\(\int \frac {(a+\frac {b}{x})^n (e x)^m}{(c+d x)^2} \, dx\) [298]

Optimal result

Integrand size = 22, antiderivative size = 76 \[ \int \frac {\left (a+\frac {b}{x}\right )^n (e x)^m}{(c+d x)^2} \, dx=-\frac {\left (a+\frac {b}{x}\right )^n \left (1+\frac {b}{a x}\right )^{-n} (e x)^m \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m) x} \] Output:

-(a+b/x)^n*(e*x)^m*AppellF1(1-m,-n,2,2-m,-b/a/x,-c/d/x)/d^2/(1-m)/((1+b/a/ 
x)^n)/x
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1018, 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.

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

\(\Big \downarrow \) 1018

\(\displaystyle x^{-m} (e x)^m \int \frac {\left (a+\frac {b}{x}\right )^n x^m}{(c+d x)^2}dx\)

\(\Big \downarrow \) 1016

\(\displaystyle x^{-m} (e x)^m \int \frac {\left (a+\frac {b}{x}\right )^n x^{m-2}}{\left (\frac {c}{x}+d\right )^2}dx\)

\(\Big \downarrow \) 999

\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \int \frac {\left (a+\frac {b}{x}\right )^n \left (\frac {1}{x}\right )^{-m}}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}\)

\(\Big \downarrow \) 152

\(\displaystyle \left (\frac {1}{x}\right )^m (e x)^m \left (-\left (a+\frac {b}{x}\right )^n\right ) \left (\frac {b}{a x}+1\right )^{-n} \int \frac {\left (\frac {b}{a x}+1\right )^n \left (\frac {1}{x}\right )^{-m}}{\left (\frac {c}{x}+d\right )^2}d\frac {1}{x}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {(e x)^m \left (a+\frac {b}{x}\right )^n \left (\frac {b}{a x}+1\right )^{-n} \operatorname {AppellF1}\left (1-m,-n,2,2-m,-\frac {b}{a x},-\frac {c}{d x}\right )}{d^2 (1-m) x}\)

Input:

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

Output:

-(((a + b/x)^n*(e*x)^m*AppellF1[1 - m, -n, 2, 2 - m, -(b/(a*x)), -(c/(d*x) 
)])/(d^2*(1 - m)*(1 + b/(a*x))^n*x))
 

Defintions of rubi rules used

rule 150

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])
 

rule 152

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]
 

rule 999

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]
 

rule 1016

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])
 

rule 1018

Int[((e_)*(x_))^(m_)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n 
_.))^(p_.), x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m] 
)   Int[x^m*(a + b*x^n)^p*(c + d/x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p, q}, x] && EqQ[mn, -n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+\frac {b}{x}\right )^n (e x)^m}{(c+d x)^2} \, dx=\text {too large to display} \] Input:

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

Output:

(e**m*(x**m*(a*x + b)**n*b + x**n*int((x**m*(a*x + b)**n*x)/(x**n*a**2*c** 
3*m*x + 2*x**n*a**2*c**2*d*m*x**2 + x**n*a**2*c*d**2*m*x**3 + x**n*a*b*c** 
3*m + 3*x**n*a*b*c**2*d*m*x - x**n*a*b*c**2*d*n*x - x**n*a*b*c**2*d*x + 3* 
x**n*a*b*c*d**2*m*x**2 - 2*x**n*a*b*c*d**2*n*x**2 - 2*x**n*a*b*c*d**2*x**2 
 + x**n*a*b*d**3*m*x**3 - x**n*a*b*d**3*n*x**3 - x**n*a*b*d**3*x**3 + x**n 
*b**2*c**2*d*m - x**n*b**2*c**2*d*n - x**n*b**2*c**2*d + 2*x**n*b**2*c*d** 
2*m*x - 2*x**n*b**2*c*d**2*n*x - 2*x**n*b**2*c*d**2*x + x**n*b**2*d**3*m*x 
**2 - x**n*b**2*d**3*n*x**2 - x**n*b**2*d**3*x**2),x)*a**3*c**3*m**2 + x** 
n*int((x**m*(a*x + b)**n*x)/(x**n*a**2*c**3*m*x + 2*x**n*a**2*c**2*d*m*x** 
2 + x**n*a**2*c*d**2*m*x**3 + x**n*a*b*c**3*m + 3*x**n*a*b*c**2*d*m*x - x* 
*n*a*b*c**2*d*n*x - x**n*a*b*c**2*d*x + 3*x**n*a*b*c*d**2*m*x**2 - 2*x**n* 
a*b*c*d**2*n*x**2 - 2*x**n*a*b*c*d**2*x**2 + x**n*a*b*d**3*m*x**3 - x**n*a 
*b*d**3*n*x**3 - x**n*a*b*d**3*x**3 + x**n*b**2*c**2*d*m - x**n*b**2*c**2* 
d*n - x**n*b**2*c**2*d + 2*x**n*b**2*c*d**2*m*x - 2*x**n*b**2*c*d**2*n*x - 
 2*x**n*b**2*c*d**2*x + x**n*b**2*d**3*m*x**2 - x**n*b**2*d**3*n*x**2 - x* 
*n*b**2*d**3*x**2),x)*a**3*c**2*d*m**2*x + x**n*int((x**m*(a*x + b)**n*x)/ 
(x**n*a**2*c**3*m*x + 2*x**n*a**2*c**2*d*m*x**2 + x**n*a**2*c*d**2*m*x**3 
+ x**n*a*b*c**3*m + 3*x**n*a*b*c**2*d*m*x - x**n*a*b*c**2*d*n*x - x**n*a*b 
*c**2*d*x + 3*x**n*a*b*c*d**2*m*x**2 - 2*x**n*a*b*c*d**2*n*x**2 - 2*x**n*a 
*b*c*d**2*x**2 + x**n*a*b*d**3*m*x**3 - x**n*a*b*d**3*n*x**3 - x**n*a*b...