Integrand size = 28, antiderivative size = 149 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{4} a^3 A x^4+\frac {1}{5} a^3 B x^5+\frac {1}{6} a^2 (3 A b+a C) x^6+\frac {1}{7} a^2 (3 b B+a D) x^7+\frac {3}{8} a b (A b+a C) x^8+\frac {1}{3} a b (b B+a D) x^9+\frac {1}{10} b^2 (A b+3 a C) x^{10}+\frac {1}{11} b^2 (b B+3 a D) x^{11}+\frac {1}{12} b^3 C x^{12}+\frac {1}{13} b^3 D x^{13} \] Output:
1/4*a^3*A*x^4+1/5*a^3*B*x^5+1/6*a^2*(3*A*b+C*a)*x^6+1/7*a^2*(3*B*b+D*a)*x^ 7+3/8*a*b*(A*b+C*a)*x^8+1/3*a*b*(B*b+D*a)*x^9+1/10*b^2*(A*b+3*C*a)*x^10+1/ 11*b^2*(B*b+3*D*a)*x^11+1/12*b^3*C*x^12+1/13*b^3*D*x^13
Time = 0.03 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{4} a^3 A x^4+\frac {1}{5} a^3 B x^5+\frac {1}{6} a^2 (3 A b+a C) x^6+\frac {1}{7} a^2 (3 b B+a D) x^7+\frac {3}{8} a b (A b+a C) x^8+\frac {1}{3} a b (b B+a D) x^9+\frac {1}{10} b^2 (A b+3 a C) x^{10}+\frac {1}{11} b^2 (b B+3 a D) x^{11}+\frac {1}{12} b^3 C x^{12}+\frac {1}{13} b^3 D x^{13} \] Input:
Integrate[x^3*(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]
Output:
(a^3*A*x^4)/4 + (a^3*B*x^5)/5 + (a^2*(3*A*b + a*C)*x^6)/6 + (a^2*(3*b*B + a*D)*x^7)/7 + (3*a*b*(A*b + a*C)*x^8)/8 + (a*b*(b*B + a*D)*x^9)/3 + (b^2*( A*b + 3*a*C)*x^10)/10 + (b^2*(b*B + 3*a*D)*x^11)/11 + (b^3*C*x^12)/12 + (b ^3*D*x^13)/13
Time = 0.43 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2333, 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.
| \(\displaystyle \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \int \left (a^3 A x^3+a^3 B x^4+a^2 x^5 (a C+3 A b)+a^2 x^6 (a D+3 b B)+b^2 x^9 (3 a C+A b)+3 a b x^7 (a C+A b)+b^2 x^{10} (3 a D+b B)+3 a b x^8 (a D+b B)+b^3 C x^{11}+b^3 D x^{12}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} a^3 A x^4+\frac {1}{5} a^3 B x^5+\frac {1}{6} a^2 x^6 (a C+3 A b)+\frac {1}{7} a^2 x^7 (a D+3 b B)+\frac {1}{10} b^2 x^{10} (3 a C+A b)+\frac {3}{8} a b x^8 (a C+A b)+\frac {1}{11} b^2 x^{11} (3 a D+b B)+\frac {1}{3} a b x^9 (a D+b B)+\frac {1}{12} b^3 C x^{12}+\frac {1}{13} b^3 D x^{13}\) |
Input:
Int[x^3*(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]
Output:
(a^3*A*x^4)/4 + (a^3*B*x^5)/5 + (a^2*(3*A*b + a*C)*x^6)/6 + (a^2*(3*b*B + a*D)*x^7)/7 + (3*a*b*(A*b + a*C)*x^8)/8 + (a*b*(b*B + a*D)*x^9)/3 + (b^2*( A*b + 3*a*C)*x^10)/10 + (b^2*(b*B + 3*a*D)*x^11)/11 + (b^3*C*x^12)/12 + (b ^3*D*x^13)/13
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.52 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99
| method | result | size |
| norman | \(\frac {b^{3} D x^{13}}{13}+\frac {b^{3} C \,x^{12}}{12}+\left (\frac {1}{11} B \,b^{3}+\frac {3}{11} a \,b^{2} D\right ) x^{11}+\left (\frac {1}{10} b^{3} A +\frac {3}{10} a C \,b^{2}\right ) x^{10}+\left (\frac {1}{3} a \,b^{2} B +\frac {1}{3} a^{2} b D\right ) x^{9}+\left (\frac {3}{8} a \,b^{2} A +\frac {3}{8} a^{2} b C \right ) x^{8}+\left (\frac {3}{7} a^{2} b B +\frac {1}{7} a^{3} D\right ) x^{7}+\left (\frac {1}{2} a^{2} b A +\frac {1}{6} C \,a^{3}\right ) x^{6}+\frac {a^{3} B \,x^{5}}{5}+\frac {a^{3} A \,x^{4}}{4}\) | \(148\) |
| default | \(\frac {b^{3} D x^{13}}{13}+\frac {b^{3} C \,x^{12}}{12}+\frac {\left (B \,b^{3}+3 a \,b^{2} D\right ) x^{11}}{11}+\frac {\left (b^{3} A +3 a C \,b^{2}\right ) x^{10}}{10}+\frac {\left (3 a \,b^{2} B +3 a^{2} b D\right ) x^{9}}{9}+\frac {\left (3 a \,b^{2} A +3 a^{2} b C \right ) x^{8}}{8}+\frac {\left (3 a^{2} b B +a^{3} D\right ) x^{7}}{7}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{6}}{6}+\frac {a^{3} B \,x^{5}}{5}+\frac {a^{3} A \,x^{4}}{4}\) | \(150\) |
| gosper | \(\frac {1}{13} b^{3} D x^{13}+\frac {1}{12} b^{3} C \,x^{12}+\frac {1}{11} x^{11} B \,b^{3}+\frac {3}{11} x^{11} a \,b^{2} D+\frac {1}{10} x^{10} b^{3} A +\frac {3}{10} x^{10} a C \,b^{2}+\frac {1}{3} x^{9} a \,b^{2} B +\frac {1}{3} x^{9} a^{2} b D+\frac {3}{8} x^{8} a \,b^{2} A +\frac {3}{8} x^{8} a^{2} b C +\frac {3}{7} x^{7} a^{2} b B +\frac {1}{7} x^{7} a^{3} D+\frac {1}{2} x^{6} a^{2} b A +\frac {1}{6} x^{6} C \,a^{3}+\frac {1}{5} a^{3} B \,x^{5}+\frac {1}{4} a^{3} A \,x^{4}\) | \(154\) |
| parallelrisch | \(\frac {1}{13} b^{3} D x^{13}+\frac {1}{12} b^{3} C \,x^{12}+\frac {1}{11} x^{11} B \,b^{3}+\frac {3}{11} x^{11} a \,b^{2} D+\frac {1}{10} x^{10} b^{3} A +\frac {3}{10} x^{10} a C \,b^{2}+\frac {1}{3} x^{9} a \,b^{2} B +\frac {1}{3} x^{9} a^{2} b D+\frac {3}{8} x^{8} a \,b^{2} A +\frac {3}{8} x^{8} a^{2} b C +\frac {3}{7} x^{7} a^{2} b B +\frac {1}{7} x^{7} a^{3} D+\frac {1}{2} x^{6} a^{2} b A +\frac {1}{6} x^{6} C \,a^{3}+\frac {1}{5} a^{3} B \,x^{5}+\frac {1}{4} a^{3} A \,x^{4}\) | \(154\) |
| orering | \(\frac {x^{4} \left (9240 b^{3} D x^{9}+10010 C \,b^{3} x^{8}+10920 b^{3} B \,x^{7}+32760 D a \,b^{2} x^{7}+12012 A \,b^{3} x^{6}+36036 C a \,b^{2} x^{6}+40040 B a \,b^{2} x^{5}+40040 D a^{2} b \,x^{5}+45045 a A \,b^{2} x^{4}+45045 C \,a^{2} b \,x^{4}+51480 B \,a^{2} b \,x^{3}+17160 D a^{3} x^{3}+60060 a^{2} A b \,x^{2}+20020 C \,a^{3} x^{2}+24024 B \,a^{3} x +30030 a^{3} A \right )}{120120}\) | \(154\) |
Input:
int(x^3*(b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/13*b^3*D*x^13+1/12*b^3*C*x^12+(1/11*B*b^3+3/11*a*b^2*D)*x^11+(1/10*b^3*A +3/10*a*C*b^2)*x^10+(1/3*a*b^2*B+1/3*a^2*b*D)*x^9+(3/8*a*b^2*A+3/8*a^2*b*C )*x^8+(3/7*a^2*b*B+1/7*a^3*D)*x^7+(1/2*a^2*b*A+1/6*C*a^3)*x^6+1/5*a^3*B*x^ 5+1/4*a^3*A*x^4
Time = 0.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {1}{12} \, C b^{3} x^{12} + \frac {1}{11} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{11} + \frac {1}{10} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{10} + \frac {1}{3} \, {\left (D a^{2} b + B a b^{2}\right )} x^{9} + \frac {1}{5} \, B a^{3} x^{5} + \frac {3}{8} \, {\left (C a^{2} b + A a b^{2}\right )} x^{8} + \frac {1}{4} \, A a^{3} x^{4} + \frac {1}{7} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{7} + \frac {1}{6} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{6} \] Input:
integrate(x^3*(b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
1/13*D*b^3*x^13 + 1/12*C*b^3*x^12 + 1/11*(3*D*a*b^2 + B*b^3)*x^11 + 1/10*( 3*C*a*b^2 + A*b^3)*x^10 + 1/3*(D*a^2*b + B*a*b^2)*x^9 + 1/5*B*a^3*x^5 + 3/ 8*(C*a^2*b + A*a*b^2)*x^8 + 1/4*A*a^3*x^4 + 1/7*(D*a^3 + 3*B*a^2*b)*x^7 + 1/6*(C*a^3 + 3*A*a^2*b)*x^6
Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.09 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A a^{3} x^{4}}{4} + \frac {B a^{3} x^{5}}{5} + \frac {C b^{3} x^{12}}{12} + \frac {D b^{3} x^{13}}{13} + x^{11} \left (\frac {B b^{3}}{11} + \frac {3 D a b^{2}}{11}\right ) + x^{10} \left (\frac {A b^{3}}{10} + \frac {3 C a b^{2}}{10}\right ) + x^{9} \left (\frac {B a b^{2}}{3} + \frac {D a^{2} b}{3}\right ) + x^{8} \cdot \left (\frac {3 A a b^{2}}{8} + \frac {3 C a^{2} b}{8}\right ) + x^{7} \cdot \left (\frac {3 B a^{2} b}{7} + \frac {D a^{3}}{7}\right ) + x^{6} \left (\frac {A a^{2} b}{2} + \frac {C a^{3}}{6}\right ) \] Input:
integrate(x**3*(b*x**2+a)**3*(D*x**3+C*x**2+B*x+A),x)
Output:
A*a**3*x**4/4 + B*a**3*x**5/5 + C*b**3*x**12/12 + D*b**3*x**13/13 + x**11* (B*b**3/11 + 3*D*a*b**2/11) + x**10*(A*b**3/10 + 3*C*a*b**2/10) + x**9*(B* a*b**2/3 + D*a**2*b/3) + x**8*(3*A*a*b**2/8 + 3*C*a**2*b/8) + x**7*(3*B*a* *2*b/7 + D*a**3/7) + x**6*(A*a**2*b/2 + C*a**3/6)
Time = 0.03 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {1}{12} \, C b^{3} x^{12} + \frac {1}{11} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{11} + \frac {1}{10} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{10} + \frac {1}{3} \, {\left (D a^{2} b + B a b^{2}\right )} x^{9} + \frac {1}{5} \, B a^{3} x^{5} + \frac {3}{8} \, {\left (C a^{2} b + A a b^{2}\right )} x^{8} + \frac {1}{4} \, A a^{3} x^{4} + \frac {1}{7} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{7} + \frac {1}{6} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{6} \] Input:
integrate(x^3*(b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/13*D*b^3*x^13 + 1/12*C*b^3*x^12 + 1/11*(3*D*a*b^2 + B*b^3)*x^11 + 1/10*( 3*C*a*b^2 + A*b^3)*x^10 + 1/3*(D*a^2*b + B*a*b^2)*x^9 + 1/5*B*a^3*x^5 + 3/ 8*(C*a^2*b + A*a*b^2)*x^8 + 1/4*A*a^3*x^4 + 1/7*(D*a^3 + 3*B*a^2*b)*x^7 + 1/6*(C*a^3 + 3*A*a^2*b)*x^6
Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {1}{12} \, C b^{3} x^{12} + \frac {3}{11} \, D a b^{2} x^{11} + \frac {1}{11} \, B b^{3} x^{11} + \frac {3}{10} \, C a b^{2} x^{10} + \frac {1}{10} \, A b^{3} x^{10} + \frac {1}{3} \, D a^{2} b x^{9} + \frac {1}{3} \, B a b^{2} x^{9} + \frac {3}{8} \, C a^{2} b x^{8} + \frac {3}{8} \, A a b^{2} x^{8} + \frac {1}{7} \, D a^{3} x^{7} + \frac {3}{7} \, B a^{2} b x^{7} + \frac {1}{6} \, C a^{3} x^{6} + \frac {1}{2} \, A a^{2} b x^{6} + \frac {1}{5} \, B a^{3} x^{5} + \frac {1}{4} \, A a^{3} x^{4} \] Input:
integrate(x^3*(b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
1/13*D*b^3*x^13 + 1/12*C*b^3*x^12 + 3/11*D*a*b^2*x^11 + 1/11*B*b^3*x^11 + 3/10*C*a*b^2*x^10 + 1/10*A*b^3*x^10 + 1/3*D*a^2*b*x^9 + 1/3*B*a*b^2*x^9 + 3/8*C*a^2*b*x^8 + 3/8*A*a*b^2*x^8 + 1/7*D*a^3*x^7 + 3/7*B*a^2*b*x^7 + 1/6* C*a^3*x^6 + 1/2*A*a^2*b*x^6 + 1/5*B*a^3*x^5 + 1/4*A*a^3*x^4
Time = 2.52 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A\,a^3\,x^4}{4}+\frac {B\,a^3\,x^5}{5}+\frac {A\,b^3\,x^{10}}{10}+\frac {C\,a^3\,x^6}{6}+\frac {B\,b^3\,x^{11}}{11}+\frac {C\,b^3\,x^{12}}{12}+\frac {a^3\,x^7\,D}{7}+\frac {b^3\,x^{13}\,D}{13}+\frac {a^2\,b\,x^9\,D}{3}+\frac {3\,a\,b^2\,x^{11}\,D}{11}+\frac {A\,a^2\,b\,x^6}{2}+\frac {3\,A\,a\,b^2\,x^8}{8}+\frac {3\,B\,a^2\,b\,x^7}{7}+\frac {B\,a\,b^2\,x^9}{3}+\frac {3\,C\,a^2\,b\,x^8}{8}+\frac {3\,C\,a\,b^2\,x^{10}}{10} \] Input:
int(x^3*(a + b*x^2)^3*(A + B*x + C*x^2 + x^3*D),x)
Output:
(A*a^3*x^4)/4 + (B*a^3*x^5)/5 + (A*b^3*x^10)/10 + (C*a^3*x^6)/6 + (B*b^3*x ^11)/11 + (C*b^3*x^12)/12 + (a^3*x^7*D)/7 + (b^3*x^13*D)/13 + (a^2*b*x^9*D )/3 + (3*a*b^2*x^11*D)/11 + (A*a^2*b*x^6)/2 + (3*A*a*b^2*x^8)/8 + (3*B*a^2 *b*x^7)/7 + (B*a*b^2*x^9)/3 + (3*C*a^2*b*x^8)/8 + (3*C*a*b^2*x^10)/10
Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int x^3 \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {x^{4} \left (9240 b^{3} d \,x^{9}+10010 b^{3} c \,x^{8}+32760 a \,b^{2} d \,x^{7}+10920 b^{4} x^{7}+12012 a \,b^{3} x^{6}+36036 a \,b^{2} c \,x^{6}+40040 a^{2} b d \,x^{5}+40040 a \,b^{3} x^{5}+45045 a^{2} b^{2} x^{4}+45045 a^{2} b c \,x^{4}+17160 a^{3} d \,x^{3}+51480 a^{2} b^{2} x^{3}+60060 a^{3} b \,x^{2}+20020 a^{3} c \,x^{2}+24024 a^{3} b x +30030 a^{4}\right )}{120120} \] Input:
int(x^3*(b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x)
Output:
(x**4*(30030*a**4 + 60060*a**3*b*x**2 + 24024*a**3*b*x + 20020*a**3*c*x**2 + 17160*a**3*d*x**3 + 45045*a**2*b**2*x**4 + 51480*a**2*b**2*x**3 + 45045 *a**2*b*c*x**4 + 40040*a**2*b*d*x**5 + 12012*a*b**3*x**6 + 40040*a*b**3*x* *5 + 36036*a*b**2*c*x**6 + 32760*a*b**2*d*x**7 + 10920*b**4*x**7 + 10010*b **3*c*x**8 + 9240*b**3*d*x**9))/120120