Integrand size = 19, antiderivative size = 154 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=a^2 c^4 x+\frac {2}{5} a c^3 (b c+2 a d) x^5+\frac {1}{9} c^2 \left (b^2 c^2+8 a b c d+6 a^2 d^2\right ) x^9+\frac {4}{13} c d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^{13}+\frac {1}{17} d^2 \left (6 b^2 c^2+8 a b c d+a^2 d^2\right ) x^{17}+\frac {2}{21} b d^3 (2 b c+a d) x^{21}+\frac {1}{25} b^2 d^4 x^{25} \]
a^2*c^4*x+2/5*a*c^3*(2*a*d+b*c)*x^5+1/9*c^2*(6*a^2*d^2+8*a*b*c*d+b^2*c^2)* x^9+4/13*c*d*(a^2*d^2+3*a*b*c*d+b^2*c^2)*x^13+1/17*d^2*(a^2*d^2+8*a*b*c*d+ 6*b^2*c^2)*x^17+2/21*b*d^3*(a*d+2*b*c)*x^21+1/25*b^2*d^4*x^25
Time = 0.03 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=a^2 c^4 x+\frac {2}{5} a c^3 (b c+2 a d) x^5+\frac {1}{9} c^2 \left (b^2 c^2+8 a b c d+6 a^2 d^2\right ) x^9+\frac {4}{13} c d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^{13}+\frac {1}{17} d^2 \left (6 b^2 c^2+8 a b c d+a^2 d^2\right ) x^{17}+\frac {2}{21} b d^3 (2 b c+a d) x^{21}+\frac {1}{25} b^2 d^4 x^{25} \]
a^2*c^4*x + (2*a*c^3*(b*c + 2*a*d)*x^5)/5 + (c^2*(b^2*c^2 + 8*a*b*c*d + 6* a^2*d^2)*x^9)/9 + (4*c*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^13)/13 + (d^2*( 6*b^2*c^2 + 8*a*b*c*d + a^2*d^2)*x^17)/17 + (2*b*d^3*(2*b*c + a*d)*x^21)/2 1 + (b^2*d^4*x^25)/25
Time = 0.32 (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 = {897, 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.
\(\displaystyle \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx\) |
\(\Big \downarrow \) 897 |
\(\displaystyle \int \left (d^2 x^{16} \left (a^2 d^2+8 a b c d+6 b^2 c^2\right )+4 c d x^{12} \left (a^2 d^2+3 a b c d+b^2 c^2\right )+c^2 x^8 \left (6 a^2 d^2+8 a b c d+b^2 c^2\right )+a^2 c^4+2 a c^3 x^4 (2 a d+b c)+2 b d^3 x^{20} (a d+2 b c)+b^2 d^4 x^{24}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{17} d^2 x^{17} \left (a^2 d^2+8 a b c d+6 b^2 c^2\right )+\frac {4}{13} c d x^{13} \left (a^2 d^2+3 a b c d+b^2 c^2\right )+\frac {1}{9} c^2 x^9 \left (6 a^2 d^2+8 a b c d+b^2 c^2\right )+a^2 c^4 x+\frac {2}{5} a c^3 x^5 (2 a d+b c)+\frac {2}{21} b d^3 x^{21} (a d+2 b c)+\frac {1}{25} b^2 d^4 x^{25}\) |
a^2*c^4*x + (2*a*c^3*(b*c + 2*a*d)*x^5)/5 + (c^2*(b^2*c^2 + 8*a*b*c*d + 6* a^2*d^2)*x^9)/9 + (4*c*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^13)/13 + (d^2*( 6*b^2*c^2 + 8*a*b*c*d + a^2*d^2)*x^17)/17 + (2*b*d^3*(2*b*c + a*d)*x^21)/2 1 + (b^2*d^4*x^25)/25
3.2.53.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> Int[ExpandIntegrand[(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b , c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Time = 3.95 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {b^{2} d^{4} x^{25}}{25}+\left (\frac {2}{21} a b \,d^{4}+\frac {4}{21} b^{2} c \,d^{3}\right ) x^{21}+\left (\frac {1}{17} a^{2} d^{4}+\frac {8}{17} a b c \,d^{3}+\frac {6}{17} b^{2} c^{2} d^{2}\right ) x^{17}+\left (\frac {4}{13} a^{2} c \,d^{3}+\frac {12}{13} a b \,c^{2} d^{2}+\frac {4}{13} b^{2} c^{3} d \right ) x^{13}+\left (\frac {2}{3} a^{2} c^{2} d^{2}+\frac {8}{9} a b \,c^{3} d +\frac {1}{9} b^{2} c^{4}\right ) x^{9}+a^{2} c^{4} x +\left (\frac {4}{5} a^{2} c^{3} d +\frac {2}{5} a b \,c^{4}\right ) x^{5}\) | \(160\) |
default | \(\frac {b^{2} d^{4} x^{25}}{25}+\frac {\left (2 a b \,d^{4}+4 b^{2} c \,d^{3}\right ) x^{21}}{21}+\frac {\left (a^{2} d^{4}+8 a b c \,d^{3}+6 b^{2} c^{2} d^{2}\right ) x^{17}}{17}+\frac {\left (4 a^{2} c \,d^{3}+12 a b \,c^{2} d^{2}+4 b^{2} c^{3} d \right ) x^{13}}{13}+\frac {\left (6 a^{2} c^{2} d^{2}+8 a b \,c^{3} d +b^{2} c^{4}\right ) x^{9}}{9}+\frac {\left (4 a^{2} c^{3} d +2 a b \,c^{4}\right ) x^{5}}{5}+a^{2} c^{4} x\) | \(163\) |
gosper | \(\frac {1}{25} b^{2} d^{4} x^{25}+\frac {2}{21} x^{21} a b \,d^{4}+\frac {4}{21} x^{21} b^{2} c \,d^{3}+\frac {1}{17} x^{17} a^{2} d^{4}+\frac {8}{17} x^{17} a b c \,d^{3}+\frac {6}{17} x^{17} b^{2} c^{2} d^{2}+\frac {4}{13} x^{13} a^{2} c \,d^{3}+\frac {12}{13} x^{13} a b \,c^{2} d^{2}+\frac {4}{13} x^{13} b^{2} c^{3} d +\frac {2}{3} x^{9} a^{2} c^{2} d^{2}+\frac {8}{9} x^{9} a b \,c^{3} d +\frac {1}{9} x^{9} b^{2} c^{4}+a^{2} c^{4} x +\frac {4}{5} x^{5} a^{2} c^{3} d +\frac {2}{5} x^{5} a b \,c^{4}\) | \(174\) |
risch | \(\frac {1}{25} b^{2} d^{4} x^{25}+\frac {2}{21} x^{21} a b \,d^{4}+\frac {4}{21} x^{21} b^{2} c \,d^{3}+\frac {1}{17} x^{17} a^{2} d^{4}+\frac {8}{17} x^{17} a b c \,d^{3}+\frac {6}{17} x^{17} b^{2} c^{2} d^{2}+\frac {4}{13} x^{13} a^{2} c \,d^{3}+\frac {12}{13} x^{13} a b \,c^{2} d^{2}+\frac {4}{13} x^{13} b^{2} c^{3} d +\frac {2}{3} x^{9} a^{2} c^{2} d^{2}+\frac {8}{9} x^{9} a b \,c^{3} d +\frac {1}{9} x^{9} b^{2} c^{4}+a^{2} c^{4} x +\frac {4}{5} x^{5} a^{2} c^{3} d +\frac {2}{5} x^{5} a b \,c^{4}\) | \(174\) |
parallelrisch | \(\frac {1}{25} b^{2} d^{4} x^{25}+\frac {2}{21} x^{21} a b \,d^{4}+\frac {4}{21} x^{21} b^{2} c \,d^{3}+\frac {1}{17} x^{17} a^{2} d^{4}+\frac {8}{17} x^{17} a b c \,d^{3}+\frac {6}{17} x^{17} b^{2} c^{2} d^{2}+\frac {4}{13} x^{13} a^{2} c \,d^{3}+\frac {12}{13} x^{13} a b \,c^{2} d^{2}+\frac {4}{13} x^{13} b^{2} c^{3} d +\frac {2}{3} x^{9} a^{2} c^{2} d^{2}+\frac {8}{9} x^{9} a b \,c^{3} d +\frac {1}{9} x^{9} b^{2} c^{4}+a^{2} c^{4} x +\frac {4}{5} x^{5} a^{2} c^{3} d +\frac {2}{5} x^{5} a b \,c^{4}\) | \(174\) |
1/25*b^2*d^4*x^25+(2/21*a*b*d^4+4/21*b^2*c*d^3)*x^21+(1/17*a^2*d^4+8/17*a* b*c*d^3+6/17*b^2*c^2*d^2)*x^17+(4/13*a^2*c*d^3+12/13*a*b*c^2*d^2+4/13*b^2* c^3*d)*x^13+(2/3*a^2*c^2*d^2+8/9*a*b*c^3*d+1/9*b^2*c^4)*x^9+a^2*c^4*x+(4/5 *a^2*c^3*d+2/5*a*b*c^4)*x^5
Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=\frac {1}{25} \, b^{2} d^{4} x^{25} + \frac {2}{21} \, {\left (2 \, b^{2} c d^{3} + a b d^{4}\right )} x^{21} + \frac {1}{17} \, {\left (6 \, b^{2} c^{2} d^{2} + 8 \, a b c d^{3} + a^{2} d^{4}\right )} x^{17} + \frac {4}{13} \, {\left (b^{2} c^{3} d + 3 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{13} + \frac {1}{9} \, {\left (b^{2} c^{4} + 8 \, a b c^{3} d + 6 \, a^{2} c^{2} d^{2}\right )} x^{9} + a^{2} c^{4} x + \frac {2}{5} \, {\left (a b c^{4} + 2 \, a^{2} c^{3} d\right )} x^{5} \]
1/25*b^2*d^4*x^25 + 2/21*(2*b^2*c*d^3 + a*b*d^4)*x^21 + 1/17*(6*b^2*c^2*d^ 2 + 8*a*b*c*d^3 + a^2*d^4)*x^17 + 4/13*(b^2*c^3*d + 3*a*b*c^2*d^2 + a^2*c* d^3)*x^13 + 1/9*(b^2*c^4 + 8*a*b*c^3*d + 6*a^2*c^2*d^2)*x^9 + a^2*c^4*x + 2/5*(a*b*c^4 + 2*a^2*c^3*d)*x^5
Time = 0.03 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=a^{2} c^{4} x + \frac {b^{2} d^{4} x^{25}}{25} + x^{21} \cdot \left (\frac {2 a b d^{4}}{21} + \frac {4 b^{2} c d^{3}}{21}\right ) + x^{17} \left (\frac {a^{2} d^{4}}{17} + \frac {8 a b c d^{3}}{17} + \frac {6 b^{2} c^{2} d^{2}}{17}\right ) + x^{13} \cdot \left (\frac {4 a^{2} c d^{3}}{13} + \frac {12 a b c^{2} d^{2}}{13} + \frac {4 b^{2} c^{3} d}{13}\right ) + x^{9} \cdot \left (\frac {2 a^{2} c^{2} d^{2}}{3} + \frac {8 a b c^{3} d}{9} + \frac {b^{2} c^{4}}{9}\right ) + x^{5} \cdot \left (\frac {4 a^{2} c^{3} d}{5} + \frac {2 a b c^{4}}{5}\right ) \]
a**2*c**4*x + b**2*d**4*x**25/25 + x**21*(2*a*b*d**4/21 + 4*b**2*c*d**3/21 ) + x**17*(a**2*d**4/17 + 8*a*b*c*d**3/17 + 6*b**2*c**2*d**2/17) + x**13*( 4*a**2*c*d**3/13 + 12*a*b*c**2*d**2/13 + 4*b**2*c**3*d/13) + x**9*(2*a**2* c**2*d**2/3 + 8*a*b*c**3*d/9 + b**2*c**4/9) + x**5*(4*a**2*c**3*d/5 + 2*a* b*c**4/5)
Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=\frac {1}{25} \, b^{2} d^{4} x^{25} + \frac {2}{21} \, {\left (2 \, b^{2} c d^{3} + a b d^{4}\right )} x^{21} + \frac {1}{17} \, {\left (6 \, b^{2} c^{2} d^{2} + 8 \, a b c d^{3} + a^{2} d^{4}\right )} x^{17} + \frac {4}{13} \, {\left (b^{2} c^{3} d + 3 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{13} + \frac {1}{9} \, {\left (b^{2} c^{4} + 8 \, a b c^{3} d + 6 \, a^{2} c^{2} d^{2}\right )} x^{9} + a^{2} c^{4} x + \frac {2}{5} \, {\left (a b c^{4} + 2 \, a^{2} c^{3} d\right )} x^{5} \]
1/25*b^2*d^4*x^25 + 2/21*(2*b^2*c*d^3 + a*b*d^4)*x^21 + 1/17*(6*b^2*c^2*d^ 2 + 8*a*b*c*d^3 + a^2*d^4)*x^17 + 4/13*(b^2*c^3*d + 3*a*b*c^2*d^2 + a^2*c* d^3)*x^13 + 1/9*(b^2*c^4 + 8*a*b*c^3*d + 6*a^2*c^2*d^2)*x^9 + a^2*c^4*x + 2/5*(a*b*c^4 + 2*a^2*c^3*d)*x^5
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=\frac {1}{25} \, b^{2} d^{4} x^{25} + \frac {4}{21} \, b^{2} c d^{3} x^{21} + \frac {2}{21} \, a b d^{4} x^{21} + \frac {6}{17} \, b^{2} c^{2} d^{2} x^{17} + \frac {8}{17} \, a b c d^{3} x^{17} + \frac {1}{17} \, a^{2} d^{4} x^{17} + \frac {4}{13} \, b^{2} c^{3} d x^{13} + \frac {12}{13} \, a b c^{2} d^{2} x^{13} + \frac {4}{13} \, a^{2} c d^{3} x^{13} + \frac {1}{9} \, b^{2} c^{4} x^{9} + \frac {8}{9} \, a b c^{3} d x^{9} + \frac {2}{3} \, a^{2} c^{2} d^{2} x^{9} + \frac {2}{5} \, a b c^{4} x^{5} + \frac {4}{5} \, a^{2} c^{3} d x^{5} + a^{2} c^{4} x \]
1/25*b^2*d^4*x^25 + 4/21*b^2*c*d^3*x^21 + 2/21*a*b*d^4*x^21 + 6/17*b^2*c^2 *d^2*x^17 + 8/17*a*b*c*d^3*x^17 + 1/17*a^2*d^4*x^17 + 4/13*b^2*c^3*d*x^13 + 12/13*a*b*c^2*d^2*x^13 + 4/13*a^2*c*d^3*x^13 + 1/9*b^2*c^4*x^9 + 8/9*a*b *c^3*d*x^9 + 2/3*a^2*c^2*d^2*x^9 + 2/5*a*b*c^4*x^5 + 4/5*a^2*c^3*d*x^5 + a ^2*c^4*x
Time = 5.65 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^4 \, dx=x^9\,\left (\frac {2\,a^2\,c^2\,d^2}{3}+\frac {8\,a\,b\,c^3\,d}{9}+\frac {b^2\,c^4}{9}\right )+x^{17}\,\left (\frac {a^2\,d^4}{17}+\frac {8\,a\,b\,c\,d^3}{17}+\frac {6\,b^2\,c^2\,d^2}{17}\right )+a^2\,c^4\,x+\frac {b^2\,d^4\,x^{25}}{25}+\frac {2\,a\,c^3\,x^5\,\left (2\,a\,d+b\,c\right )}{5}+\frac {2\,b\,d^3\,x^{21}\,\left (a\,d+2\,b\,c\right )}{21}+\frac {4\,c\,d\,x^{13}\,\left (a^2\,d^2+3\,a\,b\,c\,d+b^2\,c^2\right )}{13} \]