3.1.100 \(\int \frac {(2+x) (d+e x+f x^2+g x^3)}{(4-5 x^2+x^4)^2} \, dx\) [100]

3.1.100.1 Optimal result
3.1.100.2 Mathematica [A] (verified)
3.1.100.3 Rubi [A] (verified)
3.1.100.4 Maple [A] (verified)
3.1.100.5 Fricas [B] (verification not implemented)
3.1.100.6 Sympy [F(-1)]
3.1.100.7 Maxima [A] (verification not implemented)
3.1.100.8 Giac [A] (verification not implemented)
3.1.100.9 Mupad [B] (verification not implemented)

3.1.100.1 Optimal result

Integrand size = 31, antiderivative size = 141 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {d+e+f+g}{12 (1-x)}+\frac {d+2 e+4 f+8 g}{36 (2-x)}-\frac {d-e+f-g}{36 (1+x)}+\frac {1}{36} (2 d+5 e+8 f+11 g) \log (1-x)-\frac {1}{432} (35 d+58 e+92 f+136 g) \log (2-x)+\frac {1}{108} (2 d+e-4 f+7 g) \log (1+x)+\frac {1}{144} (d-2 e+4 f-8 g) \log (2+x) \]

output
1/12*(d+e+f+g)/(1-x)+1/36*(d+2*e+4*f+8*g)/(2-x)+1/36*(-d+e-f+g)/(1+x)+1/36 
*(2*d+5*e+8*f+11*g)*ln(1-x)-1/432*(35*d+58*e+92*f+136*g)*ln(2-x)+1/108*(2* 
d+e-4*f+7*g)*ln(1+x)+1/144*(d-2*e+4*f-8*g)*ln(2+x)
 
3.1.100.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{432} \left (\frac {12 \left (d \left (5+6 x-5 x^2\right )+2 \left (g \left (8-5 x^2\right )+f \left (4+3 x-4 x^2\right )+e \left (5-2 x^2\right )\right )\right )}{2-x-2 x^2+x^3}+12 (2 d+5 e+8 f+11 g) \log (1-x)-(35 d+58 e+92 f+136 g) \log (2-x)+4 (2 d+e-4 f+7 g) \log (1+x)+3 (d-2 e+4 f-8 g) \log (2+x)\right ) \]

input
Integrate[((2 + x)*(d + e*x + f*x^2 + g*x^3))/(4 - 5*x^2 + x^4)^2,x]
 
output
((12*(d*(5 + 6*x - 5*x^2) + 2*(g*(8 - 5*x^2) + f*(4 + 3*x - 4*x^2) + e*(5 
- 2*x^2))))/(2 - x - 2*x^2 + x^3) + 12*(2*d + 5*e + 8*f + 11*g)*Log[1 - x] 
 - (35*d + 58*e + 92*f + 136*g)*Log[2 - x] + 4*(2*d + e - 4*f + 7*g)*Log[1 
 + x] + 3*(d - 2*e + 4*f - 8*g)*Log[2 + x])/432
 
3.1.100.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2019, 2462, 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.

\(\displaystyle \int \frac {(x+2) \left (d+e x+f x^2+g x^3\right )}{\left (x^4-5 x^2+4\right )^2} \, dx\)

\(\Big \downarrow \) 2019

\(\displaystyle \int \frac {d+e x+f x^2+g x^3}{(x+2) \left (x^3-2 x^2-x+2\right )^2}dx\)

\(\Big \downarrow \) 2462

\(\displaystyle \int \left (\frac {-35 d-58 e-92 f-136 g}{432 (x-2)}+\frac {2 d+5 e+8 f+11 g}{36 (x-1)}+\frac {2 d+e-4 f+7 g}{108 (x+1)}+\frac {d-2 e+4 f-8 g}{144 (x+2)}+\frac {d+2 e+4 f+8 g}{36 (x-2)^2}+\frac {d+e+f+g}{12 (x-1)^2}+\frac {d-e+f-g}{36 (x+1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d-e+f-g}{36 (x+1)}+\frac {d+e+f+g}{12 (1-x)}+\frac {d+2 e+4 f+8 g}{36 (2-x)}+\frac {1}{36} \log (1-x) (2 d+5 e+8 f+11 g)-\frac {1}{432} \log (2-x) (35 d+58 e+92 f+136 g)+\frac {1}{108} \log (x+1) (2 d+e-4 f+7 g)+\frac {1}{144} \log (x+2) (d-2 e+4 f-8 g)\)

input
Int[((2 + x)*(d + e*x + f*x^2 + g*x^3))/(4 - 5*x^2 + x^4)^2,x]
 
output
(d + e + f + g)/(12*(1 - x)) + (d + 2*e + 4*f + 8*g)/(36*(2 - x)) - (d - e 
 + f - g)/(36*(1 + x)) + ((2*d + 5*e + 8*f + 11*g)*Log[1 - x])/36 - ((35*d 
 + 58*e + 92*f + 136*g)*Log[2 - x])/432 + ((2*d + e - 4*f + 7*g)*Log[1 + x 
])/108 + ((d - 2*e + 4*f - 8*g)*Log[2 + x])/144
 

3.1.100.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2019
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px 
, Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && 
 EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
 

rule 2462
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u*Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ 
[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 0 
] && RationalFunctionQ[u, x]
 
3.1.100.4 Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.95

method result size
default \(\left (\frac {d}{144}-\frac {e}{72}+\frac {f}{36}-\frac {g}{18}\right ) \ln \left (x +2\right )-\frac {\frac {d}{36}-\frac {e}{36}+\frac {f}{36}-\frac {g}{36}}{x +1}+\left (\frac {d}{54}+\frac {e}{108}-\frac {f}{27}+\frac {7 g}{108}\right ) \ln \left (x +1\right )-\frac {\frac {d}{12}+\frac {e}{12}+\frac {f}{12}+\frac {g}{12}}{x -1}+\left (\frac {d}{18}+\frac {5 e}{36}+\frac {2 f}{9}+\frac {11 g}{36}\right ) \ln \left (x -1\right )+\left (-\frac {35 d}{432}-\frac {29 e}{216}-\frac {23 f}{108}-\frac {17 g}{54}\right ) \ln \left (x -2\right )-\frac {\frac {d}{36}+\frac {e}{18}+\frac {f}{9}+\frac {2 g}{9}}{x -2}\) \(134\)
norman \(\frac {\left (-\frac {5 d}{36}-\frac {e}{9}-\frac {2 f}{9}-\frac {5 g}{18}\right ) x^{3}+\left (\frac {17 d}{36}+\frac {5 e}{18}+\frac {5 f}{9}+\frac {4 g}{9}\right ) x +\left (-\frac {d}{9}-\frac {2 e}{9}-\frac {5 f}{18}-\frac {5 g}{9}\right ) x^{2}+\frac {5 d}{18}+\frac {5 e}{9}+\frac {4 f}{9}+\frac {8 g}{9}}{x^{4}-5 x^{2}+4}+\left (-\frac {35 d}{432}-\frac {29 e}{216}-\frac {23 f}{108}-\frac {17 g}{54}\right ) \ln \left (x -2\right )+\left (\frac {d}{18}+\frac {5 e}{36}+\frac {2 f}{9}+\frac {11 g}{36}\right ) \ln \left (x -1\right )+\left (\frac {d}{54}+\frac {e}{108}-\frac {f}{27}+\frac {7 g}{108}\right ) \ln \left (x +1\right )+\left (\frac {d}{144}-\frac {e}{72}+\frac {f}{36}-\frac {g}{18}\right ) \ln \left (x +2\right )\) \(149\)
risch \(\frac {\left (-\frac {5 d}{36}-\frac {e}{9}-\frac {2 f}{9}-\frac {5 g}{18}\right ) x^{2}+\left (\frac {d}{6}+\frac {f}{6}\right ) x +\frac {5 d}{36}+\frac {5 e}{18}+\frac {2 f}{9}+\frac {4 g}{9}}{x^{3}-2 x^{2}-x +2}+\frac {\ln \left (x +2\right ) d}{144}-\frac {\ln \left (x +2\right ) e}{72}+\frac {\ln \left (x +2\right ) f}{36}-\frac {\ln \left (x +2\right ) g}{18}+\frac {\ln \left (-x -1\right ) d}{54}+\frac {\ln \left (-x -1\right ) e}{108}-\frac {\ln \left (-x -1\right ) f}{27}+\frac {7 \ln \left (-x -1\right ) g}{108}-\frac {35 \ln \left (2-x \right ) d}{432}-\frac {29 \ln \left (2-x \right ) e}{216}-\frac {23 \ln \left (2-x \right ) f}{108}-\frac {17 \ln \left (2-x \right ) g}{54}+\frac {\ln \left (x -1\right ) d}{18}+\frac {5 \ln \left (x -1\right ) e}{36}+\frac {2 \ln \left (x -1\right ) f}{9}+\frac {11 \ln \left (x -1\right ) g}{36}\) \(185\)
parallelrisch \(-\frac {-96 f -192 g +60 d \,x^{2}-60 d -120 e +96 f \,x^{2}+120 g \,x^{2}-72 d x -60 \ln \left (x -1\right ) x^{3} e -8 \ln \left (x +1\right ) x^{3} d -4 \ln \left (x +1\right ) x^{3} e -3 \ln \left (x +2\right ) x^{3} d +6 \ln \left (x +2\right ) x^{3} e +70 \ln \left (x -2\right ) d +116 \ln \left (x -2\right ) e -48 \ln \left (x -1\right ) d -120 \ln \left (x -1\right ) e -136 \ln \left (x -2\right ) x g +132 \ln \left (x -1\right ) x g +28 \ln \left (x +1\right ) x g -24 \ln \left (x +2\right ) x g -24 \ln \left (x +2\right ) f +32 \ln \left (x +1\right ) f +48 e \,x^{2}+58 \ln \left (x -2\right ) x^{3} e -24 \ln \left (x -1\right ) x^{3} d -6 \ln \left (x +2\right ) x e -58 \ln \left (x -2\right ) x e +24 \ln \left (x -1\right ) x d +60 \ln \left (x -1\right ) x e +8 \ln \left (x +1\right ) x d +4 \ln \left (x +1\right ) x e +3 \ln \left (x +2\right ) x d -116 \ln \left (x -2\right ) x^{2} e +48 \ln \left (x -1\right ) x^{2} d +120 \ln \left (x -1\right ) x^{2} e +16 \ln \left (x +1\right ) x^{2} d +8 \ln \left (x +1\right ) x^{2} e +6 \ln \left (x +2\right ) x^{2} d -12 \ln \left (x +2\right ) x^{2} e -6 \ln \left (x +2\right ) d +136 \ln \left (x -2\right ) x^{3} g -92 \ln \left (x -2\right ) x f +96 \ln \left (x -1\right ) x f -16 \ln \left (x +1\right ) x f +12 \ln \left (x +2\right ) x f +12 \ln \left (x +2\right ) e -16 \ln \left (x +1\right ) d -8 \ln \left (x +1\right ) e -35 \ln \left (x -2\right ) x d -56 \ln \left (x +1\right ) g +48 \ln \left (x +2\right ) g -272 \ln \left (x -2\right ) x^{2} g +264 \ln \left (x -1\right ) x^{2} g +56 \ln \left (x +1\right ) x^{2} g -48 \ln \left (x +2\right ) x^{2} g +35 \ln \left (x -2\right ) x^{3} d +272 \ln \left (x -2\right ) g -264 \ln \left (x -1\right ) g -184 \ln \left (x -2\right ) x^{2} f +192 \ln \left (x -1\right ) x^{2} f -32 \ln \left (x +1\right ) x^{2} f +24 \ln \left (x +2\right ) x^{2} f +92 \ln \left (x -2\right ) x^{3} f -96 \ln \left (x -1\right ) x^{3} f +16 \ln \left (x +1\right ) x^{3} f -12 \ln \left (x +2\right ) x^{3} f -132 \ln \left (x -1\right ) x^{3} g -28 \ln \left (x +1\right ) x^{3} g +24 \ln \left (x +2\right ) x^{3} g +184 \ln \left (x -2\right ) f -192 \ln \left (x -1\right ) f -70 \ln \left (x -2\right ) x^{2} d -72 f x}{432 \left (x^{3}-2 x^{2}-x +2\right )}\) \(623\)

input
int((x+2)*(g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x,method=_RETURNVERBOSE)
 
output
(1/144*d-1/72*e+1/36*f-1/18*g)*ln(x+2)-(1/36*d-1/36*e+1/36*f-1/36*g)/(x+1) 
+(1/54*d+1/108*e-1/27*f+7/108*g)*ln(x+1)-(1/12*d+1/12*e+1/12*f+1/12*g)/(x- 
1)+(1/18*d+5/36*e+2/9*f+11/36*g)*ln(x-1)+(-35/432*d-29/216*e-23/108*f-17/5 
4*g)*ln(x-2)-(1/36*d+1/18*e+1/9*f+2/9*g)/(x-2)
 
3.1.100.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (123) = 246\).

Time = 0.78 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.28 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {12 \, {\left (5 \, d + 4 \, e + 8 \, f + 10 \, g\right )} x^{2} - 72 \, {\left (d + f\right )} x - 3 \, {\left ({\left (d - 2 \, e + 4 \, f - 8 \, g\right )} x^{3} - 2 \, {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} x^{2} - {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} x + 2 \, d - 4 \, e + 8 \, f - 16 \, g\right )} \log \left (x + 2\right ) - 4 \, {\left ({\left (2 \, d + e - 4 \, f + 7 \, g\right )} x^{3} - 2 \, {\left (2 \, d + e - 4 \, f + 7 \, g\right )} x^{2} - {\left (2 \, d + e - 4 \, f + 7 \, g\right )} x + 4 \, d + 2 \, e - 8 \, f + 14 \, g\right )} \log \left (x + 1\right ) - 12 \, {\left ({\left (2 \, d + 5 \, e + 8 \, f + 11 \, g\right )} x^{3} - 2 \, {\left (2 \, d + 5 \, e + 8 \, f + 11 \, g\right )} x^{2} - {\left (2 \, d + 5 \, e + 8 \, f + 11 \, g\right )} x + 4 \, d + 10 \, e + 16 \, f + 22 \, g\right )} \log \left (x - 1\right ) + {\left ({\left (35 \, d + 58 \, e + 92 \, f + 136 \, g\right )} x^{3} - 2 \, {\left (35 \, d + 58 \, e + 92 \, f + 136 \, g\right )} x^{2} - {\left (35 \, d + 58 \, e + 92 \, f + 136 \, g\right )} x + 70 \, d + 116 \, e + 184 \, f + 272 \, g\right )} \log \left (x - 2\right ) - 60 \, d - 120 \, e - 96 \, f - 192 \, g}{432 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \]

input
integrate((2+x)*(g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="fricas")
 
output
-1/432*(12*(5*d + 4*e + 8*f + 10*g)*x^2 - 72*(d + f)*x - 3*((d - 2*e + 4*f 
 - 8*g)*x^3 - 2*(d - 2*e + 4*f - 8*g)*x^2 - (d - 2*e + 4*f - 8*g)*x + 2*d 
- 4*e + 8*f - 16*g)*log(x + 2) - 4*((2*d + e - 4*f + 7*g)*x^3 - 2*(2*d + e 
 - 4*f + 7*g)*x^2 - (2*d + e - 4*f + 7*g)*x + 4*d + 2*e - 8*f + 14*g)*log( 
x + 1) - 12*((2*d + 5*e + 8*f + 11*g)*x^3 - 2*(2*d + 5*e + 8*f + 11*g)*x^2 
 - (2*d + 5*e + 8*f + 11*g)*x + 4*d + 10*e + 16*f + 22*g)*log(x - 1) + ((3 
5*d + 58*e + 92*f + 136*g)*x^3 - 2*(35*d + 58*e + 92*f + 136*g)*x^2 - (35* 
d + 58*e + 92*f + 136*g)*x + 70*d + 116*e + 184*f + 272*g)*log(x - 2) - 60 
*d - 120*e - 96*f - 192*g)/(x^3 - 2*x^2 - x + 2)
 
3.1.100.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\text {Timed out} \]

input
integrate((2+x)*(g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4)**2,x)
 
output
Timed out
 
3.1.100.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.89 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \, {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) + \frac {1}{108} \, {\left (2 \, d + e - 4 \, f + 7 \, g\right )} \log \left (x + 1\right ) + \frac {1}{36} \, {\left (2 \, d + 5 \, e + 8 \, f + 11 \, g\right )} \log \left (x - 1\right ) - \frac {1}{432} \, {\left (35 \, d + 58 \, e + 92 \, f + 136 \, g\right )} \log \left (x - 2\right ) - \frac {{\left (5 \, d + 4 \, e + 8 \, f + 10 \, g\right )} x^{2} - 6 \, {\left (d + f\right )} x - 5 \, d - 10 \, e - 8 \, f - 16 \, g}{36 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \]

input
integrate((2+x)*(g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="maxima")
 
output
1/144*(d - 2*e + 4*f - 8*g)*log(x + 2) + 1/108*(2*d + e - 4*f + 7*g)*log(x 
 + 1) + 1/36*(2*d + 5*e + 8*f + 11*g)*log(x - 1) - 1/432*(35*d + 58*e + 92 
*f + 136*g)*log(x - 2) - 1/36*((5*d + 4*e + 8*f + 10*g)*x^2 - 6*(d + f)*x 
- 5*d - 10*e - 8*f - 16*g)/(x^3 - 2*x^2 - x + 2)
 
3.1.100.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.92 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \, {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{108} \, {\left (2 \, d + e - 4 \, f + 7 \, g\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{36} \, {\left (2 \, d + 5 \, e + 8 \, f + 11 \, g\right )} \log \left ({\left | x - 1 \right |}\right ) - \frac {1}{432} \, {\left (35 \, d + 58 \, e + 92 \, f + 136 \, g\right )} \log \left ({\left | x - 2 \right |}\right ) - \frac {{\left (5 \, d + 4 \, e + 8 \, f + 10 \, g\right )} x^{2} - 6 \, {\left (d + f\right )} x - 5 \, d - 10 \, e - 8 \, f - 16 \, g}{36 \, {\left (x + 1\right )} {\left (x - 1\right )} {\left (x - 2\right )}} \]

input
integrate((2+x)*(g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="giac")
 
output
1/144*(d - 2*e + 4*f - 8*g)*log(abs(x + 2)) + 1/108*(2*d + e - 4*f + 7*g)* 
log(abs(x + 1)) + 1/36*(2*d + 5*e + 8*f + 11*g)*log(abs(x - 1)) - 1/432*(3 
5*d + 58*e + 92*f + 136*g)*log(abs(x - 2)) - 1/36*((5*d + 4*e + 8*f + 10*g 
)*x^2 - 6*(d + f)*x - 5*d - 10*e - 8*f - 16*g)/((x + 1)*(x - 1)*(x - 2))
 
3.1.100.9 Mupad [B] (verification not implemented)

Time = 8.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{\left (4-5 x^2+x^4\right )^2} \, dx=\ln \left (x-1\right )\,\left (\frac {d}{18}+\frac {5\,e}{36}+\frac {2\,f}{9}+\frac {11\,g}{36}\right )+\ln \left (x+2\right )\,\left (\frac {d}{144}-\frac {e}{72}+\frac {f}{36}-\frac {g}{18}\right )+\ln \left (x+1\right )\,\left (\frac {d}{54}+\frac {e}{108}-\frac {f}{27}+\frac {7\,g}{108}\right )-\ln \left (x-2\right )\,\left (\frac {35\,d}{432}+\frac {29\,e}{216}+\frac {23\,f}{108}+\frac {17\,g}{54}\right )-\frac {\left (-\frac {5\,d}{36}-\frac {e}{9}-\frac {2\,f}{9}-\frac {5\,g}{18}\right )\,x^2+\left (\frac {d}{6}+\frac {f}{6}\right )\,x+\frac {5\,d}{36}+\frac {5\,e}{18}+\frac {2\,f}{9}+\frac {4\,g}{9}}{-x^3+2\,x^2+x-2} \]

input
int(((x + 2)*(d + e*x + f*x^2 + g*x^3))/(x^4 - 5*x^2 + 4)^2,x)
 
output
log(x - 1)*(d/18 + (5*e)/36 + (2*f)/9 + (11*g)/36) + log(x + 2)*(d/144 - e 
/72 + f/36 - g/18) + log(x + 1)*(d/54 + e/108 - f/27 + (7*g)/108) - log(x 
- 2)*((35*d)/432 + (29*e)/216 + (23*f)/108 + (17*g)/54) - ((5*d)/36 + (5*e 
)/18 + (2*f)/9 + (4*g)/9 - x^2*((5*d)/36 + e/9 + (2*f)/9 + (5*g)/18) + x*( 
d/6 + f/6))/(x + 2*x^2 - x^3 - 2)