Integrand size = 22, antiderivative size = 77 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {(b c-a d)^2 \left (c+d x^2\right )^{7/2}}{7 d^3}-\frac {2 b (b c-a d) \left (c+d x^2\right )^{9/2}}{9 d^3}+\frac {b^2 \left (c+d x^2\right )^{11/2}}{11 d^3} \]
1/7*(-a*d+b*c)^2*(d*x^2+c)^(7/2)/d^3-2/9*b*(-a*d+b*c)*(d*x^2+c)^(9/2)/d^3+ 1/11*b^2*(d*x^2+c)^(11/2)/d^3
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {\left (c+d x^2\right )^{7/2} \left (99 a^2 d^2+22 a b d \left (-2 c+7 d x^2\right )+b^2 \left (8 c^2-28 c d x^2+63 d^2 x^4\right )\right )}{693 d^3} \]
((c + d*x^2)^(7/2)*(99*a^2*d^2 + 22*a*b*d*(-2*c + 7*d*x^2) + b^2*(8*c^2 - 28*c*d*x^2 + 63*d^2*x^4)))/(693*d^3)
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {353, 53, 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.
\(\displaystyle \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \left (b x^2+a\right )^2 \left (d x^2+c\right )^{5/2}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {b^2 \left (d x^2+c\right )^{9/2}}{d^2}-\frac {2 b (b c-a d) \left (d x^2+c\right )^{7/2}}{d^2}+\frac {(a d-b c)^2 \left (d x^2+c\right )^{5/2}}{d^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {4 b \left (c+d x^2\right )^{9/2} (b c-a d)}{9 d^3}+\frac {2 \left (c+d x^2\right )^{7/2} (b c-a d)^2}{7 d^3}+\frac {2 b^2 \left (c+d x^2\right )^{11/2}}{11 d^3}\right )\) |
((2*(b*c - a*d)^2*(c + d*x^2)^(7/2))/(7*d^3) - (4*b*(b*c - a*d)*(c + d*x^2 )^(9/2))/(9*d^3) + (2*b^2*(c + d*x^2)^(11/2))/(11*d^3))/2
3.7.27.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Time = 2.89 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {7}{11} b^{2} x^{4}+\frac {14}{9} a b \,x^{2}+a^{2}\right ) d^{2}-\frac {4 \left (\frac {7 b \,x^{2}}{11}+a \right ) b c d}{9}+\frac {8 b^{2} c^{2}}{99}\right ) \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{7 d^{3}}\) | \(60\) |
gosper | \(\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}} \left (63 b^{2} d^{2} x^{4}+154 x^{2} a b \,d^{2}-28 x^{2} b^{2} c d +99 a^{2} d^{2}-44 a b c d +8 b^{2} c^{2}\right )}{693 d^{3}}\) | \(69\) |
default | \(b^{2} \left (\frac {x^{4} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{11 d}-\frac {4 c \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{9 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{63 d^{2}}\right )}{11 d}\right )+\frac {a^{2} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{7 d}+2 a b \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{9 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{63 d^{2}}\right )\) | \(117\) |
trager | \(\frac {\left (63 b^{2} d^{5} x^{10}+154 a b \,d^{5} x^{8}+161 b^{2} c \,d^{4} x^{8}+99 a^{2} d^{5} x^{6}+418 a b c \,d^{4} x^{6}+113 b^{2} c^{2} d^{3} x^{6}+297 a^{2} c \,d^{4} x^{4}+330 a b \,c^{2} d^{3} x^{4}+3 b^{2} c^{3} d^{2} x^{4}+297 a^{2} c^{2} d^{3} x^{2}+22 a b \,c^{3} d^{2} x^{2}-4 b^{2} c^{4} d \,x^{2}+99 a^{2} c^{3} d^{2}-44 a b \,c^{4} d +8 b^{2} c^{5}\right ) \sqrt {d \,x^{2}+c}}{693 d^{3}}\) | \(190\) |
risch | \(\frac {\left (63 b^{2} d^{5} x^{10}+154 a b \,d^{5} x^{8}+161 b^{2} c \,d^{4} x^{8}+99 a^{2} d^{5} x^{6}+418 a b c \,d^{4} x^{6}+113 b^{2} c^{2} d^{3} x^{6}+297 a^{2} c \,d^{4} x^{4}+330 a b \,c^{2} d^{3} x^{4}+3 b^{2} c^{3} d^{2} x^{4}+297 a^{2} c^{2} d^{3} x^{2}+22 a b \,c^{3} d^{2} x^{2}-4 b^{2} c^{4} d \,x^{2}+99 a^{2} c^{3} d^{2}-44 a b \,c^{4} d +8 b^{2} c^{5}\right ) \sqrt {d \,x^{2}+c}}{693 d^{3}}\) | \(190\) |
1/7*((7/11*b^2*x^4+14/9*a*b*x^2+a^2)*d^2-4/9*(7/11*b*x^2+a)*b*c*d+8/99*b^2 *c^2)*(d*x^2+c)^(7/2)/d^3
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (65) = 130\).
Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.31 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {{\left (63 \, b^{2} d^{5} x^{10} + 7 \, {\left (23 \, b^{2} c d^{4} + 22 \, a b d^{5}\right )} x^{8} + 8 \, b^{2} c^{5} - 44 \, a b c^{4} d + 99 \, a^{2} c^{3} d^{2} + {\left (113 \, b^{2} c^{2} d^{3} + 418 \, a b c d^{4} + 99 \, a^{2} d^{5}\right )} x^{6} + 3 \, {\left (b^{2} c^{3} d^{2} + 110 \, a b c^{2} d^{3} + 99 \, a^{2} c d^{4}\right )} x^{4} - {\left (4 \, b^{2} c^{4} d - 22 \, a b c^{3} d^{2} - 297 \, a^{2} c^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{693 \, d^{3}} \]
1/693*(63*b^2*d^5*x^10 + 7*(23*b^2*c*d^4 + 22*a*b*d^5)*x^8 + 8*b^2*c^5 - 4 4*a*b*c^4*d + 99*a^2*c^3*d^2 + (113*b^2*c^2*d^3 + 418*a*b*c*d^4 + 99*a^2*d ^5)*x^6 + 3*(b^2*c^3*d^2 + 110*a*b*c^2*d^3 + 99*a^2*c*d^4)*x^4 - (4*b^2*c^ 4*d - 22*a*b*c^3*d^2 - 297*a^2*c^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^3
Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (66) = 132\).
Time = 0.52 (sec) , antiderivative size = 384, normalized size of antiderivative = 4.99 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\begin {cases} \frac {a^{2} c^{3} \sqrt {c + d x^{2}}}{7 d} + \frac {3 a^{2} c^{2} x^{2} \sqrt {c + d x^{2}}}{7} + \frac {3 a^{2} c d x^{4} \sqrt {c + d x^{2}}}{7} + \frac {a^{2} d^{2} x^{6} \sqrt {c + d x^{2}}}{7} - \frac {4 a b c^{4} \sqrt {c + d x^{2}}}{63 d^{2}} + \frac {2 a b c^{3} x^{2} \sqrt {c + d x^{2}}}{63 d} + \frac {10 a b c^{2} x^{4} \sqrt {c + d x^{2}}}{21} + \frac {38 a b c d x^{6} \sqrt {c + d x^{2}}}{63} + \frac {2 a b d^{2} x^{8} \sqrt {c + d x^{2}}}{9} + \frac {8 b^{2} c^{5} \sqrt {c + d x^{2}}}{693 d^{3}} - \frac {4 b^{2} c^{4} x^{2} \sqrt {c + d x^{2}}}{693 d^{2}} + \frac {b^{2} c^{3} x^{4} \sqrt {c + d x^{2}}}{231 d} + \frac {113 b^{2} c^{2} x^{6} \sqrt {c + d x^{2}}}{693} + \frac {23 b^{2} c d x^{8} \sqrt {c + d x^{2}}}{99} + \frac {b^{2} d^{2} x^{10} \sqrt {c + d x^{2}}}{11} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (\frac {a^{2} x^{2}}{2} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{6}}{6}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a**2*c**3*sqrt(c + d*x**2)/(7*d) + 3*a**2*c**2*x**2*sqrt(c + d* x**2)/7 + 3*a**2*c*d*x**4*sqrt(c + d*x**2)/7 + a**2*d**2*x**6*sqrt(c + d*x **2)/7 - 4*a*b*c**4*sqrt(c + d*x**2)/(63*d**2) + 2*a*b*c**3*x**2*sqrt(c + d*x**2)/(63*d) + 10*a*b*c**2*x**4*sqrt(c + d*x**2)/21 + 38*a*b*c*d*x**6*sq rt(c + d*x**2)/63 + 2*a*b*d**2*x**8*sqrt(c + d*x**2)/9 + 8*b**2*c**5*sqrt( c + d*x**2)/(693*d**3) - 4*b**2*c**4*x**2*sqrt(c + d*x**2)/(693*d**2) + b* *2*c**3*x**4*sqrt(c + d*x**2)/(231*d) + 113*b**2*c**2*x**6*sqrt(c + d*x**2 )/693 + 23*b**2*c*d*x**8*sqrt(c + d*x**2)/99 + b**2*d**2*x**10*sqrt(c + d* x**2)/11, Ne(d, 0)), (c**(5/2)*(a**2*x**2/2 + a*b*x**4/2 + b**2*x**6/6), T rue))
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.49 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{4}}{11 \, d} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c x^{2}}{99 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b x^{2}}{9 \, d} + \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c^{2}}{693 \, d^{3}} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b c}{63 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{7 \, d} \]
1/11*(d*x^2 + c)^(7/2)*b^2*x^4/d - 4/99*(d*x^2 + c)^(7/2)*b^2*c*x^2/d^2 + 2/9*(d*x^2 + c)^(7/2)*a*b*x^2/d + 8/693*(d*x^2 + c)^(7/2)*b^2*c^2/d^3 - 4/ 63*(d*x^2 + c)^(7/2)*a*b*c/d^2 + 1/7*(d*x^2 + c)^(7/2)*a^2/d
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.27 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {63 \, {\left (d x^{2} + c\right )}^{\frac {11}{2}} b^{2} - 154 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} c + 99 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c^{2} + 154 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} a b d - 198 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b c d + 99 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d^{2}}{693 \, d^{3}} \]
1/693*(63*(d*x^2 + c)^(11/2)*b^2 - 154*(d*x^2 + c)^(9/2)*b^2*c + 99*(d*x^2 + c)^(7/2)*b^2*c^2 + 154*(d*x^2 + c)^(9/2)*a*b*d - 198*(d*x^2 + c)^(7/2)* a*b*c*d + 99*(d*x^2 + c)^(7/2)*a^2*d^2)/d^3
Time = 5.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.27 \[ \int x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {d\,\left (\frac {2\,a\,b\,{\left (d\,x^2+c\right )}^{9/2}}{9}-\frac {2\,a\,b\,c\,{\left (d\,x^2+c\right )}^{7/2}}{7}\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{11/2}}{11}-\frac {2\,b^2\,c\,{\left (d\,x^2+c\right )}^{9/2}}{9}+\frac {a^2\,d^2\,{\left (d\,x^2+c\right )}^{7/2}}{7}+\frac {b^2\,c^2\,{\left (d\,x^2+c\right )}^{7/2}}{7}}{d^3} \]