\(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx\) [3076]

Optimal result

Integrand size = 26, antiderivative size = 158 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f) m (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {98, 133} \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+1)-b (c f m+d e)) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{m (m+1) (b e-a f)^2 (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x) (b c-a d) (d e-c f)} \]

[In]

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

[Out]

(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)*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*(d*e - c*f)*m*(1 + m))

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[(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] || Sum
SimplerQ[m, 1])

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> 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] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{(b c-a d) (d e-c f) m} \\ & = \frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)}+\frac {(a d f (1+m)-b (d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\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 (d e-c f) m (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (-\frac {d}{e+f x}+\frac {(b c-a d) (-a d f (1+m)+b (d e+c f m)) \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (1+m) (c+d x)}\right )}{(b c-a d) (-d e+c f) m} \]

[In]

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

[Out]

((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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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