\(\int \frac {(a+\frac {b}{x^2}) x^3}{(c+\frac {d}{x^2})^{3/2}} \, dx\) [181]

Optimal result

Integrand size = 22, antiderivative size = 114 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\frac {d (b c-a d)}{c^3 \sqrt {c+\frac {d}{x^2}}}+\frac {(4 b c-7 a d) \sqrt {c+\frac {d}{x^2}} x^2}{8 c^3}+\frac {a \sqrt {c+\frac {d}{x^2}} x^4}{4 c^2}-\frac {3 d (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Output:

d*(-a*d+b*c)/c^3/(c+d/x^2)^(1/2)+1/8*(-7*a*d+4*b*c)*(c+d/x^2)^(1/2)*x^2/c^ 
3+1/4*a*(c+d/x^2)^(1/2)*x^4/c^2-3/8*d*(-5*a*d+4*b*c)*arctanh((c+d/x^2)^(1/ 
2)/c^(1/2))/c^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\frac {\sqrt {c} x \left (4 b c \left (3 d+c x^2\right )+a \left (-15 d^2-5 c d x^2+2 c^2 x^4\right )\right )+24 b c d \sqrt {d+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {d}-\sqrt {d+c x^2}}\right )+30 a d^2 \sqrt {d+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {d}+\sqrt {d+c x^2}}\right )}{8 c^{7/2} \sqrt {c+\frac {d}{x^2}} x} \] Input:

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

Output:

(Sqrt[c]*x*(4*b*c*(3*d + c*x^2) + a*(-15*d^2 - 5*c*d*x^2 + 2*c^2*x^4)) + 2 
4*b*c*d*Sqrt[d + c*x^2]*ArcTanh[(Sqrt[c]*x)/(Sqrt[d] - Sqrt[d + c*x^2])] + 
 30*a*d^2*Sqrt[d + c*x^2]*ArcTanh[(Sqrt[c]*x)/(-Sqrt[d] + Sqrt[d + c*x^2]) 
])/(8*c^(7/2)*Sqrt[c + d/x^2]*x)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 87, 52, 61, 73, 221}

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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((a*x^4)/(2*c*Sqrt[c + d/x^2]) - ((4*b*c - 5*a*d)*(-(x^2/(c*Sqrt[c + d/x^2 
])) - (3*d*(2/(c*Sqrt[c + d/x^2]) - (2*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/c 
^(3/2)))/(2*c)))/(4*c))/2
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

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

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]]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\left (c \,x^{2}+d \right ) \left (2 c^{\frac {7}{2}} a \,x^{5}-5 c^{\frac {5}{2}} a d \,x^{3}+4 c^{\frac {7}{2}} b \,x^{3}-15 c^{\frac {3}{2}} a \,d^{2} x +12 c^{\frac {5}{2}} b d x +15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) \sqrt {c \,x^{2}+d}\, a c \,d^{2}-12 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) \sqrt {c \,x^{2}+d}\, b \,c^{2} d \right )}{8 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} c^{\frac {9}{2}}}\)	\(140\)
risch	\(\frac {\left (2 a c \,x^{2}-7 a d +4 c b \right ) \left (c \,x^{2}+d \right )}{8 c^{3} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}+\frac {d \left (3 c \left (5 a d -4 c b \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+d}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )}{c^{\frac {3}{2}}}\right )+\frac {7 a d x}{\sqrt {c \,x^{2}+d}}-\frac {4 b c x}{\sqrt {c \,x^{2}+d}}\right ) \sqrt {c \,x^{2}+d}}{8 c^{3} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}\)	\(151\)

Input:

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

Output:

1/8*(c*x^2+d)*(2*c^(7/2)*a*x^5-5*c^(5/2)*a*d*x^3+4*c^(7/2)*b*x^3-15*c^(3/2 
)*a*d^2*x+12*c^(5/2)*b*d*x+15*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*(c*x^2+d)^(1/2 
)*a*c*d^2-12*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*(c*x^2+d)^(1/2)*b*c^2*d)/((c*x^ 
2+d)/x^2)^(3/2)/x^3/c^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.67 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (4 \, b c d^{2} - 5 \, a d^{3} + {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left (2 \, a c^{3} x^{6} + {\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \, {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, {\left (c^{5} x^{2} + c^{4} d\right )}}, \frac {3 \, {\left (4 \, b c d^{2} - 5 \, a d^{3} + {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a c^{3} x^{6} + {\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \, {\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, {\left (c^{5} x^{2} + c^{4} d\right )}}\right ] \] Input:

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

Output:

[-1/16*(3*(4*b*c*d^2 - 5*a*d^3 + (4*b*c^2*d - 5*a*c*d^2)*x^2)*sqrt(c)*log( 
-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*(2*a*c^3*x^6 + (4* 
b*c^3 - 5*a*c^2*d)*x^4 + 3*(4*b*c^2*d - 5*a*c*d^2)*x^2)*sqrt((c*x^2 + d)/x 
^2))/(c^5*x^2 + c^4*d), 1/8*(3*(4*b*c*d^2 - 5*a*d^3 + (4*b*c^2*d - 5*a*c*d 
^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + 
 (2*a*c^3*x^6 + (4*b*c^3 - 5*a*c^2*d)*x^4 + 3*(4*b*c^2*d - 5*a*c*d^2)*x^2) 
*sqrt((c*x^2 + d)/x^2))/(c^5*x^2 + c^4*d)]
 

Sympy [A] (verification not implemented)

Time = 66.64 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=a \left (\frac {x^{5}}{4 c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {5 \sqrt {d} x^{3}}{8 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {15 d^{\frac {3}{2}} x}{8 c^{3} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {15 d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 c^{\frac {7}{2}}}\right ) + b \left (\frac {x^{3}}{2 c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 \sqrt {d} x}{2 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 c^{\frac {5}{2}}}\right ) \] Input:

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

Output:

a*(x**5/(4*c*sqrt(d)*sqrt(c*x**2/d + 1)) - 5*sqrt(d)*x**3/(8*c**2*sqrt(c*x 
**2/d + 1)) - 15*d**(3/2)*x/(8*c**3*sqrt(c*x**2/d + 1)) + 15*d**2*asinh(sq 
rt(c)*x/sqrt(d))/(8*c**(7/2))) + b*(x**3/(2*c*sqrt(d)*sqrt(c*x**2/d + 1)) 
+ 3*sqrt(d)*x/(2*c**2*sqrt(c*x**2/d + 1)) - 3*d*asinh(sqrt(c)*x/sqrt(d))/( 
2*c**(5/2)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (96) = 192\).

Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=-\frac {1}{16} \, a {\left (\frac {2 \, {\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} - 25 \, {\left (c + \frac {d}{x^{2}}\right )} c d^{2} + 8 \, c^{2} d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} - 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{4} + \sqrt {c + \frac {d}{x^{2}}} c^{5}} + \frac {15 \, d^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )} + \frac {1}{4} \, b {\left (\frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )} d - 2 \, c d\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} - \sqrt {c + \frac {d}{x^{2}}} c^{3}} + \frac {3 \, d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} \] Input:

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

Output:

-1/16*a*(2*(15*(c + d/x^2)^2*d^2 - 25*(c + d/x^2)*c*d^2 + 8*c^2*d^2)/((c + 
 d/x^2)^(5/2)*c^3 - 2*(c + d/x^2)^(3/2)*c^4 + sqrt(c + d/x^2)*c^5) + 15*d^ 
2*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(7/2)) + 
1/4*b*(2*(3*(c + d/x^2)*d - 2*c*d)/((c + d/x^2)^(3/2)*c^2 - sqrt(c + d/x^2 
)*c^3) + 3*d*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/ 
c^(5/2))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, a x^{2}}{c \mathrm {sgn}\left (x\right )} + \frac {4 \, b c^{4} \mathrm {sgn}\left (x\right ) - 5 \, a c^{3} d \mathrm {sgn}\left (x\right )}{c^{5}}\right )} + \frac {3 \, {\left (4 \, b c^{3} d \mathrm {sgn}\left (x\right ) - 5 \, a c^{2} d^{2} \mathrm {sgn}\left (x\right )\right )}}{c^{5}}\right )} x}{8 \, \sqrt {c x^{2} + d}} - \frac {3 \, {\left (4 \, b c d \log \left ({\left | d \right |}\right ) - 5 \, a d^{2} \log \left ({\left | d \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {7}{2}}} + \frac {3 \, {\left (4 \, b c d - 5 \, a d^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right )}{8 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

1/8*(x^2*(2*a*x^2/(c*sgn(x)) + (4*b*c^4*sgn(x) - 5*a*c^3*d*sgn(x))/c^5) + 
3*(4*b*c^3*d*sgn(x) - 5*a*c^2*d^2*sgn(x))/c^5)*x/sqrt(c*x^2 + d) - 3/16*(4 
*b*c*d*log(abs(d)) - 5*a*d^2*log(abs(d)))*sgn(x)/c^(7/2) + 3/8*(4*b*c*d - 
5*a*d^2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))/(c^(7/2)*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 5.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\frac {a\,x^4}{4\,c\,\sqrt {c+\frac {d}{x^2}}}-\frac {15\,a\,d^2}{8\,c^3\,\sqrt {c+\frac {d}{x^2}}}+\frac {b\,x^2}{2\,c\,\sqrt {c+\frac {d}{x^2}}}-\frac {3\,b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,c^{5/2}}+\frac {15\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,c^{7/2}}+\frac {3\,b\,d}{2\,c^2\,\sqrt {c+\frac {d}{x^2}}}-\frac {5\,a\,d\,x^2}{8\,c^2\,\sqrt {c+\frac {d}{x^2}}} \] Input:

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

Output:

(a*x^4)/(4*c*(c + d/x^2)^(1/2)) - (15*a*d^2)/(8*c^3*(c + d/x^2)^(1/2)) + ( 
b*x^2)/(2*c*(c + d/x^2)^(1/2)) - (3*b*d*atanh((c + d/x^2)^(1/2)/c^(1/2)))/ 
(2*c^(5/2)) + (15*a*d^2*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(8*c^(7/2)) + (3 
*b*d)/(2*c^2*(c + d/x^2)^(1/2)) - (5*a*d*x^2)/(8*c^2*(c + d/x^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.25 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx=\frac {2 \sqrt {c \,x^{2}+d}\, a \,c^{3} x^{5}-5 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{3}-15 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x +4 \sqrt {c \,x^{2}+d}\, b \,c^{3} x^{3}+12 \sqrt {c \,x^{2}+d}\, b \,c^{2} d x +15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a c \,d^{2} x^{2}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a \,d^{3}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) b \,c^{2} d \,x^{2}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) b c \,d^{2}-10 \sqrt {c}\, a c \,d^{2} x^{2}-10 \sqrt {c}\, a \,d^{3}+9 \sqrt {c}\, b \,c^{2} d \,x^{2}+9 \sqrt {c}\, b c \,d^{2}}{8 c^{4} \left (c \,x^{2}+d \right )} \] Input:

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

Output:

(2*sqrt(c*x**2 + d)*a*c**3*x**5 - 5*sqrt(c*x**2 + d)*a*c**2*d*x**3 - 15*sq 
rt(c*x**2 + d)*a*c*d**2*x + 4*sqrt(c*x**2 + d)*b*c**3*x**3 + 12*sqrt(c*x** 
2 + d)*b*c**2*d*x + 15*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d)) 
*a*c*d**2*x**2 + 15*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*a* 
d**3 - 12*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*b*c**2*d*x** 
2 - 12*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*b*c*d**2 - 10*s 
qrt(c)*a*c*d**2*x**2 - 10*sqrt(c)*a*d**3 + 9*sqrt(c)*b*c**2*d*x**2 + 9*sqr 
t(c)*b*c*d**2)/(8*c**4*(c*x**2 + d))