\(\int \frac {(a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2}}{x^9} \, dx\) [160]

Optimal result

Integrand size = 22, antiderivative size = 134 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=-\frac {c^3 (b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^5}+\frac {c^2 (4 b c-3 a d) \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^5}-\frac {c (2 b c-a d) \left (c+\frac {d}{x^2}\right )^{9/2}}{3 d^5}+\frac {(4 b c-a d) \left (c+\frac {d}{x^2}\right )^{11/2}}{11 d^5}-\frac {b \left (c+\frac {d}{x^2}\right )^{13/2}}{13 d^5} \] Output:

-1/5*c^3*(-a*d+b*c)*(c+d/x^2)^(5/2)/d^5+1/7*c^2*(-3*a*d+4*b*c)*(c+d/x^2)^( 
7/2)/d^5-1/3*c*(-a*d+2*b*c)*(c+d/x^2)^(9/2)/d^5+1/11*(-a*d+4*b*c)*(c+d/x^2 
)^(11/2)/d^5-1/13*b*(c+d/x^2)^(13/2)/d^5
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right )^2 \left (13 a d x^2 \left (-105 d^3+70 c d^2 x^2-40 c^2 d x^4+16 c^3 x^6\right )+b \left (-1155 d^4+840 c d^3 x^2-560 c^2 d^2 x^4+320 c^3 d x^6-128 c^4 x^8\right )\right )}{15015 d^5 x^{12}} \] Input:

Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^9,x]
 

Output:

(Sqrt[c + d/x^2]*(d + c*x^2)^2*(13*a*d*x^2*(-105*d^3 + 70*c*d^2*x^2 - 40*c 
^2*d*x^4 + 16*c^3*x^6) + b*(-1155*d^4 + 840*c*d^3*x^2 - 560*c^2*d^2*x^4 + 
320*c^3*d*x^6 - 128*c^4*x^8)))/(15015*d^5*x^12)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.

\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle -\frac {1}{2} \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^6}d\frac {1}{x^2}\)

\(\Big \downarrow \) 86

\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{11/2}}{d^4}+\frac {(a d-4 b c) \left (c+\frac {d}{x^2}\right )^{9/2}}{d^4}+\frac {3 c (2 b c-a d) \left (c+\frac {d}{x^2}\right )^{7/2}}{d^4}-\frac {c^2 (4 b c-3 a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{d^4}+\frac {c^3 (b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^4}\right )d\frac {1}{x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {2 c^3 \left (c+\frac {d}{x^2}\right )^{5/2} (b c-a d)}{5 d^5}+\frac {2 c^2 \left (c+\frac {d}{x^2}\right )^{7/2} (4 b c-3 a d)}{7 d^5}+\frac {2 \left (c+\frac {d}{x^2}\right )^{11/2} (4 b c-a d)}{11 d^5}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{9/2} (2 b c-a d)}{3 d^5}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{13/2}}{13 d^5}\right )\)

Input:

Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^9,x]
 

Output:

((-2*c^3*(b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^5) + (2*c^2*(4*b*c - 3*a*d)*( 
c + d/x^2)^(7/2))/(7*d^5) - (2*c*(2*b*c - a*d)*(c + d/x^2)^(9/2))/(3*d^5) 
+ (2*(4*b*c - a*d)*(c + d/x^2)^(11/2))/(11*d^5) - (2*b*(c + d/x^2)^(13/2)) 
/(13*d^5))/2
 

Defintions of rubi rules used

rule 86

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]))))
 

rule 948

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]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (208 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}-520 a \,c^{2} d^{2} x^{6}+320 b \,c^{3} d \,x^{6}+910 a c \,d^{3} x^{4}-560 b \,c^{2} d^{2} x^{4}-1365 a \,d^{4} x^{2}+840 b c \,d^{3} x^{2}-1155 b \,d^{4}\right ) \left (c \,x^{2}+d \right )}{15015 d^{5} x^{10}}\)	\(118\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (208 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}-520 a \,c^{2} d^{2} x^{6}+320 b \,c^{3} d \,x^{6}+910 a c \,d^{3} x^{4}-560 b \,c^{2} d^{2} x^{4}-1365 a \,d^{4} x^{2}+840 b c \,d^{3} x^{2}-1155 b \,d^{4}\right ) \left (c \,x^{2}+d \right )}{15015 d^{5} x^{10}}\)	\(118\)
orering	\(\frac {\left (208 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}-520 a \,c^{2} d^{2} x^{6}+320 b \,c^{3} d \,x^{6}+910 a c \,d^{3} x^{4}-560 b \,c^{2} d^{2} x^{4}-1365 a \,d^{4} x^{2}+840 b c \,d^{3} x^{2}-1155 b \,d^{4}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \left (c +\frac {d}{x^{2}}\right )^{\frac {3}{2}}}{15015 d^{5} \left (a \,x^{2}+b \right ) x^{8}}\)	\(130\)
risch	\(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (208 a \,c^{5} d \,x^{12}-128 b \,c^{6} x^{12}-104 a \,c^{4} d^{2} x^{10}+64 b \,c^{5} d \,x^{10}+78 a \,c^{3} d^{3} x^{8}-48 b \,c^{4} d^{2} x^{8}-65 a \,c^{2} d^{4} x^{6}+40 b \,c^{3} d^{3} x^{6}-1820 a c \,d^{5} x^{4}-35 b \,c^{2} d^{4} x^{4}-1365 a \,d^{6} x^{2}-1470 b c \,d^{5} x^{2}-1155 b \,d^{6}\right )}{15015 x^{12} d^{5}}\)	\(159\)
trager	\(\frac {\left (208 a \,c^{5} d \,x^{12}-128 b \,c^{6} x^{12}-104 a \,c^{4} d^{2} x^{10}+64 b \,c^{5} d \,x^{10}+78 a \,c^{3} d^{3} x^{8}-48 b \,c^{4} d^{2} x^{8}-65 a \,c^{2} d^{4} x^{6}+40 b \,c^{3} d^{3} x^{6}-1820 a c \,d^{5} x^{4}-35 b \,c^{2} d^{4} x^{4}-1365 a \,d^{6} x^{2}-1470 b c \,d^{5} x^{2}-1155 b \,d^{6}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{15015 x^{12} d^{5}}\)	\(163\)

Input:

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^9,x,method=_RETURNVERBOSE)
 

Output:

1/15015*((c*x^2+d)/x^2)^(3/2)*(208*a*c^3*d*x^8-128*b*c^4*x^8-520*a*c^2*d^2 
*x^6+320*b*c^3*d*x^6+910*a*c*d^3*x^4-560*b*c^2*d^2*x^4-1365*a*d^4*x^2+840* 
b*c*d^3*x^2-1155*b*d^4)*(c*x^2+d)/d^5/x^10
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=-\frac {{\left (16 \, {\left (8 \, b c^{6} - 13 \, a c^{5} d\right )} x^{12} - 8 \, {\left (8 \, b c^{5} d - 13 \, a c^{4} d^{2}\right )} x^{10} + 6 \, {\left (8 \, b c^{4} d^{2} - 13 \, a c^{3} d^{3}\right )} x^{8} + 1155 \, b d^{6} - 5 \, {\left (8 \, b c^{3} d^{3} - 13 \, a c^{2} d^{4}\right )} x^{6} + 35 \, {\left (b c^{2} d^{4} + 52 \, a c d^{5}\right )} x^{4} + 105 \, {\left (14 \, b c d^{5} + 13 \, a d^{6}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15015 \, d^{5} x^{12}} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^9,x, algorithm="fricas")
 

Output:

-1/15015*(16*(8*b*c^6 - 13*a*c^5*d)*x^12 - 8*(8*b*c^5*d - 13*a*c^4*d^2)*x^ 
10 + 6*(8*b*c^4*d^2 - 13*a*c^3*d^3)*x^8 + 1155*b*d^6 - 5*(8*b*c^3*d^3 - 13 
*a*c^2*d^4)*x^6 + 35*(b*c^2*d^4 + 52*a*c*d^5)*x^4 + 105*(14*b*c*d^5 + 13*a 
*d^6)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^5*x^12)
 

Sympy [A] (verification not implemented)

Time = 5.32 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.93 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=- \frac {a 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 {a 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} - \frac {b c \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} - \frac {b d \left (\begin {cases} \frac {2 \left (- \frac {c^{5} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + c^{4} \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}} - \frac {10 c^{3} \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7} + \frac {10 c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {9}{2}}}{9} - \frac {5 c \left (c + \frac {d}{x^{2}}\right )^{\frac {11}{2}}}{11} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {13}{2}}}{13}\right )}{d^{6}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{6 x^{12}} & \text {otherwise} \end {cases}\right )}{2} \] Input:

integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**9,x)
 

Output:

-a*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 - a*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 - b*c*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 - b*d*Piecewise((2*(-c**5*(c + d/x**2)**(3/2)/3 + c**4*(c + d/x**2)**( 
5/2) - 10*c**3*(c + d/x**2)**(7/2)/7 + 10*c**2*(c + d/x**2)**(9/2)/9 - 5*c 
*(c + d/x**2)**(11/2)/11 + (c + d/x**2)**(13/2)/13)/d**6, Ne(d, 0)), (sqrt 
(c)/(6*x**12), True))/2
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=-\frac {1}{1155} \, {\left (\frac {105 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {11}{2}}}{d^{4}} - \frac {385 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} c}{d^{4}} + \frac {495 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{2}}{d^{4}} - \frac {231 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3}}{d^{4}}\right )} a - \frac {1}{15015} \, {\left (\frac {1155 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {13}{2}}}{d^{5}} - \frac {5460 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {11}{2}} c}{d^{5}} + \frac {10010 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} c^{2}}{d^{5}} - \frac {8580 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{3}}{d^{5}} + \frac {3003 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{4}}{d^{5}}\right )} b \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^9,x, algorithm="maxima")
 

Output:

-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)*a - 1/15015*(1155 
*(c + d/x^2)^(13/2)/d^5 - 5460*(c + d/x^2)^(11/2)*c/d^5 + 10010*(c + d/x^2 
)^(9/2)*c^2/d^5 - 8580*(c + d/x^2)^(7/2)*c^3/d^5 + 3003*(c + d/x^2)^(5/2)* 
c^4/d^5)*b
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (114) = 228\).

Time = 2.35 (sec) , antiderivative size = 550, normalized size of antiderivative = 4.10 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx =\text {Too large to display} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^9,x, algorithm="giac")
 

Output:

32/15015*(15015*(sqrt(c)*x - sqrt(c*x^2 + d))^18*a*c^(11/2)*sgn(x) + 48048 
*(sqrt(c)*x - sqrt(c*x^2 + d))^16*b*c^(13/2)*sgn(x) - 3003*(sqrt(c)*x - sq 
rt(c*x^2 + d))^16*a*c^(11/2)*d*sgn(x) + 96096*(sqrt(c)*x - sqrt(c*x^2 + d) 
)^14*b*c^(13/2)*d*sgn(x) - 6006*(sqrt(c)*x - sqrt(c*x^2 + d))^14*a*c^(11/2 
)*d^2*sgn(x) + 109824*(sqrt(c)*x - sqrt(c*x^2 + d))^12*b*c^(13/2)*d^2*sgn( 
x) - 28314*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(11/2)*d^3*sgn(x) + 37752* 
(sqrt(c)*x - sqrt(c*x^2 + d))^10*b*c^(13/2)*d^3*sgn(x) + 13728*(sqrt(c)*x 
- sqrt(c*x^2 + d))^10*a*c^(11/2)*d^4*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 
 + d))^8*b*c^(13/2)*d^4*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^ 
(11/2)*d^5*sgn(x) - 2288*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(13/2)*d^5*sg 
n(x) + 3718*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(11/2)*d^6*sgn(x) + 624*(s 
qrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(13/2)*d^6*sgn(x) - 1014*(sqrt(c)*x - sq 
rt(c*x^2 + d))^4*a*c^(11/2)*d^7*sgn(x) - 104*(sqrt(c)*x - sqrt(c*x^2 + d)) 
^2*b*c^(13/2)*d^7*sgn(x) + 169*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(11/2)* 
d^8*sgn(x) + 8*b*c^(13/2)*d^8*sgn(x) - 13*a*c^(11/2)*d^9*sgn(x))/((sqrt(c) 
*x - sqrt(c*x^2 + d))^2 - d)^13
 

Mupad [B] (verification not implemented)

Time = 6.46 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=\frac {16\,a\,c^5\,\sqrt {c+\frac {d}{x^2}}}{1155\,d^4}-\frac {128\,b\,c^6\,\sqrt {c+\frac {d}{x^2}}}{15015\,d^5}-\frac {4\,a\,c\,\sqrt {c+\frac {d}{x^2}}}{33\,x^8}-\frac {a\,d\,\sqrt {c+\frac {d}{x^2}}}{11\,x^{10}}-\frac {14\,b\,c\,\sqrt {c+\frac {d}{x^2}}}{143\,x^{10}}-\frac {b\,d\,\sqrt {c+\frac {d}{x^2}}}{13\,x^{12}}-\frac {a\,c^2\,\sqrt {c+\frac {d}{x^2}}}{231\,d\,x^6}+\frac {2\,a\,c^3\,\sqrt {c+\frac {d}{x^2}}}{385\,d^2\,x^4}-\frac {8\,a\,c^4\,\sqrt {c+\frac {d}{x^2}}}{1155\,d^3\,x^2}-\frac {b\,c^2\,\sqrt {c+\frac {d}{x^2}}}{429\,d\,x^8}+\frac {8\,b\,c^3\,\sqrt {c+\frac {d}{x^2}}}{3003\,d^2\,x^6}-\frac {16\,b\,c^4\,\sqrt {c+\frac {d}{x^2}}}{5005\,d^3\,x^4}+\frac {64\,b\,c^5\,\sqrt {c+\frac {d}{x^2}}}{15015\,d^4\,x^2} \] Input:

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^9,x)
 

Output:

(16*a*c^5*(c + d/x^2)^(1/2))/(1155*d^4) - (128*b*c^6*(c + d/x^2)^(1/2))/(1 
5015*d^5) - (4*a*c*(c + d/x^2)^(1/2))/(33*x^8) - (a*d*(c + d/x^2)^(1/2))/( 
11*x^10) - (14*b*c*(c + d/x^2)^(1/2))/(143*x^10) - (b*d*(c + d/x^2)^(1/2)) 
/(13*x^12) - (a*c^2*(c + d/x^2)^(1/2))/(231*d*x^6) + (2*a*c^3*(c + d/x^2)^ 
(1/2))/(385*d^2*x^4) - (8*a*c^4*(c + d/x^2)^(1/2))/(1155*d^3*x^2) - (b*c^2 
*(c + d/x^2)^(1/2))/(429*d*x^8) + (8*b*c^3*(c + d/x^2)^(1/2))/(3003*d^2*x^ 
6) - (16*b*c^4*(c + d/x^2)^(1/2))/(5005*d^3*x^4) + (64*b*c^5*(c + d/x^2)^( 
1/2))/(15015*d^4*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.03 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^9} \, dx=\frac {208 \sqrt {c \,x^{2}+d}\, a \,c^{5} d \,x^{12}-104 \sqrt {c \,x^{2}+d}\, a \,c^{4} d^{2} x^{10}+78 \sqrt {c \,x^{2}+d}\, a \,c^{3} d^{3} x^{8}-65 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{4} x^{6}-1820 \sqrt {c \,x^{2}+d}\, a c \,d^{5} x^{4}-1365 \sqrt {c \,x^{2}+d}\, a \,d^{6} x^{2}-128 \sqrt {c \,x^{2}+d}\, b \,c^{6} x^{12}+64 \sqrt {c \,x^{2}+d}\, b \,c^{5} d \,x^{10}-48 \sqrt {c \,x^{2}+d}\, b \,c^{4} d^{2} x^{8}+40 \sqrt {c \,x^{2}+d}\, b \,c^{3} d^{3} x^{6}-35 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{4} x^{4}-1470 \sqrt {c \,x^{2}+d}\, b c \,d^{5} x^{2}-1155 \sqrt {c \,x^{2}+d}\, b \,d^{6}-208 \sqrt {c}\, a \,c^{5} d \,x^{13}+128 \sqrt {c}\, b \,c^{6} x^{13}}{15015 d^{5} x^{13}} \] Input:

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^9,x)
 

Output:

(208*sqrt(c*x**2 + d)*a*c**5*d*x**12 - 104*sqrt(c*x**2 + d)*a*c**4*d**2*x* 
*10 + 78*sqrt(c*x**2 + d)*a*c**3*d**3*x**8 - 65*sqrt(c*x**2 + d)*a*c**2*d* 
*4*x**6 - 1820*sqrt(c*x**2 + d)*a*c*d**5*x**4 - 1365*sqrt(c*x**2 + d)*a*d* 
*6*x**2 - 128*sqrt(c*x**2 + d)*b*c**6*x**12 + 64*sqrt(c*x**2 + d)*b*c**5*d 
*x**10 - 48*sqrt(c*x**2 + d)*b*c**4*d**2*x**8 + 40*sqrt(c*x**2 + d)*b*c**3 
*d**3*x**6 - 35*sqrt(c*x**2 + d)*b*c**2*d**4*x**4 - 1470*sqrt(c*x**2 + d)* 
b*c*d**5*x**2 - 1155*sqrt(c*x**2 + d)*b*d**6 - 208*sqrt(c)*a*c**5*d*x**13 
+ 128*sqrt(c)*b*c**6*x**13)/(15015*d**5*x**13)