Integrand size = 20, antiderivative size = 122 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=-\frac {a^3 (A b-a B) \left (a+b x^2\right )^6}{12 b^5}+\frac {a^2 (3 A b-4 a B) \left (a+b x^2\right )^7}{14 b^5}-\frac {3 a (A b-2 a B) \left (a+b x^2\right )^8}{16 b^5}+\frac {(A b-4 a B) \left (a+b x^2\right )^9}{18 b^5}+\frac {B \left (a+b x^2\right )^{10}}{20 b^5} \]
-1/12*a^3*(A*b-B*a)*(b*x^2+a)^6/b^5+1/14*a^2*(3*A*b-4*B*a)*(b*x^2+a)^7/b^5 -3/16*a*(A*b-2*B*a)*(b*x^2+a)^8/b^5+1/18*(A*b-4*B*a)*(b*x^2+a)^9/b^5+1/20* B*(b*x^2+a)^10/b^5
Time = 0.01 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{8} a^5 A x^8+\frac {1}{10} a^4 (5 A b+a B) x^{10}+\frac {5}{12} a^3 b (2 A b+a B) x^{12}+\frac {5}{7} a^2 b^2 (A b+a B) x^{14}+\frac {5}{16} a b^3 (A b+2 a B) x^{16}+\frac {1}{18} b^4 (A b+5 a B) x^{18}+\frac {1}{20} b^5 B x^{20} \]
(a^5*A*x^8)/8 + (a^4*(5*A*b + a*B)*x^10)/10 + (5*a^3*b*(2*A*b + a*B)*x^12) /12 + (5*a^2*b^2*(A*b + a*B)*x^14)/7 + (5*a*b^3*(A*b + 2*a*B)*x^16)/16 + ( b^4*(A*b + 5*a*B)*x^18)/18 + (b^5*B*x^20)/20
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {354, 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.
\(\displaystyle \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^6 \left (b x^2+a\right )^5 \left (B x^2+A\right )dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B \left (b x^2+a\right )^9}{b^4}+\frac {(A b-4 a B) \left (b x^2+a\right )^8}{b^4}+\frac {3 a (2 a B-A b) \left (b x^2+a\right )^7}{b^4}-\frac {a^2 (4 a B-3 A b) \left (b x^2+a\right )^6}{b^4}+\frac {a^3 (a B-A b) \left (b x^2+a\right )^5}{b^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^3 \left (a+b x^2\right )^6 (A b-a B)}{6 b^5}+\frac {a^2 \left (a+b x^2\right )^7 (3 A b-4 a B)}{7 b^5}+\frac {\left (a+b x^2\right )^9 (A b-4 a B)}{9 b^5}-\frac {3 a \left (a+b x^2\right )^8 (A b-2 a B)}{8 b^5}+\frac {B \left (a+b x^2\right )^{10}}{10 b^5}\right )\) |
(-1/6*(a^3*(A*b - a*B)*(a + b*x^2)^6)/b^5 + (a^2*(3*A*b - 4*a*B)*(a + b*x^ 2)^7)/(7*b^5) - (3*a*(A*b - 2*a*B)*(a + b*x^2)^8)/(8*b^5) + ((A*b - 4*a*B) *(a + b*x^2)^9)/(9*b^5) + (B*(a + b*x^2)^10)/(10*b^5))/2
3.1.25.3.1 Defintions of rubi rules used
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_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.52 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {a^{5} A \,x^{8}}{8}+\left (\frac {1}{2} a^{4} b A +\frac {1}{10} a^{5} B \right ) x^{10}+\left (\frac {5}{6} a^{3} b^{2} A +\frac {5}{12} a^{4} b B \right ) x^{12}+\left (\frac {5}{7} a^{2} b^{3} A +\frac {5}{7} a^{3} b^{2} B \right ) x^{14}+\left (\frac {5}{16} a \,b^{4} A +\frac {5}{8} a^{2} b^{3} B \right ) x^{16}+\left (\frac {1}{18} b^{5} A +\frac {5}{18} a \,b^{4} B \right ) x^{18}+\frac {b^{5} B \,x^{20}}{20}\) | \(121\) |
default | \(\frac {b^{5} B \,x^{20}}{20}+\frac {\left (b^{5} A +5 a \,b^{4} B \right ) x^{18}}{18}+\frac {\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{16}}{16}+\frac {\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{14}}{14}+\frac {\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{12}}{12}+\frac {\left (5 a^{4} b A +a^{5} B \right ) x^{10}}{10}+\frac {a^{5} A \,x^{8}}{8}\) | \(124\) |
gosper | \(\frac {1}{8} a^{5} A \,x^{8}+\frac {1}{2} x^{10} a^{4} b A +\frac {1}{10} x^{10} a^{5} B +\frac {5}{6} x^{12} a^{3} b^{2} A +\frac {5}{12} x^{12} a^{4} b B +\frac {5}{7} x^{14} a^{2} b^{3} A +\frac {5}{7} x^{14} a^{3} b^{2} B +\frac {5}{16} x^{16} a \,b^{4} A +\frac {5}{8} x^{16} a^{2} b^{3} B +\frac {1}{18} x^{18} b^{5} A +\frac {5}{18} x^{18} a \,b^{4} B +\frac {1}{20} b^{5} B \,x^{20}\) | \(126\) |
risch | \(\frac {1}{8} a^{5} A \,x^{8}+\frac {1}{2} x^{10} a^{4} b A +\frac {1}{10} x^{10} a^{5} B +\frac {5}{6} x^{12} a^{3} b^{2} A +\frac {5}{12} x^{12} a^{4} b B +\frac {5}{7} x^{14} a^{2} b^{3} A +\frac {5}{7} x^{14} a^{3} b^{2} B +\frac {5}{16} x^{16} a \,b^{4} A +\frac {5}{8} x^{16} a^{2} b^{3} B +\frac {1}{18} x^{18} b^{5} A +\frac {5}{18} x^{18} a \,b^{4} B +\frac {1}{20} b^{5} B \,x^{20}\) | \(126\) |
parallelrisch | \(\frac {1}{8} a^{5} A \,x^{8}+\frac {1}{2} x^{10} a^{4} b A +\frac {1}{10} x^{10} a^{5} B +\frac {5}{6} x^{12} a^{3} b^{2} A +\frac {5}{12} x^{12} a^{4} b B +\frac {5}{7} x^{14} a^{2} b^{3} A +\frac {5}{7} x^{14} a^{3} b^{2} B +\frac {5}{16} x^{16} a \,b^{4} A +\frac {5}{8} x^{16} a^{2} b^{3} B +\frac {1}{18} x^{18} b^{5} A +\frac {5}{18} x^{18} a \,b^{4} B +\frac {1}{20} b^{5} B \,x^{20}\) | \(126\) |
1/8*a^5*A*x^8+(1/2*a^4*b*A+1/10*a^5*B)*x^10+(5/6*a^3*b^2*A+5/12*a^4*b*B)*x ^12+(5/7*a^2*b^3*A+5/7*a^3*b^2*B)*x^14+(5/16*a*b^4*A+5/8*a^2*b^3*B)*x^16+( 1/18*b^5*A+5/18*a*b^4*B)*x^18+1/20*b^5*B*x^20
Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{20} \, B b^{5} x^{20} + \frac {1}{18} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{18} + \frac {5}{16} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{16} + \frac {5}{7} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{14} + \frac {1}{8} \, A a^{5} x^{8} + \frac {5}{12} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{12} + \frac {1}{10} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{10} \]
1/20*B*b^5*x^20 + 1/18*(5*B*a*b^4 + A*b^5)*x^18 + 5/16*(2*B*a^2*b^3 + A*a* b^4)*x^16 + 5/7*(B*a^3*b^2 + A*a^2*b^3)*x^14 + 1/8*A*a^5*x^8 + 5/12*(B*a^4 *b + 2*A*a^3*b^2)*x^12 + 1/10*(B*a^5 + 5*A*a^4*b)*x^10
Time = 0.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {A a^{5} x^{8}}{8} + \frac {B b^{5} x^{20}}{20} + x^{18} \left (\frac {A b^{5}}{18} + \frac {5 B a b^{4}}{18}\right ) + x^{16} \cdot \left (\frac {5 A a b^{4}}{16} + \frac {5 B a^{2} b^{3}}{8}\right ) + x^{14} \cdot \left (\frac {5 A a^{2} b^{3}}{7} + \frac {5 B a^{3} b^{2}}{7}\right ) + x^{12} \cdot \left (\frac {5 A a^{3} b^{2}}{6} + \frac {5 B a^{4} b}{12}\right ) + x^{10} \left (\frac {A a^{4} b}{2} + \frac {B a^{5}}{10}\right ) \]
A*a**5*x**8/8 + B*b**5*x**20/20 + x**18*(A*b**5/18 + 5*B*a*b**4/18) + x**1 6*(5*A*a*b**4/16 + 5*B*a**2*b**3/8) + x**14*(5*A*a**2*b**3/7 + 5*B*a**3*b* *2/7) + x**12*(5*A*a**3*b**2/6 + 5*B*a**4*b/12) + x**10*(A*a**4*b/2 + B*a* *5/10)
Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{20} \, B b^{5} x^{20} + \frac {1}{18} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{18} + \frac {5}{16} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{16} + \frac {5}{7} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{14} + \frac {1}{8} \, A a^{5} x^{8} + \frac {5}{12} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{12} + \frac {1}{10} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{10} \]
1/20*B*b^5*x^20 + 1/18*(5*B*a*b^4 + A*b^5)*x^18 + 5/16*(2*B*a^2*b^3 + A*a* b^4)*x^16 + 5/7*(B*a^3*b^2 + A*a^2*b^3)*x^14 + 1/8*A*a^5*x^8 + 5/12*(B*a^4 *b + 2*A*a^3*b^2)*x^12 + 1/10*(B*a^5 + 5*A*a^4*b)*x^10
Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{20} \, B b^{5} x^{20} + \frac {5}{18} \, B a b^{4} x^{18} + \frac {1}{18} \, A b^{5} x^{18} + \frac {5}{8} \, B a^{2} b^{3} x^{16} + \frac {5}{16} \, A a b^{4} x^{16} + \frac {5}{7} \, B a^{3} b^{2} x^{14} + \frac {5}{7} \, A a^{2} b^{3} x^{14} + \frac {5}{12} \, B a^{4} b x^{12} + \frac {5}{6} \, A a^{3} b^{2} x^{12} + \frac {1}{10} \, B a^{5} x^{10} + \frac {1}{2} \, A a^{4} b x^{10} + \frac {1}{8} \, A a^{5} x^{8} \]
1/20*B*b^5*x^20 + 5/18*B*a*b^4*x^18 + 1/18*A*b^5*x^18 + 5/8*B*a^2*b^3*x^16 + 5/16*A*a*b^4*x^16 + 5/7*B*a^3*b^2*x^14 + 5/7*A*a^2*b^3*x^14 + 5/12*B*a^ 4*b*x^12 + 5/6*A*a^3*b^2*x^12 + 1/10*B*a^5*x^10 + 1/2*A*a^4*b*x^10 + 1/8*A *a^5*x^8
Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.88 \[ \int x^7 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=x^{10}\,\left (\frac {B\,a^5}{10}+\frac {A\,b\,a^4}{2}\right )+x^{18}\,\left (\frac {A\,b^5}{18}+\frac {5\,B\,a\,b^4}{18}\right )+\frac {A\,a^5\,x^8}{8}+\frac {B\,b^5\,x^{20}}{20}+\frac {5\,a^2\,b^2\,x^{14}\,\left (A\,b+B\,a\right )}{7}+\frac {5\,a^3\,b\,x^{12}\,\left (2\,A\,b+B\,a\right )}{12}+\frac {5\,a\,b^3\,x^{16}\,\left (A\,b+2\,B\,a\right )}{16} \]