\(\int \frac {(a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2}}{x^4} \, dx\) [169]

Optimal result

Integrand size = 22, antiderivative size = 159 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {c (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {c^2 (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{128 d^2 x}-\frac {c^3 (3 b c-8 a d) \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{128 d^{5/2}} \] Output:

1/64*c*(-8*a*d+3*b*c)*(c+d/x^2)^(1/2)/d/x^3+1/48*(-8*a*d+3*b*c)*(c+d/x^2)^ 
(3/2)/d/x^3-1/8*b*(c+d/x^2)^(5/2)/d/x^3+1/128*c^2*(-8*a*d+3*b*c)*(c+d/x^2) 
^(1/2)/d^2/x-1/128*c^3*(-8*a*d+3*b*c)*arctanh(d^(1/2)/(c+d/x^2)^(1/2)/x)/d 
^(5/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {\sqrt {c+\frac {d}{x^2}} \left (\sqrt {d} \sqrt {d+c x^2} \left (8 a d x^2 \left (8 d^2+14 c d x^2+3 c^2 x^4\right )+b \left (48 d^3+72 c d^2 x^2+6 c^2 d x^4-9 c^3 x^6\right )\right )+3 c^3 (3 b c-8 a d) x^8 \text {arctanh}\left (\frac {\sqrt {d+c x^2}}{\sqrt {d}}\right )\right )}{384 d^{5/2} x^7 \sqrt {d+c x^2}} \] Input:

Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x]
 

Output:

-1/384*(Sqrt[c + d/x^2]*(Sqrt[d]*Sqrt[d + c*x^2]*(8*a*d*x^2*(8*d^2 + 14*c* 
d*x^2 + 3*c^2*x^4) + b*(48*d^3 + 72*c*d^2*x^2 + 6*c^2*d*x^4 - 9*c^3*x^6)) 
+ 3*c^3*(3*b*c - 8*a*d)*x^8*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]]))/(d^(5/2)*x^ 
7*Sqrt[d + c*x^2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {959, 858, 248, 248, 262, 224, 219}

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.

\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle -\frac {(3 b c-8 a d) \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{x^4}dx}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {(3 b c-8 a d) \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{x^2}d\frac {1}{x}}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {(3 b c-8 a d) \left (\frac {1}{2} c \int \frac {\sqrt {c+\frac {d}{x^2}}}{x^2}d\frac {1}{x}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}\right )}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {(3 b c-8 a d) \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {1}{\sqrt {c+\frac {d}{x^2}} x^2}d\frac {1}{x}+\frac {\sqrt {c+\frac {d}{x^2}}}{4 x^3}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}\right )}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {(3 b c-8 a d) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}}{2 d}\right )+\frac {\sqrt {c+\frac {d}{x^2}}}{4 x^3}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}\right )}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(3 b c-8 a d) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}}{2 d}\right )+\frac {\sqrt {c+\frac {d}{x^2}}}{4 x^3}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}\right )}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(3 b c-8 a d) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {\sqrt {c+\frac {d}{x^2}}}{2 d x}-\frac {c \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{3/2}}\right )+\frac {\sqrt {c+\frac {d}{x^2}}}{4 x^3}\right )+\frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{6 x^3}\right )}{8 d}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}\)

Input:

Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x]
 

Output:

-1/8*(b*(c + d/x^2)^(5/2))/(d*x^3) + ((3*b*c - 8*a*d)*((c + d/x^2)^(3/2)/( 
6*x^3) + (c*(Sqrt[c + d/x^2]/(4*x^3) + (c*(Sqrt[c + d/x^2]/(2*d*x) - (c*Ar 
cTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(2*d^(3/2))))/4))/2))/(8*d)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 248

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) 
  Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ 
p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 959

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {\left (24 a \,c^{2} d \,x^{6}-9 b \,c^{3} x^{6}+112 a c \,d^{2} x^{4}+6 b \,c^{2} d \,x^{4}+64 a \,d^{3} x^{2}+72 b c \,d^{2} x^{2}+48 b \,d^{3}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{384 x^{7} d^{2}}+\frac {c^{3} \left (8 a d -3 c b \right ) \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}{128 d^{\frac {5}{2}} \sqrt {c \,x^{2}+d}}\)	\(151\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (24 d^{\frac {5}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) a \,c^{3} x^{8}-9 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) b \,c^{4} x^{8}-8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{3} d \,x^{8}+3 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{4} x^{8}+8 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{2} d \,x^{6}-3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,c^{3} x^{6}-24 \sqrt {c \,x^{2}+d}\, a \,c^{3} d^{2} x^{8}+9 \sqrt {c \,x^{2}+d}\, b \,c^{4} d \,x^{8}+16 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a c \,d^{2} x^{4}-6 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,c^{2} d \,x^{4}-64 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,d^{3} x^{2}+24 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b c \,d^{2} x^{2}-48 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,d^{3}\right )}{384 x^{5} \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{4}}\)	\(302\)

Input:

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/384*(24*a*c^2*d*x^6-9*b*c^3*x^6+112*a*c*d^2*x^4+6*b*c^2*d*x^4+64*a*d^3* 
x^2+72*b*c*d^2*x^2+48*b*d^3)/x^7/d^2*((c*x^2+d)/x^2)^(1/2)+1/128*c^3*(8*a* 
d-3*b*c)/d^(5/2)*ln((2*d+2*d^(1/2)*(c*x^2+d)^(1/2))/x)*((c*x^2+d)/x^2)^(1/ 
2)*x/(c*x^2+d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=\left [-\frac {3 \, {\left (3 \, b c^{4} - 8 \, a c^{3} d\right )} \sqrt {d} x^{7} \log \left (-\frac {c x^{2} + 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (3 \, b c^{3} d - 8 \, a c^{2} d^{2}\right )} x^{6} - 48 \, b d^{4} - 2 \, {\left (3 \, b c^{2} d^{2} + 56 \, a c d^{3}\right )} x^{4} - 8 \, {\left (9 \, b c d^{3} + 8 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{768 \, d^{3} x^{7}}, \frac {3 \, {\left (3 \, b c^{4} - 8 \, a c^{3} d\right )} \sqrt {-d} x^{7} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{d}\right ) + {\left (3 \, {\left (3 \, b c^{3} d - 8 \, a c^{2} d^{2}\right )} x^{6} - 48 \, b d^{4} - 2 \, {\left (3 \, b c^{2} d^{2} + 56 \, a c d^{3}\right )} x^{4} - 8 \, {\left (9 \, b c d^{3} + 8 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{384 \, d^{3} x^{7}}\right ] \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="fricas")
 

Output:

[-1/768*(3*(3*b*c^4 - 8*a*c^3*d)*sqrt(d)*x^7*log(-(c*x^2 + 2*sqrt(d)*x*sqr 
t((c*x^2 + d)/x^2) + 2*d)/x^2) - 2*(3*(3*b*c^3*d - 8*a*c^2*d^2)*x^6 - 48*b 
*d^4 - 2*(3*b*c^2*d^2 + 56*a*c*d^3)*x^4 - 8*(9*b*c*d^3 + 8*a*d^4)*x^2)*sqr 
t((c*x^2 + d)/x^2))/(d^3*x^7), 1/384*(3*(3*b*c^4 - 8*a*c^3*d)*sqrt(-d)*x^7 
*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/d) + (3*(3*b*c^3*d - 8*a*c^2*d^2) 
*x^6 - 48*b*d^4 - 2*(3*b*c^2*d^2 + 56*a*c*d^3)*x^4 - 8*(9*b*c*d^3 + 8*a*d^ 
4)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^7)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (141) = 282\).

Time = 43.04 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.81 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=- \frac {a c^{\frac {5}{2}}}{16 d x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {17 a c^{\frac {3}{2}}}{48 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {11 a \sqrt {c} d}{24 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {a c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{16 d^{\frac {3}{2}}} - \frac {a d^{2}}{6 \sqrt {c} x^{7} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {3 b c^{\frac {7}{2}}}{128 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{\frac {5}{2}}}{128 d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {13 b c^{\frac {3}{2}}}{64 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {5 b \sqrt {c} d}{16 x^{7} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{128 d^{\frac {5}{2}}} - \frac {b d^{2}}{8 \sqrt {c} x^{9} \sqrt {1 + \frac {d}{c x^{2}}}} \] Input:

integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**4,x)
 

Output:

-a*c**(5/2)/(16*d*x*sqrt(1 + d/(c*x**2))) - 17*a*c**(3/2)/(48*x**3*sqrt(1 
+ d/(c*x**2))) - 11*a*sqrt(c)*d/(24*x**5*sqrt(1 + d/(c*x**2))) + a*c**3*as 
inh(sqrt(d)/(sqrt(c)*x))/(16*d**(3/2)) - a*d**2/(6*sqrt(c)*x**7*sqrt(1 + d 
/(c*x**2))) + 3*b*c**(7/2)/(128*d**2*x*sqrt(1 + d/(c*x**2))) + b*c**(5/2)/ 
(128*d*x**3*sqrt(1 + d/(c*x**2))) - 13*b*c**(3/2)/(64*x**5*sqrt(1 + d/(c*x 
**2))) - 5*b*sqrt(c)*d/(16*x**7*sqrt(1 + d/(c*x**2))) - 3*b*c**4*asinh(sqr 
t(d)/(sqrt(c)*x))/(128*d**(5/2)) - b*d**2/(8*sqrt(c)*x**9*sqrt(1 + d/(c*x* 
*2)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (135) = 270\).

Time = 0.12 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.23 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=-\frac {1}{96} \, {\left (\frac {3 \, c^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} x^{5} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3} d x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{2} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} d x^{6} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} x^{4} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} d^{3} x^{2} - d^{4}}\right )} a + \frac {1}{256} \, {\left (\frac {3 \, c^{4} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{4} x^{7} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{4} d x^{5} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{4} d^{2} x^{3} + 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{4} d^{3} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{4} d^{2} x^{8} - 4 \, {\left (c + \frac {d}{x^{2}}\right )}^{3} d^{3} x^{6} + 6 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{4} x^{4} - 4 \, {\left (c + \frac {d}{x^{2}}\right )} d^{5} x^{2} + d^{6}}\right )} b \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="maxima")
 

Output:

-1/96*(3*c^3*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d 
)))/d^(3/2) + 2*(3*(c + d/x^2)^(5/2)*c^3*x^5 + 8*(c + d/x^2)^(3/2)*c^3*d*x 
^3 - 3*sqrt(c + d/x^2)*c^3*d^2*x)/((c + d/x^2)^3*d*x^6 - 3*(c + d/x^2)^2*d 
^2*x^4 + 3*(c + d/x^2)*d^3*x^2 - d^4))*a + 1/256*(3*c^4*log((sqrt(c + d/x^ 
2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(5/2) + 2*(3*(c + d/x^2)^ 
(7/2)*c^4*x^7 - 11*(c + d/x^2)^(5/2)*c^4*d*x^5 - 11*(c + d/x^2)^(3/2)*c^4* 
d^2*x^3 + 3*sqrt(c + d/x^2)*c^4*d^3*x)/((c + d/x^2)^4*d^2*x^8 - 4*(c + d/x 
^2)^3*d^3*x^6 + 6*(c + d/x^2)^2*d^4*x^4 - 4*(c + d/x^2)*d^5*x^2 + d^6))*b
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {\frac {3 \, {\left (3 \, b c^{5} \mathrm {sgn}\left (x\right ) - 8 \, a c^{4} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d} d^{2}} + \frac {9 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} b c^{5} \mathrm {sgn}\left (x\right ) - 24 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a c^{4} d \mathrm {sgn}\left (x\right ) - 33 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c^{5} d \mathrm {sgn}\left (x\right ) - 40 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{4} d^{2} \mathrm {sgn}\left (x\right ) - 33 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{5} d^{2} \mathrm {sgn}\left (x\right ) + 88 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{4} d^{3} \mathrm {sgn}\left (x\right ) + 9 \, \sqrt {c x^{2} + d} b c^{5} d^{3} \mathrm {sgn}\left (x\right ) - 24 \, \sqrt {c x^{2} + d} a c^{4} d^{4} \mathrm {sgn}\left (x\right )}{c^{4} d^{2} x^{8}}}{384 \, c} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="giac")
 

Output:

1/384*(3*(3*b*c^5*sgn(x) - 8*a*c^4*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(- 
d))/(sqrt(-d)*d^2) + (9*(c*x^2 + d)^(7/2)*b*c^5*sgn(x) - 24*(c*x^2 + d)^(7 
/2)*a*c^4*d*sgn(x) - 33*(c*x^2 + d)^(5/2)*b*c^5*d*sgn(x) - 40*(c*x^2 + d)^ 
(5/2)*a*c^4*d^2*sgn(x) - 33*(c*x^2 + d)^(3/2)*b*c^5*d^2*sgn(x) + 88*(c*x^2 
 + d)^(3/2)*a*c^4*d^3*sgn(x) + 9*sqrt(c*x^2 + d)*b*c^5*d^3*sgn(x) - 24*sqr 
t(c*x^2 + d)*a*c^4*d^4*sgn(x))/(c^4*d^2*x^8))/c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=\int \frac {\left (a+\frac {b}{x^2}\right )\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{x^4} \,d x \] Input:

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x)
 

Output:

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx=\frac {-24 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{2} x^{6}-112 \sqrt {c \,x^{2}+d}\, a c \,d^{3} x^{4}-64 \sqrt {c \,x^{2}+d}\, a \,d^{4} x^{2}+9 \sqrt {c \,x^{2}+d}\, b \,c^{3} d \,x^{6}-6 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{2} x^{4}-72 \sqrt {c \,x^{2}+d}\, b c \,d^{3} x^{2}-48 \sqrt {c \,x^{2}+d}\, b \,d^{4}-24 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) a \,c^{3} d \,x^{8}+9 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) b \,c^{4} x^{8}+24 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) a \,c^{3} d \,x^{8}-9 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) b \,c^{4} x^{8}}{384 d^{3} x^{8}} \] Input:

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x)
 

Output:

( - 24*sqrt(c*x**2 + d)*a*c**2*d**2*x**6 - 112*sqrt(c*x**2 + d)*a*c*d**3*x 
**4 - 64*sqrt(c*x**2 + d)*a*d**4*x**2 + 9*sqrt(c*x**2 + d)*b*c**3*d*x**6 - 
 6*sqrt(c*x**2 + d)*b*c**2*d**2*x**4 - 72*sqrt(c*x**2 + d)*b*c*d**3*x**2 - 
 48*sqrt(c*x**2 + d)*b*d**4 - 24*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x 
 - sqrt(d))/sqrt(d))*a*c**3*d*x**8 + 9*sqrt(d)*log((sqrt(c*x**2 + d) + sqr 
t(c)*x - sqrt(d))/sqrt(d))*b*c**4*x**8 + 24*sqrt(d)*log((sqrt(c*x**2 + d) 
+ sqrt(c)*x + sqrt(d))/sqrt(d))*a*c**3*d*x**8 - 9*sqrt(d)*log((sqrt(c*x**2 
 + d) + sqrt(c)*x + sqrt(d))/sqrt(d))*b*c**4*x**8)/(384*d**3*x**8)