Integrand size = 13, antiderivative size = 72 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=-\frac {a^3 \left (a+b x^3\right )^9}{27 b^4}+\frac {a^2 \left (a+b x^3\right )^{10}}{10 b^4}-\frac {a \left (a+b x^3\right )^{11}}{11 b^4}+\frac {\left (a+b x^3\right )^{12}}{36 b^4} \] Output:
-1/27*a^3*(b*x^3+a)^9/b^4+1/10*a^2*(b*x^3+a)^10/b^4-1/11*a*(b*x^3+a)^11/b^ 4+1/36*(b*x^3+a)^12/b^4
Time = 0.00 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.50 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {a^8 x^{12}}{12}+\frac {8}{15} a^7 b x^{15}+\frac {14}{9} a^6 b^2 x^{18}+\frac {8}{3} a^5 b^3 x^{21}+\frac {35}{12} a^4 b^4 x^{24}+\frac {56}{27} a^3 b^5 x^{27}+\frac {14}{15} a^2 b^6 x^{30}+\frac {8}{33} a b^7 x^{33}+\frac {b^8 x^{36}}{36} \] Input:
Integrate[x^11*(a + b*x^3)^8,x]
Output:
(a^8*x^12)/12 + (8*a^7*b*x^15)/15 + (14*a^6*b^2*x^18)/9 + (8*a^5*b^3*x^21) /3 + (35*a^4*b^4*x^24)/12 + (56*a^3*b^5*x^27)/27 + (14*a^2*b^6*x^30)/15 + (8*a*b^7*x^33)/33 + (b^8*x^36)/36
Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {798, 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.
\(\displaystyle \int x^{11} \left (a+b x^3\right )^8 \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int x^9 \left (b x^3+a\right )^8dx^3\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{3} \int \left (\frac {\left (b x^3+a\right )^{11}}{b^3}-\frac {3 a \left (b x^3+a\right )^{10}}{b^3}+\frac {3 a^2 \left (b x^3+a\right )^9}{b^3}-\frac {a^3 \left (b x^3+a\right )^8}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {a^3 \left (a+b x^3\right )^9}{9 b^4}+\frac {3 a^2 \left (a+b x^3\right )^{10}}{10 b^4}+\frac {\left (a+b x^3\right )^{12}}{12 b^4}-\frac {3 a \left (a+b x^3\right )^{11}}{11 b^4}\right )\) |
Input:
Int[x^11*(a + b*x^3)^8,x]
Output:
(-1/9*(a^3*(a + b*x^3)^9)/b^4 + (3*a^2*(a + b*x^3)^10)/(10*b^4) - (3*a*(a + b*x^3)^11)/(11*b^4) + (a + b*x^3)^12/(12*b^4))/3
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.42 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.26
method | result | size |
gosper | \(\frac {14}{15} a^{2} b^{6} x^{30}+\frac {14}{9} a^{6} b^{2} x^{18}+\frac {1}{36} b^{8} x^{36}+\frac {1}{12} a^{8} x^{12}+\frac {56}{27} a^{3} b^{5} x^{27}+\frac {35}{12} a^{4} b^{4} x^{24}+\frac {8}{33} a \,b^{7} x^{33}+\frac {8}{15} a^{7} b \,x^{15}+\frac {8}{3} a^{5} b^{3} x^{21}\) | \(91\) |
default | \(\frac {14}{15} a^{2} b^{6} x^{30}+\frac {14}{9} a^{6} b^{2} x^{18}+\frac {1}{36} b^{8} x^{36}+\frac {1}{12} a^{8} x^{12}+\frac {56}{27} a^{3} b^{5} x^{27}+\frac {35}{12} a^{4} b^{4} x^{24}+\frac {8}{33} a \,b^{7} x^{33}+\frac {8}{15} a^{7} b \,x^{15}+\frac {8}{3} a^{5} b^{3} x^{21}\) | \(91\) |
norman | \(\frac {14}{15} a^{2} b^{6} x^{30}+\frac {14}{9} a^{6} b^{2} x^{18}+\frac {1}{36} b^{8} x^{36}+\frac {1}{12} a^{8} x^{12}+\frac {56}{27} a^{3} b^{5} x^{27}+\frac {35}{12} a^{4} b^{4} x^{24}+\frac {8}{33} a \,b^{7} x^{33}+\frac {8}{15} a^{7} b \,x^{15}+\frac {8}{3} a^{5} b^{3} x^{21}\) | \(91\) |
risch | \(\frac {14}{15} a^{2} b^{6} x^{30}+\frac {14}{9} a^{6} b^{2} x^{18}+\frac {1}{36} b^{8} x^{36}+\frac {1}{12} a^{8} x^{12}+\frac {56}{27} a^{3} b^{5} x^{27}+\frac {35}{12} a^{4} b^{4} x^{24}+\frac {8}{33} a \,b^{7} x^{33}+\frac {8}{15} a^{7} b \,x^{15}+\frac {8}{3} a^{5} b^{3} x^{21}\) | \(91\) |
parallelrisch | \(\frac {14}{15} a^{2} b^{6} x^{30}+\frac {14}{9} a^{6} b^{2} x^{18}+\frac {1}{36} b^{8} x^{36}+\frac {1}{12} a^{8} x^{12}+\frac {56}{27} a^{3} b^{5} x^{27}+\frac {35}{12} a^{4} b^{4} x^{24}+\frac {8}{33} a \,b^{7} x^{33}+\frac {8}{15} a^{7} b \,x^{15}+\frac {8}{3} a^{5} b^{3} x^{21}\) | \(91\) |
orering | \(\frac {x^{12} \left (165 b^{8} x^{24}+1440 a \,b^{7} x^{21}+5544 a^{2} b^{6} x^{18}+12320 a^{3} b^{5} x^{15}+17325 a^{4} b^{4} x^{12}+15840 a^{5} b^{3} x^{9}+9240 a^{6} b^{2} x^{6}+3168 a^{7} b \,x^{3}+495 a^{8}\right )}{5940}\) | \(93\) |
Input:
int(x^11*(b*x^3+a)^8,x,method=_RETURNVERBOSE)
Output:
14/15*a^2*b^6*x^30+14/9*a^6*b^2*x^18+1/36*b^8*x^36+1/12*a^8*x^12+56/27*a^3 *b^5*x^27+35/12*a^4*b^4*x^24+8/33*a*b^7*x^33+8/15*a^7*b*x^15+8/3*a^5*b^3*x ^21
Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {1}{36} \, b^{8} x^{36} + \frac {8}{33} \, a b^{7} x^{33} + \frac {14}{15} \, a^{2} b^{6} x^{30} + \frac {56}{27} \, a^{3} b^{5} x^{27} + \frac {35}{12} \, a^{4} b^{4} x^{24} + \frac {8}{3} \, a^{5} b^{3} x^{21} + \frac {14}{9} \, a^{6} b^{2} x^{18} + \frac {8}{15} \, a^{7} b x^{15} + \frac {1}{12} \, a^{8} x^{12} \] Input:
integrate(x^11*(b*x^3+a)^8,x, algorithm="fricas")
Output:
1/36*b^8*x^36 + 8/33*a*b^7*x^33 + 14/15*a^2*b^6*x^30 + 56/27*a^3*b^5*x^27 + 35/12*a^4*b^4*x^24 + 8/3*a^5*b^3*x^21 + 14/9*a^6*b^2*x^18 + 8/15*a^7*b*x ^15 + 1/12*a^8*x^12
Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.49 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {a^{8} x^{12}}{12} + \frac {8 a^{7} b x^{15}}{15} + \frac {14 a^{6} b^{2} x^{18}}{9} + \frac {8 a^{5} b^{3} x^{21}}{3} + \frac {35 a^{4} b^{4} x^{24}}{12} + \frac {56 a^{3} b^{5} x^{27}}{27} + \frac {14 a^{2} b^{6} x^{30}}{15} + \frac {8 a b^{7} x^{33}}{33} + \frac {b^{8} x^{36}}{36} \] Input:
integrate(x**11*(b*x**3+a)**8,x)
Output:
a**8*x**12/12 + 8*a**7*b*x**15/15 + 14*a**6*b**2*x**18/9 + 8*a**5*b**3*x** 21/3 + 35*a**4*b**4*x**24/12 + 56*a**3*b**5*x**27/27 + 14*a**2*b**6*x**30/ 15 + 8*a*b**7*x**33/33 + b**8*x**36/36
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {1}{36} \, b^{8} x^{36} + \frac {8}{33} \, a b^{7} x^{33} + \frac {14}{15} \, a^{2} b^{6} x^{30} + \frac {56}{27} \, a^{3} b^{5} x^{27} + \frac {35}{12} \, a^{4} b^{4} x^{24} + \frac {8}{3} \, a^{5} b^{3} x^{21} + \frac {14}{9} \, a^{6} b^{2} x^{18} + \frac {8}{15} \, a^{7} b x^{15} + \frac {1}{12} \, a^{8} x^{12} \] Input:
integrate(x^11*(b*x^3+a)^8,x, algorithm="maxima")
Output:
1/36*b^8*x^36 + 8/33*a*b^7*x^33 + 14/15*a^2*b^6*x^30 + 56/27*a^3*b^5*x^27 + 35/12*a^4*b^4*x^24 + 8/3*a^5*b^3*x^21 + 14/9*a^6*b^2*x^18 + 8/15*a^7*b*x ^15 + 1/12*a^8*x^12
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {1}{36} \, b^{8} x^{36} + \frac {8}{33} \, a b^{7} x^{33} + \frac {14}{15} \, a^{2} b^{6} x^{30} + \frac {56}{27} \, a^{3} b^{5} x^{27} + \frac {35}{12} \, a^{4} b^{4} x^{24} + \frac {8}{3} \, a^{5} b^{3} x^{21} + \frac {14}{9} \, a^{6} b^{2} x^{18} + \frac {8}{15} \, a^{7} b x^{15} + \frac {1}{12} \, a^{8} x^{12} \] Input:
integrate(x^11*(b*x^3+a)^8,x, algorithm="giac")
Output:
1/36*b^8*x^36 + 8/33*a*b^7*x^33 + 14/15*a^2*b^6*x^30 + 56/27*a^3*b^5*x^27 + 35/12*a^4*b^4*x^24 + 8/3*a^5*b^3*x^21 + 14/9*a^6*b^2*x^18 + 8/15*a^7*b*x ^15 + 1/12*a^8*x^12
Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {a^8\,x^{12}}{12}+\frac {8\,a^7\,b\,x^{15}}{15}+\frac {14\,a^6\,b^2\,x^{18}}{9}+\frac {8\,a^5\,b^3\,x^{21}}{3}+\frac {35\,a^4\,b^4\,x^{24}}{12}+\frac {56\,a^3\,b^5\,x^{27}}{27}+\frac {14\,a^2\,b^6\,x^{30}}{15}+\frac {8\,a\,b^7\,x^{33}}{33}+\frac {b^8\,x^{36}}{36} \] Input:
int(x^11*(a + b*x^3)^8,x)
Output:
(a^8*x^12)/12 + (b^8*x^36)/36 + (8*a^7*b*x^15)/15 + (8*a*b^7*x^33)/33 + (1 4*a^6*b^2*x^18)/9 + (8*a^5*b^3*x^21)/3 + (35*a^4*b^4*x^24)/12 + (56*a^3*b^ 5*x^27)/27 + (14*a^2*b^6*x^30)/15
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28 \[ \int x^{11} \left (a+b x^3\right )^8 \, dx=\frac {x^{12} \left (165 b^{8} x^{24}+1440 a \,b^{7} x^{21}+5544 a^{2} b^{6} x^{18}+12320 a^{3} b^{5} x^{15}+17325 a^{4} b^{4} x^{12}+15840 a^{5} b^{3} x^{9}+9240 a^{6} b^{2} x^{6}+3168 a^{7} b \,x^{3}+495 a^{8}\right )}{5940} \] Input:
int(x^11*(b*x^3+a)^8,x)
Output:
(x**12*(495*a**8 + 3168*a**7*b*x**3 + 9240*a**6*b**2*x**6 + 15840*a**5*b** 3*x**9 + 17325*a**4*b**4*x**12 + 12320*a**3*b**5*x**15 + 5544*a**2*b**6*x* *18 + 1440*a*b**7*x**21 + 165*b**8*x**24))/5940