Integrand size = 21, antiderivative size = 67 \[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {c (d x)^{1+m} \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3 d (1+m)} \] Output:
c*(d*x)^(1+m)*(c/(b*x^2+a)^2)^(1/2)*(b*x^2+a)*hypergeom([3, 1/2+1/2*m],[3/ 2+1/2*m],-b*x^2/a)/a^3/d/(1+m)
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\frac {c x (d x)^m \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \left (a+b x^2\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3 (1+m)} \] Input:
Integrate[(d*x)^m*(c/(a + b*x^2)^2)^(3/2),x]
Output:
(c*x*(d*x)^m*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*Hypergeometric2F1[3, (1 + m )/2, (3 + m)/2, -((b*x^2)/a)])/(a^3*(1 + m))
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2045, 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.
\(\displaystyle \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {a^3 (d x)^m}{\left (b x^2+a\right )^3}dx}{a^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (a+b x^2\right ) \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \int \frac {(d x)^m}{\left (b x^2+a\right )^3}dx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {c \left (a+b x^2\right ) (d x)^{m+1} \sqrt {\frac {c}{\left (a+b x^2\right )^2}} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a^3 d (m+1)}\) |
Input:
Int[(d*x)^m*(c/(a + b*x^2)^2)^(3/2),x]
Output:
(c*(d*x)^(1 + m)*Sqrt[c/(a + b*x^2)^2]*(a + b*x^2)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^3*d*(1 + m))
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_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
\[\int \left (x d \right )^{m} {\left (\frac {c}{\left (b \,x^{2}+a \right )^{2}}\right )}^{\frac {3}{2}}d x\]
Input:
int((x*d)^m*(c/(b*x^2+a)^2)^(3/2),x)
Output:
int((x*d)^m*(c/(b*x^2+a)^2)^(3/2),x)
\[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int { \left (d x\right )^{m} \left (\frac {c}{{\left (b x^{2} + a\right )}^{2}}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((d*x)^m*(c/(b*x^2+a)^2)^(3/2),x, algorithm="fricas")
Output:
integral((d*x)^m*c*sqrt(c/(b^2*x^4 + 2*a*b*x^2 + a^2))/(b^2*x^4 + 2*a*b*x^ 2 + a^2), x)
Timed out. \[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate((d*x)**m*(c/(b*x**2+a)**2)**(3/2),x)
Output:
Timed out
\[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int { \left (d x\right )^{m} \left (\frac {c}{{\left (b x^{2} + a\right )}^{2}}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((d*x)^m*(c/(b*x^2+a)^2)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x)^m*(c/(b*x^2 + a)^2)^(3/2), x)
\[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int { \left (d x\right )^{m} \left (\frac {c}{{\left (b x^{2} + a\right )}^{2}}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((d*x)^m*(c/(b*x^2+a)^2)^(3/2),x, algorithm="giac")
Output:
integrate((d*x)^m*(c/(b*x^2 + a)^2)^(3/2), x)
Timed out. \[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=\int {\left (d\,x\right )}^m\,{\left (\frac {c}{{\left (b\,x^2+a\right )}^2}\right )}^{3/2} \,d x \] Input:
int((d*x)^m*(c/(a + b*x^2)^2)^(3/2),x)
Output:
int((d*x)^m*(c/(a + b*x^2)^2)^(3/2), x)
\[ \int (d x)^m \left (\frac {c}{\left (a+b x^2\right )^2}\right )^{3/2} \, dx=d^{m} \sqrt {c}\, \left (\int \frac {x^{m}}{b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}}d x \right ) c \] Input:
int((d*x)^m*(c/(b*x^2+a)^2)^(3/2),x)
Output:
d**m*sqrt(c)*int(x**m/(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6),x )*c