\(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^3 x^5} \, dx\) [33]

Optimal result

Integrand size = 18, antiderivative size = 103 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=\frac {2 a+b x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {3 b (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 b c \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \] Output:

1/2*(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2-3/2*b*(2*c*x+b)/(-4*a*c+b^2)^2/ 
(c*x^2+b*x+a)+6*b*c*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/ 
2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=\frac {\frac {\left (b^2-4 a c\right ) (2 a+b x)}{(a+x (b+c x))^2}-\frac {3 b (b+2 c x)}{a+x (b+c x)}-\frac {12 b c \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{2 \left (b^2-4 a c\right )^2} \] Input:

Integrate[1/((c + a/x^2 + b/x)^3*x^5),x]
 

Output:

(((b^2 - 4*a*c)*(2*a + b*x))/(a + x*(b + c*x))^2 - (3*b*(b + 2*c*x))/(a + 
x*(b + c*x)) - (12*b*c*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 
 4*a*c])/(2*(b^2 - 4*a*c)^2)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1692, 1159, 1086, 1083, 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.

\(\displaystyle \int \frac {1}{x^5 \left (\frac {a}{x^2}+\frac {b}{x}+c\right )^3} \, dx\)

\(\Big \downarrow \) 1692

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

\(\Big \downarrow \) 1159

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

\(\Big \downarrow \) 1086

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

Input:

Int[1/((c + a/x^2 + b/x)^3*x^5),x]
 

Output:

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

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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1086

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && ILtQ[p, -1]
 

rule 1159

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* 
x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* 
c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & 
& LtQ[p, -1] && NeQ[p, -3/2]
 

rule 1692

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
 :> Int[x^(m + 2*n*p)*(c + b/x^n + a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n 
}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {-b x -2 a}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}}-\frac {3 b \left (\frac {2 c x +b}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {4 c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (4 a c -b^{2}\right )}\)	\(118\)
risch	\(\frac {-\frac {3 b \,c^{2} x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {9 b^{2} c \,x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a c +b^{2}\right ) b x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {a \left (8 a c +b^{2}\right )}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {3 c b \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 c b \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\)	\(289\)

Input:

int(1/(c+a/x^2+b/x)^3/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/2*(-b*x-2*a)/(4*a*c-b^2)/(c*x^2+b*x+a)^2-3/2*b/(4*a*c-b^2)*((2*c*x+b)/(4 
*a*c-b^2)/(c*x^2+b*x+a)+4*c/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2) 
^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (95) = 190\).

Time = 0.08 (sec) , antiderivative size = 788, normalized size of antiderivative = 7.65 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx =\text {Too large to display} \] Input:

integrate(1/(c+a/x^2+b/x)^3/x^5,x, algorithm="fricas")
 

Output:

[-1/2*(a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9* 
(b^4*c - 4*a*b^2*c^2)*x^2 - 6*(b*c^3*x^4 + 2*b^2*c^2*x^3 + 2*a*b^2*c*x + a 
^2*b*c + (b^3*c + 2*a*b*c^2)*x^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c 
*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 2*( 
b^5 + a*b^3*c - 20*a^2*b*c^2)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 
- 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 
+ 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10 
*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 
 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x), -1/2*(a*b^4 + 4*a^2*b^2 
*c - 32*a^3*c^2 + 6*(b^3*c^2 - 4*a*b*c^3)*x^3 + 9*(b^4*c - 4*a*b^2*c^2)*x^ 
2 - 12*(b*c^3*x^4 + 2*b^2*c^2*x^3 + 2*a*b^2*c*x + a^2*b*c + (b^3*c + 2*a*b 
*c^2)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 
- 4*a*c)) + 2*(b^5 + a*b^3*c - 20*a^2*b*c^2)*x)/(a^2*b^6 - 12*a^3*b^4*c + 
48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 6 
4*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)* 
x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x 
^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (95) = 190\).

Time = 0.65 (sec) , antiderivative size = 481, normalized size of antiderivative = 4.67 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {- 192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} - 3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )} + \frac {- 8 a^{2} c - a b^{2} - 9 b^{2} c x^{2} - 6 b c^{2} x^{3} + x \left (- 10 a b c - 2 b^{3}\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \cdot \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \cdot \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \cdot \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \] Input:

integrate(1/(c+a/x**2+b/x)**3/x**5,x)
 

Output:

3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-192*a**3*b*c**4*sqrt(-1/(4*a*c 
- b**2)**5) + 144*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2)**5) - 36*a*b**5*c* 
*2*sqrt(-1/(4*a*c - b**2)**5) + 3*b**7*c*sqrt(-1/(4*a*c - b**2)**5) + 3*b* 
*2*c)/(6*b*c**2)) - 3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x + (192*a**3*b*c 
**4*sqrt(-1/(4*a*c - b**2)**5) - 144*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2) 
**5) + 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) - 3*b**7*c*sqrt(-1/(4*a*c 
 - b**2)**5) + 3*b**2*c)/(6*b*c**2)) + (-8*a**2*c - a*b**2 - 9*b**2*c*x**2 
 - 6*b*c**2*x**3 + x*(-10*a*b*c - 2*b**3))/(32*a**4*c**2 - 16*a**3*b**2*c 
+ 2*a**2*b**4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 + 2*b**4*c**2) + x**3* 
(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) + x**2*(64*a**3*c**3 - 12*a*b 
**4*c + 2*b**6) + x*(64*a**3*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(c+a/x^2+b/x)^3/x^5,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=-\frac {6 \, b c \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {6 \, b c^{2} x^{3} + 9 \, b^{2} c x^{2} + 2 \, b^{3} x + 10 \, a b c x + a b^{2} + 8 \, a^{2} c}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \] Input:

integrate(1/(c+a/x^2+b/x)^3/x^5,x, algorithm="giac")
 

Output:

-6*b*c*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c 
^2)*sqrt(-b^2 + 4*a*c)) - 1/2*(6*b*c^2*x^3 + 9*b^2*c*x^2 + 2*b^3*x + 10*a* 
b*c*x + a*b^2 + 8*a^2*c)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x + a) 
^2)
 

Mupad [B] (verification not implemented)

Time = 18.94 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.46 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=-\frac {\frac {8\,c\,a^2+a\,b^2}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b^2\,c\,x^2}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,b\,c^2\,x^3}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {b\,x\,\left (b^2+5\,a\,c\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}-\frac {6\,b\,c\,\mathrm {atan}\left (\frac {\left (\frac {3\,b^2\,c}{{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {6\,b\,c^2\,x}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{3\,b\,c}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \] Input:

int(1/(x^5*(c + a/x^2 + b/x)^3),x)
 

Output:

- ((a*b^2 + 8*a^2*c)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (9*b^2*c*x^2)/(2 
*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (3*b*c^2*x^3)/(b^4 + 16*a^2*c^2 - 8*a*b 
^2*c) + (b*x*(5*a*c + b^2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(x^2*(2*a*c + 
b^2) + a^2 + c^2*x^4 + 2*a*b*x + 2*b*c*x^3) - (6*b*c*atan((((3*b^2*c)/(4*a 
*c - b^2)^(5/2) + (6*b*c^2*x)/(4*a*c - b^2)^(5/2))*(b^4 + 16*a^2*c^2 - 8*a 
*b^2*c))/(3*b*c)))/(4*a*c - b^2)^(5/2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 575, normalized size of antiderivative = 5.58 \[ \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^5} \, dx=\frac {-12 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a^{2} b c -24 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c x -24 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a b \,c^{2} x^{2}-12 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{3} c \,x^{2}-24 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c^{2} x^{3}-12 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b \,c^{3} x^{4}-20 a^{3} c^{2}+a^{2} b^{2} c -16 a^{2} b \,c^{2} x +24 a^{2} c^{3} x^{2}+a \,b^{4}-4 a \,b^{3} c x -30 a \,b^{2} c^{2} x^{2}+12 a \,c^{4} x^{4}+2 b^{5} x +6 b^{4} c \,x^{2}-3 b^{2} c^{3} x^{4}}{128 a^{3} c^{5} x^{4}-96 a^{2} b^{2} c^{4} x^{4}+24 a \,b^{4} c^{3} x^{4}-2 b^{6} c^{2} x^{4}+256 a^{3} b \,c^{4} x^{3}-192 a^{2} b^{3} c^{3} x^{3}+48 a \,b^{5} c^{2} x^{3}-4 b^{7} c \,x^{3}+256 a^{4} c^{4} x^{2}-64 a^{3} b^{2} c^{3} x^{2}-48 a^{2} b^{4} c^{2} x^{2}+20 a \,b^{6} c \,x^{2}-2 b^{8} x^{2}+256 a^{4} b \,c^{3} x -192 a^{3} b^{3} c^{2} x +48 a^{2} b^{5} c x -4 a \,b^{7} x +128 a^{5} c^{3}-96 a^{4} b^{2} c^{2}+24 a^{3} b^{4} c -2 a^{2} b^{6}} \] Input:

int(1/(c+a/x^2+b/x)^3/x^5,x)
 

Output:

( - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b*c - 
24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c*x - 24 
*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c**2*x**2 - 1 
2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*c*x**2 - 24 
*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c**2*x**3 - 
12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**3*x**4 - 2 
0*a**3*c**2 + a**2*b**2*c - 16*a**2*b*c**2*x + 24*a**2*c**3*x**2 + a*b**4 
- 4*a*b**3*c*x - 30*a*b**2*c**2*x**2 + 12*a*c**4*x**4 + 2*b**5*x + 6*b**4* 
c*x**2 - 3*b**2*c**3*x**4)/(2*(64*a**5*c**3 - 48*a**4*b**2*c**2 + 128*a**4 
*b*c**3*x + 128*a**4*c**4*x**2 + 12*a**3*b**4*c - 96*a**3*b**3*c**2*x - 32 
*a**3*b**2*c**3*x**2 + 128*a**3*b*c**4*x**3 + 64*a**3*c**5*x**4 - a**2*b** 
6 + 24*a**2*b**5*c*x - 24*a**2*b**4*c**2*x**2 - 96*a**2*b**3*c**3*x**3 - 4 
8*a**2*b**2*c**4*x**4 - 2*a*b**7*x + 10*a*b**6*c*x**2 + 24*a*b**5*c**2*x** 
3 + 12*a*b**4*c**3*x**4 - b**8*x**2 - 2*b**7*c*x**3 - b**6*c**2*x**4))