Integrand size = 27, antiderivative size = 136 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A b+a B) x^3+\frac {1}{5} a \left (3 A b^2+a (3 b B+a C)\right ) x^5+\frac {1}{7} \left (A b^3+a \left (3 b^2 B+3 a b C+a^2 D\right )\right ) x^7+\frac {1}{9} b \left (b^2 B+3 a b C+3 a^2 D\right ) x^9+\frac {1}{11} b^2 (b C+3 a D) x^{11}+\frac {1}{13} b^3 D x^{13} \] Output:
a^3*A*x+1/3*a^2*(3*A*b+B*a)*x^3+1/5*a*(3*A*b^2+a*(3*B*b+C*a))*x^5+1/7*(A*b ^3+a*(3*B*b^2+3*C*a*b+D*a^2))*x^7+1/9*b*(B*b^2+3*C*a*b+3*D*a^2)*x^9+1/11*b ^2*(C*b+3*D*a)*x^11+1/13*b^3*D*x^13
Time = 0.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A b+a B) x^3+\frac {1}{5} a \left (3 A b^2+3 a b B+a^2 C\right ) x^5+\frac {1}{7} \left (A b^3+3 a b^2 B+3 a^2 b C+a^3 D\right ) x^7+\frac {1}{9} b \left (b^2 B+3 a b C+3 a^2 D\right ) x^9+\frac {1}{11} b^2 (b C+3 a D) x^{11}+\frac {1}{13} b^3 D x^{13} \] Input:
Integrate[(a + b*x^2)^3*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
a^3*A*x + (a^2*(3*A*b + a*B)*x^3)/3 + (a*(3*A*b^2 + 3*a*b*B + a^2*C)*x^5)/ 5 + ((A*b^3 + 3*a*b^2*B + 3*a^2*b*C + a^3*D)*x^7)/7 + (b*(b^2*B + 3*a*b*C + 3*a^2*D)*x^9)/9 + (b^2*(b*C + 3*a*D)*x^11)/11 + (b^3*D*x^13)/13
Time = 0.37 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.
\(\displaystyle \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (a^3 A+x^6 \left (a \left (a^2 D+3 a b C+3 b^2 B\right )+A b^3\right )+a^2 x^2 (a B+3 A b)+b x^8 \left (3 a^2 D+3 a b C+b^2 B\right )+a x^4 \left (a (a C+3 b B)+3 A b^2\right )+b^2 x^{10} (3 a D+b C)+b^3 D x^{12}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 A x+\frac {1}{7} x^7 \left (a \left (a^2 D+3 a b C+3 b^2 B\right )+A b^3\right )+\frac {1}{3} a^2 x^3 (a B+3 A b)+\frac {1}{9} b x^9 \left (3 a^2 D+3 a b C+b^2 B\right )+\frac {1}{5} a x^5 \left (a (a C+3 b B)+3 A b^2\right )+\frac {1}{11} b^2 x^{11} (3 a D+b C)+\frac {1}{13} b^3 D x^{13}\) |
Input:
Int[(a + b*x^2)^3*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
a^3*A*x + (a^2*(3*A*b + a*B)*x^3)/3 + (a*(3*A*b^2 + a*(3*b*B + a*C))*x^5)/ 5 + ((A*b^3 + a*(3*b^2*B + 3*a*b*C + a^2*D))*x^7)/7 + (b*(b^2*B + 3*a*b*C + 3*a^2*D)*x^9)/9 + (b^2*(b*C + 3*a*D)*x^11)/11 + (b^3*D*x^13)/13
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 = 0.50 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {b^{3} D x^{13}}{13}+\frac {\left (b^{3} C +3 a \,b^{2} D\right ) x^{11}}{11}+\frac {\left (B \,b^{3}+3 a C \,b^{2}+3 a^{2} b D\right ) x^{9}}{9}+\frac {\left (b^{3} A +3 a \,b^{2} B +3 a^{2} b C +a^{3} D\right ) x^{7}}{7}+\frac {\left (3 a \,b^{2} A +3 a^{2} b B +C \,a^{3}\right ) x^{5}}{5}+\frac {\left (3 a^{2} b A +a^{3} B \right ) x^{3}}{3}+a^{3} A x\) | \(133\) |
norman | \(\frac {b^{3} D x^{13}}{13}+\left (\frac {1}{11} b^{3} C +\frac {3}{11} a \,b^{2} D\right ) x^{11}+\left (\frac {1}{9} B \,b^{3}+\frac {1}{3} a C \,b^{2}+\frac {1}{3} a^{2} b D\right ) x^{9}+\left (\frac {1}{7} b^{3} A +\frac {3}{7} a \,b^{2} B +\frac {3}{7} a^{2} b C +\frac {1}{7} a^{3} D\right ) x^{7}+\left (\frac {3}{5} a \,b^{2} A +\frac {3}{5} a^{2} b B +\frac {1}{5} C \,a^{3}\right ) x^{5}+\left (a^{2} b A +\frac {1}{3} a^{3} B \right ) x^{3}+a^{3} A x\) | \(133\) |
gosper | \(\frac {1}{13} b^{3} D x^{13}+\frac {1}{11} b^{3} C \,x^{11}+\frac {3}{11} x^{11} a \,b^{2} D+\frac {1}{9} b^{3} B \,x^{9}+\frac {1}{3} x^{9} a C \,b^{2}+\frac {1}{3} x^{9} a^{2} b D+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a \,b^{2} B +\frac {3}{7} x^{7} a^{2} b C +\frac {1}{7} x^{7} a^{3} D+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b B +\frac {1}{5} x^{5} C \,a^{3}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} a^{3} B +a^{3} A x\) | \(150\) |
parallelrisch | \(\frac {1}{13} b^{3} D x^{13}+\frac {1}{11} b^{3} C \,x^{11}+\frac {3}{11} x^{11} a \,b^{2} D+\frac {1}{9} b^{3} B \,x^{9}+\frac {1}{3} x^{9} a C \,b^{2}+\frac {1}{3} x^{9} a^{2} b D+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a \,b^{2} B +\frac {3}{7} x^{7} a^{2} b C +\frac {1}{7} x^{7} a^{3} D+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b B +\frac {1}{5} x^{5} C \,a^{3}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} a^{3} B +a^{3} A x\) | \(150\) |
orering | \(\frac {x \left (3465 b^{3} D x^{12}+4095 b^{3} C \,x^{10}+12285 D a \,b^{2} x^{10}+5005 b^{3} B \,x^{8}+15015 C a \,b^{2} x^{8}+15015 D a^{2} b \,x^{8}+6435 A \,b^{3} x^{6}+19305 B a \,b^{2} x^{6}+19305 C \,a^{2} b \,x^{6}+6435 D a^{3} x^{6}+27027 a A \,b^{2} x^{4}+27027 B \,a^{2} b \,x^{4}+9009 C \,a^{3} x^{4}+45045 a^{2} A b \,x^{2}+15015 B \,a^{3} x^{2}+45045 a^{3} A \right )}{45045}\) | \(154\) |
Input:
int((b*x^2+a)^3*(D*x^6+C*x^4+B*x^2+A),x,method=_RETURNVERBOSE)
Output:
1/13*b^3*D*x^13+1/11*(C*b^3+3*D*a*b^2)*x^11+1/9*(B*b^3+3*C*a*b^2+3*D*a^2*b )*x^9+1/7*(A*b^3+3*B*a*b^2+3*C*a^2*b+D*a^3)*x^7+1/5*(3*A*a*b^2+3*B*a^2*b+C *a^3)*x^5+1/3*(3*A*a^2*b+B*a^3)*x^3+a^3*A*x
Time = 0.06 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {1}{11} \, {\left (3 \, D a b^{2} + C b^{3}\right )} x^{11} + \frac {1}{9} \, {\left (3 \, D a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} x^{9} + \frac {1}{7} \, {\left (D a^{3} + 3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{5} \, {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(D*x^6+C*x^4+B*x^2+A),x, algorithm="fricas")
Output:
1/13*D*b^3*x^13 + 1/11*(3*D*a*b^2 + C*b^3)*x^11 + 1/9*(3*D*a^2*b + 3*C*a*b ^2 + B*b^3)*x^9 + 1/7*(D*a^3 + 3*C*a^2*b + 3*B*a*b^2 + A*b^3)*x^7 + 1/5*(C *a^3 + 3*B*a^2*b + 3*A*a*b^2)*x^5 + A*a^3*x + 1/3*(B*a^3 + 3*A*a^2*b)*x^3
Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=A a^{3} x + \frac {D b^{3} x^{13}}{13} + x^{11} \left (\frac {C b^{3}}{11} + \frac {3 D a b^{2}}{11}\right ) + x^{9} \left (\frac {B b^{3}}{9} + \frac {C a b^{2}}{3} + \frac {D a^{2} b}{3}\right ) + x^{7} \left (\frac {A b^{3}}{7} + \frac {3 B a b^{2}}{7} + \frac {3 C a^{2} b}{7} + \frac {D a^{3}}{7}\right ) + x^{5} \cdot \left (\frac {3 A a b^{2}}{5} + \frac {3 B a^{2} b}{5} + \frac {C a^{3}}{5}\right ) + x^{3} \left (A a^{2} b + \frac {B a^{3}}{3}\right ) \] Input:
integrate((b*x**2+a)**3*(D*x**6+C*x**4+B*x**2+A),x)
Output:
A*a**3*x + D*b**3*x**13/13 + x**11*(C*b**3/11 + 3*D*a*b**2/11) + x**9*(B*b **3/9 + C*a*b**2/3 + D*a**2*b/3) + x**7*(A*b**3/7 + 3*B*a*b**2/7 + 3*C*a** 2*b/7 + D*a**3/7) + x**5*(3*A*a*b**2/5 + 3*B*a**2*b/5 + C*a**3/5) + x**3*( A*a**2*b + B*a**3/3)
Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {1}{11} \, {\left (3 \, D a b^{2} + C b^{3}\right )} x^{11} + \frac {1}{9} \, {\left (3 \, D a^{2} b + 3 \, C a b^{2} + B b^{3}\right )} x^{9} + \frac {1}{7} \, {\left (D a^{3} + 3 \, C a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{5} \, {\left (C a^{3} + 3 \, B a^{2} b + 3 \, A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(D*x^6+C*x^4+B*x^2+A),x, algorithm="maxima")
Output:
1/13*D*b^3*x^13 + 1/11*(3*D*a*b^2 + C*b^3)*x^11 + 1/9*(3*D*a^2*b + 3*C*a*b ^2 + B*b^3)*x^9 + 1/7*(D*a^3 + 3*C*a^2*b + 3*B*a*b^2 + A*b^3)*x^7 + 1/5*(C *a^3 + 3*B*a^2*b + 3*A*a*b^2)*x^5 + A*a^3*x + 1/3*(B*a^3 + 3*A*a^2*b)*x^3
Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{13} \, D b^{3} x^{13} + \frac {3}{11} \, D a b^{2} x^{11} + \frac {1}{11} \, C b^{3} x^{11} + \frac {1}{3} \, D a^{2} b x^{9} + \frac {1}{3} \, C a b^{2} x^{9} + \frac {1}{9} \, B b^{3} x^{9} + \frac {1}{7} \, D a^{3} x^{7} + \frac {3}{7} \, C a^{2} b x^{7} + \frac {3}{7} \, B a b^{2} x^{7} + \frac {1}{7} \, A b^{3} x^{7} + \frac {1}{5} \, C a^{3} x^{5} + \frac {3}{5} \, B a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {1}{3} \, B a^{3} x^{3} + A a^{2} b x^{3} + A a^{3} x \] Input:
integrate((b*x^2+a)^3*(D*x^6+C*x^4+B*x^2+A),x, algorithm="giac")
Output:
1/13*D*b^3*x^13 + 3/11*D*a*b^2*x^11 + 1/11*C*b^3*x^11 + 1/3*D*a^2*b*x^9 + 1/3*C*a*b^2*x^9 + 1/9*B*b^3*x^9 + 1/7*D*a^3*x^7 + 3/7*C*a^2*b*x^7 + 3/7*B* a*b^2*x^7 + 1/7*A*b^3*x^7 + 1/5*C*a^3*x^5 + 3/5*B*a^2*b*x^5 + 3/5*A*a*b^2* x^5 + 1/3*B*a^3*x^3 + A*a^2*b*x^3 + A*a^3*x
Time = 0.73 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {B\,a^3\,x^3}{3}+\frac {A\,b^3\,x^7}{7}+\frac {C\,a^3\,x^5}{5}+\frac {B\,b^3\,x^9}{9}+\frac {C\,b^3\,x^{11}}{11}+\frac {a^3\,x^7\,D}{7}+\frac {b^3\,x^{13}\,D}{13}+A\,a^3\,x+\frac {a^2\,b\,x^9\,D}{3}+\frac {3\,a\,b^2\,x^{11}\,D}{11}+A\,a^2\,b\,x^3+\frac {3\,A\,a\,b^2\,x^5}{5}+\frac {3\,B\,a^2\,b\,x^5}{5}+\frac {3\,B\,a\,b^2\,x^7}{7}+\frac {3\,C\,a^2\,b\,x^7}{7}+\frac {C\,a\,b^2\,x^9}{3} \] Input:
int((a + b*x^2)^3*(A + B*x^2 + C*x^4 + x^6*D),x)
Output:
(B*a^3*x^3)/3 + (A*b^3*x^7)/7 + (C*a^3*x^5)/5 + (B*b^3*x^9)/9 + (C*b^3*x^1 1)/11 + (a^3*x^7*D)/7 + (b^3*x^13*D)/13 + A*a^3*x + (a^2*b*x^9*D)/3 + (3*a *b^2*x^11*D)/11 + A*a^2*b*x^3 + (3*A*a*b^2*x^5)/5 + (3*B*a^2*b*x^5)/5 + (3 *B*a*b^2*x^7)/7 + (3*C*a^2*b*x^7)/7 + (C*a*b^2*x^9)/3
Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^2\right )^3 \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {x \left (3465 b^{3} d \,x^{12}+12285 a \,b^{2} d \,x^{10}+4095 b^{3} c \,x^{10}+15015 a^{2} b d \,x^{8}+15015 a \,b^{2} c \,x^{8}+5005 b^{4} x^{8}+6435 a^{3} d \,x^{6}+19305 a^{2} b c \,x^{6}+25740 a \,b^{3} x^{6}+9009 a^{3} c \,x^{4}+54054 a^{2} b^{2} x^{4}+60060 a^{3} b \,x^{2}+45045 a^{4}\right )}{45045} \] Input:
int((b*x^2+a)^3*(D*x^6+C*x^4+B*x^2+A),x)
Output:
(x*(45045*a**4 + 60060*a**3*b*x**2 + 9009*a**3*c*x**4 + 6435*a**3*d*x**6 + 54054*a**2*b**2*x**4 + 19305*a**2*b*c*x**6 + 15015*a**2*b*d*x**8 + 25740* a*b**3*x**6 + 15015*a*b**2*c*x**8 + 12285*a*b**2*d*x**10 + 5005*b**4*x**8 + 4095*b**3*c*x**10 + 3465*b**3*d*x**12))/45045