Integrand size = 24, antiderivative size = 157 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {c^2 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}{3 d^5}-\frac {2 c (b c-a d) (2 b c-a d) \left (c+d x^2\right )^{5/2}}{5 d^5}+\frac {\left (6 b^2 c^2-6 a b c d+a^2 d^2\right ) \left (c+d x^2\right )^{7/2}}{7 d^5}-\frac {2 b (2 b c-a d) \left (c+d x^2\right )^{9/2}}{9 d^5}+\frac {b^2 \left (c+d x^2\right )^{11/2}}{11 d^5} \] Output:
1/3*c^2*(-a*d+b*c)^2*(d*x^2+c)^(3/2)/d^5-2/5*c*(-a*d+b*c)*(-a*d+2*b*c)*(d* x^2+c)^(5/2)/d^5+1/7*(a^2*d^2-6*a*b*c*d+6*b^2*c^2)*(d*x^2+c)^(7/2)/d^5-2/9 *b*(-a*d+2*b*c)*(d*x^2+c)^(9/2)/d^5+1/11*b^2*(d*x^2+c)^(11/2)/d^5
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\left (c+d x^2\right )^{3/2} \left (33 a^2 d^2 \left (8 c^2-12 c d x^2+15 d^2 x^4\right )+22 a b d \left (-16 c^3+24 c^2 d x^2-30 c d^2 x^4+35 d^3 x^6\right )+b^2 \left (128 c^4-192 c^3 d x^2+240 c^2 d^2 x^4-280 c d^3 x^6+315 d^4 x^8\right )\right )}{3465 d^5} \] Input:
Integrate[x^5*(a + b*x^2)^2*Sqrt[c + d*x^2],x]
Output:
((c + d*x^2)^(3/2)*(33*a^2*d^2*(8*c^2 - 12*c*d*x^2 + 15*d^2*x^4) + 22*a*b* d*(-16*c^3 + 24*c^2*d*x^2 - 30*c*d^2*x^4 + 35*d^3*x^6) + b^2*(128*c^4 - 19 2*c^3*d*x^2 + 240*c^2*d^2*x^4 - 280*c*d^3*x^6 + 315*d^4*x^8)))/(3465*d^5)
Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {354, 99, 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.
| \(\displaystyle \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int x^4 \left (b x^2+a\right )^2 \sqrt {d x^2+c}dx^2\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{2} \int \left (\frac {b^2 \left (d x^2+c\right )^{9/2}}{d^4}-\frac {2 b (2 b c-a d) \left (d x^2+c\right )^{7/2}}{d^4}+\frac {\left (6 b^2 c^2-6 a b d c+a^2 d^2\right ) \left (d x^2+c\right )^{5/2}}{d^4}+\frac {2 c (b c-a d) (a d-2 b c) \left (d x^2+c\right )^{3/2}}{d^4}+\frac {c^2 (b c-a d)^2 \sqrt {d x^2+c}}{d^4}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+d x^2\right )^{7/2} \left (a^2 d^2-6 a b c d+6 b^2 c^2\right )}{7 d^5}+\frac {2 c^2 \left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^5}-\frac {4 b \left (c+d x^2\right )^{9/2} (2 b c-a d)}{9 d^5}-\frac {4 c \left (c+d x^2\right )^{5/2} (b c-a d) (2 b c-a d)}{5 d^5}+\frac {2 b^2 \left (c+d x^2\right )^{11/2}}{11 d^5}\right )\) |
Input:
Int[x^5*(a + b*x^2)^2*Sqrt[c + d*x^2],x]
Output:
((2*c^2*(b*c - a*d)^2*(c + d*x^2)^(3/2))/(3*d^5) - (4*c*(b*c - a*d)*(2*b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^5) + (2*(6*b^2*c^2 - 6*a*b*c*d + a^2*d^2)* (c + d*x^2)^(7/2))/(7*d^5) - (4*b*(2*b*c - a*d)*(c + d*x^2)^(9/2))/(9*d^5) + (2*b^2*(c + d*x^2)^(11/2))/(11*d^5))/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76
| method | result | size |
| pseudoelliptic | \(\frac {8 \left (\frac {15 \left (\frac {7}{11} b^{2} x^{4}+\frac {14}{9} a b \,x^{2}+a^{2}\right ) x^{4} d^{4}}{8}-\frac {3 c \left (\frac {70}{99} b^{2} x^{4}+\frac {5}{3} a b \,x^{2}+a^{2}\right ) x^{2} d^{3}}{2}+c^{2} \left (\frac {10}{11} b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) d^{2}-\frac {4 c^{3} \left (\frac {6 b \,x^{2}}{11}+a \right ) b d}{3}+\frac {16 b^{2} c^{4}}{33}\right ) \left (x^{2} d +c \right )^{\frac {3}{2}}}{105 d^{5}}\) | \(120\) |
| gosper | \(\frac {\left (x^{2} d +c \right )^{\frac {3}{2}} \left (315 b^{2} x^{8} d^{4}+770 a b \,d^{4} x^{6}-280 b^{2} c \,d^{3} x^{6}+495 a^{2} d^{4} x^{4}-660 a b c \,d^{3} x^{4}+240 b^{2} c^{2} d^{2} x^{4}-396 a^{2} c \,d^{3} x^{2}+528 a b \,c^{2} d^{2} x^{2}-192 b^{2} c^{3} d \,x^{2}+264 a^{2} c^{2} d^{2}-352 a b \,c^{3} d +128 b^{2} c^{4}\right )}{3465 d^{5}}\) | \(149\) |
| orering | \(\frac {\left (x^{2} d +c \right )^{\frac {3}{2}} \left (315 b^{2} x^{8} d^{4}+770 a b \,d^{4} x^{6}-280 b^{2} c \,d^{3} x^{6}+495 a^{2} d^{4} x^{4}-660 a b c \,d^{3} x^{4}+240 b^{2} c^{2} d^{2} x^{4}-396 a^{2} c \,d^{3} x^{2}+528 a b \,c^{2} d^{2} x^{2}-192 b^{2} c^{3} d \,x^{2}+264 a^{2} c^{2} d^{2}-352 a b \,c^{3} d +128 b^{2} c^{4}\right )}{3465 d^{5}}\) | \(149\) |
| trager | \(\frac {\left (315 b^{2} d^{5} x^{10}+770 a b \,d^{5} x^{8}+35 b^{2} c \,d^{4} x^{8}+495 a^{2} d^{5} x^{6}+110 a b c \,d^{4} x^{6}-40 b^{2} c^{2} d^{3} x^{6}+99 a^{2} c \,d^{4} x^{4}-132 a b \,c^{2} d^{3} x^{4}+48 b^{2} c^{3} d^{2} x^{4}-132 a^{2} c^{2} d^{3} x^{2}+176 a b \,c^{3} d^{2} x^{2}-64 b^{2} c^{4} d \,x^{2}+264 a^{2} c^{3} d^{2}-352 a b \,c^{4} d +128 b^{2} c^{5}\right ) \sqrt {x^{2} d +c}}{3465 d^{5}}\) | \(190\) |
| risch | \(\frac {\left (315 b^{2} d^{5} x^{10}+770 a b \,d^{5} x^{8}+35 b^{2} c \,d^{4} x^{8}+495 a^{2} d^{5} x^{6}+110 a b c \,d^{4} x^{6}-40 b^{2} c^{2} d^{3} x^{6}+99 a^{2} c \,d^{4} x^{4}-132 a b \,c^{2} d^{3} x^{4}+48 b^{2} c^{3} d^{2} x^{4}-132 a^{2} c^{2} d^{3} x^{2}+176 a b \,c^{3} d^{2} x^{2}-64 b^{2} c^{4} d \,x^{2}+264 a^{2} c^{3} d^{2}-352 a b \,c^{4} d +128 b^{2} c^{5}\right ) \sqrt {x^{2} d +c}}{3465 d^{5}}\) | \(190\) |
| default | \(a^{2} \left (\frac {x^{4} \left (x^{2} d +c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (x^{2} d +c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )+b^{2} \left (\frac {x^{8} \left (x^{2} d +c \right )^{\frac {3}{2}}}{11 d}-\frac {8 c \left (\frac {x^{6} \left (x^{2} d +c \right )^{\frac {3}{2}}}{9 d}-\frac {2 c \left (\frac {x^{4} \left (x^{2} d +c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (x^{2} d +c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )}{3 d}\right )}{11 d}\right )+2 a b \left (\frac {x^{6} \left (x^{2} d +c \right )^{\frac {3}{2}}}{9 d}-\frac {2 c \left (\frac {x^{4} \left (x^{2} d +c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (x^{2} d +c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )}{3 d}\right )\) | \(257\) |
Input:
int(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
8/105*(15/8*(7/11*b^2*x^4+14/9*a*b*x^2+a^2)*x^4*d^4-3/2*c*(70/99*b^2*x^4+5 /3*a*b*x^2+a^2)*x^2*d^3+c^2*(10/11*b^2*x^4+2*a*b*x^2+a^2)*d^2-4/3*c^3*(6/1 1*b*x^2+a)*b*d+16/33*b^2*c^4)*(d*x^2+c)^(3/2)/d^5
Time = 0.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.14 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (315 \, b^{2} d^{5} x^{10} + 35 \, {\left (b^{2} c d^{4} + 22 \, a b d^{5}\right )} x^{8} + 128 \, b^{2} c^{5} - 352 \, a b c^{4} d + 264 \, a^{2} c^{3} d^{2} - 5 \, {\left (8 \, b^{2} c^{2} d^{3} - 22 \, a b c d^{4} - 99 \, a^{2} d^{5}\right )} x^{6} + 3 \, {\left (16 \, b^{2} c^{3} d^{2} - 44 \, a b c^{2} d^{3} + 33 \, a^{2} c d^{4}\right )} x^{4} - 4 \, {\left (16 \, b^{2} c^{4} d - 44 \, a b c^{3} d^{2} + 33 \, a^{2} c^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3465 \, d^{5}} \] Input:
integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="fricas")
Output:
1/3465*(315*b^2*d^5*x^10 + 35*(b^2*c*d^4 + 22*a*b*d^5)*x^8 + 128*b^2*c^5 - 352*a*b*c^4*d + 264*a^2*c^3*d^2 - 5*(8*b^2*c^2*d^3 - 22*a*b*c*d^4 - 99*a^ 2*d^5)*x^6 + 3*(16*b^2*c^3*d^2 - 44*a*b*c^2*d^3 + 33*a^2*c*d^4)*x^4 - 4*(1 6*b^2*c^4*d - 44*a*b*c^3*d^2 + 33*a^2*c^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^5
Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (148) = 296\).
Time = 0.35 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.48 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} \frac {8 a^{2} c^{3} \sqrt {c + d x^{2}}}{105 d^{3}} - \frac {4 a^{2} c^{2} x^{2} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {a^{2} c x^{4} \sqrt {c + d x^{2}}}{35 d} + \frac {a^{2} x^{6} \sqrt {c + d x^{2}}}{7} - \frac {32 a b c^{4} \sqrt {c + d x^{2}}}{315 d^{4}} + \frac {16 a b c^{3} x^{2} \sqrt {c + d x^{2}}}{315 d^{3}} - \frac {4 a b c^{2} x^{4} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {2 a b c x^{6} \sqrt {c + d x^{2}}}{63 d} + \frac {2 a b x^{8} \sqrt {c + d x^{2}}}{9} + \frac {128 b^{2} c^{5} \sqrt {c + d x^{2}}}{3465 d^{5}} - \frac {64 b^{2} c^{4} x^{2} \sqrt {c + d x^{2}}}{3465 d^{4}} + \frac {16 b^{2} c^{3} x^{4} \sqrt {c + d x^{2}}}{1155 d^{3}} - \frac {8 b^{2} c^{2} x^{6} \sqrt {c + d x^{2}}}{693 d^{2}} + \frac {b^{2} c x^{8} \sqrt {c + d x^{2}}}{99 d} + \frac {b^{2} x^{10} \sqrt {c + d x^{2}}}{11} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} x^{6}}{6} + \frac {a b x^{8}}{4} + \frac {b^{2} x^{10}}{10}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(b*x**2+a)**2*(d*x**2+c)**(1/2),x)
Output:
Piecewise((8*a**2*c**3*sqrt(c + d*x**2)/(105*d**3) - 4*a**2*c**2*x**2*sqrt (c + d*x**2)/(105*d**2) + a**2*c*x**4*sqrt(c + d*x**2)/(35*d) + a**2*x**6* sqrt(c + d*x**2)/7 - 32*a*b*c**4*sqrt(c + d*x**2)/(315*d**4) + 16*a*b*c**3 *x**2*sqrt(c + d*x**2)/(315*d**3) - 4*a*b*c**2*x**4*sqrt(c + d*x**2)/(105* d**2) + 2*a*b*c*x**6*sqrt(c + d*x**2)/(63*d) + 2*a*b*x**8*sqrt(c + d*x**2) /9 + 128*b**2*c**5*sqrt(c + d*x**2)/(3465*d**5) - 64*b**2*c**4*x**2*sqrt(c + d*x**2)/(3465*d**4) + 16*b**2*c**3*x**4*sqrt(c + d*x**2)/(1155*d**3) - 8*b**2*c**2*x**6*sqrt(c + d*x**2)/(693*d**2) + b**2*c*x**8*sqrt(c + d*x**2 )/(99*d) + b**2*x**10*sqrt(c + d*x**2)/11, Ne(d, 0)), (sqrt(c)*(a**2*x**6/ 6 + a*b*x**8/4 + b**2*x**10/10), True))
Time = 0.06 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.59 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{8}}{11 \, d} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{6}}{99 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{6}}{9 \, d} + \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x^{4}}{231 \, d^{3}} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x^{4}}{21 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x^{4}}{7 \, d} - \frac {64 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} x^{2}}{1155 \, d^{4}} + \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} x^{2}}{105 \, d^{3}} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x^{2}}{35 \, d^{2}} + \frac {128 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{4}}{3465 \, d^{5}} - \frac {32 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{3}}{315 \, d^{4}} + \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c^{2}}{105 \, d^{3}} \] Input:
integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="maxima")
Output:
1/11*(d*x^2 + c)^(3/2)*b^2*x^8/d - 8/99*(d*x^2 + c)^(3/2)*b^2*c*x^6/d^2 + 2/9*(d*x^2 + c)^(3/2)*a*b*x^6/d + 16/231*(d*x^2 + c)^(3/2)*b^2*c^2*x^4/d^3 - 4/21*(d*x^2 + c)^(3/2)*a*b*c*x^4/d^2 + 1/7*(d*x^2 + c)^(3/2)*a^2*x^4/d - 64/1155*(d*x^2 + c)^(3/2)*b^2*c^3*x^2/d^4 + 16/105*(d*x^2 + c)^(3/2)*a*b *c^2*x^2/d^3 - 4/35*(d*x^2 + c)^(3/2)*a^2*c*x^2/d^2 + 128/3465*(d*x^2 + c) ^(3/2)*b^2*c^4/d^5 - 32/315*(d*x^2 + c)^(3/2)*a*b*c^3/d^4 + 8/105*(d*x^2 + c)^(3/2)*a^2*c^2/d^3
Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.30 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {315 \, {\left (d x^{2} + c\right )}^{\frac {11}{2}} b^{2} - 1540 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} c + 2970 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c^{2} - 2772 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{3} + 1155 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{4} + 770 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} a b d - 2970 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b c d + 4158 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c^{2} d - 2310 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{3} d + 495 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d^{2} - 1386 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} c d^{2} + 1155 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c^{2} d^{2}}{3465 \, d^{5}} \] Input:
integrate(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="giac")
Output:
1/3465*(315*(d*x^2 + c)^(11/2)*b^2 - 1540*(d*x^2 + c)^(9/2)*b^2*c + 2970*( d*x^2 + c)^(7/2)*b^2*c^2 - 2772*(d*x^2 + c)^(5/2)*b^2*c^3 + 1155*(d*x^2 + c)^(3/2)*b^2*c^4 + 770*(d*x^2 + c)^(9/2)*a*b*d - 2970*(d*x^2 + c)^(7/2)*a* b*c*d + 4158*(d*x^2 + c)^(5/2)*a*b*c^2*d - 2310*(d*x^2 + c)^(3/2)*a*b*c^3* d + 495*(d*x^2 + c)^(7/2)*a^2*d^2 - 1386*(d*x^2 + c)^(5/2)*a^2*c*d^2 + 115 5*(d*x^2 + c)^(3/2)*a^2*c^2*d^2)/d^5
Time = 0.93 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {264\,a^2\,c^3\,d^2-352\,a\,b\,c^4\,d+128\,b^2\,c^5}{3465\,d^5}+\frac {b^2\,x^{10}}{11}+\frac {x^6\,\left (495\,a^2\,d^5+110\,a\,b\,c\,d^4-40\,b^2\,c^2\,d^3\right )}{3465\,d^5}+\frac {b\,x^8\,\left (22\,a\,d+b\,c\right )}{99\,d}+\frac {c\,x^4\,\left (33\,a^2\,d^2-44\,a\,b\,c\,d+16\,b^2\,c^2\right )}{1155\,d^3}-\frac {4\,c^2\,x^2\,\left (33\,a^2\,d^2-44\,a\,b\,c\,d+16\,b^2\,c^2\right )}{3465\,d^4}\right ) \] Input:
int(x^5*(a + b*x^2)^2*(c + d*x^2)^(1/2),x)
Output:
(c + d*x^2)^(1/2)*((128*b^2*c^5 + 264*a^2*c^3*d^2 - 352*a*b*c^4*d)/(3465*d ^5) + (b^2*x^10)/11 + (x^6*(495*a^2*d^5 - 40*b^2*c^2*d^3 + 110*a*b*c*d^4)) /(3465*d^5) + (b*x^8*(22*a*d + b*c))/(99*d) + (c*x^4*(33*a^2*d^2 + 16*b^2* c^2 - 44*a*b*c*d))/(1155*d^3) - (4*c^2*x^2*(33*a^2*d^2 + 16*b^2*c^2 - 44*a *b*c*d))/(3465*d^4))
Time = 0.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int x^5 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\sqrt {d \,x^{2}+c}\, \left (315 b^{2} d^{5} x^{10}+770 a b \,d^{5} x^{8}+35 b^{2} c \,d^{4} x^{8}+495 a^{2} d^{5} x^{6}+110 a b c \,d^{4} x^{6}-40 b^{2} c^{2} d^{3} x^{6}+99 a^{2} c \,d^{4} x^{4}-132 a b \,c^{2} d^{3} x^{4}+48 b^{2} c^{3} d^{2} x^{4}-132 a^{2} c^{2} d^{3} x^{2}+176 a b \,c^{3} d^{2} x^{2}-64 b^{2} c^{4} d \,x^{2}+264 a^{2} c^{3} d^{2}-352 a b \,c^{4} d +128 b^{2} c^{5}\right )}{3465 d^{5}} \] Input:
int(x^5*(b*x^2+a)^2*(d*x^2+c)^(1/2),x)
Output:
(sqrt(c + d*x**2)*(264*a**2*c**3*d**2 - 132*a**2*c**2*d**3*x**2 + 99*a**2* c*d**4*x**4 + 495*a**2*d**5*x**6 - 352*a*b*c**4*d + 176*a*b*c**3*d**2*x**2 - 132*a*b*c**2*d**3*x**4 + 110*a*b*c*d**4*x**6 + 770*a*b*d**5*x**8 + 128* b**2*c**5 - 64*b**2*c**4*d*x**2 + 48*b**2*c**3*d**2*x**4 - 40*b**2*c**2*d* *3*x**6 + 35*b**2*c*d**4*x**8 + 315*b**2*d**5*x**10))/(3465*d**5)