Integrand size = 19, antiderivative size = 82 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=a^2 c^2 x+\frac {1}{2} a c (b c+a d) x^4+\frac {1}{7} \left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^7+\frac {1}{5} b d (b c+a d) x^{10}+\frac {1}{13} b^2 d^2 x^{13} \] Output:
a^2*c^2*x+1/2*a*c*(a*d+b*c)*x^4+1/7*(a^2*d^2+4*a*b*c*d+b^2*c^2)*x^7+1/5*b* d*(a*d+b*c)*x^10+1/13*b^2*d^2*x^13
Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=a^2 c^2 x+\frac {1}{2} a c (b c+a d) x^4+\frac {1}{7} \left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^7+\frac {1}{5} b d (b c+a d) x^{10}+\frac {1}{13} b^2 d^2 x^{13} \] Input:
Integrate[(a + b*x^3)^2*(c + d*x^3)^2,x]
Output:
a^2*c^2*x + (a*c*(b*c + a*d)*x^4)/2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^7 )/7 + (b*d*(b*c + a*d)*x^10)/5 + (b^2*d^2*x^13)/13
Time = 0.40 (sec) , antiderivative size = 82, 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.
\(\displaystyle \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx\) |
\(\Big \downarrow \) 897 |
\(\displaystyle \int \left (x^6 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+a^2 c^2+2 b d x^9 (a d+b c)+2 a c x^3 (a d+b c)+b^2 d^2 x^{12}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} x^7 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+a^2 c^2 x+\frac {1}{5} b d x^{10} (a d+b c)+\frac {1}{2} a c x^4 (a d+b c)+\frac {1}{13} b^2 d^2 x^{13}\) |
Input:
Int[(a + b*x^3)^2*(c + d*x^3)^2,x]
Output:
a^2*c^2*x + (a*c*(b*c + a*d)*x^4)/2 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^7 )/7 + (b*d*(b*c + a*d)*x^10)/5 + (b^2*d^2*x^13)/13
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 = 0.75 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {b^{2} d^{2} x^{13}}{13}+\left (\frac {1}{5} a b \,d^{2}+\frac {1}{5} b^{2} c d \right ) x^{10}+\left (\frac {1}{7} a^{2} d^{2}+\frac {4}{7} a b c d +\frac {1}{7} b^{2} c^{2}\right ) x^{7}+\left (\frac {1}{2} a^{2} c d +\frac {1}{2} b \,c^{2} a \right ) x^{4}+a^{2} c^{2} x\) | \(86\) |
default | \(\frac {b^{2} d^{2} x^{13}}{13}+\frac {\left (2 a b \,d^{2}+2 b^{2} c d \right ) x^{10}}{10}+\frac {\left (a^{2} d^{2}+4 a b c d +b^{2} c^{2}\right ) x^{7}}{7}+\frac {\left (2 a^{2} c d +2 b \,c^{2} a \right ) x^{4}}{4}+a^{2} c^{2} x\) | \(87\) |
gosper | \(\frac {1}{13} b^{2} d^{2} x^{13}+\frac {1}{5} x^{10} a b \,d^{2}+\frac {1}{5} x^{10} b^{2} c d +\frac {1}{7} x^{7} a^{2} d^{2}+\frac {4}{7} x^{7} a b c d +\frac {1}{7} x^{7} b^{2} c^{2}+\frac {1}{2} x^{4} a^{2} c d +\frac {1}{2} x^{4} b \,c^{2} a +a^{2} c^{2} x\) | \(92\) |
risch | \(\frac {1}{13} b^{2} d^{2} x^{13}+\frac {1}{5} x^{10} a b \,d^{2}+\frac {1}{5} x^{10} b^{2} c d +\frac {1}{7} x^{7} a^{2} d^{2}+\frac {4}{7} x^{7} a b c d +\frac {1}{7} x^{7} b^{2} c^{2}+\frac {1}{2} x^{4} a^{2} c d +\frac {1}{2} x^{4} b \,c^{2} a +a^{2} c^{2} x\) | \(92\) |
parallelrisch | \(\frac {1}{13} b^{2} d^{2} x^{13}+\frac {1}{5} x^{10} a b \,d^{2}+\frac {1}{5} x^{10} b^{2} c d +\frac {1}{7} x^{7} a^{2} d^{2}+\frac {4}{7} x^{7} a b c d +\frac {1}{7} x^{7} b^{2} c^{2}+\frac {1}{2} x^{4} a^{2} c d +\frac {1}{2} x^{4} b \,c^{2} a +a^{2} c^{2} x\) | \(92\) |
orering | \(\frac {x \left (70 b^{2} d^{2} x^{12}+182 a b \,d^{2} x^{9}+182 b^{2} c d \,x^{9}+130 a^{2} d^{2} x^{6}+520 a b c d \,x^{6}+130 b^{2} c^{2} x^{6}+455 a^{2} c d \,x^{3}+455 a b \,c^{2} x^{3}+910 a^{2} c^{2}\right )}{910}\) | \(95\) |
Input:
int((b*x^3+a)^2*(d*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
1/13*b^2*d^2*x^13+(1/5*a*b*d^2+1/5*b^2*c*d)*x^10+(1/7*a^2*d^2+4/7*a*b*c*d+ 1/7*b^2*c^2)*x^7+(1/2*a^2*c*d+1/2*b*c^2*a)*x^4+a^2*c^2*x
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=\frac {1}{13} \, b^{2} d^{2} x^{13} + \frac {1}{5} \, {\left (b^{2} c d + a b d^{2}\right )} x^{10} + \frac {1}{7} \, {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{7} + a^{2} c^{2} x + \frac {1}{2} \, {\left (a b c^{2} + a^{2} c d\right )} x^{4} \] Input:
integrate((b*x^3+a)^2*(d*x^3+c)^2,x, algorithm="fricas")
Output:
1/13*b^2*d^2*x^13 + 1/5*(b^2*c*d + a*b*d^2)*x^10 + 1/7*(b^2*c^2 + 4*a*b*c* d + a^2*d^2)*x^7 + a^2*c^2*x + 1/2*(a*b*c^2 + a^2*c*d)*x^4
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=a^{2} c^{2} x + \frac {b^{2} d^{2} x^{13}}{13} + x^{10} \left (\frac {a b d^{2}}{5} + \frac {b^{2} c d}{5}\right ) + x^{7} \left (\frac {a^{2} d^{2}}{7} + \frac {4 a b c d}{7} + \frac {b^{2} c^{2}}{7}\right ) + x^{4} \left (\frac {a^{2} c d}{2} + \frac {a b c^{2}}{2}\right ) \] Input:
integrate((b*x**3+a)**2*(d*x**3+c)**2,x)
Output:
a**2*c**2*x + b**2*d**2*x**13/13 + x**10*(a*b*d**2/5 + b**2*c*d/5) + x**7* (a**2*d**2/7 + 4*a*b*c*d/7 + b**2*c**2/7) + x**4*(a**2*c*d/2 + a*b*c**2/2)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=\frac {1}{13} \, b^{2} d^{2} x^{13} + \frac {1}{5} \, {\left (b^{2} c d + a b d^{2}\right )} x^{10} + \frac {1}{7} \, {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{7} + a^{2} c^{2} x + \frac {1}{2} \, {\left (a b c^{2} + a^{2} c d\right )} x^{4} \] Input:
integrate((b*x^3+a)^2*(d*x^3+c)^2,x, algorithm="maxima")
Output:
1/13*b^2*d^2*x^13 + 1/5*(b^2*c*d + a*b*d^2)*x^10 + 1/7*(b^2*c^2 + 4*a*b*c* d + a^2*d^2)*x^7 + a^2*c^2*x + 1/2*(a*b*c^2 + a^2*c*d)*x^4
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=\frac {1}{13} \, b^{2} d^{2} x^{13} + \frac {1}{5} \, b^{2} c d x^{10} + \frac {1}{5} \, a b d^{2} x^{10} + \frac {1}{7} \, b^{2} c^{2} x^{7} + \frac {4}{7} \, a b c d x^{7} + \frac {1}{7} \, a^{2} d^{2} x^{7} + \frac {1}{2} \, a b c^{2} x^{4} + \frac {1}{2} \, a^{2} c d x^{4} + a^{2} c^{2} x \] Input:
integrate((b*x^3+a)^2*(d*x^3+c)^2,x, algorithm="giac")
Output:
1/13*b^2*d^2*x^13 + 1/5*b^2*c*d*x^10 + 1/5*a*b*d^2*x^10 + 1/7*b^2*c^2*x^7 + 4/7*a*b*c*d*x^7 + 1/7*a^2*d^2*x^7 + 1/2*a*b*c^2*x^4 + 1/2*a^2*c*d*x^4 + a^2*c^2*x
Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=x^7\,\left (\frac {a^2\,d^2}{7}+\frac {4\,a\,b\,c\,d}{7}+\frac {b^2\,c^2}{7}\right )+a^2\,c^2\,x+\frac {b^2\,d^2\,x^{13}}{13}+\frac {a\,c\,x^4\,\left (a\,d+b\,c\right )}{2}+\frac {b\,d\,x^{10}\,\left (a\,d+b\,c\right )}{5} \] Input:
int((a + b*x^3)^2*(c + d*x^3)^2,x)
Output:
x^7*((a^2*d^2)/7 + (b^2*c^2)/7 + (4*a*b*c*d)/7) + a^2*c^2*x + (b^2*d^2*x^1 3)/13 + (a*c*x^4*(a*d + b*c))/2 + (b*d*x^10*(a*d + b*c))/5
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^2 \, dx=\frac {x \left (70 b^{2} d^{2} x^{12}+182 a b \,d^{2} x^{9}+182 b^{2} c d \,x^{9}+130 a^{2} d^{2} x^{6}+520 a b c d \,x^{6}+130 b^{2} c^{2} x^{6}+455 a^{2} c d \,x^{3}+455 a b \,c^{2} x^{3}+910 a^{2} c^{2}\right )}{910} \] Input:
int((b*x^3+a)^2*(d*x^3+c)^2,x)
Output:
(x*(910*a**2*c**2 + 455*a**2*c*d*x**3 + 130*a**2*d**2*x**6 + 455*a*b*c**2* x**3 + 520*a*b*c*d*x**6 + 182*a*b*d**2*x**9 + 130*b**2*c**2*x**6 + 182*b** 2*c*d*x**9 + 70*b**2*d**2*x**12))/910