Integrand size = 19, antiderivative size = 154 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=a^3 c^3 x+a^2 c^2 (b c+a d) x^3+\frac {3}{5} a c \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^5+\frac {1}{7} (b c+a d) \left (b^2 c^2+8 a b c d+a^2 d^2\right ) x^7+\frac {1}{3} b d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^9+\frac {3}{11} b^2 d^2 (b c+a d) x^{11}+\frac {1}{13} b^3 d^3 x^{13} \]
a^3*c^3*x+a^2*c^2*(a*d+b*c)*x^3+3/5*a*c*(a^2*d^2+3*a*b*c*d+b^2*c^2)*x^5+1/ 7*(a*d+b*c)*(a^2*d^2+8*a*b*c*d+b^2*c^2)*x^7+1/3*b*d*(a^2*d^2+3*a*b*c*d+b^2 *c^2)*x^9+3/11*b^2*d^2*(a*d+b*c)*x^11+1/13*b^3*d^3*x^13
Time = 0.02 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=a^3 c^3 x+a^2 c^2 (b c+a d) x^3+\frac {3}{5} a c \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^5+\frac {1}{7} \left (b^3 c^3+9 a b^2 c^2 d+9 a^2 b c d^2+a^3 d^3\right ) x^7+\frac {1}{3} b d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^9+\frac {3}{11} b^2 d^2 (b c+a d) x^{11}+\frac {1}{13} b^3 d^3 x^{13} \]
a^3*c^3*x + a^2*c^2*(b*c + a*d)*x^3 + (3*a*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^ 2)*x^5)/5 + ((b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*x^7)/7 + (b*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^9)/3 + (3*b^2*d^2*(b*c + a*d)*x^11) /11 + (b^3*d^3*x^13)/13
Time = 0.34 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {290, 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.
\(\displaystyle \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \int \left (a^3 c^3+3 b d x^8 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+x^6 (a d+b c) \left (a^2 d^2+8 a b c d+b^2 c^2\right )+3 a c x^4 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+3 a^2 c^2 x^2 (a d+b c)+3 b^2 d^2 x^{10} (a d+b c)+b^3 d^3 x^{12}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 c^3 x+\frac {1}{3} b d x^9 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+\frac {1}{7} x^7 (a d+b c) \left (a^2 d^2+8 a b c d+b^2 c^2\right )+\frac {3}{5} a c x^5 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a^2 c^2 x^3 (a d+b c)+\frac {3}{11} b^2 d^2 x^{11} (a d+b c)+\frac {1}{13} b^3 d^3 x^{13}\) |
a^3*c^3*x + a^2*c^2*(b*c + a*d)*x^3 + (3*a*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^ 2)*x^5)/5 + ((b*c + a*d)*(b^2*c^2 + 8*a*b*c*d + a^2*d^2)*x^7)/7 + (b*d*(b^ 2*c^2 + 3*a*b*c*d + a^2*d^2)*x^9)/3 + (3*b^2*d^2*(b*c + a*d)*x^11)/11 + (b ^3*d^3*x^13)/13
3.1.14.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Time = 2.26 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11
method | result | size |
norman | \(\frac {b^{3} d^{3} x^{13}}{13}+\left (\frac {3}{11} a \,b^{2} d^{3}+\frac {3}{11} b^{3} c \,d^{2}\right ) x^{11}+\left (\frac {1}{3} a^{2} b \,d^{3}+a \,b^{2} c \,d^{2}+\frac {1}{3} b^{3} c^{2} d \right ) x^{9}+\left (\frac {1}{7} a^{3} d^{3}+\frac {9}{7} a^{2} b c \,d^{2}+\frac {9}{7} a \,b^{2} c^{2} d +\frac {1}{7} b^{3} c^{3}\right ) x^{7}+\left (\frac {3}{5} a^{3} c \,d^{2}+\frac {9}{5} a^{2} b \,c^{2} d +\frac {3}{5} b^{2} c^{3} a \right ) x^{5}+\left (a^{3} c^{2} d +a^{2} b \,c^{3}\right ) x^{3}+a^{3} c^{3} x\) | \(171\) |
default | \(\frac {b^{3} d^{3} x^{13}}{13}+\frac {\left (3 a \,b^{2} d^{3}+3 b^{3} c \,d^{2}\right ) x^{11}}{11}+\frac {\left (3 a^{2} b \,d^{3}+9 a \,b^{2} c \,d^{2}+3 b^{3} c^{2} d \right ) x^{9}}{9}+\frac {\left (a^{3} d^{3}+9 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) x^{7}}{7}+\frac {\left (3 a^{3} c \,d^{2}+9 a^{2} b \,c^{2} d +3 b^{2} c^{3} a \right ) x^{5}}{5}+\frac {\left (3 a^{3} c^{2} d +3 a^{2} b \,c^{3}\right ) x^{3}}{3}+a^{3} c^{3} x\) | \(177\) |
gosper | \(\frac {1}{13} b^{3} d^{3} x^{13}+\frac {3}{11} x^{11} a \,b^{2} d^{3}+\frac {3}{11} x^{11} b^{3} c \,d^{2}+\frac {1}{3} x^{9} a^{2} b \,d^{3}+x^{9} a \,b^{2} c \,d^{2}+\frac {1}{3} x^{9} b^{3} c^{2} d +\frac {1}{7} x^{7} a^{3} d^{3}+\frac {9}{7} x^{7} a^{2} b c \,d^{2}+\frac {9}{7} x^{7} a \,b^{2} c^{2} d +\frac {1}{7} x^{7} b^{3} c^{3}+\frac {3}{5} x^{5} a^{3} c \,d^{2}+\frac {9}{5} x^{5} a^{2} b \,c^{2} d +\frac {3}{5} x^{5} b^{2} c^{3} a +a^{3} c^{2} d \,x^{3}+a^{2} b \,c^{3} x^{3}+a^{3} c^{3} x\) | \(188\) |
risch | \(\frac {1}{13} b^{3} d^{3} x^{13}+\frac {3}{11} x^{11} a \,b^{2} d^{3}+\frac {3}{11} x^{11} b^{3} c \,d^{2}+\frac {1}{3} x^{9} a^{2} b \,d^{3}+x^{9} a \,b^{2} c \,d^{2}+\frac {1}{3} x^{9} b^{3} c^{2} d +\frac {1}{7} x^{7} a^{3} d^{3}+\frac {9}{7} x^{7} a^{2} b c \,d^{2}+\frac {9}{7} x^{7} a \,b^{2} c^{2} d +\frac {1}{7} x^{7} b^{3} c^{3}+\frac {3}{5} x^{5} a^{3} c \,d^{2}+\frac {9}{5} x^{5} a^{2} b \,c^{2} d +\frac {3}{5} x^{5} b^{2} c^{3} a +a^{3} c^{2} d \,x^{3}+a^{2} b \,c^{3} x^{3}+a^{3} c^{3} x\) | \(188\) |
parallelrisch | \(\frac {1}{13} b^{3} d^{3} x^{13}+\frac {3}{11} x^{11} a \,b^{2} d^{3}+\frac {3}{11} x^{11} b^{3} c \,d^{2}+\frac {1}{3} x^{9} a^{2} b \,d^{3}+x^{9} a \,b^{2} c \,d^{2}+\frac {1}{3} x^{9} b^{3} c^{2} d +\frac {1}{7} x^{7} a^{3} d^{3}+\frac {9}{7} x^{7} a^{2} b c \,d^{2}+\frac {9}{7} x^{7} a \,b^{2} c^{2} d +\frac {1}{7} x^{7} b^{3} c^{3}+\frac {3}{5} x^{5} a^{3} c \,d^{2}+\frac {9}{5} x^{5} a^{2} b \,c^{2} d +\frac {3}{5} x^{5} b^{2} c^{3} a +a^{3} c^{2} d \,x^{3}+a^{2} b \,c^{3} x^{3}+a^{3} c^{3} x\) | \(188\) |
1/13*b^3*d^3*x^13+(3/11*a*b^2*d^3+3/11*b^3*c*d^2)*x^11+(1/3*a^2*b*d^3+a*b^ 2*c*d^2+1/3*b^3*c^2*d)*x^9+(1/7*a^3*d^3+9/7*a^2*b*c*d^2+9/7*a*b^2*c^2*d+1/ 7*b^3*c^3)*x^7+(3/5*a^3*c*d^2+9/5*a^2*b*c^2*d+3/5*b^2*c^3*a)*x^5+(a^3*c^2* d+a^2*b*c^3)*x^3+a^3*c^3*x
Time = 0.26 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=\frac {1}{13} \, b^{3} d^{3} x^{13} + \frac {3}{11} \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{11} + \frac {1}{3} \, {\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{9} + \frac {1}{7} \, {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{7} + a^{3} c^{3} x + \frac {3}{5} \, {\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{5} + {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x^{3} \]
1/13*b^3*d^3*x^13 + 3/11*(b^3*c*d^2 + a*b^2*d^3)*x^11 + 1/3*(b^3*c^2*d + 3 *a*b^2*c*d^2 + a^2*b*d^3)*x^9 + 1/7*(b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d ^2 + a^3*d^3)*x^7 + a^3*c^3*x + 3/5*(a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2 )*x^5 + (a^2*b*c^3 + a^3*c^2*d)*x^3
Time = 0.03 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.23 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=a^{3} c^{3} x + \frac {b^{3} d^{3} x^{13}}{13} + x^{11} \cdot \left (\frac {3 a b^{2} d^{3}}{11} + \frac {3 b^{3} c d^{2}}{11}\right ) + x^{9} \left (\frac {a^{2} b d^{3}}{3} + a b^{2} c d^{2} + \frac {b^{3} c^{2} d}{3}\right ) + x^{7} \left (\frac {a^{3} d^{3}}{7} + \frac {9 a^{2} b c d^{2}}{7} + \frac {9 a b^{2} c^{2} d}{7} + \frac {b^{3} c^{3}}{7}\right ) + x^{5} \cdot \left (\frac {3 a^{3} c d^{2}}{5} + \frac {9 a^{2} b c^{2} d}{5} + \frac {3 a b^{2} c^{3}}{5}\right ) + x^{3} \left (a^{3} c^{2} d + a^{2} b c^{3}\right ) \]
a**3*c**3*x + b**3*d**3*x**13/13 + x**11*(3*a*b**2*d**3/11 + 3*b**3*c*d**2 /11) + x**9*(a**2*b*d**3/3 + a*b**2*c*d**2 + b**3*c**2*d/3) + x**7*(a**3*d **3/7 + 9*a**2*b*c*d**2/7 + 9*a*b**2*c**2*d/7 + b**3*c**3/7) + x**5*(3*a** 3*c*d**2/5 + 9*a**2*b*c**2*d/5 + 3*a*b**2*c**3/5) + x**3*(a**3*c**2*d + a* *2*b*c**3)
Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.08 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=\frac {1}{13} \, b^{3} d^{3} x^{13} + \frac {3}{11} \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{11} + \frac {1}{3} \, {\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{9} + \frac {1}{7} \, {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{7} + a^{3} c^{3} x + \frac {3}{5} \, {\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{5} + {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x^{3} \]
1/13*b^3*d^3*x^13 + 3/11*(b^3*c*d^2 + a*b^2*d^3)*x^11 + 1/3*(b^3*c^2*d + 3 *a*b^2*c*d^2 + a^2*b*d^3)*x^9 + 1/7*(b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d ^2 + a^3*d^3)*x^7 + a^3*c^3*x + 3/5*(a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2 )*x^5 + (a^2*b*c^3 + a^3*c^2*d)*x^3
Time = 0.26 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=\frac {1}{13} \, b^{3} d^{3} x^{13} + \frac {3}{11} \, b^{3} c d^{2} x^{11} + \frac {3}{11} \, a b^{2} d^{3} x^{11} + \frac {1}{3} \, b^{3} c^{2} d x^{9} + a b^{2} c d^{2} x^{9} + \frac {1}{3} \, a^{2} b d^{3} x^{9} + \frac {1}{7} \, b^{3} c^{3} x^{7} + \frac {9}{7} \, a b^{2} c^{2} d x^{7} + \frac {9}{7} \, a^{2} b c d^{2} x^{7} + \frac {1}{7} \, a^{3} d^{3} x^{7} + \frac {3}{5} \, a b^{2} c^{3} x^{5} + \frac {9}{5} \, a^{2} b c^{2} d x^{5} + \frac {3}{5} \, a^{3} c d^{2} x^{5} + a^{2} b c^{3} x^{3} + a^{3} c^{2} d x^{3} + a^{3} c^{3} x \]
1/13*b^3*d^3*x^13 + 3/11*b^3*c*d^2*x^11 + 3/11*a*b^2*d^3*x^11 + 1/3*b^3*c^ 2*d*x^9 + a*b^2*c*d^2*x^9 + 1/3*a^2*b*d^3*x^9 + 1/7*b^3*c^3*x^7 + 9/7*a*b^ 2*c^2*d*x^7 + 9/7*a^2*b*c*d^2*x^7 + 1/7*a^3*d^3*x^7 + 3/5*a*b^2*c^3*x^5 + 9/5*a^2*b*c^2*d*x^5 + 3/5*a^3*c*d^2*x^5 + a^2*b*c^3*x^3 + a^3*c^2*d*x^3 + a^3*c^3*x
Time = 4.84 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99 \[ \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx=x^7\,\left (\frac {a^3\,d^3}{7}+\frac {9\,a^2\,b\,c\,d^2}{7}+\frac {9\,a\,b^2\,c^2\,d}{7}+\frac {b^3\,c^3}{7}\right )+a^3\,c^3\,x+\frac {b^3\,d^3\,x^{13}}{13}+\frac {3\,a\,c\,x^5\,\left (a^2\,d^2+3\,a\,b\,c\,d+b^2\,c^2\right )}{5}+\frac {b\,d\,x^9\,\left (a^2\,d^2+3\,a\,b\,c\,d+b^2\,c^2\right )}{3}+a^2\,c^2\,x^3\,\left (a\,d+b\,c\right )+\frac {3\,b^2\,d^2\,x^{11}\,\left (a\,d+b\,c\right )}{11} \]