Integrand size = 22, antiderivative size = 295 \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,2,2+m,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,2,2+m,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)} \]
1/4*(g*x)^(1+m)*(e*x+d)^n*AppellF1(1+m,-n,1,2+m,-e*x/d,-x*c^(1/2)/(-a)^(1/ 2))/a^2/g/(1+m)/((1+e*x/d)^n)+1/4*(g*x)^(1+m)*(e*x+d)^n*AppellF1(1+m,1,-n, 2+m,x*c^(1/2)/(-a)^(1/2),-e*x/d)/a^2/g/(1+m)/((1+e*x/d)^n)+1/4*(g*x)^(1+m) *(e*x+d)^n*AppellF1(1+m,-n,2,2+m,-e*x/d,-x*c^(1/2)/(-a)^(1/2))/a^2/g/(1+m) /((1+e*x/d)^n)+1/4*(g*x)^(1+m)*(e*x+d)^n*AppellF1(1+m,2,-n,2+m,x*c^(1/2)/( -a)^(1/2),-e*x/d)/a^2/g/(1+m)/((1+e*x/d)^n)
\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx \]
Time = 0.55 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {615, 2009}
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.
\(\displaystyle \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \int \left (-\frac {c (g x)^m (d+e x)^n}{2 a \left (-a c-c^2 x^2\right )}-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}-c x\right )^2}-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}+c x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,1,m+2,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,1,m+2,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,2,m+2,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,2,m+2,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}\) |
((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((e*x)/d), -((Sq rt[c]*x)/Sqrt[-a])])/(4*a^2*g*(1 + m)*(1 + (e*x)/d)^n) + ((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((e*x)/d), (Sqrt[c]*x)/Sqrt[-a]]) /(4*a^2*g*(1 + m)*(1 + (e*x)/d)^n) + ((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 2, 2 + m, -((e*x)/d), -((Sqrt[c]*x)/Sqrt[-a])])/(4*a^2*g*(1 + m) *(1 + (e*x)/d)^n) + ((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 2, 2 + m, -((e*x)/d), (Sqrt[c]*x)/Sqrt[-a]])/(4*a^2*g*(1 + m)*(1 + (e*x)/d)^n)
3.4.80.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
\[\int \frac {\left (g x \right )^{m} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]
\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {{\left (g\,x\right )}^m\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]