Integrand size = 26, antiderivative size = 64 \[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\frac {x^2 \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {2}{n},\frac {2+n}{n},-\frac {b x^n}{a}\right )}{2 a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \] Output:
1/2*x^2*(a+b*x^n)*hypergeom([3, 2/n],[(2+n)/n],-b*x^n/a)/a^3/(a^2+2*a*b*x^ n+b^2*x^(2*n))^(1/2)
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\frac {x^2 \left (a+b x^n\right )^3 \operatorname {Hypergeometric2F1}\left (3,\frac {2}{n},1+\frac {2}{n},-\frac {b x^n}{a}\right )}{2 a^3 \left (\left (a+b x^n\right )^2\right )^{3/2}} \] Input:
Integrate[x/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]
Output:
(x^2*(a + b*x^n)^3*Hypergeometric2F1[3, 2/n, 1 + 2/n, -((b*x^n)/a)])/(2*a^ 3*((a + b*x^n)^2)^(3/2))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1384, 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.
\(\displaystyle \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b^3+b^4 x^n\right ) \int \frac {x}{\left (b^2 x^n+a b\right )^3}dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^2 \left (a b^3+b^4 x^n\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {2}{n},\frac {n+2}{n},-\frac {b x^n}{a}\right )}{2 a^3 b^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
Input:
Int[x/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]
Output:
(x^2*(a*b^3 + b^4*x^n)*Hypergeometric2F1[3, 2/n, (2 + n)/n, -((b*x^n)/a)]) /(2*a^3*b^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])
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])
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)])
\[\int \frac {x}{\left (a^{2}+2 x^{n} a b +b^{2} x^{2 n}\right )^{\frac {3}{2}}}d x\]
Input:
int(x/(a^2+2*x^n*a*b+b^2*x^(2*n))^(3/2),x)
Output:
int(x/(a^2+2*x^n*a*b+b^2*x^(2*n))^(3/2),x)
\[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="fricas")
Output:
integral(sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2)*x/(b^4*x^(4*n) + 4*a^2*b^2*x^ (2*n) + 4*a^3*b*x^n + a^4 + 2*(2*a*b^3*x^n + a^2*b^2)*x^(2*n)), x)
\[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int \frac {x}{\left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x/(a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2),x)
Output:
Integral(x/((a + b*x**n)**2)**(3/2), x)
\[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="maxima")
Output:
(n^2 - 3*n + 2)*integrate(x/(a^2*b*n^2*x^n + a^3*n^2), x) + 1/2*(2*b*(n - 1)*x^2*x^n + a*(3*n - 2)*x^2)/(a^2*b^2*n^2*x^(2*n) + 2*a^3*b*n^2*x^n + a^4 *n^2)
\[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="giac")
Output:
integrate(x/(b^2*x^(2*n) + 2*a*b*x^n + a^2)^(3/2), x)
Timed out. \[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int \frac {x}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}} \,d x \] Input:
int(x/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2),x)
Output:
int(x/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2), x)
\[ \int \frac {x}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx=\int \frac {x}{x^{3 n} b^{3}+3 x^{2 n} a \,b^{2}+3 x^{n} a^{2} b +a^{3}}d x \] Input:
int(x/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x)
Output:
int(x/(x**(3*n)*b**3 + 3*x**(2*n)*a*b**2 + 3*x**n*a**2*b + a**3),x)