\(\int \frac {d+e x^n}{(a+c x^{2 n})^3} \, dx\) [39]

Optimal result

Integrand size = 19, antiderivative size = 184 \[ \int \frac {d+e x^n}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2}-\frac {x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac {d \left (1-6 n+8 n^2\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 n^2}+\frac {e (1-3 n) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 n^2 (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45 \[ \int \frac {d+e x^n}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {d x \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3}+\frac {e x^{1+n} \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a^3 (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1761, 1761, 1748, 778, 888}

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.

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

\(\Big \downarrow \) 1761

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

\(\Big \downarrow \) 1761

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

\(\Big \downarrow \) 1748

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

\(\Big \downarrow \) 778

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 1748

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Simp[d 
   Int[1/(a + c*x^(2*n)), x], x] + Simp[e   Int[x^n/(a + c*x^(2*n)), x], x] 
 /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && ( 
PosQ[a*c] ||  !IntegerQ[n])
 

rule 1761

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> S 
imp[(-x)*(d + e*x^n)*((a + c*x^(2*n))^(p + 1)/(2*a*n*(p + 1))), x] + Simp[1 
/(2*a*n*(p + 1))   Int[(d*(2*n*p + 2*n + 1) + e*(2*n*p + 3*n + 1)*x^n)*(a + 
 c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && 
 ILtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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