Integrand size = 20, antiderivative size = 67 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=-\frac {a (A b-a B) \left (a+b x^2\right )^6}{12 b^3}+\frac {(A b-2 a B) \left (a+b x^2\right )^7}{14 b^3}+\frac {B \left (a+b x^2\right )^8}{16 b^3} \] Output:
-1/12*a*(A*b-B*a)*(b*x^2+a)^6/b^3+1/14*(A*b-2*B*a)*(b*x^2+a)^7/b^3+1/16*B* (b*x^2+a)^8/b^3
Time = 0.01 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{4} a^5 A x^4+\frac {1}{6} a^4 (5 A b+a B) x^6+\frac {5}{8} a^3 b (2 A b+a B) x^8+a^2 b^2 (A b+a B) x^{10}+\frac {5}{12} a b^3 (A b+2 a B) x^{12}+\frac {1}{14} b^4 (A b+5 a B) x^{14}+\frac {1}{16} b^5 B x^{16} \] Input:
Integrate[x^3*(a + b*x^2)^5*(A + B*x^2),x]
Output:
(a^5*A*x^4)/4 + (a^4*(5*A*b + a*B)*x^6)/6 + (5*a^3*b*(2*A*b + a*B)*x^8)/8 + a^2*b^2*(A*b + a*B)*x^10 + (5*a*b^3*(A*b + 2*a*B)*x^12)/12 + (b^4*(A*b + 5*a*B)*x^14)/14 + (b^5*B*x^16)/16
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.
\(\displaystyle \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b x^2+a\right )^5 \left (B x^2+A\right )dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B \left (b x^2+a\right )^7}{b^2}+\frac {(A b-2 a B) \left (b x^2+a\right )^6}{b^2}+\frac {a (a B-A b) \left (b x^2+a\right )^5}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^7 (A b-2 a B)}{7 b^3}-\frac {a \left (a+b x^2\right )^6 (A b-a B)}{6 b^3}+\frac {B \left (a+b x^2\right )^8}{8 b^3}\right )\) |
Input:
Int[x^3*(a + b*x^2)^5*(A + B*x^2),x]
Output:
(-1/6*(a*(A*b - a*B)*(a + b*x^2)^6)/b^3 + ((A*b - 2*a*B)*(a + b*x^2)^7)/(7 *b^3) + (B*(a + b*x^2)^8)/(8*b^3))/2
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 = 0.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.78
method | result | size |
norman | \(\frac {a^{5} A \,x^{4}}{4}+\left (\frac {5}{6} a^{4} b A +\frac {1}{6} a^{5} B \right ) x^{6}+\left (\frac {5}{4} a^{3} b^{2} A +\frac {5}{8} a^{4} b B \right ) x^{8}+\left (a^{2} b^{3} A +a^{3} b^{2} B \right ) x^{10}+\left (\frac {5}{12} a \,b^{4} A +\frac {5}{6} a^{2} b^{3} B \right ) x^{12}+\left (\frac {1}{14} b^{5} A +\frac {5}{14} a \,b^{4} B \right ) x^{14}+\frac {b^{5} B \,x^{16}}{16}\) | \(119\) |
gosper | \(\frac {1}{4} a^{5} A \,x^{4}+\frac {5}{6} x^{6} a^{4} b A +\frac {1}{6} x^{6} a^{5} B +\frac {5}{4} x^{8} a^{3} b^{2} A +\frac {5}{8} x^{8} a^{4} b B +A \,a^{2} b^{3} x^{10}+B \,a^{3} b^{2} x^{10}+\frac {5}{12} x^{12} a \,b^{4} A +\frac {5}{6} x^{12} a^{2} b^{3} B +\frac {1}{14} x^{14} b^{5} A +\frac {5}{14} x^{14} a \,b^{4} B +\frac {1}{16} b^{5} B \,x^{16}\) | \(124\) |
default | \(\frac {b^{5} B \,x^{16}}{16}+\frac {\left (b^{5} A +5 a \,b^{4} B \right ) x^{14}}{14}+\frac {\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{12}}{12}+\frac {\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{10}}{10}+\frac {\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{8}}{8}+\frac {\left (5 a^{4} b A +a^{5} B \right ) x^{6}}{6}+\frac {a^{5} A \,x^{4}}{4}\) | \(124\) |
risch | \(\frac {1}{4} a^{5} A \,x^{4}+\frac {5}{6} x^{6} a^{4} b A +\frac {1}{6} x^{6} a^{5} B +\frac {5}{4} x^{8} a^{3} b^{2} A +\frac {5}{8} x^{8} a^{4} b B +A \,a^{2} b^{3} x^{10}+B \,a^{3} b^{2} x^{10}+\frac {5}{12} x^{12} a \,b^{4} A +\frac {5}{6} x^{12} a^{2} b^{3} B +\frac {1}{14} x^{14} b^{5} A +\frac {5}{14} x^{14} a \,b^{4} B +\frac {1}{16} b^{5} B \,x^{16}\) | \(124\) |
parallelrisch | \(\frac {1}{4} a^{5} A \,x^{4}+\frac {5}{6} x^{6} a^{4} b A +\frac {1}{6} x^{6} a^{5} B +\frac {5}{4} x^{8} a^{3} b^{2} A +\frac {5}{8} x^{8} a^{4} b B +A \,a^{2} b^{3} x^{10}+B \,a^{3} b^{2} x^{10}+\frac {5}{12} x^{12} a \,b^{4} A +\frac {5}{6} x^{12} a^{2} b^{3} B +\frac {1}{14} x^{14} b^{5} A +\frac {5}{14} x^{14} a \,b^{4} B +\frac {1}{16} b^{5} B \,x^{16}\) | \(124\) |
orering | \(\frac {x^{4} \left (21 b^{5} B \,x^{12}+24 A \,b^{5} x^{10}+120 B a \,b^{4} x^{10}+140 a A \,b^{4} x^{8}+280 B \,a^{2} b^{3} x^{8}+336 a^{2} A \,b^{3} x^{6}+336 B \,a^{3} b^{2} x^{6}+420 a^{3} A \,b^{2} x^{4}+210 B \,a^{4} b \,x^{4}+280 a^{4} A b \,x^{2}+56 B \,a^{5} x^{2}+84 a^{5} A \right )}{336}\) | \(128\) |
Input:
int(x^3*(b*x^2+a)^5*(B*x^2+A),x,method=_RETURNVERBOSE)
Output:
1/4*a^5*A*x^4+(5/6*a^4*b*A+1/6*a^5*B)*x^6+(5/4*a^3*b^2*A+5/8*a^4*b*B)*x^8+ (A*a^2*b^3+B*a^3*b^2)*x^10+(5/12*a*b^4*A+5/6*a^2*b^3*B)*x^12+(1/14*b^5*A+5 /14*a*b^4*B)*x^14+1/16*b^5*B*x^16
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.76 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{16} \, B b^{5} x^{16} + \frac {1}{14} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{14} + \frac {5}{12} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{10} + \frac {1}{4} \, A a^{5} x^{4} + \frac {5}{8} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{8} + \frac {1}{6} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{6} \] Input:
integrate(x^3*(b*x^2+a)^5*(B*x^2+A),x, algorithm="fricas")
Output:
1/16*B*b^5*x^16 + 1/14*(5*B*a*b^4 + A*b^5)*x^14 + 5/12*(2*B*a^2*b^3 + A*a* b^4)*x^12 + (B*a^3*b^2 + A*a^2*b^3)*x^10 + 1/4*A*a^5*x^4 + 5/8*(B*a^4*b + 2*A*a^3*b^2)*x^8 + 1/6*(B*a^5 + 5*A*a^4*b)*x^6
Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (58) = 116\).
Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.96 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {A a^{5} x^{4}}{4} + \frac {B b^{5} x^{16}}{16} + x^{14} \left (\frac {A b^{5}}{14} + \frac {5 B a b^{4}}{14}\right ) + x^{12} \cdot \left (\frac {5 A a b^{4}}{12} + \frac {5 B a^{2} b^{3}}{6}\right ) + x^{10} \left (A a^{2} b^{3} + B a^{3} b^{2}\right ) + x^{8} \cdot \left (\frac {5 A a^{3} b^{2}}{4} + \frac {5 B a^{4} b}{8}\right ) + x^{6} \cdot \left (\frac {5 A a^{4} b}{6} + \frac {B a^{5}}{6}\right ) \] Input:
integrate(x**3*(b*x**2+a)**5*(B*x**2+A),x)
Output:
A*a**5*x**4/4 + B*b**5*x**16/16 + x**14*(A*b**5/14 + 5*B*a*b**4/14) + x**1 2*(5*A*a*b**4/12 + 5*B*a**2*b**3/6) + x**10*(A*a**2*b**3 + B*a**3*b**2) + x**8*(5*A*a**3*b**2/4 + 5*B*a**4*b/8) + x**6*(5*A*a**4*b/6 + B*a**5/6)
Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.76 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{16} \, B b^{5} x^{16} + \frac {1}{14} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{14} + \frac {5}{12} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{10} + \frac {1}{4} \, A a^{5} x^{4} + \frac {5}{8} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{8} + \frac {1}{6} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{6} \] Input:
integrate(x^3*(b*x^2+a)^5*(B*x^2+A),x, algorithm="maxima")
Output:
1/16*B*b^5*x^16 + 1/14*(5*B*a*b^4 + A*b^5)*x^14 + 5/12*(2*B*a^2*b^3 + A*a* b^4)*x^12 + (B*a^3*b^2 + A*a^2*b^3)*x^10 + 1/4*A*a^5*x^4 + 5/8*(B*a^4*b + 2*A*a^3*b^2)*x^8 + 1/6*(B*a^5 + 5*A*a^4*b)*x^6
Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.84 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{16} \, B b^{5} x^{16} + \frac {5}{14} \, B a b^{4} x^{14} + \frac {1}{14} \, A b^{5} x^{14} + \frac {5}{6} \, B a^{2} b^{3} x^{12} + \frac {5}{12} \, A a b^{4} x^{12} + B a^{3} b^{2} x^{10} + A a^{2} b^{3} x^{10} + \frac {5}{8} \, B a^{4} b x^{8} + \frac {5}{4} \, A a^{3} b^{2} x^{8} + \frac {1}{6} \, B a^{5} x^{6} + \frac {5}{6} \, A a^{4} b x^{6} + \frac {1}{4} \, A a^{5} x^{4} \] Input:
integrate(x^3*(b*x^2+a)^5*(B*x^2+A),x, algorithm="giac")
Output:
1/16*B*b^5*x^16 + 5/14*B*a*b^4*x^14 + 1/14*A*b^5*x^14 + 5/6*B*a^2*b^3*x^12 + 5/12*A*a*b^4*x^12 + B*a^3*b^2*x^10 + A*a^2*b^3*x^10 + 5/8*B*a^4*b*x^8 + 5/4*A*a^3*b^2*x^8 + 1/6*B*a^5*x^6 + 5/6*A*a^4*b*x^6 + 1/4*A*a^5*x^4
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.58 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=x^6\,\left (\frac {B\,a^5}{6}+\frac {5\,A\,b\,a^4}{6}\right )+x^{14}\,\left (\frac {A\,b^5}{14}+\frac {5\,B\,a\,b^4}{14}\right )+\frac {A\,a^5\,x^4}{4}+\frac {B\,b^5\,x^{16}}{16}+a^2\,b^2\,x^{10}\,\left (A\,b+B\,a\right )+\frac {5\,a^3\,b\,x^8\,\left (2\,A\,b+B\,a\right )}{8}+\frac {5\,a\,b^3\,x^{12}\,\left (A\,b+2\,B\,a\right )}{12} \] Input:
int(x^3*(A + B*x^2)*(a + b*x^2)^5,x)
Output:
x^6*((B*a^5)/6 + (5*A*a^4*b)/6) + x^14*((A*b^5)/14 + (5*B*a*b^4)/14) + (A* a^5*x^4)/4 + (B*b^5*x^16)/16 + a^2*b^2*x^10*(A*b + B*a) + (5*a^3*b*x^8*(2* A*b + B*a))/8 + (5*a*b^3*x^12*(A*b + 2*B*a))/12
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int x^3 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {x^{4} \left (7 b^{6} x^{12}+48 a \,b^{5} x^{10}+140 a^{2} b^{4} x^{8}+224 a^{3} b^{3} x^{6}+210 a^{4} b^{2} x^{4}+112 a^{5} b \,x^{2}+28 a^{6}\right )}{112} \] Input:
int(x^3*(b*x^2+a)^5*(B*x^2+A),x)
Output:
(x**4*(28*a**6 + 112*a**5*b*x**2 + 210*a**4*b**2*x**4 + 224*a**3*b**3*x**6 + 140*a**2*b**4*x**8 + 48*a*b**5*x**10 + 7*b**6*x**12))/112