Integrand size = 24, antiderivative size = 139 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2}{9} a^2 c^3 x^{9/2}+\frac {2}{13} a c^2 (2 b c+3 a d) x^{13/2}+\frac {2}{17} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^{17/2}+\frac {2}{21} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{21/2}+\frac {2}{25} b d^2 (3 b c+2 a d) x^{25/2}+\frac {2}{29} b^2 d^3 x^{29/2} \]
2/9*a^2*c^3*x^(9/2)+2/13*a*c^2*(3*a*d+2*b*c)*x^(13/2)+2/17*c*(3*a^2*d^2+6* a*b*c*d+b^2*c^2)*x^(17/2)+2/21*d*(a^2*d^2+6*a*b*c*d+3*b^2*c^2)*x^(21/2)+2/ 25*b*d^2*(2*a*d+3*b*c)*x^(25/2)+2/29*b^2*d^3*x^(29/2)
Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.91 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2 x^{9/2} \left (725 a^2 \left (1547 c^3+3213 c^2 d x^2+2457 c d^2 x^4+663 d^3 x^6\right )+522 a b x^2 \left (2975 c^3+6825 c^2 d x^2+5525 c d^2 x^4+1547 d^3 x^6\right )+117 b^2 x^4 \left (5075 c^3+12325 c^2 d x^2+10353 c d^2 x^4+2975 d^3 x^6\right )\right )}{10094175} \]
(2*x^(9/2)*(725*a^2*(1547*c^3 + 3213*c^2*d*x^2 + 2457*c*d^2*x^4 + 663*d^3* x^6) + 522*a*b*x^2*(2975*c^3 + 6825*c^2*d*x^2 + 5525*c*d^2*x^4 + 1547*d^3* x^6) + 117*b^2*x^4*(5075*c^3 + 12325*c^2*d*x^2 + 10353*c*d^2*x^4 + 2975*d^ 3*x^6)))/10094175
Time = 0.27 (sec) , antiderivative size = 139, 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.
\(\displaystyle \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (d x^{19/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+c x^{15/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+a^2 c^3 x^{7/2}+a c^2 x^{11/2} (3 a d+2 b c)+b d^2 x^{23/2} (2 a d+3 b c)+b^2 d^3 x^{27/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{21} d x^{21/2} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {2}{17} c x^{17/2} \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {2}{9} a^2 c^3 x^{9/2}+\frac {2}{13} a c^2 x^{13/2} (3 a d+2 b c)+\frac {2}{25} b d^2 x^{25/2} (2 a d+3 b c)+\frac {2}{29} b^2 d^3 x^{29/2}\) |
(2*a^2*c^3*x^(9/2))/9 + (2*a*c^2*(2*b*c + 3*a*d)*x^(13/2))/13 + (2*c*(b^2* c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^(17/2))/17 + (2*d*(3*b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^(21/2))/21 + (2*b*d^2*(3*b*c + 2*a*d)*x^(25/2))/25 + (2*b^2*d^3 *x^(29/2))/29
3.5.7.3.1 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 = 3.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {2 b^{2} d^{3} x^{\frac {29}{2}}}{29}+\frac {2 \left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{\frac {25}{2}}}{25}+\frac {2 \left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 a^{2} c^{3} x^{\frac {9}{2}}}{9}\) | \(128\) |
default | \(\frac {2 b^{2} d^{3} x^{\frac {29}{2}}}{29}+\frac {2 \left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{\frac {25}{2}}}{25}+\frac {2 \left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{\frac {21}{2}}}{21}+\frac {2 \left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{\frac {17}{2}}}{17}+\frac {2 \left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 a^{2} c^{3} x^{\frac {9}{2}}}{9}\) | \(128\) |
gosper | \(\frac {2 x^{\frac {9}{2}} \left (348075 b^{2} d^{3} x^{10}+807534 a b \,d^{3} x^{8}+1211301 b^{2} c \,d^{2} x^{8}+480675 a^{2} d^{3} x^{6}+2884050 x^{6} d^{2} a b c +1442025 b^{2} c^{2} d \,x^{6}+1781325 a^{2} c \,d^{2} x^{4}+3562650 a b \,c^{2} d \,x^{4}+593775 b^{2} c^{3} x^{4}+2329425 a^{2} c^{2} d \,x^{2}+1552950 a b \,c^{3} x^{2}+1121575 a^{2} c^{3}\right )}{10094175}\) | \(138\) |
trager | \(\frac {2 x^{\frac {9}{2}} \left (348075 b^{2} d^{3} x^{10}+807534 a b \,d^{3} x^{8}+1211301 b^{2} c \,d^{2} x^{8}+480675 a^{2} d^{3} x^{6}+2884050 x^{6} d^{2} a b c +1442025 b^{2} c^{2} d \,x^{6}+1781325 a^{2} c \,d^{2} x^{4}+3562650 a b \,c^{2} d \,x^{4}+593775 b^{2} c^{3} x^{4}+2329425 a^{2} c^{2} d \,x^{2}+1552950 a b \,c^{3} x^{2}+1121575 a^{2} c^{3}\right )}{10094175}\) | \(138\) |
risch | \(\frac {2 x^{\frac {9}{2}} \left (348075 b^{2} d^{3} x^{10}+807534 a b \,d^{3} x^{8}+1211301 b^{2} c \,d^{2} x^{8}+480675 a^{2} d^{3} x^{6}+2884050 x^{6} d^{2} a b c +1442025 b^{2} c^{2} d \,x^{6}+1781325 a^{2} c \,d^{2} x^{4}+3562650 a b \,c^{2} d \,x^{4}+593775 b^{2} c^{3} x^{4}+2329425 a^{2} c^{2} d \,x^{2}+1552950 a b \,c^{3} x^{2}+1121575 a^{2} c^{3}\right )}{10094175}\) | \(138\) |
2/29*b^2*d^3*x^(29/2)+2/25*(2*a*b*d^3+3*b^2*c*d^2)*x^(25/2)+2/21*(a^2*d^3+ 6*a*b*c*d^2+3*b^2*c^2*d)*x^(21/2)+2/17*(3*a^2*c*d^2+6*a*b*c^2*d+b^2*c^3)*x ^(17/2)+2/13*(3*a^2*c^2*d+2*a*b*c^3)*x^(13/2)+2/9*a^2*c^3*x^(9/2)
Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.95 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2}{10094175} \, {\left (348075 \, b^{2} d^{3} x^{14} + 403767 \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{12} + 480675 \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{10} + 1121575 \, a^{2} c^{3} x^{4} + 593775 \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{8} + 776475 \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{6}\right )} \sqrt {x} \]
2/10094175*(348075*b^2*d^3*x^14 + 403767*(3*b^2*c*d^2 + 2*a*b*d^3)*x^12 + 480675*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^10 + 1121575*a^2*c^3*x^4 + 593775*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^8 + 776475*(2*a*b*c^3 + 3*a ^2*c^2*d)*x^6)*sqrt(x)
Time = 1.78 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.38 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2 a^{2} c^{3} x^{\frac {9}{2}}}{9} + \frac {6 a^{2} c^{2} d x^{\frac {13}{2}}}{13} + \frac {6 a^{2} c d^{2} x^{\frac {17}{2}}}{17} + \frac {2 a^{2} d^{3} x^{\frac {21}{2}}}{21} + \frac {4 a b c^{3} x^{\frac {13}{2}}}{13} + \frac {12 a b c^{2} d x^{\frac {17}{2}}}{17} + \frac {4 a b c d^{2} x^{\frac {21}{2}}}{7} + \frac {4 a b d^{3} x^{\frac {25}{2}}}{25} + \frac {2 b^{2} c^{3} x^{\frac {17}{2}}}{17} + \frac {2 b^{2} c^{2} d x^{\frac {21}{2}}}{7} + \frac {6 b^{2} c d^{2} x^{\frac {25}{2}}}{25} + \frac {2 b^{2} d^{3} x^{\frac {29}{2}}}{29} \]
2*a**2*c**3*x**(9/2)/9 + 6*a**2*c**2*d*x**(13/2)/13 + 6*a**2*c*d**2*x**(17 /2)/17 + 2*a**2*d**3*x**(21/2)/21 + 4*a*b*c**3*x**(13/2)/13 + 12*a*b*c**2* d*x**(17/2)/17 + 4*a*b*c*d**2*x**(21/2)/7 + 4*a*b*d**3*x**(25/2)/25 + 2*b* *2*c**3*x**(17/2)/17 + 2*b**2*c**2*d*x**(21/2)/7 + 6*b**2*c*d**2*x**(25/2) /25 + 2*b**2*d**3*x**(29/2)/29
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2}{29} \, b^{2} d^{3} x^{\frac {29}{2}} + \frac {2}{25} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{\frac {25}{2}} + \frac {2}{21} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {21}{2}} + \frac {2}{9} \, a^{2} c^{3} x^{\frac {9}{2}} + \frac {2}{17} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{\frac {17}{2}} + \frac {2}{13} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{\frac {13}{2}} \]
2/29*b^2*d^3*x^(29/2) + 2/25*(3*b^2*c*d^2 + 2*a*b*d^3)*x^(25/2) + 2/21*(3* b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^(21/2) + 2/9*a^2*c^3*x^(9/2) + 2/17*( b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(17/2) + 2/13*(2*a*b*c^3 + 3*a^2*c^ 2*d)*x^(13/2)
Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.97 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {2}{29} \, b^{2} d^{3} x^{\frac {29}{2}} + \frac {6}{25} \, b^{2} c d^{2} x^{\frac {25}{2}} + \frac {4}{25} \, a b d^{3} x^{\frac {25}{2}} + \frac {2}{7} \, b^{2} c^{2} d x^{\frac {21}{2}} + \frac {4}{7} \, a b c d^{2} x^{\frac {21}{2}} + \frac {2}{21} \, a^{2} d^{3} x^{\frac {21}{2}} + \frac {2}{17} \, b^{2} c^{3} x^{\frac {17}{2}} + \frac {12}{17} \, a b c^{2} d x^{\frac {17}{2}} + \frac {6}{17} \, a^{2} c d^{2} x^{\frac {17}{2}} + \frac {4}{13} \, a b c^{3} x^{\frac {13}{2}} + \frac {6}{13} \, a^{2} c^{2} d x^{\frac {13}{2}} + \frac {2}{9} \, a^{2} c^{3} x^{\frac {9}{2}} \]
2/29*b^2*d^3*x^(29/2) + 6/25*b^2*c*d^2*x^(25/2) + 4/25*a*b*d^3*x^(25/2) + 2/7*b^2*c^2*d*x^(21/2) + 4/7*a*b*c*d^2*x^(21/2) + 2/21*a^2*d^3*x^(21/2) + 2/17*b^2*c^3*x^(17/2) + 12/17*a*b*c^2*d*x^(17/2) + 6/17*a^2*c*d^2*x^(17/2) + 4/13*a*b*c^3*x^(13/2) + 6/13*a^2*c^2*d*x^(13/2) + 2/9*a^2*c^3*x^(9/2)
Time = 4.92 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86 \[ \int x^{7/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=x^{17/2}\,\left (\frac {6\,a^2\,c\,d^2}{17}+\frac {12\,a\,b\,c^2\,d}{17}+\frac {2\,b^2\,c^3}{17}\right )+x^{21/2}\,\left (\frac {2\,a^2\,d^3}{21}+\frac {4\,a\,b\,c\,d^2}{7}+\frac {2\,b^2\,c^2\,d}{7}\right )+\frac {2\,a^2\,c^3\,x^{9/2}}{9}+\frac {2\,b^2\,d^3\,x^{29/2}}{29}+\frac {2\,a\,c^2\,x^{13/2}\,\left (3\,a\,d+2\,b\,c\right )}{13}+\frac {2\,b\,d^2\,x^{25/2}\,\left (2\,a\,d+3\,b\,c\right )}{25} \]