Integrand size = 22, antiderivative size = 85 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{3} a^3 A x^{3/2}+\frac {2}{9} a^2 (3 A b+a B) x^{9/2}+\frac {2}{5} a b (A b+a B) x^{15/2}+\frac {2}{21} b^2 (A b+3 a B) x^{21/2}+\frac {2}{27} b^3 B x^{27/2} \] Output:
2/3*a^3*A*x^(3/2)+2/9*a^2*(3*A*b+B*a)*x^(9/2)+2/5*a*b*(A*b+B*a)*x^(15/2)+2 /21*b^2*(A*b+3*B*a)*x^(21/2)+2/27*b^3*B*x^(27/2)
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{945} x^{3/2} \left (105 a^3 \left (3 A+B x^3\right )+63 a^2 b x^3 \left (5 A+3 B x^3\right )+27 a b^2 x^6 \left (7 A+5 B x^3\right )+5 b^3 x^9 \left (9 A+7 B x^3\right )\right ) \] Input:
Integrate[Sqrt[x]*(a + b*x^3)^3*(A + B*x^3),x]
Output:
(2*x^(3/2)*(105*a^3*(3*A + B*x^3) + 63*a^2*b*x^3*(5*A + 3*B*x^3) + 27*a*b^ 2*x^6*(7*A + 5*B*x^3) + 5*b^3*x^9*(9*A + 7*B*x^3)))/945
Time = 0.36 (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.
\(\displaystyle \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (a^3 A \sqrt {x}+a^2 x^{7/2} (a B+3 A b)+b^2 x^{19/2} (3 a B+A b)+3 a b x^{13/2} (a B+A b)+b^3 B x^{25/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} a^3 A x^{3/2}+\frac {2}{9} a^2 x^{9/2} (a B+3 A b)+\frac {2}{21} b^2 x^{21/2} (3 a B+A b)+\frac {2}{5} a b x^{15/2} (a B+A b)+\frac {2}{27} b^3 B x^{27/2}\) |
Input:
Int[Sqrt[x]*(a + b*x^3)^3*(A + B*x^3),x]
Output:
(2*a^3*A*x^(3/2))/3 + (2*a^2*(3*A*b + a*B)*x^(9/2))/9 + (2*a*b*(A*b + a*B) *x^(15/2))/5 + (2*b^2*(A*b + 3*a*B)*x^(21/2))/21 + (2*b^3*B*x^(27/2))/27
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 = 0.60 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 b^{3} B \,x^{\frac {27}{2}}}{27}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\) | \(76\) |
default | \(\frac {2 b^{3} B \,x^{\frac {27}{2}}}{27}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {3}{2}} \left (35 B \,b^{3} x^{12}+45 A \,x^{9} b^{3}+135 B \,x^{9} a \,b^{2}+189 A \,x^{6} a \,b^{2}+189 B \,x^{6} a^{2} b +315 a^{2} A b \,x^{3}+105 B \,x^{3} a^{3}+315 a^{3} A \right )}{945}\) | \(80\) |
trager | \(\frac {2 x^{\frac {3}{2}} \left (35 B \,b^{3} x^{12}+45 A \,x^{9} b^{3}+135 B \,x^{9} a \,b^{2}+189 A \,x^{6} a \,b^{2}+189 B \,x^{6} a^{2} b +315 a^{2} A b \,x^{3}+105 B \,x^{3} a^{3}+315 a^{3} A \right )}{945}\) | \(80\) |
risch | \(\frac {2 x^{\frac {3}{2}} \left (35 B \,b^{3} x^{12}+45 A \,x^{9} b^{3}+135 B \,x^{9} a \,b^{2}+189 A \,x^{6} a \,b^{2}+189 B \,x^{6} a^{2} b +315 a^{2} A b \,x^{3}+105 B \,x^{3} a^{3}+315 a^{3} A \right )}{945}\) | \(80\) |
orering | \(\frac {2 x^{\frac {3}{2}} \left (35 B \,b^{3} x^{12}+45 A \,x^{9} b^{3}+135 B \,x^{9} a \,b^{2}+189 A \,x^{6} a \,b^{2}+189 B \,x^{6} a^{2} b +315 a^{2} A b \,x^{3}+105 B \,x^{3} a^{3}+315 a^{3} A \right )}{945}\) | \(80\) |
Input:
int(x^(1/2)*(b*x^3+a)^3*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
2/27*b^3*B*x^(27/2)+2/21*(A*b^3+3*B*a*b^2)*x^(21/2)+2/15*(3*A*a*b^2+3*B*a^ 2*b)*x^(15/2)+2/9*(3*A*a^2*b+B*a^3)*x^(9/2)+2/3*a^3*A*x^(3/2)
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{945} \, {\left (35 \, B b^{3} x^{13} + 45 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{10} + 189 \, {\left (B a^{2} b + A a b^{2}\right )} x^{7} + 315 \, A a^{3} x + 105 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4}\right )} \sqrt {x} \] Input:
integrate(x^(1/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="fricas")
Output:
2/945*(35*B*b^3*x^13 + 45*(3*B*a*b^2 + A*b^3)*x^10 + 189*(B*a^2*b + A*a*b^ 2)*x^7 + 315*A*a^3*x + 105*(B*a^3 + 3*A*a^2*b)*x^4)*sqrt(x)
Time = 1.50 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2 A a^{3} x^{\frac {3}{2}}}{3} + \frac {2 A a^{2} b x^{\frac {9}{2}}}{3} + \frac {2 A a b^{2} x^{\frac {15}{2}}}{5} + \frac {2 A b^{3} x^{\frac {21}{2}}}{21} + \frac {2 B a^{3} x^{\frac {9}{2}}}{9} + \frac {2 B a^{2} b x^{\frac {15}{2}}}{5} + \frac {2 B a b^{2} x^{\frac {21}{2}}}{7} + \frac {2 B b^{3} x^{\frac {27}{2}}}{27} \] Input:
integrate(x**(1/2)*(b*x**3+a)**3*(B*x**3+A),x)
Output:
2*A*a**3*x**(3/2)/3 + 2*A*a**2*b*x**(9/2)/3 + 2*A*a*b**2*x**(15/2)/5 + 2*A *b**3*x**(21/2)/21 + 2*B*a**3*x**(9/2)/9 + 2*B*a**2*b*x**(15/2)/5 + 2*B*a* b**2*x**(21/2)/7 + 2*B*b**3*x**(27/2)/27
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{27} \, B b^{3} x^{\frac {27}{2}} + \frac {2}{21} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac {21}{2}} + \frac {2}{5} \, {\left (B a^{2} b + A a b^{2}\right )} x^{\frac {15}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} + \frac {2}{9} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {9}{2}} \] Input:
integrate(x^(1/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="maxima")
Output:
2/27*B*b^3*x^(27/2) + 2/21*(3*B*a*b^2 + A*b^3)*x^(21/2) + 2/5*(B*a^2*b + A *a*b^2)*x^(15/2) + 2/3*A*a^3*x^(3/2) + 2/9*(B*a^3 + 3*A*a^2*b)*x^(9/2)
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2}{27} \, B b^{3} x^{\frac {27}{2}} + \frac {2}{7} \, B a b^{2} x^{\frac {21}{2}} + \frac {2}{21} \, A b^{3} x^{\frac {21}{2}} + \frac {2}{5} \, B a^{2} b x^{\frac {15}{2}} + \frac {2}{5} \, A a b^{2} x^{\frac {15}{2}} + \frac {2}{9} \, B a^{3} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} b x^{\frac {9}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} \] Input:
integrate(x^(1/2)*(b*x^3+a)^3*(B*x^3+A),x, algorithm="giac")
Output:
2/27*B*b^3*x^(27/2) + 2/7*B*a*b^2*x^(21/2) + 2/21*A*b^3*x^(21/2) + 2/5*B*a ^2*b*x^(15/2) + 2/5*A*a*b^2*x^(15/2) + 2/9*B*a^3*x^(9/2) + 2/3*A*a^2*b*x^( 9/2) + 2/3*A*a^3*x^(3/2)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=x^{9/2}\,\left (\frac {2\,B\,a^3}{9}+\frac {2\,A\,b\,a^2}{3}\right )+x^{21/2}\,\left (\frac {2\,A\,b^3}{21}+\frac {2\,B\,a\,b^2}{7}\right )+\frac {2\,A\,a^3\,x^{3/2}}{3}+\frac {2\,B\,b^3\,x^{27/2}}{27}+\frac {2\,a\,b\,x^{15/2}\,\left (A\,b+B\,a\right )}{5} \] Input:
int(x^(1/2)*(A + B*x^3)*(a + b*x^3)^3,x)
Output:
x^(9/2)*((2*B*a^3)/9 + (2*A*a^2*b)/3) + x^(21/2)*((2*A*b^3)/21 + (2*B*a*b^ 2)/7) + (2*A*a^3*x^(3/2))/3 + (2*B*b^3*x^(27/2))/27 + (2*a*b*x^(15/2)*(A*b + B*a))/5
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int \sqrt {x} \left (a+b x^3\right )^3 \left (A+B x^3\right ) \, dx=\frac {2 \sqrt {x}\, x \left (35 b^{4} x^{12}+180 a \,b^{3} x^{9}+378 a^{2} b^{2} x^{6}+420 a^{3} b \,x^{3}+315 a^{4}\right )}{945} \] Input:
int(x^(1/2)*(b*x^3+a)^3*(B*x^3+A),x)
Output:
(2*sqrt(x)*x*(315*a**4 + 420*a**3*b*x**3 + 378*a**2*b**2*x**6 + 180*a*b**3 *x**9 + 35*b**4*x**12))/945