Integrand size = 24, antiderivative size = 137 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=-\frac {2 a^2 c^3}{5 x^{5/2}}-\frac {2 a c^2 (2 b c+3 a d)}{\sqrt {x}}+\frac {2}{3} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^{3/2}+\frac {2}{7} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{7/2}+\frac {2}{11} b d^2 (3 b c+2 a d) x^{11/2}+\frac {2}{15} b^2 d^3 x^{15/2} \]
-2/5*a^2*c^3/x^(5/2)+2/3*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^(3/2)+2/7*d*(a^ 2*d^2+6*a*b*c*d+3*b^2*c^2)*x^(7/2)+2/11*b*d^2*(2*a*d+3*b*c)*x^(11/2)+2/15* b^2*d^3*x^(15/2)-2*a*c^2*(3*a*d+2*b*c)/x^(1/2)
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=\frac {-66 a^2 \left (7 c^3+105 c^2 d x^2-35 c d^2 x^4-5 d^3 x^6\right )+60 a b x^2 \left (-77 c^3+77 c^2 d x^2+33 c d^2 x^4+7 d^3 x^6\right )+2 b^2 x^4 \left (385 c^3+495 c^2 d x^2+315 c d^2 x^4+77 d^3 x^6\right )}{1155 x^{5/2}} \]
(-66*a^2*(7*c^3 + 105*c^2*d*x^2 - 35*c*d^2*x^4 - 5*d^3*x^6) + 60*a*b*x^2*( -77*c^3 + 77*c^2*d*x^2 + 33*c*d^2*x^4 + 7*d^3*x^6) + 2*b^2*x^4*(385*c^3 + 495*c^2*d*x^2 + 315*c*d^2*x^4 + 77*d^3*x^6))/(1155*x^(5/2))
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (d x^{5/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+c \sqrt {x} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {a^2 c^3}{x^{7/2}}+\frac {a c^2 (3 a d+2 b c)}{x^{3/2}}+b d^2 x^{9/2} (2 a d+3 b c)+b^2 d^3 x^{13/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{7} d x^{7/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {2}{3} c x^{3/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )-\frac {2 a^2 c^3}{5 x^{5/2}}-\frac {2 a c^2 (3 a d+2 b c)}{\sqrt {x}}+\frac {2}{11} b d^2 x^{11/2} (2 a d+3 b c)+\frac {2}{15} b^2 d^3 x^{15/2}\) |
(-2*a^2*c^3)/(5*x^(5/2)) - (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^(3/2))/3 + (2*d*(3*b^2*c^2 + 6*a*b*c*d + a^2 *d^2)*x^(7/2))/7 + (2*b*d^2*(3*b*c + 2*a*d)*x^(11/2))/11 + (2*b^2*d^3*x^(1 5/2))/15
3.5.14.3.1 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 = 132, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {2 b^{2} d^{3} x^{\frac {15}{2}}}{15}+\frac {4 a b \,d^{3} x^{\frac {11}{2}}}{11}+\frac {6 b^{2} c \,d^{2} x^{\frac {11}{2}}}{11}+\frac {2 a^{2} d^{3} x^{\frac {7}{2}}}{7}+\frac {12 a b c \,d^{2} x^{\frac {7}{2}}}{7}+\frac {6 b^{2} c^{2} d \,x^{\frac {7}{2}}}{7}+2 a^{2} c \,d^{2} x^{\frac {3}{2}}+4 a b \,c^{2} d \,x^{\frac {3}{2}}+\frac {2 b^{2} c^{3} x^{\frac {3}{2}}}{3}-\frac {2 a^{2} c^{3}}{5 x^{\frac {5}{2}}}-\frac {2 a \,c^{2} \left (3 a d +2 b c \right )}{\sqrt {x}}\) | \(132\) |
default | \(\frac {2 b^{2} d^{3} x^{\frac {15}{2}}}{15}+\frac {4 a b \,d^{3} x^{\frac {11}{2}}}{11}+\frac {6 b^{2} c \,d^{2} x^{\frac {11}{2}}}{11}+\frac {2 a^{2} d^{3} x^{\frac {7}{2}}}{7}+\frac {12 a b c \,d^{2} x^{\frac {7}{2}}}{7}+\frac {6 b^{2} c^{2} d \,x^{\frac {7}{2}}}{7}+2 a^{2} c \,d^{2} x^{\frac {3}{2}}+4 a b \,c^{2} d \,x^{\frac {3}{2}}+\frac {2 b^{2} c^{3} x^{\frac {3}{2}}}{3}-\frac {2 a^{2} c^{3}}{5 x^{\frac {5}{2}}}-\frac {2 a \,c^{2} \left (3 a d +2 b c \right )}{\sqrt {x}}\) | \(132\) |
gosper | \(-\frac {2 \left (-77 b^{2} d^{3} x^{10}-210 a b \,d^{3} x^{8}-315 b^{2} c \,d^{2} x^{8}-165 a^{2} d^{3} x^{6}-990 x^{6} d^{2} a b c -495 b^{2} c^{2} d \,x^{6}-1155 a^{2} c \,d^{2} x^{4}-2310 a b \,c^{2} d \,x^{4}-385 b^{2} c^{3} x^{4}+3465 a^{2} c^{2} d \,x^{2}+2310 a b \,c^{3} x^{2}+231 a^{2} c^{3}\right )}{1155 x^{\frac {5}{2}}}\) | \(138\) |
trager | \(-\frac {2 \left (-77 b^{2} d^{3} x^{10}-210 a b \,d^{3} x^{8}-315 b^{2} c \,d^{2} x^{8}-165 a^{2} d^{3} x^{6}-990 x^{6} d^{2} a b c -495 b^{2} c^{2} d \,x^{6}-1155 a^{2} c \,d^{2} x^{4}-2310 a b \,c^{2} d \,x^{4}-385 b^{2} c^{3} x^{4}+3465 a^{2} c^{2} d \,x^{2}+2310 a b \,c^{3} x^{2}+231 a^{2} c^{3}\right )}{1155 x^{\frac {5}{2}}}\) | \(138\) |
risch | \(-\frac {2 \left (-77 b^{2} d^{3} x^{10}-210 a b \,d^{3} x^{8}-315 b^{2} c \,d^{2} x^{8}-165 a^{2} d^{3} x^{6}-990 x^{6} d^{2} a b c -495 b^{2} c^{2} d \,x^{6}-1155 a^{2} c \,d^{2} x^{4}-2310 a b \,c^{2} d \,x^{4}-385 b^{2} c^{3} x^{4}+3465 a^{2} c^{2} d \,x^{2}+2310 a b \,c^{3} x^{2}+231 a^{2} c^{3}\right )}{1155 x^{\frac {5}{2}}}\) | \(138\) |
2/15*b^2*d^3*x^(15/2)+4/11*a*b*d^3*x^(11/2)+6/11*b^2*c*d^2*x^(11/2)+2/7*a^ 2*d^3*x^(7/2)+12/7*a*b*c*d^2*x^(7/2)+6/7*b^2*c^2*d*x^(7/2)+2*a^2*c*d^2*x^( 3/2)+4*a*b*c^2*d*x^(3/2)+2/3*b^2*c^3*x^(3/2)-2/5*a^2*c^3/x^(5/2)-2*a*c^2*( 3*a*d+2*b*c)/x^(1/2)
Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=\frac {2 \, {\left (77 \, b^{2} d^{3} x^{10} + 105 \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{8} + 165 \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{6} - 231 \, a^{2} c^{3} + 385 \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} - 1155 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )}}{1155 \, x^{\frac {5}{2}}} \]
2/1155*(77*b^2*d^3*x^10 + 105*(3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + 165*(3*b^2*c ^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 - 231*a^2*c^3 + 385*(b^2*c^3 + 6*a*b*c^2 *d + 3*a^2*c*d^2)*x^4 - 1155*(2*a*b*c^3 + 3*a^2*c^2*d)*x^2)/x^(5/2)
Time = 0.96 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=- \frac {2 a^{2} c^{3}}{5 x^{\frac {5}{2}}} - \frac {6 a^{2} c^{2} d}{\sqrt {x}} + 2 a^{2} c d^{2} x^{\frac {3}{2}} + \frac {2 a^{2} d^{3} x^{\frac {7}{2}}}{7} - \frac {4 a b c^{3}}{\sqrt {x}} + 4 a b c^{2} d x^{\frac {3}{2}} + \frac {12 a b c d^{2} x^{\frac {7}{2}}}{7} + \frac {4 a b d^{3} x^{\frac {11}{2}}}{11} + \frac {2 b^{2} c^{3} x^{\frac {3}{2}}}{3} + \frac {6 b^{2} c^{2} d x^{\frac {7}{2}}}{7} + \frac {6 b^{2} c d^{2} x^{\frac {11}{2}}}{11} + \frac {2 b^{2} d^{3} x^{\frac {15}{2}}}{15} \]
-2*a**2*c**3/(5*x**(5/2)) - 6*a**2*c**2*d/sqrt(x) + 2*a**2*c*d**2*x**(3/2) + 2*a**2*d**3*x**(7/2)/7 - 4*a*b*c**3/sqrt(x) + 4*a*b*c**2*d*x**(3/2) + 1 2*a*b*c*d**2*x**(7/2)/7 + 4*a*b*d**3*x**(11/2)/11 + 2*b**2*c**3*x**(3/2)/3 + 6*b**2*c**2*d*x**(7/2)/7 + 6*b**2*c*d**2*x**(11/2)/11 + 2*b**2*d**3*x** (15/2)/15
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=\frac {2}{15} \, b^{2} d^{3} x^{\frac {15}{2}} + \frac {2}{11} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{\frac {11}{2}} + \frac {2}{7} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {7}{2}} + \frac {2}{3} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{\frac {3}{2}} - \frac {2 \, {\left (a^{2} c^{3} + 5 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )}}{5 \, x^{\frac {5}{2}}} \]
2/15*b^2*d^3*x^(15/2) + 2/11*(3*b^2*c*d^2 + 2*a*b*d^3)*x^(11/2) + 2/7*(3*b ^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^(7/2) + 2/3*(b^2*c^3 + 6*a*b*c^2*d + 3 *a^2*c*d^2)*x^(3/2) - 2/5*(a^2*c^3 + 5*(2*a*b*c^3 + 3*a^2*c^2*d)*x^2)/x^(5 /2)
Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=\frac {2}{15} \, b^{2} d^{3} x^{\frac {15}{2}} + \frac {6}{11} \, b^{2} c d^{2} x^{\frac {11}{2}} + \frac {4}{11} \, a b d^{3} x^{\frac {11}{2}} + \frac {6}{7} \, b^{2} c^{2} d x^{\frac {7}{2}} + \frac {12}{7} \, a b c d^{2} x^{\frac {7}{2}} + \frac {2}{7} \, a^{2} d^{3} x^{\frac {7}{2}} + \frac {2}{3} \, b^{2} c^{3} x^{\frac {3}{2}} + 4 \, a b c^{2} d x^{\frac {3}{2}} + 2 \, a^{2} c d^{2} x^{\frac {3}{2}} - \frac {2 \, {\left (10 \, a b c^{3} x^{2} + 15 \, a^{2} c^{2} d x^{2} + a^{2} c^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]
2/15*b^2*d^3*x^(15/2) + 6/11*b^2*c*d^2*x^(11/2) + 4/11*a*b*d^3*x^(11/2) + 6/7*b^2*c^2*d*x^(7/2) + 12/7*a*b*c*d^2*x^(7/2) + 2/7*a^2*d^3*x^(7/2) + 2/3 *b^2*c^3*x^(3/2) + 4*a*b*c^2*d*x^(3/2) + 2*a^2*c*d^2*x^(3/2) - 2/5*(10*a*b *c^3*x^2 + 15*a^2*c^2*d*x^2 + a^2*c^3)/x^(5/2)
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^3}{x^{7/2}} \, dx=x^{3/2}\,\left (2\,a^2\,c\,d^2+4\,a\,b\,c^2\,d+\frac {2\,b^2\,c^3}{3}\right )+x^{7/2}\,\left (\frac {2\,a^2\,d^3}{7}+\frac {12\,a\,b\,c\,d^2}{7}+\frac {6\,b^2\,c^2\,d}{7}\right )-\frac {x^2\,\left (6\,d\,a^2\,c^2+4\,b\,a\,c^3\right )+\frac {2\,a^2\,c^3}{5}}{x^{5/2}}+\frac {2\,b^2\,d^3\,x^{15/2}}{15}+\frac {2\,b\,d^2\,x^{11/2}\,\left (2\,a\,d+3\,b\,c\right )}{11} \]