Integrand size = 24, antiderivative size = 137 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=-\frac {2 a^2 c^3}{3 x^{3/2}}+2 a c^2 (2 b c+3 a d) \sqrt {x}+\frac {2}{5} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^{5/2}+\frac {2}{9} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{9/2}+\frac {2}{13} b d^2 (3 b c+2 a d) x^{13/2}+\frac {2}{17} b^2 d^3 x^{17/2} \]
-2/3*a^2*c^3/x^(3/2)+2/5*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^(5/2)+2/9*d*(a^ 2*d^2+6*a*b*c*d+3*b^2*c^2)*x^(9/2)+2/13*b*d^2*(2*a*d+3*b*c)*x^(13/2)+2/17* b^2*d^3*x^(17/2)+2*a*c^2*(3*a*d+2*b*c)*x^(1/2)
Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=\frac {-442 a^2 \left (15 c^3-135 c^2 d x^2-27 c d^2 x^4-5 d^3 x^6\right )+204 a b x^2 \left (195 c^3+117 c^2 d x^2+65 c d^2 x^4+15 d^3 x^6\right )+6 b^2 x^4 \left (663 c^3+1105 c^2 d x^2+765 c d^2 x^4+195 d^3 x^6\right )}{9945 x^{3/2}} \]
(-442*a^2*(15*c^3 - 135*c^2*d*x^2 - 27*c*d^2*x^4 - 5*d^3*x^6) + 204*a*b*x^ 2*(195*c^3 + 117*c^2*d*x^2 + 65*c*d^2*x^4 + 15*d^3*x^6) + 6*b^2*x^4*(663*c ^3 + 1105*c^2*d*x^2 + 765*c*d^2*x^4 + 195*d^3*x^6))/(9945*x^(3/2))
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (d x^{7/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+c x^{3/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {a^2 c^3}{x^{5/2}}+\frac {a c^2 (3 a d+2 b c)}{\sqrt {x}}+b d^2 x^{11/2} (2 a d+3 b c)+b^2 d^3 x^{15/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{9} d x^{9/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {2}{5} c x^{5/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )-\frac {2 a^2 c^3}{3 x^{3/2}}+2 a c^2 \sqrt {x} (3 a d+2 b c)+\frac {2}{13} b d^2 x^{13/2} (2 a d+3 b c)+\frac {2}{17} b^2 d^3 x^{17/2}\) |
(-2*a^2*c^3)/(3*x^(3/2)) + 2*a*c^2*(2*b*c + 3*a*d)*Sqrt[x] + (2*c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^(5/2))/5 + (2*d*(3*b^2*c^2 + 6*a*b*c*d + a^2*d ^2)*x^(9/2))/9 + (2*b*d^2*(3*b*c + 2*a*d)*x^(13/2))/13 + (2*b^2*d^3*x^(17/ 2))/17
3.5.13.3.1 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 = 2.80 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {2 b^{2} d^{3} x^{\frac {17}{2}}}{17}+\frac {4 a b \,d^{3} x^{\frac {13}{2}}}{13}+\frac {6 b^{2} c \,d^{2} x^{\frac {13}{2}}}{13}+\frac {2 a^{2} d^{3} x^{\frac {9}{2}}}{9}+\frac {4 a b c \,d^{2} x^{\frac {9}{2}}}{3}+\frac {2 b^{2} c^{2} d \,x^{\frac {9}{2}}}{3}+\frac {6 a^{2} c \,d^{2} x^{\frac {5}{2}}}{5}+\frac {12 a b \,c^{2} d \,x^{\frac {5}{2}}}{5}+\frac {2 b^{2} c^{3} x^{\frac {5}{2}}}{5}+6 a^{2} c^{2} d \sqrt {x}+4 a b \,c^{3} \sqrt {x}-\frac {2 a^{2} c^{3}}{3 x^{\frac {3}{2}}}\) | \(136\) |
default | \(\frac {2 b^{2} d^{3} x^{\frac {17}{2}}}{17}+\frac {4 a b \,d^{3} x^{\frac {13}{2}}}{13}+\frac {6 b^{2} c \,d^{2} x^{\frac {13}{2}}}{13}+\frac {2 a^{2} d^{3} x^{\frac {9}{2}}}{9}+\frac {4 a b c \,d^{2} x^{\frac {9}{2}}}{3}+\frac {2 b^{2} c^{2} d \,x^{\frac {9}{2}}}{3}+\frac {6 a^{2} c \,d^{2} x^{\frac {5}{2}}}{5}+\frac {12 a b \,c^{2} d \,x^{\frac {5}{2}}}{5}+\frac {2 b^{2} c^{3} x^{\frac {5}{2}}}{5}+6 a^{2} c^{2} d \sqrt {x}+4 a b \,c^{3} \sqrt {x}-\frac {2 a^{2} c^{3}}{3 x^{\frac {3}{2}}}\) | \(136\) |
gosper | \(-\frac {2 \left (-585 b^{2} d^{3} x^{10}-1530 a b \,d^{3} x^{8}-2295 b^{2} c \,d^{2} x^{8}-1105 a^{2} d^{3} x^{6}-6630 x^{6} d^{2} a b c -3315 b^{2} c^{2} d \,x^{6}-5967 a^{2} c \,d^{2} x^{4}-11934 a b \,c^{2} d \,x^{4}-1989 b^{2} c^{3} x^{4}-29835 a^{2} c^{2} d \,x^{2}-19890 a b \,c^{3} x^{2}+3315 a^{2} c^{3}\right )}{9945 x^{\frac {3}{2}}}\) | \(138\) |
trager | \(-\frac {2 \left (-585 b^{2} d^{3} x^{10}-1530 a b \,d^{3} x^{8}-2295 b^{2} c \,d^{2} x^{8}-1105 a^{2} d^{3} x^{6}-6630 x^{6} d^{2} a b c -3315 b^{2} c^{2} d \,x^{6}-5967 a^{2} c \,d^{2} x^{4}-11934 a b \,c^{2} d \,x^{4}-1989 b^{2} c^{3} x^{4}-29835 a^{2} c^{2} d \,x^{2}-19890 a b \,c^{3} x^{2}+3315 a^{2} c^{3}\right )}{9945 x^{\frac {3}{2}}}\) | \(138\) |
risch | \(-\frac {2 \left (-585 b^{2} d^{3} x^{10}-1530 a b \,d^{3} x^{8}-2295 b^{2} c \,d^{2} x^{8}-1105 a^{2} d^{3} x^{6}-6630 x^{6} d^{2} a b c -3315 b^{2} c^{2} d \,x^{6}-5967 a^{2} c \,d^{2} x^{4}-11934 a b \,c^{2} d \,x^{4}-1989 b^{2} c^{3} x^{4}-29835 a^{2} c^{2} d \,x^{2}-19890 a b \,c^{3} x^{2}+3315 a^{2} c^{3}\right )}{9945 x^{\frac {3}{2}}}\) | \(138\) |
2/17*b^2*d^3*x^(17/2)+4/13*a*b*d^3*x^(13/2)+6/13*b^2*c*d^2*x^(13/2)+2/9*a^ 2*d^3*x^(9/2)+4/3*a*b*c*d^2*x^(9/2)+2/3*b^2*c^2*d*x^(9/2)+6/5*a^2*c*d^2*x^ (5/2)+12/5*a*b*c^2*d*x^(5/2)+2/5*b^2*c^3*x^(5/2)+6*a^2*c^2*d*x^(1/2)+4*a*b *c^3*x^(1/2)-2/3*a^2*c^3/x^(3/2)
Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=\frac {2 \, {\left (585 \, b^{2} d^{3} x^{10} + 765 \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{8} + 1105 \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{6} - 3315 \, a^{2} c^{3} + 1989 \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} + 9945 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )}}{9945 \, x^{\frac {3}{2}}} \]
2/9945*(585*b^2*d^3*x^10 + 765*(3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + 1105*(3*b^2 *c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 - 3315*a^2*c^3 + 1989*(b^2*c^3 + 6*a*b *c^2*d + 3*a^2*c*d^2)*x^4 + 9945*(2*a*b*c^3 + 3*a^2*c^2*d)*x^2)/x^(3/2)
Time = 0.81 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=- \frac {2 a^{2} c^{3}}{3 x^{\frac {3}{2}}} + 6 a^{2} c^{2} d \sqrt {x} + \frac {6 a^{2} c d^{2} x^{\frac {5}{2}}}{5} + \frac {2 a^{2} d^{3} x^{\frac {9}{2}}}{9} + 4 a b c^{3} \sqrt {x} + \frac {12 a b c^{2} d x^{\frac {5}{2}}}{5} + \frac {4 a b c d^{2} x^{\frac {9}{2}}}{3} + \frac {4 a b d^{3} x^{\frac {13}{2}}}{13} + \frac {2 b^{2} c^{3} x^{\frac {5}{2}}}{5} + \frac {2 b^{2} c^{2} d x^{\frac {9}{2}}}{3} + \frac {6 b^{2} c d^{2} x^{\frac {13}{2}}}{13} + \frac {2 b^{2} d^{3} x^{\frac {17}{2}}}{17} \]
-2*a**2*c**3/(3*x**(3/2)) + 6*a**2*c**2*d*sqrt(x) + 6*a**2*c*d**2*x**(5/2) /5 + 2*a**2*d**3*x**(9/2)/9 + 4*a*b*c**3*sqrt(x) + 12*a*b*c**2*d*x**(5/2)/ 5 + 4*a*b*c*d**2*x**(9/2)/3 + 4*a*b*d**3*x**(13/2)/13 + 2*b**2*c**3*x**(5/ 2)/5 + 2*b**2*c**2*d*x**(9/2)/3 + 6*b**2*c*d**2*x**(13/2)/13 + 2*b**2*d**3 *x**(17/2)/17
Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=\frac {2}{17} \, b^{2} d^{3} x^{\frac {17}{2}} + \frac {2}{13} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{\frac {13}{2}} + \frac {2}{9} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {9}{2}} - \frac {2 \, a^{2} c^{3}}{3 \, x^{\frac {3}{2}}} + \frac {2}{5} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{\frac {5}{2}} + 2 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} \sqrt {x} \]
2/17*b^2*d^3*x^(17/2) + 2/13*(3*b^2*c*d^2 + 2*a*b*d^3)*x^(13/2) + 2/9*(3*b ^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^(9/2) - 2/3*a^2*c^3/x^(3/2) + 2/5*(b^2 *c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(5/2) + 2*(2*a*b*c^3 + 3*a^2*c^2*d)*sq rt(x)
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=\frac {2}{17} \, b^{2} d^{3} x^{\frac {17}{2}} + \frac {6}{13} \, b^{2} c d^{2} x^{\frac {13}{2}} + \frac {4}{13} \, a b d^{3} x^{\frac {13}{2}} + \frac {2}{3} \, b^{2} c^{2} d x^{\frac {9}{2}} + \frac {4}{3} \, a b c d^{2} x^{\frac {9}{2}} + \frac {2}{9} \, a^{2} d^{3} x^{\frac {9}{2}} + \frac {2}{5} \, b^{2} c^{3} x^{\frac {5}{2}} + \frac {12}{5} \, a b c^{2} d x^{\frac {5}{2}} + \frac {6}{5} \, a^{2} c d^{2} x^{\frac {5}{2}} + 4 \, a b c^{3} \sqrt {x} + 6 \, a^{2} c^{2} d \sqrt {x} - \frac {2 \, a^{2} c^{3}}{3 \, x^{\frac {3}{2}}} \]
2/17*b^2*d^3*x^(17/2) + 6/13*b^2*c*d^2*x^(13/2) + 4/13*a*b*d^3*x^(13/2) + 2/3*b^2*c^2*d*x^(9/2) + 4/3*a*b*c*d^2*x^(9/2) + 2/9*a^2*d^3*x^(9/2) + 2/5* b^2*c^3*x^(5/2) + 12/5*a*b*c^2*d*x^(5/2) + 6/5*a^2*c*d^2*x^(5/2) + 4*a*b*c ^3*sqrt(x) + 6*a^2*c^2*d*sqrt(x) - 2/3*a^2*c^3/x^(3/2)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{5/2}} \, dx=x^{5/2}\,\left (\frac {6\,a^2\,c\,d^2}{5}+\frac {12\,a\,b\,c^2\,d}{5}+\frac {2\,b^2\,c^3}{5}\right )+x^{9/2}\,\left (\frac {2\,a^2\,d^3}{9}+\frac {4\,a\,b\,c\,d^2}{3}+\frac {2\,b^2\,c^2\,d}{3}\right )-\frac {2\,a^2\,c^3}{3\,x^{3/2}}+\frac {2\,b^2\,d^3\,x^{17/2}}{17}+2\,a\,c^2\,\sqrt {x}\,\left (3\,a\,d+2\,b\,c\right )+\frac {2\,b\,d^2\,x^{13/2}\,\left (2\,a\,d+3\,b\,c\right )}{13} \]