Integrand size = 22, antiderivative size = 85 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2}{3} a^3 A x^{3/2}+\frac {2}{7} a^2 (3 A b+a B) x^{7/2}+\frac {6}{11} a b (A b+a B) x^{11/2}+\frac {2}{15} b^2 (A b+3 a B) x^{15/2}+\frac {2}{19} b^3 B x^{19/2} \] Output:
2/3*a^3*A*x^(3/2)+2/7*a^2*(3*A*b+B*a)*x^(7/2)+6/11*a*b*(A*b+B*a)*x^(11/2)+ 2/15*b^2*(A*b+3*B*a)*x^(15/2)+2/19*b^3*B*x^(19/2)
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2 x^{3/2} \left (1045 a^3 \left (7 A+3 B x^2\right )+855 a^2 b x^2 \left (11 A+7 B x^2\right )+399 a b^2 x^4 \left (15 A+11 B x^2\right )+77 b^3 x^6 \left (19 A+15 B x^2\right )\right )}{21945} \] Input:
Integrate[Sqrt[x]*(a + b*x^2)^3*(A + B*x^2),x]
Output:
(2*x^(3/2)*(1045*a^3*(7*A + 3*B*x^2) + 855*a^2*b*x^2*(11*A + 7*B*x^2) + 39 9*a*b^2*x^4*(15*A + 11*B*x^2) + 77*b^3*x^6*(19*A + 15*B*x^2)))/21945
Time = 0.20 (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 = {355, 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.
\(\displaystyle \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (a^3 A \sqrt {x}+a^2 x^{5/2} (a B+3 A b)+b^2 x^{13/2} (3 a B+A b)+3 a b x^{9/2} (a B+A b)+b^3 B x^{17/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} a^3 A x^{3/2}+\frac {2}{7} a^2 x^{7/2} (a B+3 A b)+\frac {2}{15} b^2 x^{15/2} (3 a B+A b)+\frac {6}{11} a b x^{11/2} (a B+A b)+\frac {2}{19} b^3 B x^{19/2}\) |
Input:
Int[Sqrt[x]*(a + b*x^2)^3*(A + B*x^2),x]
Output:
(2*a^3*A*x^(3/2))/3 + (2*a^2*(3*A*b + a*B)*x^(7/2))/7 + (6*a*b*(A*b + a*B) *x^(11/2))/11 + (2*b^2*(A*b + 3*a*B)*x^(15/2))/15 + (2*b^3*B*x^(19/2))/19
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 b^{3} B \,x^{\frac {19}{2}}}{19}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\) | \(76\) |
default | \(\frac {2 b^{3} B \,x^{\frac {19}{2}}}{19}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{3} A \,x^{\frac {3}{2}}}{3}\) | \(76\) |
gosper | \(\frac {2 x^{\frac {3}{2}} \left (1155 b^{3} B \,x^{8}+1463 A \,b^{3} x^{6}+4389 B a \,b^{2} x^{6}+5985 a A \,b^{2} x^{4}+5985 B \,a^{2} b \,x^{4}+9405 a^{2} A b \,x^{2}+3135 B \,a^{3} x^{2}+7315 a^{3} A \right )}{21945}\) | \(80\) |
trager | \(\frac {2 x^{\frac {3}{2}} \left (1155 b^{3} B \,x^{8}+1463 A \,b^{3} x^{6}+4389 B a \,b^{2} x^{6}+5985 a A \,b^{2} x^{4}+5985 B \,a^{2} b \,x^{4}+9405 a^{2} A b \,x^{2}+3135 B \,a^{3} x^{2}+7315 a^{3} A \right )}{21945}\) | \(80\) |
risch | \(\frac {2 x^{\frac {3}{2}} \left (1155 b^{3} B \,x^{8}+1463 A \,b^{3} x^{6}+4389 B a \,b^{2} x^{6}+5985 a A \,b^{2} x^{4}+5985 B \,a^{2} b \,x^{4}+9405 a^{2} A b \,x^{2}+3135 B \,a^{3} x^{2}+7315 a^{3} A \right )}{21945}\) | \(80\) |
orering | \(\frac {2 x^{\frac {3}{2}} \left (1155 b^{3} B \,x^{8}+1463 A \,b^{3} x^{6}+4389 B a \,b^{2} x^{6}+5985 a A \,b^{2} x^{4}+5985 B \,a^{2} b \,x^{4}+9405 a^{2} A b \,x^{2}+3135 B \,a^{3} x^{2}+7315 a^{3} A \right )}{21945}\) | \(80\) |
Input:
int(x^(1/2)*(b*x^2+a)^3*(B*x^2+A),x,method=_RETURNVERBOSE)
Output:
2/19*b^3*B*x^(19/2)+2/15*(A*b^3+3*B*a*b^2)*x^(15/2)+2/11*(3*A*a*b^2+3*B*a^ 2*b)*x^(11/2)+2/7*(3*A*a^2*b+B*a^3)*x^(7/2)+2/3*a^3*A*x^(3/2)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2}{21945} \, {\left (1155 \, B b^{3} x^{9} + 1463 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{7} + 5985 \, {\left (B a^{2} b + A a b^{2}\right )} x^{5} + 7315 \, A a^{3} x + 3135 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )} \sqrt {x} \] Input:
integrate(x^(1/2)*(b*x^2+a)^3*(B*x^2+A),x, algorithm="fricas")
Output:
2/21945*(1155*B*b^3*x^9 + 1463*(3*B*a*b^2 + A*b^3)*x^7 + 5985*(B*a^2*b + A *a*b^2)*x^5 + 7315*A*a^3*x + 3135*(B*a^3 + 3*A*a^2*b)*x^3)*sqrt(x)
Time = 0.68 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2 A a^{3} x^{\frac {3}{2}}}{3} + \frac {2 B b^{3} x^{\frac {19}{2}}}{19} + \frac {2 x^{\frac {15}{2}} \left (A b^{3} + 3 B a b^{2}\right )}{15} + \frac {2 x^{\frac {11}{2}} \cdot \left (3 A a b^{2} + 3 B a^{2} b\right )}{11} + \frac {2 x^{\frac {7}{2}} \cdot \left (3 A a^{2} b + B a^{3}\right )}{7} \] Input:
integrate(x**(1/2)*(b*x**2+a)**3*(B*x**2+A),x)
Output:
2*A*a**3*x**(3/2)/3 + 2*B*b**3*x**(19/2)/19 + 2*x**(15/2)*(A*b**3 + 3*B*a* b**2)/15 + 2*x**(11/2)*(3*A*a*b**2 + 3*B*a**2*b)/11 + 2*x**(7/2)*(3*A*a**2 *b + B*a**3)/7
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2}{19} \, B b^{3} x^{\frac {19}{2}} + \frac {2}{15} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac {15}{2}} + \frac {6}{11} \, {\left (B a^{2} b + A a b^{2}\right )} x^{\frac {11}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} + \frac {2}{7} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {7}{2}} \] Input:
integrate(x^(1/2)*(b*x^2+a)^3*(B*x^2+A),x, algorithm="maxima")
Output:
2/19*B*b^3*x^(19/2) + 2/15*(3*B*a*b^2 + A*b^3)*x^(15/2) + 6/11*(B*a^2*b + A*a*b^2)*x^(11/2) + 2/3*A*a^3*x^(3/2) + 2/7*(B*a^3 + 3*A*a^2*b)*x^(7/2)
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2}{19} \, B b^{3} x^{\frac {19}{2}} + \frac {2}{5} \, B a b^{2} x^{\frac {15}{2}} + \frac {2}{15} \, A b^{3} x^{\frac {15}{2}} + \frac {6}{11} \, B a^{2} b x^{\frac {11}{2}} + \frac {6}{11} \, A a b^{2} x^{\frac {11}{2}} + \frac {2}{7} \, B a^{3} x^{\frac {7}{2}} + \frac {6}{7} \, A a^{2} b x^{\frac {7}{2}} + \frac {2}{3} \, A a^{3} x^{\frac {3}{2}} \] Input:
integrate(x^(1/2)*(b*x^2+a)^3*(B*x^2+A),x, algorithm="giac")
Output:
2/19*B*b^3*x^(19/2) + 2/5*B*a*b^2*x^(15/2) + 2/15*A*b^3*x^(15/2) + 6/11*B* a^2*b*x^(11/2) + 6/11*A*a*b^2*x^(11/2) + 2/7*B*a^3*x^(7/2) + 6/7*A*a^2*b*x ^(7/2) + 2/3*A*a^3*x^(3/2)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=x^{7/2}\,\left (\frac {2\,B\,a^3}{7}+\frac {6\,A\,b\,a^2}{7}\right )+x^{15/2}\,\left (\frac {2\,A\,b^3}{15}+\frac {2\,B\,a\,b^2}{5}\right )+\frac {2\,A\,a^3\,x^{3/2}}{3}+\frac {2\,B\,b^3\,x^{19/2}}{19}+\frac {6\,a\,b\,x^{11/2}\,\left (A\,b+B\,a\right )}{11} \] Input:
int(x^(1/2)*(A + B*x^2)*(a + b*x^2)^3,x)
Output:
x^(7/2)*((2*B*a^3)/7 + (6*A*a^2*b)/7) + x^(15/2)*((2*A*b^3)/15 + (2*B*a*b^ 2)/5) + (2*A*a^3*x^(3/2))/3 + (2*B*b^3*x^(19/2))/19 + (6*a*b*x^(11/2)*(A*b + B*a))/11
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int \sqrt {x} \left (a+b x^2\right )^3 \left (A+B x^2\right ) \, dx=\frac {2 \sqrt {x}\, x \left (1155 b^{4} x^{8}+5852 a \,b^{3} x^{6}+11970 a^{2} b^{2} x^{4}+12540 a^{3} b \,x^{2}+7315 a^{4}\right )}{21945} \] Input:
int(x^(1/2)*(b*x^2+a)^3*(B*x^2+A),x)
Output:
(2*sqrt(x)*x*(7315*a**4 + 12540*a**3*b*x**2 + 11970*a**2*b**2*x**4 + 5852* a*b**3*x**6 + 1155*b**4*x**8))/21945