\(\int \frac {A+B x+C x^2+D x^3}{x^2 (a+b x^2)^3} \, dx\) [59]

Optimal result

Integrand size = 28, antiderivative size = 144 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}+\frac {a (b B-a D)-b (A b-a C) x}{4 a^2 b \left (a+b x^2\right )^2}+\frac {4 a B-(7 A b-3 a C) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {B \log (x)}{a^3}-\frac {B \log \left (a+b x^2\right )}{2 a^3} \] Output:

-A/a^3/x+1/4*(a*(B*b-D*a)-b*(A*b-C*a)*x)/a^2/b/(b*x^2+a)^2+1/8*(4*B*a-(7*A 
*b-3*C*a)*x)/a^3/(b*x^2+a)-3/8*(5*A*b-C*a)*arctan(b^(1/2)*x/a^(1/2))/a^(7/ 
2)/b^(1/2)+B*ln(x)/a^3-1/2*B*ln(b*x^2+a)/a^3
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {A}{a^3 x}+\frac {a b B-a^2 D-A b^2 x+a b C x}{4 a^2 b \left (a+b x^2\right )^2}+\frac {4 a B-7 A b x+3 a C x}{8 a^3 \left (a+b x^2\right )}+\frac {3 (-5 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {B \log (x)}{a^3}-\frac {B \log \left (a+b x^2\right )}{2 a^3} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/(x^2*(a + b*x^2)^3),x]
 

Output:

-(A/(a^3*x)) + (a*b*B - a^2*D - A*b^2*x + a*b*C*x)/(4*a^2*b*(a + b*x^2)^2) 
 + (4*a*B - 7*A*b*x + 3*a*C*x)/(8*a^3*(a + b*x^2)) + (3*(-5*A*b + a*C)*Arc 
Tan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]) + (B*Log[x])/a^3 - (B*Log[a 
+ b*x^2])/(2*a^3)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2336, 25, 2336, 25, 2333, 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.

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

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}-\frac {\int -\frac {-3 \left (\frac {A b}{a}-C\right ) x^2+4 B x+4 A}{x^2 \left (b x^2+a\right )^2}dx}{4 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {-3 \left (\frac {A b}{a}-C\right ) x^2+4 B x+4 A}{x^2 \left (b x^2+a\right )^2}dx}{4 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{2 a \left (a+b x^2\right )}-\frac {\int -\frac {-\left (\left (\frac {7 A b}{a}-3 C\right ) x^2\right )+8 B x+8 A}{x^2 \left (b x^2+a\right )}dx}{2 a}}{4 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-\left (\left (\frac {7 A b}{a}-3 C\right ) x^2\right )+8 B x+8 A}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{2 a \left (a+b x^2\right )}}{4 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {\int \left (\frac {8 A}{a x^2}+\frac {8 B}{a x}+\frac {-15 A b-8 B x b+3 a C}{a \left (b x^2+a\right )}\right )dx}{2 a}+\frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{2 a \left (a+b x^2\right )}}{4 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {-\frac {3 (5 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}-\frac {8 A}{a x}-\frac {4 B \log \left (a+b x^2\right )}{a}+\frac {8 B \log (x)}{a}}{2 a}+\frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{2 a \left (a+b x^2\right )}}{4 a}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/(x^2*(a + b*x^2)^3),x]
 

Output:

(b*B - a*D - b*((A*b)/a - C)*x)/(4*a*b*(a + b*x^2)^2) + ((4*B - ((7*A*b)/a 
 - 3*C)*x)/(2*a*(a + b*x^2)) + ((-8*A)/(a*x) - (3*(5*A*b - a*C)*ArcTan[(Sq 
rt[b]*x)/Sqrt[a]])/(a^(3/2)*Sqrt[b]) + (8*B*Log[x])/a - (4*B*Log[a + b*x^2 
])/a)/(2*a))/(4*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} C a b \right ) x^{3}-\frac {B a b \,x^{2}}{2}+\frac {a \left (9 A b -5 C a \right ) x}{8}-\frac {a^{2} \left (3 B b -D a \right )}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {B \ln \left (b \,x^{2}+a \right )}{2}+\frac {\left (15 A b -3 C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{3}}-\frac {A}{a^{3} x}+\frac {B \ln \left (x \right )}{a^{3}}\)	\(125\)

Input:

int((D*x^3+C*x^2+B*x+A)/x^2/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/a^3*(((7/8*b^2*A-3/8*C*a*b)*x^3-1/2*B*a*b*x^2+1/8*a*(9*A*b-5*C*a)*x-1/4 
*a^2*(3*B*b-D*a)/b)/(b*x^2+a)^2+1/2*B*ln(b*x^2+a)+1/8*(15*A*b-3*C*a)/(a*b) 
^(1/2)*arctan(b*x/(a*b)^(1/2)))-A/a^3/x+B*ln(x)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 524, normalized size of antiderivative = 3.64 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=\left [\frac {8 \, B a^{2} b^{2} x^{3} - 16 \, A a^{3} b + 6 \, {\left (C a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 10 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (C a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (C a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 4 \, {\left (D a^{4} - 3 \, B a^{3} b\right )} x - 8 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (b x^{2} + a\right ) + 16 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (x\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, \frac {4 \, B a^{2} b^{2} x^{3} - 8 \, A a^{3} b + 3 \, {\left (C a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 5 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (C a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (C a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 2 \, {\left (D a^{4} - 3 \, B a^{3} b\right )} x - 4 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (b x^{2} + a\right ) + 8 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (x\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^2/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

[1/16*(8*B*a^2*b^2*x^3 - 16*A*a^3*b + 6*(C*a^2*b^2 - 5*A*a*b^3)*x^4 + 10*( 
C*a^3*b - 5*A*a^2*b^2)*x^2 + 3*((C*a*b^2 - 5*A*b^3)*x^5 + 2*(C*a^2*b - 5*A 
*a*b^2)*x^3 + (C*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)* 
x - a)/(b*x^2 + a)) - 4*(D*a^4 - 3*B*a^3*b)*x - 8*(B*a*b^3*x^5 + 2*B*a^2*b 
^2*x^3 + B*a^3*b*x)*log(b*x^2 + a) + 16*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B 
*a^3*b*x)*log(x))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x), 1/8*(4*B*a^2*b^ 
2*x^3 - 8*A*a^3*b + 3*(C*a^2*b^2 - 5*A*a*b^3)*x^4 + 5*(C*a^3*b - 5*A*a^2*b 
^2)*x^2 + 3*((C*a*b^2 - 5*A*b^3)*x^5 + 2*(C*a^2*b - 5*A*a*b^2)*x^3 + (C*a^ 
3 - 5*A*a^2*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - 2*(D*a^4 - 3*B*a^3*b)* 
x - 4*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B*a^3*b*x)*log(b*x^2 + a) + 8*(B*a* 
b^3*x^5 + 2*B*a^2*b^2*x^3 + B*a^3*b*x)*log(x))/(a^4*b^3*x^5 + 2*a^5*b^2*x^ 
3 + a^6*b*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((D*x**3+C*x**2+B*x+A)/x**2/(b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {4 \, B a b^{2} x^{3} + 3 \, {\left (C a b^{2} - 5 \, A b^{3}\right )} x^{4} - 8 \, A a^{2} b + 5 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 2 \, {\left (D a^{3} - 3 \, B a^{2} b\right )} x}{8 \, {\left (a^{3} b^{3} x^{5} + 2 \, a^{4} b^{2} x^{3} + a^{5} b x\right )}} - \frac {B \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {B \log \left (x\right )}{a^{3}} + \frac {3 \, {\left (C a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^2/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

1/8*(4*B*a*b^2*x^3 + 3*(C*a*b^2 - 5*A*b^3)*x^4 - 8*A*a^2*b + 5*(C*a^2*b - 
5*A*a*b^2)*x^2 - 2*(D*a^3 - 3*B*a^2*b)*x)/(a^3*b^3*x^5 + 2*a^4*b^2*x^3 + a 
^5*b*x) - 1/2*B*log(b*x^2 + a)/a^3 + B*log(x)/a^3 + 3/8*(C*a - 5*A*b)*arct 
an(b*x/sqrt(a*b))/(sqrt(a*b)*a^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=-\frac {B \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {B \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {3 \, {\left (C a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {4 \, B a b^{2} x^{3} + 3 \, {\left (C a b^{2} - 5 \, A b^{3}\right )} x^{4} - 8 \, A a^{2} b + 5 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 2 \, {\left (D a^{3} - 3 \, B a^{2} b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b x} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^2/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

-1/2*B*log(b*x^2 + a)/a^3 + B*log(abs(x))/a^3 + 3/8*(C*a - 5*A*b)*arctan(b 
*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/8*(4*B*a*b^2*x^3 + 3*(C*a*b^2 - 5*A*b^3) 
*x^4 - 8*A*a^2*b + 5*(C*a^2*b - 5*A*a*b^2)*x^2 - 2*(D*a^3 - 3*B*a^2*b)*x)/ 
((b*x^2 + a)^2*a^3*b*x)
 

Mupad [B] (verification not implemented)

Time = 2.56 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.40 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {\frac {3\,B}{4\,a}+\frac {B\,b\,x^2}{2\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {5\,C\,x}{8\,a}+\frac {3\,C\,b\,x^3}{8\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {A}{a}+\frac {25\,A\,b\,x^2}{8\,a^2}+\frac {15\,A\,b^2\,x^4}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {D}{4\,b\,{\left (b\,x^2+a\right )}^2}-\frac {B\,\ln \left (b\,x^2+a\right )}{2\,a^3}+\frac {B\,\ln \left (x\right )}{a^3}-\frac {15\,A\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}}+\frac {3\,C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {b}} \] Input:

int((A + B*x + C*x^2 + x^3*D)/(x^2*(a + b*x^2)^3),x)
 

Output:

((3*B)/(4*a) + (B*b*x^2)/(2*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) + ((5*C*x)/( 
8*a) + (3*C*b*x^3)/(8*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - (A/a + (25*A*b*x 
^2)/(8*a^2) + (15*A*b^2*x^4)/(8*a^3))/(a^2*x + b^2*x^5 + 2*a*b*x^3) - D/(4 
*b*(a + b*x^2)^2) - (B*log(a + b*x^2))/(2*a^3) + (B*log(x))/a^3 - (15*A*b^ 
(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(7/2)) + (3*C*atan((b^(1/2)*x)/a^(1/ 
2)))/(8*a^(5/2)*b^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.28 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b x +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} c x -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{3}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b c \,x^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{5}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c \,x^{5}-4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} x -8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} x^{3}-4 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{4} x^{5}+8 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x +16 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}+8 \,\mathrm {log}\left (x \right ) b^{4} x^{5}-8 a^{3} b -2 a^{3} d x -25 a^{2} b^{2} x^{2}+4 a^{2} b^{2} x +5 a^{2} b c \,x^{2}-15 a \,b^{3} x^{4}+3 a \,b^{2} c \,x^{4}-2 b^{4} x^{5}}{8 a^{3} b x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((D*x^3+C*x^2+B*x+A)/x^2/(b*x^2+a)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*x + 3*sqrt(b)* 
sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*c*x - 30*sqrt(b)*sqrt(a)*atan(( 
b*x)/(sqrt(b)*sqrt(a)))*a*b**2*x**3 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a*b*c*x**3 - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))* 
b**3*x**5 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*c*x**5 - 
4*log(a + b*x**2)*a**2*b**2*x - 8*log(a + b*x**2)*a*b**3*x**3 - 4*log(a + 
b*x**2)*b**4*x**5 + 8*log(x)*a**2*b**2*x + 16*log(x)*a*b**3*x**3 + 8*log(x 
)*b**4*x**5 - 8*a**3*b - 2*a**3*d*x - 25*a**2*b**2*x**2 + 4*a**2*b**2*x + 
5*a**2*b*c*x**2 - 15*a*b**3*x**4 + 3*a*b**2*c*x**4 - 2*b**4*x**5)/(8*a**3* 
b*x*(a**2 + 2*a*b*x**2 + b**2*x**4))