Integrand size = 24, antiderivative size = 137 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=-\frac {2 a^2 c^3}{\sqrt {x}}+\frac {2}{3} a c^2 (2 b c+3 a d) x^{3/2}+\frac {2}{7} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^{7/2}+\frac {2}{11} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{11/2}+\frac {2}{15} b d^2 (3 b c+2 a d) x^{15/2}+\frac {2}{19} b^2 d^3 x^{19/2} \]
2/3*a*c^2*(3*a*d+2*b*c)*x^(3/2)+2/7*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^(7/2 )+2/11*d*(a^2*d^2+6*a*b*c*d+3*b^2*c^2)*x^(11/2)+2/15*b*d^2*(2*a*d+3*b*c)*x ^(15/2)+2/19*b^2*d^3*x^(19/2)-2*a^2*c^3/x^(1/2)
Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=\frac {-570 a^2 \left (77 c^3-77 c^2 d x^2-33 c d^2 x^4-7 d^3 x^6\right )+76 a b x^2 \left (385 c^3+495 c^2 d x^2+315 c d^2 x^4+77 d^3 x^6\right )+6 b^2 x^4 \left (1045 c^3+1995 c^2 d x^2+1463 c d^2 x^4+385 d^3 x^6\right )}{21945 \sqrt {x}} \]
(-570*a^2*(77*c^3 - 77*c^2*d*x^2 - 33*c*d^2*x^4 - 7*d^3*x^6) + 76*a*b*x^2* (385*c^3 + 495*c^2*d*x^2 + 315*c*d^2*x^4 + 77*d^3*x^6) + 6*b^2*x^4*(1045*c ^3 + 1995*c^2*d*x^2 + 1463*c*d^2*x^4 + 385*d^3*x^6))/(21945*Sqrt[x])
Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (d x^{9/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+c x^{5/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {a^2 c^3}{x^{3/2}}+a c^2 \sqrt {x} (3 a d+2 b c)+b d^2 x^{13/2} (2 a d+3 b c)+b^2 d^3 x^{17/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{11} d x^{11/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {2}{7} c x^{7/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )-\frac {2 a^2 c^3}{\sqrt {x}}+\frac {2}{3} a c^2 x^{3/2} (3 a d+2 b c)+\frac {2}{15} b d^2 x^{15/2} (2 a d+3 b c)+\frac {2}{19} b^2 d^3 x^{19/2}\) |
(-2*a^2*c^3)/Sqrt[x] + (2*a*c^2*(2*b*c + 3*a*d)*x^(3/2))/3 + (2*c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^(7/2))/7 + (2*d*(3*b^2*c^2 + 6*a*b*c*d + a^2*d ^2)*x^(11/2))/11 + (2*b*d^2*(3*b*c + 2*a*d)*x^(15/2))/15 + (2*b^2*d^3*x^(1 9/2))/19
3.5.12.3.1 Defintions of rubi rules used
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 = 2.82 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {2 b^{2} d^{3} x^{\frac {19}{2}}}{19}+\frac {4 a b \,d^{3} x^{\frac {15}{2}}}{15}+\frac {2 b^{2} c \,d^{2} x^{\frac {15}{2}}}{5}+\frac {2 a^{2} d^{3} x^{\frac {11}{2}}}{11}+\frac {12 a b c \,d^{2} x^{\frac {11}{2}}}{11}+\frac {6 b^{2} c^{2} d \,x^{\frac {11}{2}}}{11}+\frac {6 a^{2} c \,d^{2} x^{\frac {7}{2}}}{7}+\frac {12 a b \,c^{2} d \,x^{\frac {7}{2}}}{7}+\frac {2 b^{2} c^{3} x^{\frac {7}{2}}}{7}+2 a^{2} c^{2} d \,x^{\frac {3}{2}}+\frac {4 a b \,c^{3} x^{\frac {3}{2}}}{3}-\frac {2 a^{2} c^{3}}{\sqrt {x}}\) | \(136\) |
default | \(\frac {2 b^{2} d^{3} x^{\frac {19}{2}}}{19}+\frac {4 a b \,d^{3} x^{\frac {15}{2}}}{15}+\frac {2 b^{2} c \,d^{2} x^{\frac {15}{2}}}{5}+\frac {2 a^{2} d^{3} x^{\frac {11}{2}}}{11}+\frac {12 a b c \,d^{2} x^{\frac {11}{2}}}{11}+\frac {6 b^{2} c^{2} d \,x^{\frac {11}{2}}}{11}+\frac {6 a^{2} c \,d^{2} x^{\frac {7}{2}}}{7}+\frac {12 a b \,c^{2} d \,x^{\frac {7}{2}}}{7}+\frac {2 b^{2} c^{3} x^{\frac {7}{2}}}{7}+2 a^{2} c^{2} d \,x^{\frac {3}{2}}+\frac {4 a b \,c^{3} x^{\frac {3}{2}}}{3}-\frac {2 a^{2} c^{3}}{\sqrt {x}}\) | \(136\) |
gosper | \(-\frac {2 \left (-1155 b^{2} d^{3} x^{10}-2926 a b \,d^{3} x^{8}-4389 b^{2} c \,d^{2} x^{8}-1995 a^{2} d^{3} x^{6}-11970 x^{6} d^{2} a b c -5985 b^{2} c^{2} d \,x^{6}-9405 a^{2} c \,d^{2} x^{4}-18810 a b \,c^{2} d \,x^{4}-3135 b^{2} c^{3} x^{4}-21945 a^{2} c^{2} d \,x^{2}-14630 a b \,c^{3} x^{2}+21945 a^{2} c^{3}\right )}{21945 \sqrt {x}}\) | \(138\) |
trager | \(-\frac {2 \left (-1155 b^{2} d^{3} x^{10}-2926 a b \,d^{3} x^{8}-4389 b^{2} c \,d^{2} x^{8}-1995 a^{2} d^{3} x^{6}-11970 x^{6} d^{2} a b c -5985 b^{2} c^{2} d \,x^{6}-9405 a^{2} c \,d^{2} x^{4}-18810 a b \,c^{2} d \,x^{4}-3135 b^{2} c^{3} x^{4}-21945 a^{2} c^{2} d \,x^{2}-14630 a b \,c^{3} x^{2}+21945 a^{2} c^{3}\right )}{21945 \sqrt {x}}\) | \(138\) |
risch | \(-\frac {2 \left (-1155 b^{2} d^{3} x^{10}-2926 a b \,d^{3} x^{8}-4389 b^{2} c \,d^{2} x^{8}-1995 a^{2} d^{3} x^{6}-11970 x^{6} d^{2} a b c -5985 b^{2} c^{2} d \,x^{6}-9405 a^{2} c \,d^{2} x^{4}-18810 a b \,c^{2} d \,x^{4}-3135 b^{2} c^{3} x^{4}-21945 a^{2} c^{2} d \,x^{2}-14630 a b \,c^{3} x^{2}+21945 a^{2} c^{3}\right )}{21945 \sqrt {x}}\) | \(138\) |
2/19*b^2*d^3*x^(19/2)+4/15*a*b*d^3*x^(15/2)+2/5*b^2*c*d^2*x^(15/2)+2/11*a^ 2*d^3*x^(11/2)+12/11*a*b*c*d^2*x^(11/2)+6/11*b^2*c^2*d*x^(11/2)+6/7*a^2*c* d^2*x^(7/2)+12/7*a*b*c^2*d*x^(7/2)+2/7*b^2*c^3*x^(7/2)+2*a^2*c^2*d*x^(3/2) +4/3*a*b*c^3*x^(3/2)-2*a^2*c^3/x^(1/2)
Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=\frac {2 \, {\left (1155 \, b^{2} d^{3} x^{10} + 1463 \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{8} + 1995 \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{6} - 21945 \, a^{2} c^{3} + 3135 \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} + 7315 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )}}{21945 \, \sqrt {x}} \]
2/21945*(1155*b^2*d^3*x^10 + 1463*(3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + 1995*(3* b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 - 21945*a^2*c^3 + 3135*(b^2*c^3 + 6 *a*b*c^2*d + 3*a^2*c*d^2)*x^4 + 7315*(2*a*b*c^3 + 3*a^2*c^2*d)*x^2)/sqrt(x )
Time = 0.73 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=- \frac {2 a^{2} c^{3}}{\sqrt {x}} + 2 a^{2} c^{2} d x^{\frac {3}{2}} + \frac {6 a^{2} c d^{2} x^{\frac {7}{2}}}{7} + \frac {2 a^{2} d^{3} x^{\frac {11}{2}}}{11} + \frac {4 a b c^{3} x^{\frac {3}{2}}}{3} + \frac {12 a b c^{2} d x^{\frac {7}{2}}}{7} + \frac {12 a b c d^{2} x^{\frac {11}{2}}}{11} + \frac {4 a b d^{3} x^{\frac {15}{2}}}{15} + \frac {2 b^{2} c^{3} x^{\frac {7}{2}}}{7} + \frac {6 b^{2} c^{2} d x^{\frac {11}{2}}}{11} + \frac {2 b^{2} c d^{2} x^{\frac {15}{2}}}{5} + \frac {2 b^{2} d^{3} x^{\frac {19}{2}}}{19} \]
-2*a**2*c**3/sqrt(x) + 2*a**2*c**2*d*x**(3/2) + 6*a**2*c*d**2*x**(7/2)/7 + 2*a**2*d**3*x**(11/2)/11 + 4*a*b*c**3*x**(3/2)/3 + 12*a*b*c**2*d*x**(7/2) /7 + 12*a*b*c*d**2*x**(11/2)/11 + 4*a*b*d**3*x**(15/2)/15 + 2*b**2*c**3*x* *(7/2)/7 + 6*b**2*c**2*d*x**(11/2)/11 + 2*b**2*c*d**2*x**(15/2)/5 + 2*b**2 *d**3*x**(19/2)/19
Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=\frac {2}{19} \, b^{2} d^{3} x^{\frac {19}{2}} + \frac {2}{15} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{\frac {15}{2}} + \frac {2}{11} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {11}{2}} - \frac {2 \, a^{2} c^{3}}{\sqrt {x}} + \frac {2}{7} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{\frac {7}{2}} + \frac {2}{3} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{\frac {3}{2}} \]
2/19*b^2*d^3*x^(19/2) + 2/15*(3*b^2*c*d^2 + 2*a*b*d^3)*x^(15/2) + 2/11*(3* b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^(11/2) - 2*a^2*c^3/sqrt(x) + 2/7*(b^2 *c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(7/2) + 2/3*(2*a*b*c^3 + 3*a^2*c^2*d)* x^(3/2)
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=\frac {2}{19} \, b^{2} d^{3} x^{\frac {19}{2}} + \frac {2}{5} \, b^{2} c d^{2} x^{\frac {15}{2}} + \frac {4}{15} \, a b d^{3} x^{\frac {15}{2}} + \frac {6}{11} \, b^{2} c^{2} d x^{\frac {11}{2}} + \frac {12}{11} \, a b c d^{2} x^{\frac {11}{2}} + \frac {2}{11} \, a^{2} d^{3} x^{\frac {11}{2}} + \frac {2}{7} \, b^{2} c^{3} x^{\frac {7}{2}} + \frac {12}{7} \, a b c^{2} d x^{\frac {7}{2}} + \frac {6}{7} \, a^{2} c d^{2} x^{\frac {7}{2}} + \frac {4}{3} \, a b c^{3} x^{\frac {3}{2}} + 2 \, a^{2} c^{2} d x^{\frac {3}{2}} - \frac {2 \, a^{2} c^{3}}{\sqrt {x}} \]
2/19*b^2*d^3*x^(19/2) + 2/5*b^2*c*d^2*x^(15/2) + 4/15*a*b*d^3*x^(15/2) + 6 /11*b^2*c^2*d*x^(11/2) + 12/11*a*b*c*d^2*x^(11/2) + 2/11*a^2*d^3*x^(11/2) + 2/7*b^2*c^3*x^(7/2) + 12/7*a*b*c^2*d*x^(7/2) + 6/7*a^2*c*d^2*x^(7/2) + 4 /3*a*b*c^3*x^(3/2) + 2*a^2*c^2*d*x^(3/2) - 2*a^2*c^3/sqrt(x)
Time = 0.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{3/2}} \, dx=x^{7/2}\,\left (\frac {6\,a^2\,c\,d^2}{7}+\frac {12\,a\,b\,c^2\,d}{7}+\frac {2\,b^2\,c^3}{7}\right )+x^{11/2}\,\left (\frac {2\,a^2\,d^3}{11}+\frac {12\,a\,b\,c\,d^2}{11}+\frac {6\,b^2\,c^2\,d}{11}\right )-\frac {2\,a^2\,c^3}{\sqrt {x}}+\frac {2\,b^2\,d^3\,x^{19/2}}{19}+\frac {2\,a\,c^2\,x^{3/2}\,\left (3\,a\,d+2\,b\,c\right )}{3}+\frac {2\,b\,d^2\,x^{15/2}\,\left (2\,a\,d+3\,b\,c\right )}{15} \]