Integrand size = 20, antiderivative size = 95 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {a^2 (A b-a B) \left (a+b x^3\right )^6}{18 b^4}-\frac {a (2 A b-3 a B) \left (a+b x^3\right )^7}{21 b^4}+\frac {(A b-3 a B) \left (a+b x^3\right )^8}{24 b^4}+\frac {B \left (a+b x^3\right )^9}{27 b^4} \] Output:
1/18*a^2*(A*b-B*a)*(b*x^3+a)^6/b^4-1/21*a*(2*A*b-3*B*a)*(b*x^3+a)^7/b^4+1/ 24*(A*b-3*B*a)*(b*x^3+a)^8/b^4+1/27*B*(b*x^3+a)^9/b^4
Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {x^9 \left (168 a^5 A+126 a^4 (5 A b+a B) x^3+504 a^3 b (2 A b+a B) x^6+840 a^2 b^2 (A b+a B) x^9+360 a b^3 (A b+2 a B) x^{12}+63 b^4 (A b+5 a B) x^{15}+56 b^5 B x^{18}\right )}{1512} \] Input:
Integrate[x^8*(a + b*x^3)^5*(A + B*x^3),x]
Output:
(x^9*(168*a^5*A + 126*a^4*(5*A*b + a*B)*x^3 + 504*a^3*b*(2*A*b + a*B)*x^6 + 840*a^2*b^2*(A*b + a*B)*x^9 + 360*a*b^3*(A*b + 2*a*B)*x^12 + 63*b^4*(A*b + 5*a*B)*x^15 + 56*b^5*B*x^18))/1512
Time = 0.47 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {948, 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.
\(\displaystyle \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^6 \left (b x^3+a\right )^5 \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^8}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^7}{b^3}+\frac {a (3 a B-2 A b) \left (b x^3+a\right )^6}{b^3}-\frac {a^2 (a B-A b) \left (b x^3+a\right )^5}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {a^2 \left (a+b x^3\right )^6 (A b-a B)}{6 b^4}+\frac {\left (a+b x^3\right )^8 (A b-3 a B)}{8 b^4}-\frac {a \left (a+b x^3\right )^7 (2 A b-3 a B)}{7 b^4}+\frac {B \left (a+b x^3\right )^9}{9 b^4}\right )\) |
Input:
Int[x^8*(a + b*x^3)^5*(A + B*x^3),x]
Output:
((a^2*(A*b - a*B)*(a + b*x^3)^6)/(6*b^4) - (a*(2*A*b - 3*a*B)*(a + b*x^3)^ 7)/(7*b^4) + ((A*b - 3*a*B)*(a + b*x^3)^8)/(8*b^4) + (B*(a + b*x^3)^9)/(9* b^4))/3
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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.70 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.27
method | result | size |
norman | \(\frac {b^{5} B \,x^{27}}{27}+\left (\frac {5}{9} a^{2} b^{3} A +\frac {5}{9} a^{3} b^{2} B \right ) x^{18}+\left (\frac {2}{3} a^{3} b^{2} A +\frac {1}{3} a^{4} b B \right ) x^{15}+\left (\frac {5}{12} a^{4} b A +\frac {1}{12} a^{5} B \right ) x^{12}+\left (\frac {5}{21} a \,b^{4} A +\frac {10}{21} a^{2} b^{3} B \right ) x^{21}+\left (\frac {1}{24} b^{5} A +\frac {5}{24} a \,b^{4} B \right ) x^{24}+\frac {a^{5} A \,x^{9}}{9}\) | \(121\) |
default | \(\frac {b^{5} B \,x^{27}}{27}+\frac {\left (b^{5} A +5 a \,b^{4} B \right ) x^{24}}{24}+\frac {\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{21}}{21}+\frac {\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{18}}{18}+\frac {\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{15}}{15}+\frac {\left (5 a^{4} b A +a^{5} B \right ) x^{12}}{12}+\frac {a^{5} A \,x^{9}}{9}\) | \(124\) |
gosper | \(\frac {1}{27} b^{5} B \,x^{27}+\frac {5}{9} x^{18} a^{2} b^{3} A +\frac {5}{9} x^{18} a^{3} b^{2} B +\frac {2}{3} x^{15} a^{3} b^{2} A +\frac {1}{3} x^{15} a^{4} b B +\frac {5}{12} x^{12} a^{4} b A +\frac {1}{12} x^{12} a^{5} B +\frac {5}{21} x^{21} a \,b^{4} A +\frac {10}{21} x^{21} a^{2} b^{3} B +\frac {1}{24} x^{24} b^{5} A +\frac {5}{24} x^{24} a \,b^{4} B +\frac {1}{9} a^{5} A \,x^{9}\) | \(126\) |
risch | \(\frac {1}{27} b^{5} B \,x^{27}+\frac {5}{9} x^{18} a^{2} b^{3} A +\frac {5}{9} x^{18} a^{3} b^{2} B +\frac {2}{3} x^{15} a^{3} b^{2} A +\frac {1}{3} x^{15} a^{4} b B +\frac {5}{12} x^{12} a^{4} b A +\frac {1}{12} x^{12} a^{5} B +\frac {5}{21} x^{21} a \,b^{4} A +\frac {10}{21} x^{21} a^{2} b^{3} B +\frac {1}{24} x^{24} b^{5} A +\frac {5}{24} x^{24} a \,b^{4} B +\frac {1}{9} a^{5} A \,x^{9}\) | \(126\) |
parallelrisch | \(\frac {1}{27} b^{5} B \,x^{27}+\frac {5}{9} x^{18} a^{2} b^{3} A +\frac {5}{9} x^{18} a^{3} b^{2} B +\frac {2}{3} x^{15} a^{3} b^{2} A +\frac {1}{3} x^{15} a^{4} b B +\frac {5}{12} x^{12} a^{4} b A +\frac {1}{12} x^{12} a^{5} B +\frac {5}{21} x^{21} a \,b^{4} A +\frac {10}{21} x^{21} a^{2} b^{3} B +\frac {1}{24} x^{24} b^{5} A +\frac {5}{24} x^{24} a \,b^{4} B +\frac {1}{9} a^{5} A \,x^{9}\) | \(126\) |
orering | \(\frac {x^{9} \left (56 b^{5} B \,x^{18}+63 A \,b^{5} x^{15}+315 B a \,b^{4} x^{15}+360 a A \,b^{4} x^{12}+720 B \,a^{2} b^{3} x^{12}+840 a^{2} A \,b^{3} x^{9}+840 B \,a^{3} b^{2} x^{9}+1008 a^{3} A \,b^{2} x^{6}+504 B \,a^{4} b \,x^{6}+630 a^{4} A b \,x^{3}+126 B \,a^{5} x^{3}+168 a^{5} A \right )}{1512}\) | \(128\) |
Input:
int(x^8*(b*x^3+a)^5*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
1/27*b^5*B*x^27+(5/9*a^2*b^3*A+5/9*a^3*b^2*B)*x^18+(2/3*a^3*b^2*A+1/3*a^4* b*B)*x^15+(5/12*a^4*b*A+1/12*a^5*B)*x^12+(5/21*a*b^4*A+10/21*a^2*b^3*B)*x^ 21+(1/24*b^5*A+5/24*a*b^4*B)*x^24+1/9*a^5*A*x^9
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.25 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {1}{27} \, B b^{5} x^{27} + \frac {1}{24} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{24} + \frac {5}{21} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{21} + \frac {5}{9} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{18} + \frac {1}{3} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{15} + \frac {1}{9} \, A a^{5} x^{9} + \frac {1}{12} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{12} \] Input:
integrate(x^8*(b*x^3+a)^5*(B*x^3+A),x, algorithm="fricas")
Output:
1/27*B*b^5*x^27 + 1/24*(5*B*a*b^4 + A*b^5)*x^24 + 5/21*(2*B*a^2*b^3 + A*a* b^4)*x^21 + 5/9*(B*a^3*b^2 + A*a^2*b^3)*x^18 + 1/3*(B*a^4*b + 2*A*a^3*b^2) *x^15 + 1/9*A*a^5*x^9 + 1/12*(B*a^5 + 5*A*a^4*b)*x^12
Time = 0.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.43 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {A a^{5} x^{9}}{9} + \frac {B b^{5} x^{27}}{27} + x^{24} \left (\frac {A b^{5}}{24} + \frac {5 B a b^{4}}{24}\right ) + x^{21} \cdot \left (\frac {5 A a b^{4}}{21} + \frac {10 B a^{2} b^{3}}{21}\right ) + x^{18} \cdot \left (\frac {5 A a^{2} b^{3}}{9} + \frac {5 B a^{3} b^{2}}{9}\right ) + x^{15} \cdot \left (\frac {2 A a^{3} b^{2}}{3} + \frac {B a^{4} b}{3}\right ) + x^{12} \cdot \left (\frac {5 A a^{4} b}{12} + \frac {B a^{5}}{12}\right ) \] Input:
integrate(x**8*(b*x**3+a)**5*(B*x**3+A),x)
Output:
A*a**5*x**9/9 + B*b**5*x**27/27 + x**24*(A*b**5/24 + 5*B*a*b**4/24) + x**2 1*(5*A*a*b**4/21 + 10*B*a**2*b**3/21) + x**18*(5*A*a**2*b**3/9 + 5*B*a**3* b**2/9) + x**15*(2*A*a**3*b**2/3 + B*a**4*b/3) + x**12*(5*A*a**4*b/12 + B* a**5/12)
Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.25 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {1}{27} \, B b^{5} x^{27} + \frac {1}{24} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{24} + \frac {5}{21} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{21} + \frac {5}{9} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{18} + \frac {1}{3} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{15} + \frac {1}{9} \, A a^{5} x^{9} + \frac {1}{12} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{12} \] Input:
integrate(x^8*(b*x^3+a)^5*(B*x^3+A),x, algorithm="maxima")
Output:
1/27*B*b^5*x^27 + 1/24*(5*B*a*b^4 + A*b^5)*x^24 + 5/21*(2*B*a^2*b^3 + A*a* b^4)*x^21 + 5/9*(B*a^3*b^2 + A*a^2*b^3)*x^18 + 1/3*(B*a^4*b + 2*A*a^3*b^2) *x^15 + 1/9*A*a^5*x^9 + 1/12*(B*a^5 + 5*A*a^4*b)*x^12
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {1}{27} \, B b^{5} x^{27} + \frac {5}{24} \, B a b^{4} x^{24} + \frac {1}{24} \, A b^{5} x^{24} + \frac {10}{21} \, B a^{2} b^{3} x^{21} + \frac {5}{21} \, A a b^{4} x^{21} + \frac {5}{9} \, B a^{3} b^{2} x^{18} + \frac {5}{9} \, A a^{2} b^{3} x^{18} + \frac {1}{3} \, B a^{4} b x^{15} + \frac {2}{3} \, A a^{3} b^{2} x^{15} + \frac {1}{12} \, B a^{5} x^{12} + \frac {5}{12} \, A a^{4} b x^{12} + \frac {1}{9} \, A a^{5} x^{9} \] Input:
integrate(x^8*(b*x^3+a)^5*(B*x^3+A),x, algorithm="giac")
Output:
1/27*B*b^5*x^27 + 5/24*B*a*b^4*x^24 + 1/24*A*b^5*x^24 + 10/21*B*a^2*b^3*x^ 21 + 5/21*A*a*b^4*x^21 + 5/9*B*a^3*b^2*x^18 + 5/9*A*a^2*b^3*x^18 + 1/3*B*a ^4*b*x^15 + 2/3*A*a^3*b^2*x^15 + 1/12*B*a^5*x^12 + 5/12*A*a^4*b*x^12 + 1/9 *A*a^5*x^9
Time = 0.70 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=x^{12}\,\left (\frac {B\,a^5}{12}+\frac {5\,A\,b\,a^4}{12}\right )+x^{24}\,\left (\frac {A\,b^5}{24}+\frac {5\,B\,a\,b^4}{24}\right )+\frac {A\,a^5\,x^9}{9}+\frac {B\,b^5\,x^{27}}{27}+\frac {5\,a^2\,b^2\,x^{18}\,\left (A\,b+B\,a\right )}{9}+\frac {a^3\,b\,x^{15}\,\left (2\,A\,b+B\,a\right )}{3}+\frac {5\,a\,b^3\,x^{21}\,\left (A\,b+2\,B\,a\right )}{21} \] Input:
int(x^8*(A + B*x^3)*(a + b*x^3)^5,x)
Output:
x^12*((B*a^5)/12 + (5*A*a^4*b)/12) + x^24*((A*b^5)/24 + (5*B*a*b^4)/24) + (A*a^5*x^9)/9 + (B*b^5*x^27)/27 + (5*a^2*b^2*x^18*(A*b + B*a))/9 + (a^3*b* x^15*(2*A*b + B*a))/3 + (5*a*b^3*x^21*(A*b + 2*B*a))/21
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int x^8 \left (a+b x^3\right )^5 \left (A+B x^3\right ) \, dx=\frac {x^{9} \left (28 b^{6} x^{18}+189 a \,b^{5} x^{15}+540 a^{2} b^{4} x^{12}+840 a^{3} b^{3} x^{9}+756 a^{4} b^{2} x^{6}+378 a^{5} b \,x^{3}+84 a^{6}\right )}{756} \] Input:
int(x^8*(b*x^3+a)^5*(B*x^3+A),x)
Output:
(x**9*(84*a**6 + 378*a**5*b*x**3 + 756*a**4*b**2*x**6 + 840*a**3*b**3*x**9 + 540*a**2*b**4*x**12 + 189*a*b**5*x**15 + 28*b**6*x**18))/756