Integrand size = 11, antiderivative size = 147 \[ \int x^8 (a+b x)^{10} \, dx=\frac {a^8 (a+b x)^{11}}{11 b^9}-\frac {2 a^7 (a+b x)^{12}}{3 b^9}+\frac {28 a^6 (a+b x)^{13}}{13 b^9}-\frac {4 a^5 (a+b x)^{14}}{b^9}+\frac {14 a^4 (a+b x)^{15}}{3 b^9}-\frac {7 a^3 (a+b x)^{16}}{2 b^9}+\frac {28 a^2 (a+b x)^{17}}{17 b^9}-\frac {4 a (a+b x)^{18}}{9 b^9}+\frac {(a+b x)^{19}}{19 b^9} \]
1/11*a^8*(b*x+a)^11/b^9-2/3*a^7*(b*x+a)^12/b^9+28/13*a^6*(b*x+a)^13/b^9-4* a^5*(b*x+a)^14/b^9+14/3*a^4*(b*x+a)^15/b^9-7/2*a^3*(b*x+a)^16/b^9+28/17*a^ 2*(b*x+a)^17/b^9-4/9*a*(b*x+a)^18/b^9+1/19*(b*x+a)^19/b^9
Time = 0.00 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.85 \[ \int x^8 (a+b x)^{10} \, dx=\frac {a^{10} x^9}{9}+a^9 b x^{10}+\frac {45}{11} a^8 b^2 x^{11}+10 a^7 b^3 x^{12}+\frac {210}{13} a^6 b^4 x^{13}+18 a^5 b^5 x^{14}+14 a^4 b^6 x^{15}+\frac {15}{2} a^3 b^7 x^{16}+\frac {45}{17} a^2 b^8 x^{17}+\frac {5}{9} a b^9 x^{18}+\frac {b^{10} x^{19}}{19} \]
(a^10*x^9)/9 + a^9*b*x^10 + (45*a^8*b^2*x^11)/11 + 10*a^7*b^3*x^12 + (210* a^6*b^4*x^13)/13 + 18*a^5*b^5*x^14 + 14*a^4*b^6*x^15 + (15*a^3*b^7*x^16)/2 + (45*a^2*b^8*x^17)/17 + (5*a*b^9*x^18)/9 + (b^10*x^19)/19
Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {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.
\(\displaystyle \int x^8 (a+b x)^{10} \, dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (\frac {a^8 (a+b x)^{10}}{b^8}-\frac {8 a^7 (a+b x)^{11}}{b^8}+\frac {28 a^6 (a+b x)^{12}}{b^8}-\frac {56 a^5 (a+b x)^{13}}{b^8}+\frac {70 a^4 (a+b x)^{14}}{b^8}-\frac {56 a^3 (a+b x)^{15}}{b^8}+\frac {28 a^2 (a+b x)^{16}}{b^8}+\frac {(a+b x)^{18}}{b^8}-\frac {8 a (a+b x)^{17}}{b^8}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^8 (a+b x)^{11}}{11 b^9}-\frac {2 a^7 (a+b x)^{12}}{3 b^9}+\frac {28 a^6 (a+b x)^{13}}{13 b^9}-\frac {4 a^5 (a+b x)^{14}}{b^9}+\frac {14 a^4 (a+b x)^{15}}{3 b^9}-\frac {7 a^3 (a+b x)^{16}}{2 b^9}+\frac {28 a^2 (a+b x)^{17}}{17 b^9}+\frac {(a+b x)^{19}}{19 b^9}-\frac {4 a (a+b x)^{18}}{9 b^9}\) |
(a^8*(a + b*x)^11)/(11*b^9) - (2*a^7*(a + b*x)^12)/(3*b^9) + (28*a^6*(a + b*x)^13)/(13*b^9) - (4*a^5*(a + b*x)^14)/b^9 + (14*a^4*(a + b*x)^15)/(3*b^ 9) - (7*a^3*(a + b*x)^16)/(2*b^9) + (28*a^2*(a + b*x)^17)/(17*b^9) - (4*a* (a + b*x)^18)/(9*b^9) + (a + b*x)^19/(19*b^9)
3.2.26.3.1 Defintions of rubi rules used
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]
Time = 0.18 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76
method | result | size |
gosper | \(\frac {1}{9} a^{10} x^{9}+a^{9} b \,x^{10}+\frac {45}{11} a^{8} b^{2} x^{11}+10 a^{7} b^{3} x^{12}+\frac {210}{13} a^{6} b^{4} x^{13}+18 a^{5} b^{5} x^{14}+14 a^{4} b^{6} x^{15}+\frac {15}{2} a^{3} b^{7} x^{16}+\frac {45}{17} a^{2} b^{8} x^{17}+\frac {5}{9} a \,b^{9} x^{18}+\frac {1}{19} b^{10} x^{19}\) | \(112\) |
default | \(\frac {1}{9} a^{10} x^{9}+a^{9} b \,x^{10}+\frac {45}{11} a^{8} b^{2} x^{11}+10 a^{7} b^{3} x^{12}+\frac {210}{13} a^{6} b^{4} x^{13}+18 a^{5} b^{5} x^{14}+14 a^{4} b^{6} x^{15}+\frac {15}{2} a^{3} b^{7} x^{16}+\frac {45}{17} a^{2} b^{8} x^{17}+\frac {5}{9} a \,b^{9} x^{18}+\frac {1}{19} b^{10} x^{19}\) | \(112\) |
norman | \(\frac {1}{9} a^{10} x^{9}+a^{9} b \,x^{10}+\frac {45}{11} a^{8} b^{2} x^{11}+10 a^{7} b^{3} x^{12}+\frac {210}{13} a^{6} b^{4} x^{13}+18 a^{5} b^{5} x^{14}+14 a^{4} b^{6} x^{15}+\frac {15}{2} a^{3} b^{7} x^{16}+\frac {45}{17} a^{2} b^{8} x^{17}+\frac {5}{9} a \,b^{9} x^{18}+\frac {1}{19} b^{10} x^{19}\) | \(112\) |
risch | \(\frac {1}{9} a^{10} x^{9}+a^{9} b \,x^{10}+\frac {45}{11} a^{8} b^{2} x^{11}+10 a^{7} b^{3} x^{12}+\frac {210}{13} a^{6} b^{4} x^{13}+18 a^{5} b^{5} x^{14}+14 a^{4} b^{6} x^{15}+\frac {15}{2} a^{3} b^{7} x^{16}+\frac {45}{17} a^{2} b^{8} x^{17}+\frac {5}{9} a \,b^{9} x^{18}+\frac {1}{19} b^{10} x^{19}\) | \(112\) |
parallelrisch | \(\frac {1}{9} a^{10} x^{9}+a^{9} b \,x^{10}+\frac {45}{11} a^{8} b^{2} x^{11}+10 a^{7} b^{3} x^{12}+\frac {210}{13} a^{6} b^{4} x^{13}+18 a^{5} b^{5} x^{14}+14 a^{4} b^{6} x^{15}+\frac {15}{2} a^{3} b^{7} x^{16}+\frac {45}{17} a^{2} b^{8} x^{17}+\frac {5}{9} a \,b^{9} x^{18}+\frac {1}{19} b^{10} x^{19}\) | \(112\) |
1/9*a^10*x^9+a^9*b*x^10+45/11*a^8*b^2*x^11+10*a^7*b^3*x^12+210/13*a^6*b^4* x^13+18*a^5*b^5*x^14+14*a^4*b^6*x^15+15/2*a^3*b^7*x^16+45/17*a^2*b^8*x^17+ 5/9*a*b^9*x^18+1/19*b^10*x^19
Time = 0.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int x^8 (a+b x)^{10} \, dx=\frac {1}{19} \, b^{10} x^{19} + \frac {5}{9} \, a b^{9} x^{18} + \frac {45}{17} \, a^{2} b^{8} x^{17} + \frac {15}{2} \, a^{3} b^{7} x^{16} + 14 \, a^{4} b^{6} x^{15} + 18 \, a^{5} b^{5} x^{14} + \frac {210}{13} \, a^{6} b^{4} x^{13} + 10 \, a^{7} b^{3} x^{12} + \frac {45}{11} \, a^{8} b^{2} x^{11} + a^{9} b x^{10} + \frac {1}{9} \, a^{10} x^{9} \]
1/19*b^10*x^19 + 5/9*a*b^9*x^18 + 45/17*a^2*b^8*x^17 + 15/2*a^3*b^7*x^16 + 14*a^4*b^6*x^15 + 18*a^5*b^5*x^14 + 210/13*a^6*b^4*x^13 + 10*a^7*b^3*x^12 + 45/11*a^8*b^2*x^11 + a^9*b*x^10 + 1/9*a^10*x^9
Time = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int x^8 (a+b x)^{10} \, dx=\frac {a^{10} x^{9}}{9} + a^{9} b x^{10} + \frac {45 a^{8} b^{2} x^{11}}{11} + 10 a^{7} b^{3} x^{12} + \frac {210 a^{6} b^{4} x^{13}}{13} + 18 a^{5} b^{5} x^{14} + 14 a^{4} b^{6} x^{15} + \frac {15 a^{3} b^{7} x^{16}}{2} + \frac {45 a^{2} b^{8} x^{17}}{17} + \frac {5 a b^{9} x^{18}}{9} + \frac {b^{10} x^{19}}{19} \]
a**10*x**9/9 + a**9*b*x**10 + 45*a**8*b**2*x**11/11 + 10*a**7*b**3*x**12 + 210*a**6*b**4*x**13/13 + 18*a**5*b**5*x**14 + 14*a**4*b**6*x**15 + 15*a** 3*b**7*x**16/2 + 45*a**2*b**8*x**17/17 + 5*a*b**9*x**18/9 + b**10*x**19/19
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int x^8 (a+b x)^{10} \, dx=\frac {1}{19} \, b^{10} x^{19} + \frac {5}{9} \, a b^{9} x^{18} + \frac {45}{17} \, a^{2} b^{8} x^{17} + \frac {15}{2} \, a^{3} b^{7} x^{16} + 14 \, a^{4} b^{6} x^{15} + 18 \, a^{5} b^{5} x^{14} + \frac {210}{13} \, a^{6} b^{4} x^{13} + 10 \, a^{7} b^{3} x^{12} + \frac {45}{11} \, a^{8} b^{2} x^{11} + a^{9} b x^{10} + \frac {1}{9} \, a^{10} x^{9} \]
1/19*b^10*x^19 + 5/9*a*b^9*x^18 + 45/17*a^2*b^8*x^17 + 15/2*a^3*b^7*x^16 + 14*a^4*b^6*x^15 + 18*a^5*b^5*x^14 + 210/13*a^6*b^4*x^13 + 10*a^7*b^3*x^12 + 45/11*a^8*b^2*x^11 + a^9*b*x^10 + 1/9*a^10*x^9
Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int x^8 (a+b x)^{10} \, dx=\frac {1}{19} \, b^{10} x^{19} + \frac {5}{9} \, a b^{9} x^{18} + \frac {45}{17} \, a^{2} b^{8} x^{17} + \frac {15}{2} \, a^{3} b^{7} x^{16} + 14 \, a^{4} b^{6} x^{15} + 18 \, a^{5} b^{5} x^{14} + \frac {210}{13} \, a^{6} b^{4} x^{13} + 10 \, a^{7} b^{3} x^{12} + \frac {45}{11} \, a^{8} b^{2} x^{11} + a^{9} b x^{10} + \frac {1}{9} \, a^{10} x^{9} \]
1/19*b^10*x^19 + 5/9*a*b^9*x^18 + 45/17*a^2*b^8*x^17 + 15/2*a^3*b^7*x^16 + 14*a^4*b^6*x^15 + 18*a^5*b^5*x^14 + 210/13*a^6*b^4*x^13 + 10*a^7*b^3*x^12 + 45/11*a^8*b^2*x^11 + a^9*b*x^10 + 1/9*a^10*x^9
Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.76 \[ \int x^8 (a+b x)^{10} \, dx=\frac {a^{10}\,x^9}{9}+a^9\,b\,x^{10}+\frac {45\,a^8\,b^2\,x^{11}}{11}+10\,a^7\,b^3\,x^{12}+\frac {210\,a^6\,b^4\,x^{13}}{13}+18\,a^5\,b^5\,x^{14}+14\,a^4\,b^6\,x^{15}+\frac {15\,a^3\,b^7\,x^{16}}{2}+\frac {45\,a^2\,b^8\,x^{17}}{17}+\frac {5\,a\,b^9\,x^{18}}{9}+\frac {b^{10}\,x^{19}}{19} \]