\(\int \frac {(a+\frac {b}{x})^n x^m}{(c+d x)^2} \, dx\) [292]

Optimal result

Integrand size = 20, antiderivative size = 73 \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=-\frac {\left (a+\frac {b}{x}\right )^n \left (1+\frac {b}{a x}\right )^{-n} x^{-1+m} \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m)} \]

[Out]

-(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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {528, 511, 140, 138} \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=-\frac {x^{m-1} \left (a+\frac {b}{x}\right )^n \left (\frac {b}{a x}+1\right )^{-n} \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m)} \]

[In]

Int[((a + b/x)^n*x^m)/(c + d*x)^2,x]

[Out]

-(((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)
)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> 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] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[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]

Rule 511

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(-(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]

Rule 528

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] |
|  !IntegerQ[p])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (a+\frac {b}{x}\right )^n x^{-2+m}}{\left (d+\frac {c}{x}\right )^2} \, dx \\ & = -\left (\left (\left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-m} (a+b x)^n}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )\right ) \\ & = -\left (\left (\left (a+\frac {b}{x}\right )^n \left (1+\frac {b}{a x}\right )^{-n} \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-m} \left (1+\frac {b x}{a}\right )^n}{(d+c x)^2} \, dx,x,\frac {1}{x}\right )\right ) \\ & = -\frac {\left (a+\frac {b}{x}\right )^n \left (1+\frac {b}{a x}\right )^{-n} x^{-1+m} F_1\left (1-m;-n,2;2-m;-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m)} \\ \end{align*}

Mathematica [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx \]

[In]

Integrate[((a + b/x)^n*x^m)/(c + d*x)^2,x]

[Out]

Integrate[((a + b/x)^n*x^m)/(c + d*x)^2, x]

Maple [F]

\[\int \frac {\left (a +\frac {b}{x}\right )^{n} x^{m}}{\left (d x +c \right )^{2}}d x\]

[In]

int((a+b/x)^n*x^m/(d*x+c)^2,x)

[Out]

int((a+b/x)^n*x^m/(d*x+c)^2,x)

Fricas [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate((a+b/x)^n*x^m/(d*x+c)^2,x, algorithm="fricas")

[Out]

integral(x^m*((a*x + b)/x)^n/(d^2*x^2 + 2*c*d*x + c^2), x)

Sympy [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {x^{m} \left (a + \frac {b}{x}\right )^{n}}{\left (c + d x\right )^{2}}\, dx \]

[In]

integrate((a+b/x)**n*x**m/(d*x+c)**2,x)

[Out]

Integral(x**m*(a + b/x)**n/(c + d*x)**2, x)

Maxima [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate((a+b/x)^n*x^m/(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((a + b/x)^n*x^m/(d*x + c)^2, x)

Giac [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int { \frac {{\left (a + \frac {b}{x}\right )}^{n} x^{m}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

integrate((a+b/x)^n*x^m/(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((a + b/x)^n*x^m/(d*x + c)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2} \, dx=\int \frac {x^m\,{\left (a+\frac {b}{x}\right )}^n}{{\left (c+d\,x\right )}^2} \,d x \]

[In]

int((x^m*(a + b/x)^n)/(c + d*x)^2,x)

[Out]

int((x^m*(a + b/x)^n)/(c + d*x)^2, x)