3.2.53 \(\int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx\) [153]

3.2.53.1 Optimal result

Integrand size = 16, antiderivative size = 218 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=-\frac {a^{10} A}{5 x^5}-\frac {a^9 (10 A b+a B)}{4 x^4}-\frac {5 a^8 b (9 A b+2 a B)}{3 x^3}-\frac {15 a^7 b^2 (8 A b+3 a B)}{2 x^2}-\frac {30 a^6 b^3 (7 A b+4 a B)}{x}+42 a^4 b^5 (5 A b+6 a B) x+15 a^3 b^6 (4 A b+7 a B) x^2+5 a^2 b^7 (3 A b+8 a B) x^3+\frac {5}{4} a b^8 (2 A b+9 a B) x^4+\frac {1}{5} b^9 (A b+10 a B) x^5+\frac {1}{6} b^{10} B x^6+42 a^5 b^4 (6 A b+5 a B) \log (x) \]

output

-1/5*a^10*A/x^5-1/4*a^9*(10*A*b+B*a)/x^4-5/3*a^8*b*(9*A*b+2*B*a)/x^3-15/2* 
a^7*b^2*(8*A*b+3*B*a)/x^2-30*a^6*b^3*(7*A*b+4*B*a)/x+42*a^4*b^5*(5*A*b+6*B 
*a)*x+15*a^3*b^6*(4*A*b+7*B*a)*x^2+5*a^2*b^7*(3*A*b+8*B*a)*x^3+5/4*a*b^8*( 
2*A*b+9*B*a)*x^4+1/5*b^9*(A*b+10*B*a)*x^5+1/6*b^10*B*x^6+42*a^5*b^4*(6*A*b 
+5*B*a)*ln(x)
 

3.2.53.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=-\frac {210 a^6 A b^4}{x}+252 a^5 b^5 B x+105 a^4 b^6 x (2 A+B x)-\frac {60 a^7 b^3 (A+2 B x)}{x^2}+20 a^3 b^7 x^2 (3 A+2 B x)-\frac {15 a^8 b^2 (2 A+3 B x)}{2 x^3}+\frac {15}{4} a^2 b^8 x^3 (4 A+3 B x)-\frac {5 a^9 b (3 A+4 B x)}{6 x^4}+\frac {1}{2} a b^9 x^4 (5 A+4 B x)-\frac {a^{10} (4 A+5 B x)}{20 x^5}+\frac {1}{30} b^{10} x^5 (6 A+5 B x)+42 a^5 b^4 (6 A b+5 a B) \log (x) \]

input

Integrate[((a + b*x)^10*(A + B*x))/x^6,x]
 

output

(-210*a^6*A*b^4)/x + 252*a^5*b^5*B*x + 105*a^4*b^6*x*(2*A + B*x) - (60*a^7 
*b^3*(A + 2*B*x))/x^2 + 20*a^3*b^7*x^2*(3*A + 2*B*x) - (15*a^8*b^2*(2*A + 
3*B*x))/(2*x^3) + (15*a^2*b^8*x^3*(4*A + 3*B*x))/4 - (5*a^9*b*(3*A + 4*B*x 
))/(6*x^4) + (a*b^9*x^4*(5*A + 4*B*x))/2 - (a^10*(4*A + 5*B*x))/(20*x^5) + 
 (b^10*x^5*(6*A + 5*B*x))/30 + 42*a^5*b^4*(6*A*b + 5*a*B)*Log[x]
 

3.2.53.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {85, 2009}

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.

\(\displaystyle \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx\)

\(\Big \downarrow \) 85

\(\displaystyle \int \left (\frac {a^{10} A}{x^6}+\frac {a^9 (a B+10 A b)}{x^5}+\frac {5 a^8 b (2 a B+9 A b)}{x^4}+\frac {15 a^7 b^2 (3 a B+8 A b)}{x^3}+\frac {30 a^6 b^3 (4 a B+7 A b)}{x^2}+\frac {42 a^5 b^4 (5 a B+6 A b)}{x}+42 a^4 b^5 (6 a B+5 A b)+30 a^3 b^6 x (7 a B+4 A b)+15 a^2 b^7 x^2 (8 a B+3 A b)+b^9 x^4 (10 a B+A b)+5 a b^8 x^3 (9 a B+2 A b)+b^{10} B x^5\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^{10} A}{5 x^5}-\frac {a^9 (a B+10 A b)}{4 x^4}-\frac {5 a^8 b (2 a B+9 A b)}{3 x^3}-\frac {15 a^7 b^2 (3 a B+8 A b)}{2 x^2}-\frac {30 a^6 b^3 (4 a B+7 A b)}{x}+42 a^5 b^4 \log (x) (5 a B+6 A b)+42 a^4 b^5 x (6 a B+5 A b)+15 a^3 b^6 x^2 (7 a B+4 A b)+5 a^2 b^7 x^3 (8 a B+3 A b)+\frac {1}{5} b^9 x^5 (10 a B+A b)+\frac {5}{4} a b^8 x^4 (9 a B+2 A b)+\frac {1}{6} b^{10} B x^6\)

input

Int[((a + b*x)^10*(A + B*x))/x^6,x]
 

output

-1/5*(a^10*A)/x^5 - (a^9*(10*A*b + a*B))/(4*x^4) - (5*a^8*b*(9*A*b + 2*a*B 
))/(3*x^3) - (15*a^7*b^2*(8*A*b + 3*a*B))/(2*x^2) - (30*a^6*b^3*(7*A*b + 4 
*a*B))/x + 42*a^4*b^5*(5*A*b + 6*a*B)*x + 15*a^3*b^6*(4*A*b + 7*a*B)*x^2 + 
 5*a^2*b^7*(3*A*b + 8*a*B)*x^3 + (5*a*b^8*(2*A*b + 9*a*B)*x^4)/4 + (b^9*(A 
*b + 10*a*B)*x^5)/5 + (b^10*B*x^6)/6 + 42*a^5*b^4*(6*A*b + 5*a*B)*Log[x]
 

3.2.53.3.1 Defintions of rubi rules used

rule 85

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* 
f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n 
 + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 
1])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.2.53.4 Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {b^{10} B \,x^{6}}{6}+\frac {A \,b^{10} x^{5}}{5}+2 B a \,b^{9} x^{5}+\frac {5 A a \,b^{9} x^{4}}{2}+\frac {45 B \,a^{2} b^{8} x^{4}}{4}+15 A \,a^{2} b^{8} x^{3}+40 B \,a^{3} b^{7} x^{3}+60 A \,a^{3} b^{7} x^{2}+105 B \,a^{4} b^{6} x^{2}+210 A \,a^{4} b^{6} x +252 B \,a^{5} b^{5} x +42 a^{5} b^{4} \left (6 A b +5 B a \right ) \ln \left (x \right )-\frac {5 a^{8} b \left (9 A b +2 B a \right )}{3 x^{3}}-\frac {30 a^{6} b^{3} \left (7 A b +4 B a \right )}{x}-\frac {15 a^{7} b^{2} \left (8 A b +3 B a \right )}{2 x^{2}}-\frac {a^{9} \left (10 A b +B a \right )}{4 x^{4}}-\frac {a^{10} A}{5 x^{5}}\)	\(222\)
norman	\(\frac {\left (\frac {1}{5} b^{10} A +2 a \,b^{9} B \right ) x^{10}+\left (\frac {5}{2} a \,b^{9} A +\frac {45}{4} a^{2} b^{8} B \right ) x^{9}+\left (-60 a^{7} b^{3} A -\frac {45}{2} a^{8} b^{2} B \right ) x^{3}+\left (-15 a^{8} b^{2} A -\frac {10}{3} a^{9} b B \right ) x^{2}+\left (-\frac {5}{2} a^{9} b A -\frac {1}{4} a^{10} B \right ) x +\left (15 a^{2} b^{8} A +40 a^{3} b^{7} B \right ) x^{8}+\left (60 a^{3} b^{7} A +105 a^{4} b^{6} B \right ) x^{7}+\left (210 a^{4} b^{6} A +252 a^{5} b^{5} B \right ) x^{6}+\left (-210 a^{6} b^{4} A -120 a^{7} b^{3} B \right ) x^{4}-\frac {a^{10} A}{5}+\frac {b^{10} B \,x^{11}}{6}}{x^{5}}+\left (252 a^{5} b^{5} A +210 a^{6} b^{4} B \right ) \ln \left (x \right )\)	\(235\)
risch	\(\frac {b^{10} B \,x^{6}}{6}+\frac {A \,b^{10} x^{5}}{5}+2 B a \,b^{9} x^{5}+\frac {5 A a \,b^{9} x^{4}}{2}+\frac {45 B \,a^{2} b^{8} x^{4}}{4}+15 A \,a^{2} b^{8} x^{3}+40 B \,a^{3} b^{7} x^{3}+60 A \,a^{3} b^{7} x^{2}+105 B \,a^{4} b^{6} x^{2}+210 A \,a^{4} b^{6} x +252 B \,a^{5} b^{5} x +\frac {\left (-210 a^{6} b^{4} A -120 a^{7} b^{3} B \right ) x^{4}+\left (-60 a^{7} b^{3} A -\frac {45}{2} a^{8} b^{2} B \right ) x^{3}+\left (-15 a^{8} b^{2} A -\frac {10}{3} a^{9} b B \right ) x^{2}+\left (-\frac {5}{2} a^{9} b A -\frac {1}{4} a^{10} B \right ) x -\frac {a^{10} A}{5}}{x^{5}}+252 A \ln \left (x \right ) a^{5} b^{5}+210 B \ln \left (x \right ) a^{6} b^{4}\)	\(236\)
parallelrisch	\(\frac {10 b^{10} B \,x^{11}+12 A \,b^{10} x^{10}+120 B a \,b^{9} x^{10}+150 a A \,b^{9} x^{9}+675 B \,a^{2} b^{8} x^{9}+900 a^{2} A \,b^{8} x^{8}+2400 B \,a^{3} b^{7} x^{8}+3600 a^{3} A \,b^{7} x^{7}+6300 B \,a^{4} b^{6} x^{7}+15120 A \ln \left (x \right ) x^{5} a^{5} b^{5}+12600 a^{4} A \,b^{6} x^{6}+12600 B \ln \left (x \right ) x^{5} a^{6} b^{4}+15120 B \,a^{5} b^{5} x^{6}-12600 a^{6} A \,b^{4} x^{4}-7200 B \,a^{7} b^{3} x^{4}-3600 a^{7} A \,b^{3} x^{3}-1350 B \,a^{8} b^{2} x^{3}-900 a^{8} A \,b^{2} x^{2}-200 B \,a^{9} b \,x^{2}-150 a^{9} A b x -15 a^{10} B x -12 a^{10} A}{60 x^{5}}\)	\(248\)

input

int((b*x+a)^10*(B*x+A)/x^6,x,method=_RETURNVERBOSE)
 

output

1/6*b^10*B*x^6+1/5*A*b^10*x^5+2*B*a*b^9*x^5+5/2*A*a*b^9*x^4+45/4*B*a^2*b^8 
*x^4+15*A*a^2*b^8*x^3+40*B*a^3*b^7*x^3+60*A*a^3*b^7*x^2+105*B*a^4*b^6*x^2+ 
210*A*a^4*b^6*x+252*B*a^5*b^5*x+42*a^5*b^4*(6*A*b+5*B*a)*ln(x)-5/3*a^8*b*( 
9*A*b+2*B*a)/x^3-30*a^6*b^3*(7*A*b+4*B*a)/x-15/2*a^7*b^2*(8*A*b+3*B*a)/x^2 
-1/4*a^9*(10*A*b+B*a)/x^4-1/5*a^10*A/x^5
 

3.2.53.5 Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=\frac {10 \, B b^{10} x^{11} - 12 \, A a^{10} + 12 \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + 75 \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + 300 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + 900 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} + 2520 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + 2520 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} \log \left (x\right ) - 1800 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} - 450 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} - 100 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} - 15 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{5}} \]

input

integrate((b*x+a)^10*(B*x+A)/x^6,x, algorithm="fricas")
 

output

1/60*(10*B*b^10*x^11 - 12*A*a^10 + 12*(10*B*a*b^9 + A*b^10)*x^10 + 75*(9*B 
*a^2*b^8 + 2*A*a*b^9)*x^9 + 300*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^8 + 900*(7*B 
*a^4*b^6 + 4*A*a^3*b^7)*x^7 + 2520*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x^6 + 2520* 
(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5*log(x) - 1800*(4*B*a^7*b^3 + 7*A*a^6*b^4)* 
x^4 - 450*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 - 100*(2*B*a^9*b + 9*A*a^8*b^2)* 
x^2 - 15*(B*a^10 + 10*A*a^9*b)*x)/x^5
 

3.2.53.6 Sympy [A] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=\frac {B b^{10} x^{6}}{6} + 42 a^{5} b^{4} \cdot \left (6 A b + 5 B a\right ) \log {\left (x \right )} + x^{5} \left (\frac {A b^{10}}{5} + 2 B a b^{9}\right ) + x^{4} \cdot \left (\frac {5 A a b^{9}}{2} + \frac {45 B a^{2} b^{8}}{4}\right ) + x^{3} \cdot \left (15 A a^{2} b^{8} + 40 B a^{3} b^{7}\right ) + x^{2} \cdot \left (60 A a^{3} b^{7} + 105 B a^{4} b^{6}\right ) + x \left (210 A a^{4} b^{6} + 252 B a^{5} b^{5}\right ) + \frac {- 12 A a^{10} + x^{4} \left (- 12600 A a^{6} b^{4} - 7200 B a^{7} b^{3}\right ) + x^{3} \left (- 3600 A a^{7} b^{3} - 1350 B a^{8} b^{2}\right ) + x^{2} \left (- 900 A a^{8} b^{2} - 200 B a^{9} b\right ) + x \left (- 150 A a^{9} b - 15 B a^{10}\right )}{60 x^{5}} \]

input

integrate((b*x+a)**10*(B*x+A)/x**6,x)
 

output

B*b**10*x**6/6 + 42*a**5*b**4*(6*A*b + 5*B*a)*log(x) + x**5*(A*b**10/5 + 2 
*B*a*b**9) + x**4*(5*A*a*b**9/2 + 45*B*a**2*b**8/4) + x**3*(15*A*a**2*b**8 
 + 40*B*a**3*b**7) + x**2*(60*A*a**3*b**7 + 105*B*a**4*b**6) + x*(210*A*a* 
*4*b**6 + 252*B*a**5*b**5) + (-12*A*a**10 + x**4*(-12600*A*a**6*b**4 - 720 
0*B*a**7*b**3) + x**3*(-3600*A*a**7*b**3 - 1350*B*a**8*b**2) + x**2*(-900* 
A*a**8*b**2 - 200*B*a**9*b) + x*(-150*A*a**9*b - 15*B*a**10))/(60*x**5)
 

3.2.53.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=\frac {1}{6} \, B b^{10} x^{6} + \frac {1}{5} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{5} + \frac {5}{4} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{4} + 5 \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{3} + 15 \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{2} + 42 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x + 42 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} \log \left (x\right ) - \frac {12 \, A a^{10} + 1800 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 450 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 100 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 15 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{5}} \]

input

integrate((b*x+a)^10*(B*x+A)/x^6,x, algorithm="maxima")
 

output

1/6*B*b^10*x^6 + 1/5*(10*B*a*b^9 + A*b^10)*x^5 + 5/4*(9*B*a^2*b^8 + 2*A*a* 
b^9)*x^4 + 5*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^3 + 15*(7*B*a^4*b^6 + 4*A*a^3*b 
^7)*x^2 + 42*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x + 42*(5*B*a^6*b^4 + 6*A*a^5*b^5 
)*log(x) - 1/60*(12*A*a^10 + 1800*(4*B*a^7*b^3 + 7*A*a^6*b^4)*x^4 + 450*(3 
*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 100*(2*B*a^9*b + 9*A*a^8*b^2)*x^2 + 15*(B* 
a^10 + 10*A*a^9*b)*x)/x^5
 

3.2.53.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=\frac {1}{6} \, B b^{10} x^{6} + 2 \, B a b^{9} x^{5} + \frac {1}{5} \, A b^{10} x^{5} + \frac {45}{4} \, B a^{2} b^{8} x^{4} + \frac {5}{2} \, A a b^{9} x^{4} + 40 \, B a^{3} b^{7} x^{3} + 15 \, A a^{2} b^{8} x^{3} + 105 \, B a^{4} b^{6} x^{2} + 60 \, A a^{3} b^{7} x^{2} + 252 \, B a^{5} b^{5} x + 210 \, A a^{4} b^{6} x + 42 \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{10} + 1800 \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 450 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + 100 \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + 15 \, {\left (B a^{10} + 10 \, A a^{9} b\right )} x}{60 \, x^{5}} \]

input

integrate((b*x+a)^10*(B*x+A)/x^6,x, algorithm="giac")
 

output

1/6*B*b^10*x^6 + 2*B*a*b^9*x^5 + 1/5*A*b^10*x^5 + 45/4*B*a^2*b^8*x^4 + 5/2 
*A*a*b^9*x^4 + 40*B*a^3*b^7*x^3 + 15*A*a^2*b^8*x^3 + 105*B*a^4*b^6*x^2 + 6 
0*A*a^3*b^7*x^2 + 252*B*a^5*b^5*x + 210*A*a^4*b^6*x + 42*(5*B*a^6*b^4 + 6* 
A*a^5*b^5)*log(abs(x)) - 1/60*(12*A*a^10 + 1800*(4*B*a^7*b^3 + 7*A*a^6*b^4 
)*x^4 + 450*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 100*(2*B*a^9*b + 9*A*a^8*b^2 
)*x^2 + 15*(B*a^10 + 10*A*a^9*b)*x)/x^5
 

3.2.53.9 Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^{10} (A+B x)}{x^6} \, dx=x^5\,\left (\frac {A\,b^{10}}{5}+2\,B\,a\,b^9\right )-\frac {x\,\left (\frac {B\,a^{10}}{4}+\frac {5\,A\,b\,a^9}{2}\right )+\frac {A\,a^{10}}{5}+x^2\,\left (\frac {10\,B\,a^9\,b}{3}+15\,A\,a^8\,b^2\right )+x^3\,\left (\frac {45\,B\,a^8\,b^2}{2}+60\,A\,a^7\,b^3\right )+x^4\,\left (120\,B\,a^7\,b^3+210\,A\,a^6\,b^4\right )}{x^5}+\ln \left (x\right )\,\left (210\,B\,a^6\,b^4+252\,A\,a^5\,b^5\right )+\frac {B\,b^{10}\,x^6}{6}+15\,a^3\,b^6\,x^2\,\left (4\,A\,b+7\,B\,a\right )+5\,a^2\,b^7\,x^3\,\left (3\,A\,b+8\,B\,a\right )+42\,a^4\,b^5\,x\,\left (5\,A\,b+6\,B\,a\right )+\frac {5\,a\,b^8\,x^4\,\left (2\,A\,b+9\,B\,a\right )}{4} \]

input

int(((A + B*x)*(a + b*x)^10)/x^6,x)
 

output

x^5*((A*b^10)/5 + 2*B*a*b^9) - (x*((B*a^10)/4 + (5*A*a^9*b)/2) + (A*a^10)/ 
5 + x^2*(15*A*a^8*b^2 + (10*B*a^9*b)/3) + x^3*(60*A*a^7*b^3 + (45*B*a^8*b^ 
2)/2) + x^4*(210*A*a^6*b^4 + 120*B*a^7*b^3))/x^5 + log(x)*(252*A*a^5*b^5 + 
 210*B*a^6*b^4) + (B*b^10*x^6)/6 + 15*a^3*b^6*x^2*(4*A*b + 7*B*a) + 5*a^2* 
b^7*x^3*(3*A*b + 8*B*a) + 42*a^4*b^5*x*(5*A*b + 6*B*a) + (5*a*b^8*x^4*(2*A 
*b + 9*B*a))/4