\(\int \frac {A+B x^2}{x (b x^2+c x^4)^3} \, dx\) [92]

Optimal result

Integrand size = 24, antiderivative size = 148 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{6 b^3 x^6}-\frac {b B-3 A c}{4 b^4 x^4}+\frac {3 c (b B-2 A c)}{2 b^5 x^2}+\frac {c^2 (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}+\frac {c^2 (3 b B-4 A c)}{2 b^5 \left (b+c x^2\right )}+\frac {2 c^2 (3 b B-5 A c) \log (x)}{b^6}-\frac {c^2 (3 b B-5 A c) \log \left (b+c x^2\right )}{b^6} \] Output:

-1/6*A/b^3/x^6-1/4*(-3*A*c+B*b)/b^4/x^4+3/2*c*(-2*A*c+B*b)/b^5/x^2+1/4*c^2 
*(-A*c+B*b)/b^4/(c*x^2+b)^2+1/2*c^2*(-4*A*c+3*B*b)/b^5/(c*x^2+b)+2*c^2*(-5 
*A*c+3*B*b)*ln(x)/b^6-c^2*(-5*A*c+3*B*b)*ln(c*x^2+b)/b^6
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {2 A b^3}{x^6}-\frac {3 b^2 (b B-3 A c)}{x^4}+\frac {18 b c (b B-2 A c)}{x^2}+\frac {3 b^2 c^2 (b B-A c)}{\left (b+c x^2\right )^2}+\frac {6 b c^2 (3 b B-4 A c)}{b+c x^2}+24 c^2 (3 b B-5 A c) \log (x)+12 c^2 (-3 b B+5 A c) \log \left (b+c x^2\right )}{12 b^6} \] Input:

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

Output:

((-2*A*b^3)/x^6 - (3*b^2*(b*B - 3*A*c))/x^4 + (18*b*c*(b*B - 2*A*c))/x^2 + 
 (3*b^2*c^2*(b*B - A*c))/(b + c*x^2)^2 + (6*b*c^2*(3*b*B - 4*A*c))/(b + c* 
x^2) + 24*c^2*(3*b*B - 5*A*c)*Log[x] + 12*c^2*(-3*b*B + 5*A*c)*Log[b + c*x 
^2])/(12*b^6)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 354, 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.

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

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {A+B x^2}{x^7 \left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^8 \left (c x^2+b\right )^3}dx^2\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {1}{2} \int \left (-\frac {2 (3 b B-5 A c) c^3}{b^6 \left (c x^2+b\right )}-\frac {(3 b B-4 A c) c^3}{b^5 \left (c x^2+b\right )^2}-\frac {(b B-A c) c^3}{b^4 \left (c x^2+b\right )^3}+\frac {2 (3 b B-5 A c) c^2}{b^6 x^2}-\frac {3 (b B-2 A c) c}{b^5 x^4}+\frac {b B-3 A c}{b^4 x^6}+\frac {A}{b^3 x^8}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2 c^2 \log \left (x^2\right ) (3 b B-5 A c)}{b^6}-\frac {2 c^2 (3 b B-5 A c) \log \left (b+c x^2\right )}{b^6}+\frac {c^2 (3 b B-4 A c)}{b^5 \left (b+c x^2\right )}+\frac {3 c (b B-2 A c)}{b^5 x^2}+\frac {c^2 (b B-A c)}{2 b^4 \left (b+c x^2\right )^2}-\frac {b B-3 A c}{2 b^4 x^4}-\frac {A}{3 b^3 x^6}\right )\)

Input:

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

Output:

(-1/3*A/(b^3*x^6) - (b*B - 3*A*c)/(2*b^4*x^4) + (3*c*(b*B - 2*A*c))/(b^5*x 
^2) + (c^2*(b*B - A*c))/(2*b^4*(b + c*x^2)^2) + (c^2*(3*b*B - 4*A*c))/(b^5 
*(b + c*x^2)) + (2*c^2*(3*b*B - 5*A*c)*Log[x^2])/b^6 - (2*c^2*(3*b*B - 5*A 
*c)*Log[b + c*x^2])/b^6)/2
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {A}{6 b^{3} x^{6}}-\frac {-3 A c +B b}{4 b^{4} x^{4}}-\frac {3 c \left (2 A c -B b \right )}{2 b^{5} x^{2}}-\frac {2 c^{2} \left (5 A c -3 B b \right ) \ln \left (x \right )}{b^{6}}+\frac {c^{3} \left (\frac {\left (10 A c -6 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b \left (4 A c -3 B b \right )}{c \left (c \,x^{2}+b \right )}-\frac {b^{2} \left (A c -B b \right )}{2 c \left (c \,x^{2}+b \right )^{2}}\right )}{2 b^{6}}\)	\(143\)
norman	\(\frac {-\frac {A}{6 b}+\frac {\left (5 A c -3 B b \right ) x^{2}}{12 b^{2}}-\frac {c \left (5 A c -3 B b \right ) x^{4}}{3 b^{3}}+\frac {2 c \left (5 A \,c^{3}-3 B b \,c^{2}\right ) x^{8}}{b^{5}}+\frac {c^{2} \left (15 A \,c^{3}-9 B b \,c^{2}\right ) x^{10}}{2 b^{6}}}{x^{6} \left (c \,x^{2}+b \right )^{2}}+\frac {c^{2} \left (5 A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{b^{6}}-\frac {2 c^{2} \left (5 A c -3 B b \right ) \ln \left (x \right )}{b^{6}}\)	\(148\)
risch	\(\frac {-\frac {c^{3} \left (5 A c -3 B b \right ) x^{8}}{b^{5}}-\frac {3 c^{2} \left (5 A c -3 B b \right ) x^{6}}{2 b^{4}}-\frac {c \left (5 A c -3 B b \right ) x^{4}}{3 b^{3}}+\frac {\left (5 A c -3 B b \right ) x^{2}}{12 b^{2}}-\frac {A}{6 b}}{x^{6} \left (c \,x^{2}+b \right )^{2}}-\frac {10 c^{3} \ln \left (x \right ) A}{b^{6}}+\frac {6 c^{2} \ln \left (x \right ) B}{b^{5}}+\frac {5 c^{3} \ln \left (-c \,x^{2}-b \right ) A}{b^{6}}-\frac {3 c^{2} \ln \left (-c \,x^{2}-b \right ) B}{b^{5}}\)	\(159\)
parallelrisch	\(-\frac {120 A \ln \left (x \right ) x^{10} c^{5}-60 A \ln \left (c \,x^{2}+b \right ) x^{10} c^{5}-72 B \ln \left (x \right ) x^{10} b \,c^{4}+36 B \ln \left (c \,x^{2}+b \right ) x^{10} b \,c^{4}-90 A \,x^{10} c^{5}+54 B \,x^{10} b \,c^{4}+240 A \ln \left (x \right ) x^{8} b \,c^{4}-120 A \ln \left (c \,x^{2}+b \right ) x^{8} b \,c^{4}-144 B \ln \left (x \right ) x^{8} b^{2} c^{3}+72 B \ln \left (c \,x^{2}+b \right ) x^{8} b^{2} c^{3}-120 A \,x^{8} b \,c^{4}+72 B \,x^{8} b^{2} c^{3}+120 A \ln \left (x \right ) x^{6} b^{2} c^{3}-60 A \ln \left (c \,x^{2}+b \right ) x^{6} b^{2} c^{3}-72 B \ln \left (x \right ) x^{6} b^{3} c^{2}+36 B \ln \left (c \,x^{2}+b \right ) x^{6} b^{3} c^{2}+20 A \,x^{4} b^{3} c^{2}-12 B \,x^{4} b^{4} c -5 A \,x^{2} b^{4} c +3 B \,b^{5} x^{2}+2 A \,b^{5}}{12 b^{6} x^{6} \left (c \,x^{2}+b \right )^{2}}\)	\(297\)

Input:

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

Output:

-1/6*A/b^3/x^6-1/4*(-3*A*c+B*b)/b^4/x^4-3/2*c*(2*A*c-B*b)/b^5/x^2-2*c^2*(5 
*A*c-3*B*b)/b^6*ln(x)+1/2/b^6*c^3*((10*A*c-6*B*b)/c*ln(c*x^2+b)-b*(4*A*c-3 
*B*b)/c/(c*x^2+b)-1/2*b^2*(A*c-B*b)/c/(c*x^2+b)^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {12 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + 18 \, {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6} - 2 \, A b^{5} + 4 \, {\left (3 \, B b^{4} c - 5 \, A b^{3} c^{2}\right )} x^{4} - {\left (3 \, B b^{5} - 5 \, A b^{4} c\right )} x^{2} - 12 \, {\left ({\left (3 \, B b c^{4} - 5 \, A c^{5}\right )} x^{10} + 2 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6}\right )} \log \left (c x^{2} + b\right ) + 24 \, {\left ({\left (3 \, B b c^{4} - 5 \, A c^{5}\right )} x^{10} + 2 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6}\right )} \log \left (x\right )}{12 \, {\left (b^{6} c^{2} x^{10} + 2 \, b^{7} c x^{8} + b^{8} x^{6}\right )}} \] Input:

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

Output:

1/12*(12*(3*B*b^2*c^3 - 5*A*b*c^4)*x^8 + 18*(3*B*b^3*c^2 - 5*A*b^2*c^3)*x^ 
6 - 2*A*b^5 + 4*(3*B*b^4*c - 5*A*b^3*c^2)*x^4 - (3*B*b^5 - 5*A*b^4*c)*x^2 
- 12*((3*B*b*c^4 - 5*A*c^5)*x^10 + 2*(3*B*b^2*c^3 - 5*A*b*c^4)*x^8 + (3*B* 
b^3*c^2 - 5*A*b^2*c^3)*x^6)*log(c*x^2 + b) + 24*((3*B*b*c^4 - 5*A*c^5)*x^1 
0 + 2*(3*B*b^2*c^3 - 5*A*b*c^4)*x^8 + (3*B*b^3*c^2 - 5*A*b^2*c^3)*x^6)*log 
(x))/(b^6*c^2*x^10 + 2*b^7*c*x^8 + b^8*x^6)
 

Sympy [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {- 2 A b^{4} + x^{8} \left (- 60 A c^{4} + 36 B b c^{3}\right ) + x^{6} \left (- 90 A b c^{3} + 54 B b^{2} c^{2}\right ) + x^{4} \left (- 20 A b^{2} c^{2} + 12 B b^{3} c\right ) + x^{2} \cdot \left (5 A b^{3} c - 3 B b^{4}\right )}{12 b^{7} x^{6} + 24 b^{6} c x^{8} + 12 b^{5} c^{2} x^{10}} + \frac {2 c^{2} \left (- 5 A c + 3 B b\right ) \log {\left (x \right )}}{b^{6}} - \frac {c^{2} \left (- 5 A c + 3 B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{b^{6}} \] Input:

integrate((B*x**2+A)/x/(c*x**4+b*x**2)**3,x)
 

Output:

(-2*A*b**4 + x**8*(-60*A*c**4 + 36*B*b*c**3) + x**6*(-90*A*b*c**3 + 54*B*b 
**2*c**2) + x**4*(-20*A*b**2*c**2 + 12*B*b**3*c) + x**2*(5*A*b**3*c - 3*B* 
b**4))/(12*b**7*x**6 + 24*b**6*c*x**8 + 12*b**5*c**2*x**10) + 2*c**2*(-5*A 
*c + 3*B*b)*log(x)/b**6 - c**2*(-5*A*c + 3*B*b)*log(b/c + x**2)/b**6
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {12 \, {\left (3 \, B b c^{3} - 5 \, A c^{4}\right )} x^{8} + 18 \, {\left (3 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{6} - 2 \, A b^{4} + 4 \, {\left (3 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{4} - {\left (3 \, B b^{4} - 5 \, A b^{3} c\right )} x^{2}}{12 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )}} - \frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (c x^{2} + b\right )}{b^{6}} + \frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (x^{2}\right )}{b^{6}} \] Input:

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

Output:

1/12*(12*(3*B*b*c^3 - 5*A*c^4)*x^8 + 18*(3*B*b^2*c^2 - 5*A*b*c^3)*x^6 - 2* 
A*b^4 + 4*(3*B*b^3*c - 5*A*b^2*c^2)*x^4 - (3*B*b^4 - 5*A*b^3*c)*x^2)/(b^5* 
c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6) - (3*B*b*c^2 - 5*A*c^3)*log(c*x^2 + b)/b 
^6 + (3*B*b*c^2 - 5*A*c^3)*log(x^2)/b^6
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (x^{2}\right )}{b^{6}} - \frac {{\left (3 \, B b c^{3} - 5 \, A c^{4}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{b^{6} c} + \frac {18 \, B b c^{4} x^{4} - 30 \, A c^{5} x^{4} + 42 \, B b^{2} c^{3} x^{2} - 68 \, A b c^{4} x^{2} + 25 \, B b^{3} c^{2} - 39 \, A b^{2} c^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{6}} - \frac {66 \, B b c^{2} x^{6} - 110 \, A c^{3} x^{6} - 18 \, B b^{2} c x^{4} + 36 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} - 9 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, b^{6} x^{6}} \] Input:

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

Output:

(3*B*b*c^2 - 5*A*c^3)*log(x^2)/b^6 - (3*B*b*c^3 - 5*A*c^4)*log(abs(c*x^2 + 
 b))/(b^6*c) + 1/4*(18*B*b*c^4*x^4 - 30*A*c^5*x^4 + 42*B*b^2*c^3*x^2 - 68* 
A*b*c^4*x^2 + 25*B*b^3*c^2 - 39*A*b^2*c^3)/((c*x^2 + b)^2*b^6) - 1/12*(66* 
B*b*c^2*x^6 - 110*A*c^3*x^6 - 18*B*b^2*c*x^4 + 36*A*b*c^2*x^4 + 3*B*b^3*x^ 
2 - 9*A*b^2*c*x^2 + 2*A*b^3)/(b^6*x^6)
 

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {\ln \left (c\,x^2+b\right )\,\left (5\,A\,c^3-3\,B\,b\,c^2\right )}{b^6}-\frac {\frac {A}{6\,b}-\frac {x^2\,\left (5\,A\,c-3\,B\,b\right )}{12\,b^2}+\frac {3\,c^2\,x^6\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^4}+\frac {c^3\,x^8\,\left (5\,A\,c-3\,B\,b\right )}{b^5}+\frac {c\,x^4\,\left (5\,A\,c-3\,B\,b\right )}{3\,b^3}}{b^2\,x^6+2\,b\,c\,x^8+c^2\,x^{10}}-\frac {\ln \left (x\right )\,\left (10\,A\,c^3-6\,B\,b\,c^2\right )}{b^6} \] Input:

int((A + B*x^2)/(x*(b*x^2 + c*x^4)^3),x)
 

Output:

(log(b + c*x^2)*(5*A*c^3 - 3*B*b*c^2))/b^6 - (A/(6*b) - (x^2*(5*A*c - 3*B* 
b))/(12*b^2) + (3*c^2*x^6*(5*A*c - 3*B*b))/(2*b^4) + (c^3*x^8*(5*A*c - 3*B 
*b))/b^5 + (c*x^4*(5*A*c - 3*B*b))/(3*b^3))/(b^2*x^6 + c^2*x^10 + 2*b*c*x^ 
8) - (log(x)*(10*A*c^3 - 6*B*b*c^2))/b^6
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.06 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {60 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{2} c^{3} x^{6}+120 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{4} x^{8}+60 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{5} x^{10}-36 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{4} c^{2} x^{6}-72 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c^{3} x^{8}-36 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{4} x^{10}-120 \,\mathrm {log}\left (x \right ) a \,b^{2} c^{3} x^{6}-240 \,\mathrm {log}\left (x \right ) a b \,c^{4} x^{8}-120 \,\mathrm {log}\left (x \right ) a \,c^{5} x^{10}+72 \,\mathrm {log}\left (x \right ) b^{4} c^{2} x^{6}+144 \,\mathrm {log}\left (x \right ) b^{3} c^{3} x^{8}+72 \,\mathrm {log}\left (x \right ) b^{2} c^{4} x^{10}-2 a \,b^{5}+5 a \,b^{4} c \,x^{2}-20 a \,b^{3} c^{2} x^{4}-60 a \,b^{2} c^{3} x^{6}+30 a \,c^{5} x^{10}-3 b^{6} x^{2}+12 b^{5} c \,x^{4}+36 b^{4} c^{2} x^{6}-18 b^{2} c^{4} x^{10}}{12 b^{6} x^{6} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

int((B*x^2+A)/x/(c*x^4+b*x^2)^3,x)
 

Output:

(60*log(b + c*x**2)*a*b**2*c**3*x**6 + 120*log(b + c*x**2)*a*b*c**4*x**8 + 
 60*log(b + c*x**2)*a*c**5*x**10 - 36*log(b + c*x**2)*b**4*c**2*x**6 - 72* 
log(b + c*x**2)*b**3*c**3*x**8 - 36*log(b + c*x**2)*b**2*c**4*x**10 - 120* 
log(x)*a*b**2*c**3*x**6 - 240*log(x)*a*b*c**4*x**8 - 120*log(x)*a*c**5*x** 
10 + 72*log(x)*b**4*c**2*x**6 + 144*log(x)*b**3*c**3*x**8 + 72*log(x)*b**2 
*c**4*x**10 - 2*a*b**5 + 5*a*b**4*c*x**2 - 20*a*b**3*c**2*x**4 - 60*a*b**2 
*c**3*x**6 + 30*a*c**5*x**10 - 3*b**6*x**2 + 12*b**5*c*x**4 + 36*b**4*c**2 
*x**6 - 18*b**2*c**4*x**10)/(12*b**6*x**6*(b**2 + 2*b*c*x**2 + c**2*x**4))