Integrand size = 22, antiderivative size = 104 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {c^2 (b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^4}-\frac {c (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^4}+\frac {(3 b c-a d) \left (c+\frac {d}{x^2}\right )^{9/2}}{9 d^4}-\frac {b \left (c+\frac {d}{x^2}\right )^{11/2}}{11 d^4} \] Output:
1/5*c^2*(-a*d+b*c)*(c+d/x^2)^(5/2)/d^4-1/7*c*(-2*a*d+3*b*c)*(c+d/x^2)^(7/2 )/d^4+1/9*(-a*d+3*b*c)*(c+d/x^2)^(9/2)/d^4-1/11*b*(c+d/x^2)^(11/2)/d^4
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right )^2 \left (-11 a d x^2 \left (35 d^2-20 c d x^2+8 c^2 x^4\right )-3 b \left (105 d^3-70 c d^2 x^2+40 c^2 d x^4-16 c^3 x^6\right )\right )}{3465 d^4 x^{10}} \] Input:
Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^7,x]
Output:
(Sqrt[c + d/x^2]*(d + c*x^2)^2*(-11*a*d*x^2*(35*d^2 - 20*c*d*x^2 + 8*c^2*x ^4) - 3*b*(105*d^3 - 70*c*d^2*x^2 + 40*c^2*d*x^4 - 16*c^3*x^6)))/(3465*d^4 *x^10)
Time = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 86, 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.
| \(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle -\frac {1}{2} \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4}d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{9/2}}{d^3}+\frac {(a d-3 b c) \left (c+\frac {d}{x^2}\right )^{7/2}}{d^3}+\frac {c (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{d^3}-\frac {c^2 (b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^3}\right )d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 c^2 \left (c+\frac {d}{x^2}\right )^{5/2} (b c-a d)}{5 d^4}+\frac {2 \left (c+\frac {d}{x^2}\right )^{9/2} (3 b c-a d)}{9 d^4}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{7/2} (3 b c-2 a d)}{7 d^4}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{11/2}}{11 d^4}\right )\) |
Input:
Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^7,x]
Output:
((2*c^2*(b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^4) - (2*c*(3*b*c - 2*a*d)*(c + d/x^2)^(7/2))/(7*d^4) + (2*(3*b*c - a*d)*(c + d/x^2)^(9/2))/(9*d^4) - (2* b*(c + d/x^2)^(11/2))/(11*d^4))/2
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90
| method | result | size |
| gosper | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (88 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-220 a c \,d^{2} x^{4}+120 b \,c^{2} d \,x^{4}+385 a \,d^{3} x^{2}-210 b c \,d^{2} x^{2}+315 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{3465 d^{4} x^{8}}\) | \(94\) |
| default | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (88 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-220 a c \,d^{2} x^{4}+120 b \,c^{2} d \,x^{4}+385 a \,d^{3} x^{2}-210 b c \,d^{2} x^{2}+315 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{3465 d^{4} x^{8}}\) | \(94\) |
| orering | \(-\frac {\left (88 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-220 a c \,d^{2} x^{4}+120 b \,c^{2} d \,x^{4}+385 a \,d^{3} x^{2}-210 b c \,d^{2} x^{2}+315 b \,d^{3}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \left (c +\frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3465 d^{4} \left (a \,x^{2}+b \right ) x^{6}}\) | \(106\) |
| risch | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (88 a \,c^{4} d \,x^{10}-48 b \,c^{5} x^{10}-44 a \,c^{3} d^{2} x^{8}+24 b \,c^{4} d \,x^{8}+33 a \,c^{2} d^{3} x^{6}-18 b \,c^{3} d^{2} x^{6}+550 a c \,d^{4} x^{4}+15 b \,c^{2} d^{3} x^{4}+385 a \,d^{5} x^{2}+420 b c \,d^{4} x^{2}+315 b \,d^{5}\right )}{3465 x^{10} d^{4}}\) | \(135\) |
| trager | \(-\frac {\left (88 a \,c^{4} d \,x^{10}-48 b \,c^{5} x^{10}-44 a \,c^{3} d^{2} x^{8}+24 b \,c^{4} d \,x^{8}+33 a \,c^{2} d^{3} x^{6}-18 b \,c^{3} d^{2} x^{6}+550 a c \,d^{4} x^{4}+15 b \,c^{2} d^{3} x^{4}+385 a \,d^{5} x^{2}+420 b c \,d^{4} x^{2}+315 b \,d^{5}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{3465 x^{10} d^{4}}\) | \(139\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/3465*((c*x^2+d)/x^2)^(3/2)*(88*a*c^2*d*x^6-48*b*c^3*x^6-220*a*c*d^2*x^4 +120*b*c^2*d*x^4+385*a*d^3*x^2-210*b*c*d^2*x^2+315*b*d^3)*(c*x^2+d)/d^4/x^ 8
Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {{\left (8 \, {\left (6 \, b c^{5} - 11 \, a c^{4} d\right )} x^{10} - 4 \, {\left (6 \, b c^{4} d - 11 \, a c^{3} d^{2}\right )} x^{8} + 3 \, {\left (6 \, b c^{3} d^{2} - 11 \, a c^{2} d^{3}\right )} x^{6} - 315 \, b d^{5} - 5 \, {\left (3 \, b c^{2} d^{3} + 110 \, a c d^{4}\right )} x^{4} - 35 \, {\left (12 \, b c d^{4} + 11 \, a d^{5}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3465 \, d^{4} x^{10}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^7,x, algorithm="fricas")
Output:
1/3465*(8*(6*b*c^5 - 11*a*c^4*d)*x^10 - 4*(6*b*c^4*d - 11*a*c^3*d^2)*x^8 + 3*(6*b*c^3*d^2 - 11*a*c^2*d^3)*x^6 - 315*b*d^5 - 5*(3*b*c^2*d^3 + 110*a*c *d^4)*x^4 - 35*(12*b*c*d^4 + 11*a*d^5)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^4*x^1 0)
Time = 5.42 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.13 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=- \frac {a c \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{6}} & \text {otherwise} \end {cases}\right )}{2} - \frac {a d \left (\begin {cases} \frac {2 \left (- \frac {c^{3} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {3 c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} - \frac {3 c \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {9}{2}}}{9}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{8}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b c \left (\begin {cases} \frac {2 \left (- \frac {c^{3} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {3 c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} - \frac {3 c \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {9}{2}}}{9}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{8}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b d \left (\begin {cases} \frac {2 \left (\frac {c^{4} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {4 c^{3} \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {6 c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7} - \frac {4 c \left (c + \frac {d}{x^{2}}\right )^{\frac {9}{2}}}{9} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {11}{2}}}{11}\right )}{d^{5}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{5 x^{10}} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**7,x)
Output:
-a*c*Piecewise((2*(c**2*(c + d/x**2)**(3/2)/3 - 2*c*(c + d/x**2)**(5/2)/5 + (c + d/x**2)**(7/2)/7)/d**3, Ne(d, 0)), (sqrt(c)/(3*x**6), True))/2 - a* d*Piecewise((2*(-c**3*(c + d/x**2)**(3/2)/3 + 3*c**2*(c + d/x**2)**(5/2)/5 - 3*c*(c + d/x**2)**(7/2)/7 + (c + d/x**2)**(9/2)/9)/d**4, Ne(d, 0)), (sq rt(c)/(4*x**8), True))/2 - b*c*Piecewise((2*(-c**3*(c + d/x**2)**(3/2)/3 + 3*c**2*(c + d/x**2)**(5/2)/5 - 3*c*(c + d/x**2)**(7/2)/7 + (c + d/x**2)** (9/2)/9)/d**4, Ne(d, 0)), (sqrt(c)/(4*x**8), True))/2 - b*d*Piecewise((2*( c**4*(c + d/x**2)**(3/2)/3 - 4*c**3*(c + d/x**2)**(5/2)/5 + 6*c**2*(c + d/ x**2)**(7/2)/7 - 4*c*(c + d/x**2)**(9/2)/9 + (c + d/x**2)**(11/2)/11)/d**5 , Ne(d, 0)), (sqrt(c)/(5*x**10), True))/2
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=-\frac {1}{315} \, {\left (\frac {35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}}}{d^{3}} - \frac {90 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c}{d^{3}} + \frac {63 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{2}}{d^{3}}\right )} a - \frac {1}{1155} \, {\left (\frac {105 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {11}{2}}}{d^{4}} - \frac {385 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} c}{d^{4}} + \frac {495 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{2}}{d^{4}} - \frac {231 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3}}{d^{4}}\right )} b \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^7,x, algorithm="maxima")
Output:
-1/315*(35*(c + d/x^2)^(9/2)/d^3 - 90*(c + d/x^2)^(7/2)*c/d^3 + 63*(c + d/ x^2)^(5/2)*c^2/d^3)*a - 1/1155*(105*(c + d/x^2)^(11/2)/d^4 - 385*(c + d/x^ 2)^(9/2)*c/d^4 + 495*(c + d/x^2)^(7/2)*c^2/d^4 - 231*(c + d/x^2)^(5/2)*c^3 /d^4)*b
Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (88) = 176\).
Time = 1.44 (sec) , antiderivative size = 490, normalized size of antiderivative = 4.71 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {16 \, {\left (2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{16} a c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 6930 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{14} b c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{14} a c^{\frac {9}{2}} d \mathrm {sgn}\left (x\right ) + 12474 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} b c^{\frac {11}{2}} d \mathrm {sgn}\left (x\right ) + 231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} a c^{\frac {9}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 15246 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} b c^{\frac {11}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 4851 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} a c^{\frac {9}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 4950 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {11}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 2475 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {9}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 990 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {11}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 495 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {9}{2}} d^{5} \mathrm {sgn}\left (x\right ) - 330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {11}{2}} d^{5} \mathrm {sgn}\left (x\right ) + 605 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {9}{2}} d^{6} \mathrm {sgn}\left (x\right ) + 66 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {11}{2}} d^{6} \mathrm {sgn}\left (x\right ) - 121 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {9}{2}} d^{7} \mathrm {sgn}\left (x\right ) - 6 \, b c^{\frac {11}{2}} d^{7} \mathrm {sgn}\left (x\right ) + 11 \, a c^{\frac {9}{2}} d^{8} \mathrm {sgn}\left (x\right )\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{11}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^7,x, algorithm="giac")
Output:
16/3465*(2310*(sqrt(c)*x - sqrt(c*x^2 + d))^16*a*c^(9/2)*sgn(x) + 6930*(sq rt(c)*x - sqrt(c*x^2 + d))^14*b*c^(11/2)*sgn(x) - 1155*(sqrt(c)*x - sqrt(c *x^2 + d))^14*a*c^(9/2)*d*sgn(x) + 12474*(sqrt(c)*x - sqrt(c*x^2 + d))^12* b*c^(11/2)*d*sgn(x) + 231*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(9/2)*d^2*s gn(x) + 15246*(sqrt(c)*x - sqrt(c*x^2 + d))^10*b*c^(11/2)*d^2*sgn(x) - 485 1*(sqrt(c)*x - sqrt(c*x^2 + d))^10*a*c^(9/2)*d^3*sgn(x) + 4950*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(11/2)*d^3*sgn(x) + 2475*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(9/2)*d^4*sgn(x) + 990*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(11 /2)*d^4*sgn(x) + 495*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(9/2)*d^5*sgn(x) - 330*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(11/2)*d^5*sgn(x) + 605*(sqrt(c) *x - sqrt(c*x^2 + d))^4*a*c^(9/2)*d^6*sgn(x) + 66*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(11/2)*d^6*sgn(x) - 121*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(9 /2)*d^7*sgn(x) - 6*b*c^(11/2)*d^7*sgn(x) + 11*a*c^(9/2)*d^8*sgn(x))/((sqrt (c)*x - sqrt(c*x^2 + d))^2 - d)^11
Time = 5.49 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {16\,b\,c^5\,\sqrt {c+\frac {d}{x^2}}}{1155\,d^4}-\frac {8\,a\,c^4\,\sqrt {c+\frac {d}{x^2}}}{315\,d^3}-\frac {10\,a\,c\,\sqrt {c+\frac {d}{x^2}}}{63\,x^6}-\frac {a\,d\,\sqrt {c+\frac {d}{x^2}}}{9\,x^8}-\frac {4\,b\,c\,\sqrt {c+\frac {d}{x^2}}}{33\,x^8}-\frac {b\,d\,\sqrt {c+\frac {d}{x^2}}}{11\,x^{10}}-\frac {a\,c^2\,\sqrt {c+\frac {d}{x^2}}}{105\,d\,x^4}+\frac {4\,a\,c^3\,\sqrt {c+\frac {d}{x^2}}}{315\,d^2\,x^2}-\frac {b\,c^2\,\sqrt {c+\frac {d}{x^2}}}{231\,d\,x^6}+\frac {2\,b\,c^3\,\sqrt {c+\frac {d}{x^2}}}{385\,d^2\,x^4}-\frac {8\,b\,c^4\,\sqrt {c+\frac {d}{x^2}}}{1155\,d^3\,x^2} \] Input:
int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^7,x)
Output:
(16*b*c^5*(c + d/x^2)^(1/2))/(1155*d^4) - (8*a*c^4*(c + d/x^2)^(1/2))/(315 *d^3) - (10*a*c*(c + d/x^2)^(1/2))/(63*x^6) - (a*d*(c + d/x^2)^(1/2))/(9*x ^8) - (4*b*c*(c + d/x^2)^(1/2))/(33*x^8) - (b*d*(c + d/x^2)^(1/2))/(11*x^1 0) - (a*c^2*(c + d/x^2)^(1/2))/(105*d*x^4) + (4*a*c^3*(c + d/x^2)^(1/2))/( 315*d^2*x^2) - (b*c^2*(c + d/x^2)^(1/2))/(231*d*x^6) + (2*b*c^3*(c + d/x^2 )^(1/2))/(385*d^2*x^4) - (8*b*c^4*(c + d/x^2)^(1/2))/(1155*d^3*x^2)
Time = 0.21 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.23 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^7} \, dx=\frac {-88 \sqrt {c \,x^{2}+d}\, a \,c^{4} d \,x^{10}+44 \sqrt {c \,x^{2}+d}\, a \,c^{3} d^{2} x^{8}-33 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{3} x^{6}-550 \sqrt {c \,x^{2}+d}\, a c \,d^{4} x^{4}-385 \sqrt {c \,x^{2}+d}\, a \,d^{5} x^{2}+48 \sqrt {c \,x^{2}+d}\, b \,c^{5} x^{10}-24 \sqrt {c \,x^{2}+d}\, b \,c^{4} d \,x^{8}+18 \sqrt {c \,x^{2}+d}\, b \,c^{3} d^{2} x^{6}-15 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{3} x^{4}-420 \sqrt {c \,x^{2}+d}\, b c \,d^{4} x^{2}-315 \sqrt {c \,x^{2}+d}\, b \,d^{5}+88 \sqrt {c}\, a \,c^{4} d \,x^{11}-48 \sqrt {c}\, b \,c^{5} x^{11}}{3465 d^{4} x^{11}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)/x^7,x)
Output:
( - 88*sqrt(c*x**2 + d)*a*c**4*d*x**10 + 44*sqrt(c*x**2 + d)*a*c**3*d**2*x **8 - 33*sqrt(c*x**2 + d)*a*c**2*d**3*x**6 - 550*sqrt(c*x**2 + d)*a*c*d**4 *x**4 - 385*sqrt(c*x**2 + d)*a*d**5*x**2 + 48*sqrt(c*x**2 + d)*b*c**5*x**1 0 - 24*sqrt(c*x**2 + d)*b*c**4*d*x**8 + 18*sqrt(c*x**2 + d)*b*c**3*d**2*x* *6 - 15*sqrt(c*x**2 + d)*b*c**2*d**3*x**4 - 420*sqrt(c*x**2 + d)*b*c*d**4* x**2 - 315*sqrt(c*x**2 + d)*b*d**5 + 88*sqrt(c)*a*c**4*d*x**11 - 48*sqrt(c )*b*c**5*x**11)/(3465*d**4*x**11)