3.31.94 \(\int \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx\) [3094]

3.31.94.1 Optimal result

Integrand size = 26, antiderivative size = 384 \[ \int \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx=-\frac {d (a d f (3+m)-b (d e+c f (2+m))) (a+b x)^{1+m} (c+d x)^{-2-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (2+m)}-\frac {d \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )-a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d)^2 (b e-a f) (d e-c f)^3 (1+m) (2+m)}-\frac {f (a+b x)^{1+m} (c+d x)^{-2-m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {f^2 (a d f (3+m)-b (3 d e+c f m)) (a+b x)^m (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{(b e-a f) (d e-c f)^4 m} \]

output

-d*(a*d*f*(3+m)-b*(d*e+c*f*(2+m)))*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/(-a*d+b*c) 
/(-a*f+b*e)/(-c*f+d*e)^2/(2+m)-d*(a^2*d^2*f^2*(m^2+5*m+6)-b^2*(d^2*e^2-c*d 
*e*f*(5+2*m)-c^2*f^2*(m^2+3*m+2))-a*b*d*f*(d*e*(3+2*m)+c*f*(2*m^2+8*m+9))) 
*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/(-a*d+b*c)^2/(-a*f+b*e)/(-c*f+d*e)^3/(1+m)/( 
2+m)-f*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)+f^2*(a*d 
*f*(3+m)-b*(c*f*m+3*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)^4/m/((d*x+c)^m)
 

3.31.94.2 Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 333, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx=-\frac {(a+b x)^{1+m} (c+d x)^{-2-m} \left (\frac {d}{e+f x}+\frac {f (b d e+b c f (2+m)-a d f (3+m)) (c+d x)}{(b e-a f) (d e-c f) (e+f x)}+\frac {(c+d x) \left (d (b e-a f) (1+m) \left (-a^2 d^2 f^2 \left (6+5 m+m^2\right )+b^2 \left (d^2 e^2-c d e f (5+2 m)-c^2 f^2 \left (2+3 m+m^2\right )\right )+a b d f \left (d e (3+2 m)+c f \left (9+8 m+2 m^2\right )\right )\right )+(b c-a d)^2 f^2 \left (2+3 m+m^2\right ) (-a d f (3+m)+b (3 d e+c f m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(b c-a d) (b e-a f)^2 (d e-c f)^2 (1+m)^2}\right )}{(b c-a d) (-d e+c f) (2+m)} \]

input

Integrate[((a + b*x)^m*(c + d*x)^(-3 - m))/(e + f*x)^2,x]
 

output

-(((a + b*x)^(1 + m)*(c + d*x)^(-2 - m)*(d/(e + f*x) + (f*(b*d*e + b*c*f*( 
2 + m) - a*d*f*(3 + m))*(c + d*x))/((b*e - a*f)*(d*e - c*f)*(e + f*x)) + ( 
(c + d*x)*(d*(b*e - a*f)*(1 + m)*(-(a^2*d^2*f^2*(6 + 5*m + m^2)) + b^2*(d^ 
2*e^2 - c*d*e*f*(5 + 2*m) - c^2*f^2*(2 + 3*m + m^2)) + a*b*d*f*(d*e*(3 + 2 
*m) + c*f*(9 + 8*m + 2*m^2))) + (b*c - a*d)^2*f^2*(2 + 3*m + m^2)*(-(a*d*f 
*(3 + m)) + b*(3*d*e + c*f*m))*Hypergeometric2F1[1, 1 + m, 2 + m, ((d*e - 
c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/((b*c - a*d)*(b*e - a*f)^2*(d*e 
 - c*f)^2*(1 + m)^2)))/((b*c - a*d)*(-(d*e) + c*f)*(2 + m)))
 

3.31.94.3 Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {114, 172, 172, 27, 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.

\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{-m-3}}{(e+f x)^2} \, dx\)

\(\Big \downarrow \) 114

\(\displaystyle -\frac {\int \frac {(a+b x)^m (c+d x)^{-m-3} (a d f (m+3)-b (d e+c f m)+2 b d f x)}{e+f x}dx}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 172

\(\displaystyle -\frac {\frac {\int \frac {(a+b x)^m (c+d x)^{-m-2} \left (-\left (\left (d^2 e^2-2 c d f (m+2) e-c^2 f^2 m (m+2)\right ) b^2\right )-a d f (2 m+3) (d e+c f (m+2)) b+d f (a d f (m+3)-b (d e+c f (m+2))) x b+a^2 d^2 f^2 \left (m^2+5 m+6\right )\right )}{e+f x}dx}{(m+2) (b c-a d) (d e-c f)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (b c-a d) (d e-c f)}}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 172

\(\displaystyle -\frac {\frac {\frac {\int \frac {(b c-a d)^2 f^2 \left (m^2+3 m+2\right ) (a d f (m+3)-b (3 d e+c f m)) (a+b x)^m (c+d x)^{-m-1}}{e+f x}dx}{(m+1) (b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )-\left (b^2 \left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )\right )}{(m+1) (b c-a d) (d e-c f)}}{(m+2) (b c-a d) (d e-c f)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (b c-a d) (d e-c f)}}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {f^2 \left (m^2+3 m+2\right ) (b c-a d) (a d f (m+3)-b (c f m+3 d e)) \int \frac {(a+b x)^m (c+d x)^{-m-1}}{e+f x}dx}{(m+1) (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )-\left (b^2 \left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )\right )}{(m+1) (b c-a d) (d e-c f)}}{(m+2) (b c-a d) (d e-c f)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (b c-a d) (d e-c f)}}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 141

\(\displaystyle -\frac {\frac {\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (c f \left (2 m^2+8 m+9\right )+d e (2 m+3)\right )-\left (b^2 \left (-c^2 f^2 \left (m^2+3 m+2\right )-c d e f (2 m+5)+d^2 e^2\right )\right )\right )}{(m+1) (b c-a d) (d e-c f)}-\frac {f^2 \left (m^2+3 m+2\right ) (b c-a d) (a+b x)^m (c+d x)^{-m} (a d f (m+3)-b (c f m+3 d e)) \operatorname {Hypergeometric2F1}\left (1,-m,1-m,\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (m+1) (d e-c f)^2}}{(m+2) (b c-a d) (d e-c f)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+2)+b d e)}{(m+2) (b c-a d) (d e-c f)}}{(b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{-m-2}}{(e+f x) (b e-a f) (d e-c f)}\)

input

Int[((a + b*x)^m*(c + d*x)^(-3 - m))/(e + f*x)^2,x]
 

output

-((f*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*e - a*f)*(d*e - c*f)*(e + f 
*x))) - (-((d*(b*d*e + b*c*f*(2 + m) - a*d*f*(3 + m))*(a + b*x)^(1 + m)*(c 
 + d*x)^(-2 - m))/((b*c - a*d)*(d*e - c*f)*(2 + m))) + ((d*(a^2*d^2*f^2*(6 
 + 5*m + m^2) - b^2*(d^2*e^2 - c*d*e*f*(5 + 2*m) - c^2*f^2*(2 + 3*m + m^2) 
) - a*b*d*f*(d*e*(3 + 2*m) + c*f*(9 + 8*m + 2*m^2)))*(a + b*x)^(1 + m)*(c 
+ d*x)^(-1 - m))/((b*c - a*d)*(d*e - c*f)*(1 + m)) - ((b*c - a*d)*f^2*(2 + 
 3*m + m^2)*(a*d*f*(3 + m) - b*(3*d*e + c*f*m))*(a + b*x)^m*Hypergeometric 
2F1[1, -m, 1 - m, ((b*e - a*f)*(c + d*x))/((d*e - c*f)*(a + b*x))])/((d*e 
- c*f)^2*m*(1 + m)*(c + d*x)^m))/((b*c - a*d)*(d*e - c*f)*(2 + m)))/((b*e 
- a*f)*(d*e - c*f))
 

3.31.94.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 114

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[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*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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]
 

rule 172

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> With[{mnp = Simplify[m + n + p]}, 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] + Simp[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)*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1 
])))] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && NeQ[m, -1]
 

3.31.94.4 Maple [F]

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

input

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

output

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

3.31.94.5 Fricas [F]

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

input

integrate((b*x+a)^m*(d*x+c)^(-3-m)/(f*x+e)^2,x, algorithm="fricas")
 

output

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

3.31.94.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{-3-m}}{(e+f x)^2} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.31.94.7 Maxima [F]

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

input

integrate((b*x+a)^m*(d*x+c)^(-3-m)/(f*x+e)^2,x, algorithm="maxima")
 

output

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

3.31.94.8 Giac [F]

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

input

integrate((b*x+a)^m*(d*x+c)^(-3-m)/(f*x+e)^2,x, algorithm="giac")
 

output

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

3.31.94.9 Mupad [F(-1)]

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

input

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

output

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