\(\int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx\) [1766]

Optimal result

Integrand size = 26, antiderivative size = 541 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{2-m}}{5 (b e-a f) (d e-c f) (e+f x)^5}-\frac {f (7 b d e-b c f (4-m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{20 (b e-a f)^2 (d e-c f)^2 (e+f x)^4}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (3 d e (7+4 m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (27 d^2 e^2-3 c d e f (11-4 m)+c^2 f^2 \left (12-7 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{2-m}}{60 (b e-a f)^3 (d e-c f)^3 (e+f x)^3}+\frac {(b c-a d)^2 \left (3 a^2 b d^2 f^2 (5 d e-c f (2-m)) \left (2+3 m+m^2\right )-a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-3 a b^2 d f (1+m) \left (20 d^2 e^2-10 c d e f (2-m)+c^2 f^2 \left (6-5 m+m^2\right )\right )+b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (2-m)+15 c^2 d e f^2 \left (6-5 m+m^2\right )-c^3 f^3 \left (24-26 m+9 m^2-m^3\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{60 (b e-a f)^6 (d e-c f)^3 (1+m)} \] Output:

-1/5*f*(b*x+a)^(1+m)*(d*x+c)^(2-m)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)^5-1/20*f* 
(7*b*d*e-b*c*f*(4-m)-a*d*f*(3+m))*(b*x+a)^(1+m)*(d*x+c)^(2-m)/(-a*f+b*e)^2 
/(-c*f+d*e)^2/(f*x+e)^4-1/60*f*(a^2*d^2*f^2*(m^2+5*m+6)-a*b*d*f*(3*d*e*(7+ 
4*m)-c*f*(-2*m^2+2*m+9))+b^2*(27*d^2*e^2-3*c*d*e*f*(11-4*m)+c^2*f^2*(m^2-7 
*m+12)))*(b*x+a)^(1+m)*(d*x+c)^(2-m)/(-a*f+b*e)^3/(-c*f+d*e)^3/(f*x+e)^3+1 
/60*(-a*d+b*c)^2*(3*a^2*b*d^2*f^2*(5*d*e-c*f*(2-m))*(m^2+3*m+2)-a^3*d^3*f^ 
3*(m^3+6*m^2+11*m+6)-3*a*b^2*d*f*(1+m)*(20*d^2*e^2-10*c*d*e*f*(2-m)+c^2*f^ 
2*(m^2-5*m+6))+b^3*(60*d^3*e^3-60*c*d^2*e^2*f*(2-m)+15*c^2*d*e*f^2*(m^2-5* 
m+6)-c^3*f^3*(-m^3+9*m^2-26*m+24)))*(b*x+a)^(1+m)*(d*x+c)^(-1-m)*hypergeom 
([3, 1+m],[2+m],(-c*f+d*e)*(b*x+a)/(-a*f+b*e)/(d*x+c))/(-a*f+b*e)^6/(-c*f+ 
d*e)^3/(1+m)
 

Mathematica [A] (verified)

Time = 1.88 (sec) , antiderivative size = 484, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (-12 f (c+d x)^3+\frac {3 f (-b (7 d e+c f (-4+m))+a d f (3+m)) (c+d x)^3 (e+f x)}{(b e-a f) (d e-c f)}-\frac {(e+f x)^2 \left (f (b e-a f)^3 (1+m) \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )+a b d f \left (-3 d e (7+4 m)+c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (27 d^2 e^2+3 c d e f (-11+4 m)+c^2 f^2 \left (12-7 m+m^2\right )\right )\right ) (c+d x)^3-(b c-a d)^2 \left (3 a^2 b d^2 f^2 (5 d e+c f (-2+m)) \left (2+3 m+m^2\right )-a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-3 a b^2 d f (1+m) \left (20 d^2 e^2+10 c d e f (-2+m)+c^2 f^2 \left (6-5 m+m^2\right )\right )+b^3 \left (60 d^3 e^3+60 c d^2 e^2 f (-2+m)+15 c^2 d e f^2 \left (6-5 m+m^2\right )+c^3 f^3 \left (-24+26 m-9 m^2+m^3\right )\right )\right ) (e+f x)^3 \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(b e-a f)^5 (d e-c f)^2 (1+m)}\right )}{60 (b e-a f) (d e-c f) (e+f x)^5} \] Input:

Integrate[((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^6,x]
 

Output:

((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*(-12*f*(c + d*x)^3 + (3*f*(-(b*(7*d* 
e + c*f*(-4 + m))) + a*d*f*(3 + m))*(c + d*x)^3*(e + f*x))/((b*e - a*f)*(d 
*e - c*f)) - ((e + f*x)^2*(f*(b*e - a*f)^3*(1 + m)*(a^2*d^2*f^2*(6 + 5*m + 
 m^2) + a*b*d*f*(-3*d*e*(7 + 4*m) + c*f*(9 + 2*m - 2*m^2)) + b^2*(27*d^2*e 
^2 + 3*c*d*e*f*(-11 + 4*m) + c^2*f^2*(12 - 7*m + m^2)))*(c + d*x)^3 - (b*c 
 - a*d)^2*(3*a^2*b*d^2*f^2*(5*d*e + c*f*(-2 + m))*(2 + 3*m + m^2) - a^3*d^ 
3*f^3*(6 + 11*m + 6*m^2 + m^3) - 3*a*b^2*d*f*(1 + m)*(20*d^2*e^2 + 10*c*d* 
e*f*(-2 + m) + c^2*f^2*(6 - 5*m + m^2)) + b^3*(60*d^3*e^3 + 60*c*d^2*e^2*f 
*(-2 + m) + 15*c^2*d*e*f^2*(6 - 5*m + m^2) + c^3*f^3*(-24 + 26*m - 9*m^2 + 
 m^3)))*(e + f*x)^3*Hypergeometric2F1[3, 1 + m, 2 + m, ((d*e - c*f)*(a + b 
*x))/((b*e - a*f)*(c + d*x))]))/((b*e - a*f)^5*(d*e - c*f)^2*(1 + m))))/(6 
0*(b*e - a*f)*(d*e - c*f)*(e + f*x)^5)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 591, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {114, 25, 168, 25, 168, 25, 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.

\(\displaystyle \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx\)

\(\Big \downarrow \) 114

\(\displaystyle -\frac {\int -\frac {(a+b x)^m (c+d x)^{1-m} (b (5 d e-c f (4-m))-a d f (m+3)-2 b d f x)}{(e+f x)^5}dx}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(a+b x)^m (c+d x)^{1-m} (5 b d e-b c f (4-m)-a d f (m+3)-2 b d f x)}{(e+f x)^5}dx}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {-\frac {\int -\frac {(a+b x)^m (c+d x)^{1-m} \left (\left (20 d^2 e^2-c d f (29-11 m) e+c^2 f^2 \left (m^2-7 m+12\right )\right ) b^2-a d f \left (d e (11 m+18)-c f \left (-2 m^2+2 m+9\right )\right ) b-d f (7 b d e-b c f (4-m)-a d f (m+3)) x b+a^2 d^2 f^2 \left (m^2+5 m+6\right )\right )}{(e+f x)^4}dx}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {(a+b x)^m (c+d x)^{1-m} \left (\left (20 d^2 e^2-c d f (29-11 m) e+c^2 f^2 \left (m^2-7 m+12\right )\right ) b^2-a d f \left (d e (11 m+18)-c f \left (-2 m^2+2 m+9\right )\right ) b-d f (7 b d e-b c f (4-m)-a d f (m+3)) x b+a^2 d^2 f^2 \left (m^2+5 m+6\right )\right )}{(e+f x)^4}dx}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\frac {-\frac {\int -\frac {\left (\left (60 d^3 e^3-60 c d^2 f (2-m) e^2+15 c^2 d f^2 \left (m^2-5 m+6\right ) e-c^3 f^3 \left (-m^3+9 m^2-26 m+24\right )\right ) b^3-3 a d f (m+1) \left (20 d^2 e^2-10 c d f (2-m) e+c^2 f^2 \left (m^2-5 m+6\right )\right ) b^2+3 a^2 d^2 f^2 (5 d e-c f (2-m)) \left (m^2+3 m+2\right ) b-a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )\right ) (a+b x)^m (c+d x)^{1-m}}{(e+f x)^3}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (3 d e (4 m+7)-c f \left (-2 m^2+2 m+9\right )\right )+b^2 \left (c^2 f^2 \left (m^2-7 m+12\right )-3 c d e f (11-4 m)+27 d^2 e^2\right )\right )}{3 (e+f x)^3 (b e-a f) (d e-c f)}}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {\left (\left (60 d^3 e^3-60 c d^2 f (2-m) e^2+15 c^2 d f^2 \left (m^2-5 m+6\right ) e-c^3 f^3 \left (-m^3+9 m^2-26 m+24\right )\right ) b^3-3 a d f (m+1) \left (20 d^2 e^2-10 c d f (2-m) e+c^2 f^2 \left (m^2-5 m+6\right )\right ) b^2+3 a^2 d^2 f^2 (5 d e-c f (2-m)) \left (m^2+3 m+2\right ) b-a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )\right ) (a+b x)^m (c+d x)^{1-m}}{(e+f x)^3}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (3 d e (4 m+7)-c f \left (-2 m^2+2 m+9\right )\right )+b^2 \left (c^2 f^2 \left (m^2-7 m+12\right )-3 c d e f (11-4 m)+27 d^2 e^2\right )\right )}{3 (e+f x)^3 (b e-a f) (d e-c f)}}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\left (-a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )+3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (5 d e-c f (2-m))-3 a b^2 d f (m+1) \left (c^2 f^2 \left (m^2-5 m+6\right )-10 c d e f (2-m)+20 d^2 e^2\right )+b^3 \left (-c^3 f^3 \left (-m^3+9 m^2-26 m+24\right )+15 c^2 d e f^2 \left (m^2-5 m+6\right )-60 c d^2 e^2 f (2-m)+60 d^3 e^3\right )\right ) \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^3}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (3 d e (4 m+7)-c f \left (-2 m^2+2 m+9\right )\right )+b^2 \left (c^2 f^2 \left (m^2-7 m+12\right )-3 c d e f (11-4 m)+27 d^2 e^2\right )\right )}{3 (e+f x)^3 (b e-a f) (d e-c f)}}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

\(\Big \downarrow \) 141

\(\displaystyle \frac {\frac {\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )+3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (5 d e-c f (2-m))-3 a b^2 d f (m+1) \left (c^2 f^2 \left (m^2-5 m+6\right )-10 c d e f (2-m)+20 d^2 e^2\right )+b^3 \left (-c^3 f^3 \left (-m^3+9 m^2-26 m+24\right )+15 c^2 d e f^2 \left (m^2-5 m+6\right )-60 c d^2 e^2 f (2-m)+60 d^3 e^3\right )\right ) \operatorname {Hypergeometric2F1}\left (3,m+1,m+2,\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{3 (m+1) (b e-a f)^4 (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (3 d e (4 m+7)-c f \left (-2 m^2+2 m+9\right )\right )+b^2 \left (c^2 f^2 \left (m^2-7 m+12\right )-3 c d e f (11-4 m)+27 d^2 e^2\right )\right )}{3 (e+f x)^3 (b e-a f) (d e-c f)}}{4 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (m+3)-b c f (4-m)+7 b d e)}{4 (e+f x)^4 (b e-a f) (d e-c f)}}{5 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{5 (e+f x)^5 (b e-a f) (d e-c f)}\)

Input:

Int[((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^6,x]
 

Output:

-1/5*(f*(a + b*x)^(1 + m)*(c + d*x)^(2 - m))/((b*e - a*f)*(d*e - c*f)*(e + 
 f*x)^5) + (-1/4*(f*(7*b*d*e - b*c*f*(4 - m) - a*d*f*(3 + m))*(a + b*x)^(1 
 + m)*(c + d*x)^(2 - m))/((b*e - a*f)*(d*e - c*f)*(e + f*x)^4) + (-1/3*(f* 
(a^2*d^2*f^2*(6 + 5*m + m^2) - a*b*d*f*(3*d*e*(7 + 4*m) - c*f*(9 + 2*m - 2 
*m^2)) + b^2*(27*d^2*e^2 - 3*c*d*e*f*(11 - 4*m) + c^2*f^2*(12 - 7*m + m^2) 
))*(a + b*x)^(1 + m)*(c + d*x)^(2 - m))/((b*e - a*f)*(d*e - c*f)*(e + f*x) 
^3) + ((b*c - a*d)^2*(3*a^2*b*d^2*f^2*(5*d*e - c*f*(2 - m))*(2 + 3*m + m^2 
) - a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) - 3*a*b^2*d*f*(1 + m)*(20*d^2*e^2 
 - 10*c*d*e*f*(2 - m) + c^2*f^2*(6 - 5*m + m^2)) + b^3*(60*d^3*e^3 - 60*c* 
d^2*e^2*f*(2 - m) + 15*c^2*d*e*f^2*(6 - 5*m + m^2) - c^3*f^3*(24 - 26*m + 
9*m^2 - m^3)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[3, 1 
 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(3*(b*e - a 
*f)^4*(d*e - c*f)*(1 + m)))/(4*(b*e - a*f)*(d*e - c*f)))/(5*(b*e - a*f)*(d 
*e - c*f))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S 
imp[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]
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (x d +c \right )^{-m +1}}{\left (f x +e \right )^{6}}d x\]

Input:

int((b*x+a)^m*(d*x+c)^(-m+1)/(f*x+e)^6,x)
 

Output:

int((b*x+a)^m*(d*x+c)^(-m+1)/(f*x+e)^6,x)
 

Fricas [F]

\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{6}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^6,x, algorithm="fricas")
 

Output:

integral((b*x + a)^m*(d*x + c)^(-m + 1)/(f^6*x^6 + 6*e*f^5*x^5 + 15*e^2*f^ 
4*x^4 + 20*e^3*f^3*x^3 + 15*e^4*f^2*x^2 + 6*e^5*f*x + e^6), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\text {Timed out} \] Input:

integrate((b*x+a)**m*(d*x+c)**(1-m)/(f*x+e)**6,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{6}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^6,x, algorithm="maxima")
 

Output:

integrate((b*x + a)^m*(d*x + c)^(-m + 1)/(f*x + e)^6, x)
 

Giac [F]

\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}}{{\left (f x + e\right )}^{6}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^6,x, algorithm="giac")
 

Output:

integrate((b*x + a)^m*(d*x + c)^(-m + 1)/(f*x + e)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m}}{{\left (e+f\,x\right )}^6} \,d x \] Input:

int(((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^6,x)
 

Output:

int(((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^6, x)
 

Reduce [F]

\[ \int \frac {(a+b x)^m (c+d x)^{1-m}}{(e+f x)^6} \, dx=\left (\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{m} e^{6}+6 \left (d x +c \right )^{m} e^{5} f x +15 \left (d x +c \right )^{m} e^{4} f^{2} x^{2}+20 \left (d x +c \right )^{m} e^{3} f^{3} x^{3}+15 \left (d x +c \right )^{m} e^{2} f^{4} x^{4}+6 \left (d x +c \right )^{m} e \,f^{5} x^{5}+\left (d x +c \right )^{m} f^{6} x^{6}}d x \right ) c +\left (\int \frac {\left (b x +a \right )^{m} x}{\left (d x +c \right )^{m} e^{6}+6 \left (d x +c \right )^{m} e^{5} f x +15 \left (d x +c \right )^{m} e^{4} f^{2} x^{2}+20 \left (d x +c \right )^{m} e^{3} f^{3} x^{3}+15 \left (d x +c \right )^{m} e^{2} f^{4} x^{4}+6 \left (d x +c \right )^{m} e \,f^{5} x^{5}+\left (d x +c \right )^{m} f^{6} x^{6}}d x \right ) d \] Input:

int((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^6,x)
 

Output:

int((a + b*x)**m/((c + d*x)**m*e**6 + 6*(c + d*x)**m*e**5*f*x + 15*(c + d* 
x)**m*e**4*f**2*x**2 + 20*(c + d*x)**m*e**3*f**3*x**3 + 15*(c + d*x)**m*e* 
*2*f**4*x**4 + 6*(c + d*x)**m*e*f**5*x**5 + (c + d*x)**m*f**6*x**6),x)*c + 
 int(((a + b*x)**m*x)/((c + d*x)**m*e**6 + 6*(c + d*x)**m*e**5*f*x + 15*(c 
 + d*x)**m*e**4*f**2*x**2 + 20*(c + d*x)**m*e**3*f**3*x**3 + 15*(c + d*x)* 
*m*e**2*f**4*x**4 + 6*(c + d*x)**m*e*f**5*x**5 + (c + d*x)**m*f**6*x**6),x 
)*d