Integrand size = 30, antiderivative size = 76 \[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {(d x)^{1+m} \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{a d (1+m) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \] Output:
(d*x)^(1+m)*(a+b*x^n)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-b*x^n/a)/a/d/(1+ m)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82 \[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x (d x)^m \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},1+\frac {1+m}{n},-\frac {b x^n}{a}\right )}{a (1+m) \sqrt {\left (a+b x^n\right )^2}} \] Input:
Integrate[(d*x)^m/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]
Output:
(x*(d*x)^m*(a + b*x^n)*Hypergeometric2F1[1, (1 + m)/n, 1 + (1 + m)/n, -((b *x^n)/a)])/(a*(1 + m)*Sqrt[(a + b*x^n)^2])
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
\(\displaystyle \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b+b^2 x^n\right ) \int \frac {(d x)^m}{b^2 x^n+a b}dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(d x)^{m+1} \left (a b+b^2 x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right )}{a b d (m+1) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\) |
Input:
Int[(d*x)^m/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]
Output:
((d*x)^(1 + m)*(a*b + b^2*x^n)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n) /n, -((b*x^n)/a)])/(a*b*d*(1 + m)*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)])
\[\int \frac {\left (d x \right )^{m}}{\sqrt {a^{2}+2 x^{n} a b +b^{2} x^{2 n}}}d x\]
Input:
int((d*x)^m/(a^2+2*x^n*a*b+b^2*x^(2*n))^(1/2),x)
Output:
int((d*x)^m/(a^2+2*x^n*a*b+b^2*x^(2*n))^(1/2),x)
\[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \] Input:
integrate((d*x)^m/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="fricas")
Output:
integral((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
\[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {\left (d x\right )^{m}}{\sqrt {\left (a + b x^{n}\right )^{2}}}\, dx \] Input:
integrate((d*x)**m/(a**2+2*a*b*x**n+b**2*x**(2*n))**(1/2),x)
Output:
Integral((d*x)**m/sqrt((a + b*x**n)**2), x)
\[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \] Input:
integrate((d*x)^m/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="maxima")
Output:
integrate((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
\[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {\left (d x\right )^{m}}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \] Input:
integrate((d*x)^m/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x, algorithm="giac")
Output:
integrate((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
Timed out. \[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {{\left (d\,x\right )}^m}{\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n}} \,d x \] Input:
int((d*x)^m/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(1/2),x)
Output:
int((d*x)^m/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(1/2), x)
\[ \int \frac {(d x)^m}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=d^{m} \left (\int \frac {x^{m}}{x^{n} b +a}d x \right ) \] Input:
int((d*x)^m/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x)
Output:
d**m*int(x**m/(x**n*b + a),x)