Integrand size = 20, antiderivative size = 73 \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=-\frac {\left (a+\frac {b}{x}\right )^n \left (1+\frac {b}{a x}\right )^{-n} x^{-1+m} \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m)} \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx \]
Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1016, 999, 152, 150}
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.
\(\displaystyle \int \frac {x^m \left (a+\frac {b}{x}\right )^n}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 1016 |
\(\displaystyle \int \frac {x^{m-2} \left (a+\frac {b}{x}\right )^n}{\left (\frac {c}{x}+d\right )^2}dx\) |
\(\Big \downarrow \) 999 |
\(\displaystyle -\left (\frac {1}{x}\right )^m x^m \int \frac {\left (a+\frac {b}{x}\right )^n \left (\frac {1}{x}\right )^{-m}}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle \left (\frac {1}{x}\right )^m x^m \left (-\left (a+\frac {b}{x}\right )^n\right ) \left (\frac {b}{a x}+1\right )^{-n} \int \frac {\left (\frac {b}{a x}+1\right )^n \left (\frac {1}{x}\right )^{-m}}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {x^{m-1} \left (a+\frac {b}{x}\right )^n \left (\frac {b}{a x}+1\right )^{-n} \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m)}\) |
-(((a + b/x)^n*x^(-1 + m)*AppellF1[1 - m, -n, 2, 2 - m, -(b/(a*x)), -(c/(d *x))])/(d^2*(1 - m)*(1 + b/(a*x))^n))
3.3.92.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> Simp[(-(e*x)^m)*(x^(-1))^m Subst[Int[(a + b/x^n)^p*( (c + d/x^n)^q/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, b, c, d, e, m, p, q} , x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0] && !RationalQ[m]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^( p_.), x_Symbol] :> Int[x^(m - n*q)*(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ [{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] || !I ntegerQ[p])
\[\int \frac {\left (a +\frac {b}{x}\right )^{n} x^{m}}{\left (d x +c \right )^{2}}d x\]
\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {x^{m} \left (a + \frac {b}{x}\right )^{n}}{\left (c + d x\right )^{2}}\, dx \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]
\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {x^m\,{\left (a+\frac {b}{x}\right )}^n}{{\left (c+d\,x\right )}^2} \,d x \]