Integrand size = 28, antiderivative size = 73 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {(d x)^{1+m} \left (a+b x^2\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^5 d (1+m) \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
(d*x)^(1+m)*(b*x^2+a)*hypergeom([5, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^5/d /(1+m)/((b*x^2+a)^2)^(1/2)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x (d x)^m \left (a+b x^2\right ) \operatorname {Hypergeometric2F1}\left (5,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^5 (1+m) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[(d*x)^m/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(x*(d*x)^m*(a + b*x^2)*Hypergeometric2F1[5, (1 + m)/2, (3 + m)/2, -((b*x^2 )/a)])/(a^5*(1 + m)*Sqrt[(a + b*x^2)^2])
Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1384, 27, 278}
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.
\(\displaystyle \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {(d x)^m}{b^5 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {(d x)^m}{\left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {\left (a+b x^2\right ) (d x)^{m+1} \operatorname {Hypergeometric2F1}\left (5,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a^5 d (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[(d*x)^m/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
((d*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[5, (1 + m)/2, (3 + m)/2, -((b *x^2)/a)])/(a^5*d*(1 + m)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[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)])
\[\int \frac {\left (d x \right )^{m}}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}d x\]
Input:
int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
integral(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(d*x)^m/(b^6*x^12 + 6*a*b^5*x^10 + 15*a^2*b^4*x^8 + 20*a^3*b^3*x^6 + 15*a^4*b^2*x^4 + 6*a^5*b*x^2 + a^6), x )
\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {\left (d x\right )^{m}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((d*x)**m/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral((d*x)**m/((a + b*x**2)**2)**(5/2), x)
\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
integrate((d*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(5/2), x)
\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
integrate((d*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(5/2), x)
Timed out. \[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:
int((d*x)^m/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int((d*x)^m/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
\[ \int \frac {(d x)^m}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=d^{m} \left (\int \frac {x^{m}}{b^{5} x^{10}+5 a \,b^{4} x^{8}+10 a^{2} b^{3} x^{6}+10 a^{3} b^{2} x^{4}+5 a^{4} b \,x^{2}+a^{5}}d x \right ) \] Input:
int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
d**m*int(x**m/(a**5 + 5*a**4*b*x**2 + 10*a**3*b**2*x**4 + 10*a**2*b**3*x** 6 + 5*a*b**4*x**8 + b**5*x**10),x)