Integrand size = 26, antiderivative size = 138 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a^3 B x^3+\frac {1}{4} a^3 C x^4+\frac {1}{5} a^2 (3 b B+a D) x^5+\frac {1}{2} a^2 b C x^6+\frac {3}{7} a b (b B+a D) x^7+\frac {3}{8} a b^2 C x^8+\frac {1}{9} b^2 (b B+3 a D) x^9+\frac {1}{10} b^3 C x^{10}+\frac {1}{11} b^3 D x^{11}+\frac {A \left (a+b x^2\right )^4}{8 b} \]
1/3*a^3*B*x^3+1/4*a^3*C*x^4+1/5*a^2*(3*B*b+D*a)*x^5+1/2*a^2*b*C*x^6+3/7*a* b*(B*b+D*a)*x^7+3/8*a*b^2*C*x^8+1/9*b^2*(B*b+3*D*a)*x^9+1/10*b^3*C*x^10+1/ 11*b^3*D*x^11+1/8*A*(b*x^2+a)^4/b
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {7 b^3 x^8 \left (495 A+4 x \left (110 B+99 C x+90 D x^2\right )\right )+462 a^3 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))+198 a^2 b x^4 (105 A+2 x (42 B+5 x (7 C+6 D x)))+165 a b^2 x^6 (84 A+x (72 B+7 x (9 C+8 D x)))}{27720} \]
(7*b^3*x^8*(495*A + 4*x*(110*B + 99*C*x + 90*D*x^2)) + 462*a^3*x^2*(30*A + x*(20*B + 3*x*(5*C + 4*D*x))) + 198*a^2*b*x^4*(105*A + 2*x*(42*B + 5*x*(7 *C + 6*D*x))) + 165*a*b^2*x^6*(84*A + x*(72*B + 7*x*(9*C + 8*D*x))))/27720
Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2017, 2341, 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.
\(\displaystyle \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^3 \left (x \left (D x^3+C x^2+B x+A\right )-A x\right )dx+\frac {A \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (b^3 D x^{10}+b^3 C x^9+b^2 (b B+3 a D) x^8+3 a b^2 C x^7+3 a b (b B+a D) x^6+3 a^2 b C x^5+a^2 (3 b B+a D) x^4+a^3 C x^3+a^3 B x^2\right )dx+\frac {A \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} a^3 B x^3+\frac {1}{4} a^3 C x^4+\frac {1}{5} a^2 x^5 (a D+3 b B)+\frac {1}{2} a^2 b C x^6+\frac {A \left (a+b x^2\right )^4}{8 b}+\frac {1}{9} b^2 x^9 (3 a D+b B)+\frac {3}{8} a b^2 C x^8+\frac {3}{7} a b x^7 (a D+b B)+\frac {1}{10} b^3 C x^{10}+\frac {1}{11} b^3 D x^{11}\) |
(a^3*B*x^3)/3 + (a^3*C*x^4)/4 + (a^2*(3*b*B + a*D)*x^5)/5 + (a^2*b*C*x^6)/ 2 + (3*a*b*(b*B + a*D)*x^7)/7 + (3*a*b^2*C*x^8)/8 + (b^2*(b*B + 3*a*D)*x^9 )/9 + (b^3*C*x^10)/10 + (b^3*D*x^11)/11 + (A*(a + b*x^2)^4)/(8*b)
3.1.80.3.1 Defintions of rubi rules used
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] *x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p , 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] && !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ [{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 3.42 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\frac {b^{3} D x^{11}}{11}+\frac {b^{3} C \,x^{10}}{10}+\left (\frac {1}{9} B \,b^{3}+\frac {1}{3} a \,b^{2} D\right ) x^{9}+\left (\frac {1}{8} b^{3} A +\frac {3}{8} C \,b^{2} a \right ) x^{8}+\left (\frac {3}{7} a \,b^{2} B +\frac {3}{7} D a^{2} b \right ) x^{7}+\left (\frac {1}{2} a \,b^{2} A +\frac {1}{2} C \,a^{2} b \right ) x^{6}+\left (\frac {3}{5} a^{2} b B +\frac {1}{5} D a^{3}\right ) x^{5}+\left (\frac {3}{4} a^{2} b A +\frac {1}{4} C \,a^{3}\right ) x^{4}+\frac {a^{3} B \,x^{3}}{3}+\frac {a^{3} A \,x^{2}}{2}\) | \(148\) |
default | \(\frac {b^{3} D x^{11}}{11}+\frac {b^{3} C \,x^{10}}{10}+\frac {\left (B \,b^{3}+3 a \,b^{2} D\right ) x^{9}}{9}+\frac {\left (b^{3} A +3 C \,b^{2} a \right ) x^{8}}{8}+\frac {\left (3 a \,b^{2} B +3 D a^{2} b \right ) x^{7}}{7}+\frac {\left (3 a \,b^{2} A +3 C \,a^{2} b \right ) x^{6}}{6}+\frac {\left (3 a^{2} b B +D a^{3}\right ) x^{5}}{5}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{4}}{4}+\frac {a^{3} B \,x^{3}}{3}+\frac {a^{3} A \,x^{2}}{2}\) | \(150\) |
gosper | \(\frac {1}{11} b^{3} D x^{11}+\frac {1}{10} b^{3} C \,x^{10}+\frac {1}{9} b^{3} B \,x^{9}+\frac {1}{3} x^{9} a \,b^{2} D+\frac {1}{8} x^{8} b^{3} A +\frac {3}{8} a \,b^{2} C \,x^{8}+\frac {3}{7} x^{7} a \,b^{2} B +\frac {3}{7} x^{7} D a^{2} b +\frac {1}{2} x^{6} a \,b^{2} A +\frac {1}{2} a^{2} b C \,x^{6}+\frac {3}{5} x^{5} a^{2} b B +\frac {1}{5} x^{5} D a^{3}+\frac {3}{4} x^{4} a^{2} b A +\frac {1}{4} a^{3} C \,x^{4}+\frac {1}{3} a^{3} B \,x^{3}+\frac {1}{2} a^{3} A \,x^{2}\) | \(154\) |
parallelrisch | \(\frac {1}{11} b^{3} D x^{11}+\frac {1}{10} b^{3} C \,x^{10}+\frac {1}{9} b^{3} B \,x^{9}+\frac {1}{3} x^{9} a \,b^{2} D+\frac {1}{8} x^{8} b^{3} A +\frac {3}{8} a \,b^{2} C \,x^{8}+\frac {3}{7} x^{7} a \,b^{2} B +\frac {3}{7} x^{7} D a^{2} b +\frac {1}{2} x^{6} a \,b^{2} A +\frac {1}{2} a^{2} b C \,x^{6}+\frac {3}{5} x^{5} a^{2} b B +\frac {1}{5} x^{5} D a^{3}+\frac {3}{4} x^{4} a^{2} b A +\frac {1}{4} a^{3} C \,x^{4}+\frac {1}{3} a^{3} B \,x^{3}+\frac {1}{2} a^{3} A \,x^{2}\) | \(154\) |
1/11*b^3*D*x^11+1/10*b^3*C*x^10+(1/9*B*b^3+1/3*a*b^2*D)*x^9+(1/8*b^3*A+3/8 *C*b^2*a)*x^8+(3/7*a*b^2*B+3/7*D*a^2*b)*x^7+(1/2*a*b^2*A+1/2*C*a^2*b)*x^6+ (3/5*a^2*b*B+1/5*D*a^3)*x^5+(3/4*a^2*b*A+1/4*C*a^3)*x^4+1/3*a^3*B*x^3+1/2* a^3*A*x^2
Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.05 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{11} \, D b^{3} x^{11} + \frac {1}{10} \, C b^{3} x^{10} + \frac {1}{9} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{9} + \frac {1}{8} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (D a^{2} b + B a b^{2}\right )} x^{7} + \frac {1}{3} \, B a^{3} x^{3} + \frac {1}{2} \, {\left (C a^{2} b + A a b^{2}\right )} x^{6} + \frac {1}{2} \, A a^{3} x^{2} + \frac {1}{5} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{5} + \frac {1}{4} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{4} \]
1/11*D*b^3*x^11 + 1/10*C*b^3*x^10 + 1/9*(3*D*a*b^2 + B*b^3)*x^9 + 1/8*(3*C *a*b^2 + A*b^3)*x^8 + 3/7*(D*a^2*b + B*a*b^2)*x^7 + 1/3*B*a^3*x^3 + 1/2*(C *a^2*b + A*a*b^2)*x^6 + 1/2*A*a^3*x^2 + 1/5*(D*a^3 + 3*B*a^2*b)*x^5 + 1/4* (C*a^3 + 3*A*a^2*b)*x^4
Time = 0.03 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.18 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A a^{3} x^{2}}{2} + \frac {B a^{3} x^{3}}{3} + \frac {C b^{3} x^{10}}{10} + \frac {D b^{3} x^{11}}{11} + x^{9} \left (\frac {B b^{3}}{9} + \frac {D a b^{2}}{3}\right ) + x^{8} \left (\frac {A b^{3}}{8} + \frac {3 C a b^{2}}{8}\right ) + x^{7} \cdot \left (\frac {3 B a b^{2}}{7} + \frac {3 D a^{2} b}{7}\right ) + x^{6} \left (\frac {A a b^{2}}{2} + \frac {C a^{2} b}{2}\right ) + x^{5} \cdot \left (\frac {3 B a^{2} b}{5} + \frac {D a^{3}}{5}\right ) + x^{4} \cdot \left (\frac {3 A a^{2} b}{4} + \frac {C a^{3}}{4}\right ) \]
A*a**3*x**2/2 + B*a**3*x**3/3 + C*b**3*x**10/10 + D*b**3*x**11/11 + x**9*( B*b**3/9 + D*a*b**2/3) + x**8*(A*b**3/8 + 3*C*a*b**2/8) + x**7*(3*B*a*b**2 /7 + 3*D*a**2*b/7) + x**6*(A*a*b**2/2 + C*a**2*b/2) + x**5*(3*B*a**2*b/5 + D*a**3/5) + x**4*(3*A*a**2*b/4 + C*a**3/4)
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.05 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{11} \, D b^{3} x^{11} + \frac {1}{10} \, C b^{3} x^{10} + \frac {1}{9} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{9} + \frac {1}{8} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (D a^{2} b + B a b^{2}\right )} x^{7} + \frac {1}{3} \, B a^{3} x^{3} + \frac {1}{2} \, {\left (C a^{2} b + A a b^{2}\right )} x^{6} + \frac {1}{2} \, A a^{3} x^{2} + \frac {1}{5} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{5} + \frac {1}{4} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{4} \]
1/11*D*b^3*x^11 + 1/10*C*b^3*x^10 + 1/9*(3*D*a*b^2 + B*b^3)*x^9 + 1/8*(3*C *a*b^2 + A*b^3)*x^8 + 3/7*(D*a^2*b + B*a*b^2)*x^7 + 1/3*B*a^3*x^3 + 1/2*(C *a^2*b + A*a*b^2)*x^6 + 1/2*A*a^3*x^2 + 1/5*(D*a^3 + 3*B*a^2*b)*x^5 + 1/4* (C*a^3 + 3*A*a^2*b)*x^4
Time = 0.37 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{11} \, D b^{3} x^{11} + \frac {1}{10} \, C b^{3} x^{10} + \frac {1}{3} \, D a b^{2} x^{9} + \frac {1}{9} \, B b^{3} x^{9} + \frac {3}{8} \, C a b^{2} x^{8} + \frac {1}{8} \, A b^{3} x^{8} + \frac {3}{7} \, D a^{2} b x^{7} + \frac {3}{7} \, B a b^{2} x^{7} + \frac {1}{2} \, C a^{2} b x^{6} + \frac {1}{2} \, A a b^{2} x^{6} + \frac {1}{5} \, D a^{3} x^{5} + \frac {3}{5} \, B a^{2} b x^{5} + \frac {1}{4} \, C a^{3} x^{4} + \frac {3}{4} \, A a^{2} b x^{4} + \frac {1}{3} \, B a^{3} x^{3} + \frac {1}{2} \, A a^{3} x^{2} \]
1/11*D*b^3*x^11 + 1/10*C*b^3*x^10 + 1/3*D*a*b^2*x^9 + 1/9*B*b^3*x^9 + 3/8* C*a*b^2*x^8 + 1/8*A*b^3*x^8 + 3/7*D*a^2*b*x^7 + 3/7*B*a*b^2*x^7 + 1/2*C*a^ 2*b*x^6 + 1/2*A*a*b^2*x^6 + 1/5*D*a^3*x^5 + 3/5*B*a^2*b*x^5 + 1/4*C*a^3*x^ 4 + 3/4*A*a^2*b*x^4 + 1/3*B*a^3*x^3 + 1/2*A*a^3*x^2
Time = 5.79 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int x \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A\,a^3\,x^2}{2}+\frac {B\,a^3\,x^3}{3}+\frac {A\,b^3\,x^8}{8}+\frac {C\,a^3\,x^4}{4}+\frac {B\,b^3\,x^9}{9}+\frac {C\,b^3\,x^{10}}{10}+\frac {a^3\,x^5\,D}{5}+\frac {b^3\,x^{11}\,D}{11}+\frac {3\,a^2\,b\,x^7\,D}{7}+\frac {a\,b^2\,x^9\,D}{3}+\frac {3\,A\,a^2\,b\,x^4}{4}+\frac {A\,a\,b^2\,x^6}{2}+\frac {3\,B\,a^2\,b\,x^5}{5}+\frac {3\,B\,a\,b^2\,x^7}{7}+\frac {C\,a^2\,b\,x^6}{2}+\frac {3\,C\,a\,b^2\,x^8}{8} \]