c + a))