\(\int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx\) [3117]

Optimal result

Integrand size = 26, antiderivative size = 190 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {(a d f (1-m)-b (d e-c f m)) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^2 (b e-a f) m}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,1+m,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {132, 72, 71, 156, 12, 133} \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=\frac {(a+b x)^m (c+d x)^{-m} (a d f (1-m)-b (d e-c f m)) \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^2 m (b e-a f)}-\frac {(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac {d (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,m+1,-\frac {d (a+b x)}{b c-a d}\right )}{f^2 m} \]

[In]

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

[Out]

-(((a + b*x)^m*(c + d*x)^(1 - m))/(f*(e + f*x))) + ((a*d*f*(1 - m) - b*(d*e - c*f*m))*(a + b*x)^m*Hypergeometr
ic2F1[1, m, 1 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(f^2*(b*e - a*f)*m*(c + d*x)^m) + (d*(a +
b*x)^m*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, m, 1 + m, -((d*(a + b*x))/(b*c - a*d))])/(f^2*m*(c
+ d*x)^m)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 132

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

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.26 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-m} \left (d (e+f x) \left (\frac {b (e+f x)}{b e-a f}\right )^m \operatorname {AppellF1}\left (1+m,m,1,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )+(-d e+c f) \operatorname {Hypergeometric2F1}\left (m,1+m,2+m,\frac {(-d e+c f) (a+b x)}{(b c-a d) (e+f x)}\right )\right )}{f (-b e+a f) (1+m) (e+f x)} \]

[In]

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

[Out]

-(((a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(d*(e + f*x)*((b*(e + f*x))/(b*e - a*f))^m*AppellF1[1 + m,
m, 1, 2 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)] + (-(d*e) + c*f)*Hypergeometric2F1[m,
1 + m, 2 + m, ((-(d*e) + c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))]))/(f*(-(b*e) + a*f)*(1 + m)*(c + d*x)^m*(e
+ f*x)*((b*(e + f*x))/(b*e - a*f))^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(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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

[In]

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

[Out]

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