Integrand size = 22, antiderivative size = 85 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{5} a^3 A x^{5/2}+\frac {2}{11} a^2 (3 A b+a B) x^{11/2}+\frac {6}{17} a b (A b+a B) x^{17/2}+\frac {2}{23} b^2 (A b+3 a B) x^{23/2}+\frac {2}{29} b^3 B x^{29/2} \] Output:
2/5*a^3*A*x^(5/2)+2/11*a^2*(3*A*b+B*a)*x^(11/2)+6/17*a*b*(A*b+B*a)*x^(17/2 )+2/23*b^2*(A*b+3*B*a)*x^(23/2)+2/29*b^3*B*x^(29/2)
Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2 x^{5/2} \left (11339 a^3 \left (11 A+5 B x^3\right )+10005 a^2 b x^3 \left (17 A+11 B x^3\right )+4785 a b^2 x^6 \left (23 A+17 B x^3\right )+935 b^3 x^9 \left (29 A+23 B x^3\right )\right )}{623645} \] Input:
Integrate[x^(3/2)*(a + b*x^3)^3*(A + B*x^3),x]
Output:
(2*x^(5/2)*(11339*a^3*(11*A + 5*B*x^3) + 10005*a^2*b*x^3*(17*A + 11*B*x^3) + 4785*a*b^2*x^6*(23*A + 17*B*x^3) + 935*b^3*x^9*(29*A + 23*B*x^3)))/6236 45
Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {950, 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.
\(\displaystyle \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (a^3 A x^{3/2}+a^2 x^{9/2} (a B+3 A b)+b^2 x^{21/2} (3 a B+A b)+3 a b x^{15/2} (a B+A b)+b^3 B x^{27/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} a^3 A x^{5/2}+\frac {2}{11} a^2 x^{11/2} (a B+3 A b)+\frac {2}{23} b^2 x^{23/2} (3 a B+A b)+\frac {6}{17} a b x^{17/2} (a B+A b)+\frac {2}{29} b^3 B x^{29/2}\) |
Input:
Int[x^(3/2)*(a + b*x^3)^3*(A + B*x^3),x]
Output:
(2*a^3*A*x^(5/2))/5 + (2*a^2*(3*A*b + a*B)*x^(11/2))/11 + (6*a*b*(A*b + a* B)*x^(17/2))/17 + (2*b^2*(A*b + 3*a*B)*x^(23/2))/23 + (2*b^3*B*x^(29/2))/2 9
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^ n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGt Q[p, 0] && IGtQ[q, 0]
Time = 0.58 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 b^{3} B \,x^{\frac {29}{2}}}{29}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{3} A \,x^{\frac {5}{2}}}{5}\) | \(76\) |
default | \(\frac {2 b^{3} B \,x^{\frac {29}{2}}}{29}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {23}{2}}}{23}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {11}{2}}}{11}+\frac {2 a^{3} A \,x^{\frac {5}{2}}}{5}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {5}{2}} \left (21505 B \,b^{3} x^{12}+27115 A \,x^{9} b^{3}+81345 B \,x^{9} a \,b^{2}+110055 A \,x^{6} a \,b^{2}+110055 B \,x^{6} a^{2} b +170085 a^{2} A b \,x^{3}+56695 B \,x^{3} a^{3}+124729 a^{3} A \right )}{623645}\) | \(80\) |
trager | \(\frac {2 x^{\frac {5}{2}} \left (21505 B \,b^{3} x^{12}+27115 A \,x^{9} b^{3}+81345 B \,x^{9} a \,b^{2}+110055 A \,x^{6} a \,b^{2}+110055 B \,x^{6} a^{2} b +170085 a^{2} A b \,x^{3}+56695 B \,x^{3} a^{3}+124729 a^{3} A \right )}{623645}\) | \(80\) |
risch | \(\frac {2 x^{\frac {5}{2}} \left (21505 B \,b^{3} x^{12}+27115 A \,x^{9} b^{3}+81345 B \,x^{9} a \,b^{2}+110055 A \,x^{6} a \,b^{2}+110055 B \,x^{6} a^{2} b +170085 a^{2} A b \,x^{3}+56695 B \,x^{3} a^{3}+124729 a^{3} A \right )}{623645}\) | \(80\) |
orering | \(\frac {2 x^{\frac {5}{2}} \left (21505 B \,b^{3} x^{12}+27115 A \,x^{9} b^{3}+81345 B \,x^{9} a \,b^{2}+110055 A \,x^{6} a \,b^{2}+110055 B \,x^{6} a^{2} b +170085 a^{2} A b \,x^{3}+56695 B \,x^{3} a^{3}+124729 a^{3} A \right )}{623645}\) | \(80\) |
Input:
int(x^(3/2)*(b*x^3+a)^3*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
2/29*b^3*B*x^(29/2)+2/23*(A*b^3+3*B*a*b^2)*x^(23/2)+2/17*(3*A*a*b^2+3*B*a^ 2*b)*x^(17/2)+2/11*(3*A*a^2*b+B*a^3)*x^(11/2)+2/5*a^3*A*x^(5/2)
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{623645} \, {\left (21505 \, B b^{3} x^{14} + 27115 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{11} + 110055 \, {\left (B a^{2} b + A a b^{2}\right )} x^{8} + 124729 \, A a^{3} x^{2} + 56695 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5}\right )} \sqrt {x} \] Input:
integrate(x^(3/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="fricas")
Output:
2/623645*(21505*B*b^3*x^14 + 27115*(3*B*a*b^2 + A*b^3)*x^11 + 110055*(B*a^ 2*b + A*a*b^2)*x^8 + 124729*A*a^3*x^2 + 56695*(B*a^3 + 3*A*a^2*b)*x^5)*sqr t(x)
Time = 1.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2 A a^{3} x^{\frac {5}{2}}}{5} + \frac {6 A a^{2} b x^{\frac {11}{2}}}{11} + \frac {6 A a b^{2} x^{\frac {17}{2}}}{17} + \frac {2 A b^{3} x^{\frac {23}{2}}}{23} + \frac {2 B a^{3} x^{\frac {11}{2}}}{11} + \frac {6 B a^{2} b x^{\frac {17}{2}}}{17} + \frac {6 B a b^{2} x^{\frac {23}{2}}}{23} + \frac {2 B b^{3} x^{\frac {29}{2}}}{29} \] Input:
integrate(x**(3/2)*(b*x**3+a)**3*(B*x**3+A),x)
Output:
2*A*a**3*x**(5/2)/5 + 6*A*a**2*b*x**(11/2)/11 + 6*A*a*b**2*x**(17/2)/17 + 2*A*b**3*x**(23/2)/23 + 2*B*a**3*x**(11/2)/11 + 6*B*a**2*b*x**(17/2)/17 + 6*B*a*b**2*x**(23/2)/23 + 2*B*b**3*x**(29/2)/29
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{29} \, B b^{3} x^{\frac {29}{2}} + \frac {2}{23} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac {23}{2}} + \frac {6}{17} \, {\left (B a^{2} b + A a b^{2}\right )} x^{\frac {17}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} + \frac {2}{11} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {11}{2}} \] Input:
integrate(x^(3/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="maxima")
Output:
2/29*B*b^3*x^(29/2) + 2/23*(3*B*a*b^2 + A*b^3)*x^(23/2) + 6/17*(B*a^2*b + A*a*b^2)*x^(17/2) + 2/5*A*a^3*x^(5/2) + 2/11*(B*a^3 + 3*A*a^2*b)*x^(11/2)
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{29} \, B b^{3} x^{\frac {29}{2}} + \frac {6}{23} \, B a b^{2} x^{\frac {23}{2}} + \frac {2}{23} \, A b^{3} x^{\frac {23}{2}} + \frac {6}{17} \, B a^{2} b x^{\frac {17}{2}} + \frac {6}{17} \, A a b^{2} x^{\frac {17}{2}} + \frac {2}{11} \, B a^{3} x^{\frac {11}{2}} + \frac {6}{11} \, A a^{2} b x^{\frac {11}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} \] Input:
integrate(x^(3/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="giac")
Output:
2/29*B*b^3*x^(29/2) + 6/23*B*a*b^2*x^(23/2) + 2/23*A*b^3*x^(23/2) + 6/17*B *a^2*b*x^(17/2) + 6/17*A*a*b^2*x^(17/2) + 2/11*B*a^3*x^(11/2) + 6/11*A*a^2 *b*x^(11/2) + 2/5*A*a^3*x^(5/2)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=x^{11/2}\,\left (\frac {2\,B\,a^3}{11}+\frac {6\,A\,b\,a^2}{11}\right )+x^{23/2}\,\left (\frac {2\,A\,b^3}{23}+\frac {6\,B\,a\,b^2}{23}\right )+\frac {2\,A\,a^3\,x^{5/2}}{5}+\frac {2\,B\,b^3\,x^{29/2}}{29}+\frac {6\,a\,b\,x^{17/2}\,\left (A\,b+B\,a\right )}{17} \] Input:
int(x^(3/2)*(A + B*x^3)*(a + b*x^3)^3,x)
Output:
x^(11/2)*((2*B*a^3)/11 + (6*A*a^2*b)/11) + x^(23/2)*((2*A*b^3)/23 + (6*B*a *b^2)/23) + (2*A*a^3*x^(5/2))/5 + (2*B*b^3*x^(29/2))/29 + (6*a*b*x^(17/2)* (A*b + B*a))/17
Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.59 \[ \int x^{3/2} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2 \sqrt {x}\, x^{2} \left (21505 b^{4} x^{12}+108460 a \,b^{3} x^{9}+220110 a^{2} b^{2} x^{6}+226780 a^{3} b \,x^{3}+124729 a^{4}\right )}{623645} \] Input:
int(x^(3/2)*(b*x^3+a)^3*(B*x^3+A),x)
Output:
(2*sqrt(x)*x**2*(124729*a**4 + 226780*a**3*b*x**3 + 220110*a**2*b**2*x**6 + 108460*a*b**3*x**9 + 21505*b**4*x**12))/623645