3.2.48 \(\int \frac {(a+b x)^{10} (A+B x)}{x} \, dx\) [148]

3.2.48.1 Optimal result
3.2.48.2 Mathematica [A] (verified)
3.2.48.3 Rubi [A] (verified)
3.2.48.4 Maple [A] (verified)
3.2.48.5 Fricas [A] (verification not implemented)
3.2.48.6 Sympy [A] (verification not implemented)
3.2.48.7 Maxima [A] (verification not implemented)
3.2.48.8 Giac [A] (verification not implemented)
3.2.48.9 Mupad [B] (verification not implemented)

3.2.48.1 Optimal result

Integrand size = 16, antiderivative size = 148 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=10 a^9 A b x+\frac {45}{2} a^8 A b^2 x^2+40 a^7 A b^3 x^3+\frac {105}{2} a^6 A b^4 x^4+\frac {252}{5} a^5 A b^5 x^5+35 a^4 A b^6 x^6+\frac {120}{7} a^3 A b^7 x^7+\frac {45}{8} a^2 A b^8 x^8+\frac {10}{9} a A b^9 x^9+\frac {1}{10} A b^{10} x^{10}+\frac {B (a+b x)^{11}}{11 b}+a^{10} A \log (x) \]

output
10*a^9*A*b*x+45/2*a^8*A*b^2*x^2+40*a^7*A*b^3*x^3+105/2*a^6*A*b^4*x^4+252/5 
*a^5*A*b^5*x^5+35*a^4*A*b^6*x^6+120/7*a^3*A*b^7*x^7+45/8*a^2*A*b^8*x^8+10/ 
9*a*A*b^9*x^9+1/10*A*b^10*x^10+1/11*B*(b*x+a)^11/b+a^10*A*ln(x)
 
3.2.48.2 Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=a^{10} B x+5 a^9 b x (2 A+B x)+\frac {15}{2} a^8 b^2 x^2 (3 A+2 B x)+10 a^7 b^3 x^3 (4 A+3 B x)+\frac {21}{2} a^6 b^4 x^4 (5 A+4 B x)+\frac {42}{5} a^5 b^5 x^5 (6 A+5 B x)+5 a^4 b^6 x^6 (7 A+6 B x)+\frac {15}{7} a^3 b^7 x^7 (8 A+7 B x)+\frac {5}{8} a^2 b^8 x^8 (9 A+8 B x)+\frac {1}{9} a b^9 x^9 (10 A+9 B x)+\frac {1}{110} b^{10} x^{10} (11 A+10 B x)+a^{10} A \log (x) \]

input
Integrate[((a + b*x)^10*(A + B*x))/x,x]
 
output
a^10*B*x + 5*a^9*b*x*(2*A + B*x) + (15*a^8*b^2*x^2*(3*A + 2*B*x))/2 + 10*a 
^7*b^3*x^3*(4*A + 3*B*x) + (21*a^6*b^4*x^4*(5*A + 4*B*x))/2 + (42*a^5*b^5* 
x^5*(6*A + 5*B*x))/5 + 5*a^4*b^6*x^6*(7*A + 6*B*x) + (15*a^3*b^7*x^7*(8*A 
+ 7*B*x))/7 + (5*a^2*b^8*x^8*(9*A + 8*B*x))/8 + (a*b^9*x^9*(10*A + 9*B*x)) 
/9 + (b^10*x^10*(11*A + 10*B*x))/110 + a^10*A*Log[x]
 
3.2.48.3 Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {90, 49, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle A \int \frac {(a+b x)^{10}}{x}dx+\frac {B (a+b x)^{11}}{11 b}\)

\(\Big \downarrow \) 49

\(\displaystyle A \int \left (\frac {a^{10}}{x}+10 b a^9+45 b^2 x a^8+120 b^3 x^2 a^7+210 b^4 x^3 a^6+252 b^5 x^4 a^5+210 b^6 x^5 a^4+120 b^7 x^6 a^3+45 b^8 x^7 a^2+10 b^9 x^8 a+b^{10} x^9\right )dx+\frac {B (a+b x)^{11}}{11 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle A \left (a^{10} \log (x)+10 a^9 b x+\frac {45}{2} a^8 b^2 x^2+40 a^7 b^3 x^3+\frac {105}{2} a^6 b^4 x^4+\frac {252}{5} a^5 b^5 x^5+35 a^4 b^6 x^6+\frac {120}{7} a^3 b^7 x^7+\frac {45}{8} a^2 b^8 x^8+\frac {10}{9} a b^9 x^9+\frac {b^{10} x^{10}}{10}\right )+\frac {B (a+b x)^{11}}{11 b}\)

input
Int[((a + b*x)^10*(A + B*x))/x,x]
 
output
(B*(a + b*x)^11)/(11*b) + A*(10*a^9*b*x + (45*a^8*b^2*x^2)/2 + 40*a^7*b^3* 
x^3 + (105*a^6*b^4*x^4)/2 + (252*a^5*b^5*x^5)/5 + 35*a^4*b^6*x^6 + (120*a^ 
3*b^7*x^7)/7 + (45*a^2*b^8*x^8)/8 + (10*a*b^9*x^9)/9 + (b^10*x^10)/10 + a^ 
10*Log[x])
 

3.2.48.3.1 Defintions of rubi rules used

rule 49
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 90
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.2.48.4 Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.55

method result size
norman \(\left (\frac {1}{10} b^{10} A +a \,b^{9} B \right ) x^{10}+\left (\frac {10}{9} a \,b^{9} A +5 a^{2} b^{8} B \right ) x^{9}+\left (\frac {45}{8} a^{2} b^{8} A +15 a^{3} b^{7} B \right ) x^{8}+\left (\frac {120}{7} a^{3} b^{7} A +30 a^{4} b^{6} B \right ) x^{7}+\left (\frac {252}{5} a^{5} b^{5} A +42 a^{6} b^{4} B \right ) x^{5}+\left (\frac {105}{2} a^{6} b^{4} A +30 a^{7} b^{3} B \right ) x^{4}+\left (\frac {45}{2} a^{8} b^{2} A +5 a^{9} b B \right ) x^{2}+\left (35 a^{4} b^{6} A +42 a^{5} b^{5} B \right ) x^{6}+\left (40 a^{7} b^{3} A +15 a^{8} b^{2} B \right ) x^{3}+\left (10 a^{9} b A +a^{10} B \right ) x +\frac {b^{10} B \,x^{11}}{11}+a^{10} A \ln \left (x \right )\) \(230\)
default \(\frac {b^{10} B \,x^{11}}{11}+\frac {A \,b^{10} x^{10}}{10}+B a \,b^{9} x^{10}+\frac {10 a A \,b^{9} x^{9}}{9}+5 B \,a^{2} b^{8} x^{9}+\frac {45 a^{2} A \,b^{8} x^{8}}{8}+15 B \,a^{3} b^{7} x^{8}+\frac {120 a^{3} A \,b^{7} x^{7}}{7}+30 B \,a^{4} b^{6} x^{7}+35 a^{4} A \,b^{6} x^{6}+42 B \,a^{5} b^{5} x^{6}+\frac {252 a^{5} A \,b^{5} x^{5}}{5}+42 B \,a^{6} b^{4} x^{5}+\frac {105 a^{6} A \,b^{4} x^{4}}{2}+30 B \,a^{7} b^{3} x^{4}+40 a^{7} A \,b^{3} x^{3}+15 B \,a^{8} b^{2} x^{3}+\frac {45 a^{8} A \,b^{2} x^{2}}{2}+5 B \,a^{9} b \,x^{2}+10 a^{9} A b x +a^{10} B x +a^{10} A \ln \left (x \right )\) \(238\)
risch \(\frac {b^{10} B \,x^{11}}{11}+\frac {A \,b^{10} x^{10}}{10}+B a \,b^{9} x^{10}+\frac {10 a A \,b^{9} x^{9}}{9}+5 B \,a^{2} b^{8} x^{9}+\frac {45 a^{2} A \,b^{8} x^{8}}{8}+15 B \,a^{3} b^{7} x^{8}+\frac {120 a^{3} A \,b^{7} x^{7}}{7}+30 B \,a^{4} b^{6} x^{7}+35 a^{4} A \,b^{6} x^{6}+42 B \,a^{5} b^{5} x^{6}+\frac {252 a^{5} A \,b^{5} x^{5}}{5}+42 B \,a^{6} b^{4} x^{5}+\frac {105 a^{6} A \,b^{4} x^{4}}{2}+30 B \,a^{7} b^{3} x^{4}+40 a^{7} A \,b^{3} x^{3}+15 B \,a^{8} b^{2} x^{3}+\frac {45 a^{8} A \,b^{2} x^{2}}{2}+5 B \,a^{9} b \,x^{2}+10 a^{9} A b x +a^{10} B x +a^{10} A \ln \left (x \right )\) \(238\)
parallelrisch \(\frac {b^{10} B \,x^{11}}{11}+\frac {A \,b^{10} x^{10}}{10}+B a \,b^{9} x^{10}+\frac {10 a A \,b^{9} x^{9}}{9}+5 B \,a^{2} b^{8} x^{9}+\frac {45 a^{2} A \,b^{8} x^{8}}{8}+15 B \,a^{3} b^{7} x^{8}+\frac {120 a^{3} A \,b^{7} x^{7}}{7}+30 B \,a^{4} b^{6} x^{7}+35 a^{4} A \,b^{6} x^{6}+42 B \,a^{5} b^{5} x^{6}+\frac {252 a^{5} A \,b^{5} x^{5}}{5}+42 B \,a^{6} b^{4} x^{5}+\frac {105 a^{6} A \,b^{4} x^{4}}{2}+30 B \,a^{7} b^{3} x^{4}+40 a^{7} A \,b^{3} x^{3}+15 B \,a^{8} b^{2} x^{3}+\frac {45 a^{8} A \,b^{2} x^{2}}{2}+5 B \,a^{9} b \,x^{2}+10 a^{9} A b x +a^{10} B x +a^{10} A \ln \left (x \right )\) \(238\)

input
int((b*x+a)^10*(B*x+A)/x,x,method=_RETURNVERBOSE)
 
output
(1/10*b^10*A+a*b^9*B)*x^10+(10/9*a*b^9*A+5*a^2*b^8*B)*x^9+(45/8*a^2*b^8*A+ 
15*a^3*b^7*B)*x^8+(120/7*a^3*b^7*A+30*a^4*b^6*B)*x^7+(252/5*a^5*b^5*A+42*a 
^6*b^4*B)*x^5+(105/2*a^6*b^4*A+30*a^7*b^3*B)*x^4+(45/2*a^8*b^2*A+5*a^9*b*B 
)*x^2+(35*A*a^4*b^6+42*B*a^5*b^5)*x^6+(40*A*a^7*b^3+15*B*a^8*b^2)*x^3+(10* 
A*a^9*b+B*a^10)*x+1/11*b^10*B*x^11+a^10*A*ln(x)
 
3.2.48.5 Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=\frac {1}{11} \, B b^{10} x^{11} + A a^{10} \log \left (x\right ) + \frac {1}{10} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + \frac {5}{9} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + \frac {15}{8} \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + \frac {30}{7} \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} + 7 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + \frac {42}{5} \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + \frac {15}{2} \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 5 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + \frac {5}{2} \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + {\left (B a^{10} + 10 \, A a^{9} b\right )} x \]

input
integrate((b*x+a)^10*(B*x+A)/x,x, algorithm="fricas")
 
output
1/11*B*b^10*x^11 + A*a^10*log(x) + 1/10*(10*B*a*b^9 + A*b^10)*x^10 + 5/9*( 
9*B*a^2*b^8 + 2*A*a*b^9)*x^9 + 15/8*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^8 + 30/7 
*(7*B*a^4*b^6 + 4*A*a^3*b^7)*x^7 + 7*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x^6 + 42/ 
5*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 15/2*(4*B*a^7*b^3 + 7*A*a^6*b^4)*x^4 + 
 5*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 5/2*(2*B*a^9*b + 9*A*a^8*b^2)*x^2 + ( 
B*a^10 + 10*A*a^9*b)*x
 
3.2.48.6 Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=A a^{10} \log {\left (x \right )} + \frac {B b^{10} x^{11}}{11} + x^{10} \left (\frac {A b^{10}}{10} + B a b^{9}\right ) + x^{9} \cdot \left (\frac {10 A a b^{9}}{9} + 5 B a^{2} b^{8}\right ) + x^{8} \cdot \left (\frac {45 A a^{2} b^{8}}{8} + 15 B a^{3} b^{7}\right ) + x^{7} \cdot \left (\frac {120 A a^{3} b^{7}}{7} + 30 B a^{4} b^{6}\right ) + x^{6} \cdot \left (35 A a^{4} b^{6} + 42 B a^{5} b^{5}\right ) + x^{5} \cdot \left (\frac {252 A a^{5} b^{5}}{5} + 42 B a^{6} b^{4}\right ) + x^{4} \cdot \left (\frac {105 A a^{6} b^{4}}{2} + 30 B a^{7} b^{3}\right ) + x^{3} \cdot \left (40 A a^{7} b^{3} + 15 B a^{8} b^{2}\right ) + x^{2} \cdot \left (\frac {45 A a^{8} b^{2}}{2} + 5 B a^{9} b\right ) + x \left (10 A a^{9} b + B a^{10}\right ) \]

input
integrate((b*x+a)**10*(B*x+A)/x,x)
 
output
A*a**10*log(x) + B*b**10*x**11/11 + x**10*(A*b**10/10 + B*a*b**9) + x**9*( 
10*A*a*b**9/9 + 5*B*a**2*b**8) + x**8*(45*A*a**2*b**8/8 + 15*B*a**3*b**7) 
+ x**7*(120*A*a**3*b**7/7 + 30*B*a**4*b**6) + x**6*(35*A*a**4*b**6 + 42*B* 
a**5*b**5) + x**5*(252*A*a**5*b**5/5 + 42*B*a**6*b**4) + x**4*(105*A*a**6* 
b**4/2 + 30*B*a**7*b**3) + x**3*(40*A*a**7*b**3 + 15*B*a**8*b**2) + x**2*( 
45*A*a**8*b**2/2 + 5*B*a**9*b) + x*(10*A*a**9*b + B*a**10)
 
3.2.48.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=\frac {1}{11} \, B b^{10} x^{11} + A a^{10} \log \left (x\right ) + \frac {1}{10} \, {\left (10 \, B a b^{9} + A b^{10}\right )} x^{10} + \frac {5}{9} \, {\left (9 \, B a^{2} b^{8} + 2 \, A a b^{9}\right )} x^{9} + \frac {15}{8} \, {\left (8 \, B a^{3} b^{7} + 3 \, A a^{2} b^{8}\right )} x^{8} + \frac {30}{7} \, {\left (7 \, B a^{4} b^{6} + 4 \, A a^{3} b^{7}\right )} x^{7} + 7 \, {\left (6 \, B a^{5} b^{5} + 5 \, A a^{4} b^{6}\right )} x^{6} + \frac {42}{5} \, {\left (5 \, B a^{6} b^{4} + 6 \, A a^{5} b^{5}\right )} x^{5} + \frac {15}{2} \, {\left (4 \, B a^{7} b^{3} + 7 \, A a^{6} b^{4}\right )} x^{4} + 5 \, {\left (3 \, B a^{8} b^{2} + 8 \, A a^{7} b^{3}\right )} x^{3} + \frac {5}{2} \, {\left (2 \, B a^{9} b + 9 \, A a^{8} b^{2}\right )} x^{2} + {\left (B a^{10} + 10 \, A a^{9} b\right )} x \]

input
integrate((b*x+a)^10*(B*x+A)/x,x, algorithm="maxima")
 
output
1/11*B*b^10*x^11 + A*a^10*log(x) + 1/10*(10*B*a*b^9 + A*b^10)*x^10 + 5/9*( 
9*B*a^2*b^8 + 2*A*a*b^9)*x^9 + 15/8*(8*B*a^3*b^7 + 3*A*a^2*b^8)*x^8 + 30/7 
*(7*B*a^4*b^6 + 4*A*a^3*b^7)*x^7 + 7*(6*B*a^5*b^5 + 5*A*a^4*b^6)*x^6 + 42/ 
5*(5*B*a^6*b^4 + 6*A*a^5*b^5)*x^5 + 15/2*(4*B*a^7*b^3 + 7*A*a^6*b^4)*x^4 + 
 5*(3*B*a^8*b^2 + 8*A*a^7*b^3)*x^3 + 5/2*(2*B*a^9*b + 9*A*a^8*b^2)*x^2 + ( 
B*a^10 + 10*A*a^9*b)*x
 
3.2.48.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=\frac {1}{11} \, B b^{10} x^{11} + B a b^{9} x^{10} + \frac {1}{10} \, A b^{10} x^{10} + 5 \, B a^{2} b^{8} x^{9} + \frac {10}{9} \, A a b^{9} x^{9} + 15 \, B a^{3} b^{7} x^{8} + \frac {45}{8} \, A a^{2} b^{8} x^{8} + 30 \, B a^{4} b^{6} x^{7} + \frac {120}{7} \, A a^{3} b^{7} x^{7} + 42 \, B a^{5} b^{5} x^{6} + 35 \, A a^{4} b^{6} x^{6} + 42 \, B a^{6} b^{4} x^{5} + \frac {252}{5} \, A a^{5} b^{5} x^{5} + 30 \, B a^{7} b^{3} x^{4} + \frac {105}{2} \, A a^{6} b^{4} x^{4} + 15 \, B a^{8} b^{2} x^{3} + 40 \, A a^{7} b^{3} x^{3} + 5 \, B a^{9} b x^{2} + \frac {45}{2} \, A a^{8} b^{2} x^{2} + B a^{10} x + 10 \, A a^{9} b x + A a^{10} \log \left ({\left | x \right |}\right ) \]

input
integrate((b*x+a)^10*(B*x+A)/x,x, algorithm="giac")
 
output
1/11*B*b^10*x^11 + B*a*b^9*x^10 + 1/10*A*b^10*x^10 + 5*B*a^2*b^8*x^9 + 10/ 
9*A*a*b^9*x^9 + 15*B*a^3*b^7*x^8 + 45/8*A*a^2*b^8*x^8 + 30*B*a^4*b^6*x^7 + 
 120/7*A*a^3*b^7*x^7 + 42*B*a^5*b^5*x^6 + 35*A*a^4*b^6*x^6 + 42*B*a^6*b^4* 
x^5 + 252/5*A*a^5*b^5*x^5 + 30*B*a^7*b^3*x^4 + 105/2*A*a^6*b^4*x^4 + 15*B* 
a^8*b^2*x^3 + 40*A*a^7*b^3*x^3 + 5*B*a^9*b*x^2 + 45/2*A*a^8*b^2*x^2 + B*a^ 
10*x + 10*A*a^9*b*x + A*a^10*log(abs(x))
 
3.2.48.9 Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^{10} (A+B x)}{x} \, dx=x\,\left (B\,a^{10}+10\,A\,b\,a^9\right )+x^{10}\,\left (\frac {A\,b^{10}}{10}+B\,a\,b^9\right )+\frac {B\,b^{10}\,x^{11}}{11}+A\,a^{10}\,\ln \left (x\right )+5\,a^7\,b^2\,x^3\,\left (8\,A\,b+3\,B\,a\right )+\frac {15\,a^6\,b^3\,x^4\,\left (7\,A\,b+4\,B\,a\right )}{2}+\frac {42\,a^5\,b^4\,x^5\,\left (6\,A\,b+5\,B\,a\right )}{5}+7\,a^4\,b^5\,x^6\,\left (5\,A\,b+6\,B\,a\right )+\frac {30\,a^3\,b^6\,x^7\,\left (4\,A\,b+7\,B\,a\right )}{7}+\frac {15\,a^2\,b^7\,x^8\,\left (3\,A\,b+8\,B\,a\right )}{8}+\frac {5\,a^8\,b\,x^2\,\left (9\,A\,b+2\,B\,a\right )}{2}+\frac {5\,a\,b^8\,x^9\,\left (2\,A\,b+9\,B\,a\right )}{9} \]

input
int(((A + B*x)*(a + b*x)^10)/x,x)
 
output
x*(B*a^10 + 10*A*a^9*b) + x^10*((A*b^10)/10 + B*a*b^9) + (B*b^10*x^11)/11 
+ A*a^10*log(x) + 5*a^7*b^2*x^3*(8*A*b + 3*B*a) + (15*a^6*b^3*x^4*(7*A*b + 
 4*B*a))/2 + (42*a^5*b^4*x^5*(6*A*b + 5*B*a))/5 + 7*a^4*b^5*x^6*(5*A*b + 6 
*B*a) + (30*a^3*b^6*x^7*(4*A*b + 7*B*a))/7 + (15*a^2*b^7*x^8*(3*A*b + 8*B* 
a))/8 + (5*a^8*b*x^2*(9*A*b + 2*B*a))/2 + (5*a*b^8*x^9*(2*A*b + 9*B*a))/9