\(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx\) [1726]

Optimal result

Integrand size = 26, antiderivative size = 149 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{(b e-a f) (d e-c f)^2 m} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (-\frac {d}{e+f x}+\frac {(b c-a d) (-a d f (1+m)+b (d e+c f m)) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (1+m) (c+d x)}\right )}{(b c-a d) (-d e+c f) m} \] Input:

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

Output:

((a + b*x)^(1 + m)*(-(d/(e + f*x)) + ((b*c - a*d)*(-(a*d*f*(1 + m)) + b*(d 
*e + c*f*m))*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/(( 
b*e - a*f)*(c + d*x))])/((b*e - a*f)^2*(1 + m)*(c + d*x))))/((b*c - a*d)*( 
-(d*e) + c*f)*m*(c + d*x)^m)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {107, 141}

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.

\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m-1}}{(e+f x)^2} \, dx\)

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 141

\(\displaystyle \frac {(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+1)-b (c f m+d e)) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{m (m+1) (b e-a f)^2 (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x) (b c-a d) (d e-c f)}\)

Input:

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

Output:

(d*(a + b*x)^(1 + m))/((b*c - a*d)*(d*e - c*f)*m*(c + d*x)^m*(e + f*x)) + 
((a*d*f*(1 + m) - b*(d*e + c*f*m))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hy 
pergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + 
 d*x))])/((b*e - a*f)^2*(d*e - c*f)*m*(1 + m))
 

Defintions of rubi rules used

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 141

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( 
n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f 
))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, 
p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !Su 
mSimplerQ[p, 1]) &&  !ILtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\text {Timed out} \] Input:

integrate((b*x+a)**m*(d*x+c)**(-1-m)/(f*x+e)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^{m+1}} \,d x \] Input:

int((a + b*x)^m/((e + f*x)^2*(c + d*x)^(m + 1)),x)
 

Output:

int((a + b*x)^m/((e + f*x)^2*(c + d*x)^(m + 1)), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{m} c \,e^{2}+2 \left (d x +c \right )^{m} c e f x +\left (d x +c \right )^{m} c \,f^{2} x^{2}+\left (d x +c \right )^{m} d \,e^{2} x +2 \left (d x +c \right )^{m} d e f \,x^{2}+\left (d x +c \right )^{m} d \,f^{2} x^{3}}d x \] Input:

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

Output:

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