\(\int \frac {1}{(a^2+2 a c x^n+c^2 x^{2 n})^{5/2}} \, dx\) [117]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {1}{\left (a^2+2 a c x^n+c^2 x^{2 n}\right )^{5/2}} \, dx=\frac {x \left (a+c x^n\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1}{n},1+\frac {1}{n},-\frac {c x^n}{a}\right )}{a^5 \sqrt {a^2+2 a c x^n+c^2 x^{2 n}}} \] Output:

x*(a+c*x^n)*hypergeom([5, 1/n],[1+1/n],-c*x^n/a)/a^5/(a^2+2*a*c*x^n+c^2*x^ 
(2*n))^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a^2+2 a c x^n+c^2 x^{2 n}\right )^{5/2}} \, dx=\frac {x \left (a+c x^n\right )^5 \operatorname {Hypergeometric2F1}\left (5,\frac {1}{n},1+\frac {1}{n},-\frac {c x^n}{a}\right )}{a^5 \left (\left (a+c x^n\right )^2\right )^{5/2}} \] Input:

Integrate[(a^2 + 2*a*c*x^n + c^2*x^(2*n))^(-5/2),x]
 

Output:

(x*(a + c*x^n)^5*Hypergeometric2F1[5, n^(-1), 1 + n^(-1), -((c*x^n)/a)])/( 
a^5*((a + c*x^n)^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1384, 778}

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.

\(\displaystyle \int \frac {1}{\left (a^2+2 a c x^n+c^2 x^{2 n}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {\left (a c^5+c^6 x^n\right ) \int \frac {1}{\left (c^2 x^n+a c\right )^5}dx}{\sqrt {a^2+2 a c x^n+c^2 x^{2 n}}}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {x \left (a c^5+c^6 x^n\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1}{n},1+\frac {1}{n},-\frac {c x^n}{a}\right )}{a^5 c^5 \sqrt {a^2+2 a c x^n+c^2 x^{2 n}}}\)

Input:

Int[(a^2 + 2*a*c*x^n + c^2*x^(2*n))^(-5/2),x]
 

Output:

(x*(a*c^5 + c^6*x^n)*Hypergeometric2F1[5, n^(-1), 1 + n^(-1), -((c*x^n)/a) 
])/(a^5*c^5*Sqrt[a^2 + 2*a*c*x^n + c^2*x^(2*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 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

Maple [F]

\[\int \frac {1}{\left (a^{2}+2 a c \,x^{n}+c^{2} x^{2 n}\right )^{\frac {5}{2}}}d x\]

Input:

int(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x)
 

Output:

int(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x)
 

Fricas [F]

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

integrate(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(c^2*x^(2*n) + 2*a*c*x^n + a^2)/(c^6*x^(6*n) + 8*a^3*c^3*x^(3 
*n) + 12*a^4*c^2*x^(2*n) + 6*a^5*c*x^n + a^6 + 3*(2*a*c^5*x^n + a^2*c^4)*x 
^(4*n) + 3*(4*a^2*c^4*x^(2*n) + 4*a^3*c^3*x^n + a^4*c^2)*x^(2*n)), x)
 

Sympy [F]

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

integrate(1/(a**2+2*a*c*x**n+c**2*x**(2*n))**(5/2),x)
 

Output:

Integral((a**2 + 2*a*c*x**n + c**2*x**(2*n))**(-5/2), x)
 

Maxima [F]

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

integrate(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x, algorithm="maxima")
 

Output:

(24*n^4 - 50*n^3 + 35*n^2 - 10*n + 1)*integrate(1/24/(a^4*c*n^4*x^n + a^5* 
n^4), x) + 1/24*((24*n^3 - 26*n^2 + 9*n - 1)*c^3*x*x^(3*n) + (84*n^3 - 85* 
n^2 + 28*n - 3)*a*c^2*x*x^(2*n) + (104*n^3 - 94*n^2 + 29*n - 3)*a^2*c*x*x^ 
n + (50*n^3 - 35*n^2 + 10*n - 1)*a^3*x)/(a^4*c^4*n^4*x^(4*n) + 4*a^5*c^3*n 
^4*x^(3*n) + 6*a^6*c^2*n^4*x^(2*n) + 4*a^7*c*n^4*x^n + a^8*n^4)
 

Giac [F]

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

integrate(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x, algorithm="giac")
 

Output:

integrate((c^2*x^(2*n) + 2*a*c*x^n + a^2)^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a^2+2 a c x^n+c^2 x^{2 n}\right )^{5/2}} \, dx=\int \frac {1}{{\left (a^2+c^2\,x^{2\,n}+2\,a\,c\,x^n\right )}^{5/2}} \,d x \] Input:

int(1/(a^2 + c^2*x^(2*n) + 2*a*c*x^n)^(5/2),x)
 

Output:

int(1/(a^2 + c^2*x^(2*n) + 2*a*c*x^n)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a^2+2 a c x^n+c^2 x^{2 n}\right )^{5/2}} \, dx=\int \frac {1}{x^{5 n} c^{5}+5 x^{4 n} a \,c^{4}+10 x^{3 n} a^{2} c^{3}+10 x^{2 n} a^{3} c^{2}+5 x^{n} a^{4} c +a^{5}}d x \] Input:

int(1/(a^2+2*a*c*x^n+c^2*x^(2*n))^(5/2),x)
 

Output:

int(1/(x**(5*n)*c**5 + 5*x**(4*n)*a*c**4 + 10*x**(3*n)*a**2*c**3 + 10*x**( 
2*n)*a**3*c**2 + 5*x**n*a**4*c + a**5),x)