\(\int \frac {(f+g x)^2}{(d+e x) (d^2-e^2 x^2)^2} \, dx\) [565]

Optimal result

Integrand size = 29, antiderivative size = 121 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {(e f+d g)^2}{8 d^3 e^3 (d-e x)}-\frac {(e f-d g)^2}{8 d^2 e^3 (d+e x)^2}-\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d+e x)}+\frac {(3 e f-d g) (e f+d g) \text {arctanh}\left (\frac {e x}{d}\right )}{8 d^4 e^3} \]

[Out]

1/8*(d*g+e*f)^2/d^3/e^3/(-e*x+d)-1/8*(-d*g+e*f)^2/d^2/e^3/(e*x+d)^2+1/4*(d^2*g^2-e^2*f^2)/d^3/e^3/(e*x+d)+1/8*
(-d*g+3*e*f)*(d*g+e*f)*arctanh(e*x/d)/d^4/e^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {862, 90, 214} \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right ) (3 e f-d g) (d g+e f)}{8 d^4 e^3}+\frac {(d g+e f)^2}{8 d^3 e^3 (d-e x)}-\frac {(e f-d g)^2}{8 d^2 e^3 (d+e x)^2}-\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d+e x)} \]

[In]

Int[(f + g*x)^2/((d + e*x)*(d^2 - e^2*x^2)^2),x]

[Out]

(e*f + d*g)^2/(8*d^3*e^3*(d - e*x)) - (e*f - d*g)^2/(8*d^2*e^3*(d + e*x)^2) - (e^2*f^2 - d^2*g^2)/(4*d^3*e^3*(
d + e*x)) + ((3*e*f - d*g)*(e*f + d*g)*ArcTanh[(e*x)/d])/(8*d^4*e^3)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 214

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 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {(f+g x)^2}{(d-e x)^2 (d+e x)^3} \, dx \\ & = \int \left (\frac {(e f+d g)^2}{8 d^3 e^2 (d-e x)^2}+\frac {(-e f+d g)^2}{4 d^2 e^2 (d+e x)^3}+\frac {e^2 f^2-d^2 g^2}{4 d^3 e^2 (d+e x)^2}+\frac {(3 e f-d g) (e f+d g)}{8 d^3 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx \\ & = \frac {(e f+d g)^2}{8 d^3 e^3 (d-e x)}-\frac {(e f-d g)^2}{8 d^2 e^3 (d+e x)^2}-\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d+e x)}+\frac {((3 e f-d g) (e f+d g)) \int \frac {1}{d^2-e^2 x^2} \, dx}{8 d^3 e^2} \\ & = \frac {(e f+d g)^2}{8 d^3 e^3 (d-e x)}-\frac {(e f-d g)^2}{8 d^2 e^3 (d+e x)^2}-\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d+e x)}+\frac {(3 e f-d g) (e f+d g) \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^4 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.15 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {\frac {2 d (e f+d g)^2}{d-e x}-\frac {2 d^2 (e f-d g)^2}{(d+e x)^2}+\frac {4 d \left (-e^2 f^2+d^2 g^2\right )}{d+e x}+\left (-3 e^2 f^2-2 d e f g+d^2 g^2\right ) \log (d-e x)+\left (3 e^2 f^2+2 d e f g-d^2 g^2\right ) \log (d+e x)}{16 d^4 e^3} \]

[In]

Integrate[(f + g*x)^2/((d + e*x)*(d^2 - e^2*x^2)^2),x]

[Out]

((2*d*(e*f + d*g)^2)/(d - e*x) - (2*d^2*(e*f - d*g)^2)/(d + e*x)^2 + (4*d*(-(e^2*f^2) + d^2*g^2))/(d + e*x) +
(-3*e^2*f^2 - 2*d*e*f*g + d^2*g^2)*Log[d - e*x] + (3*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*Log[d + e*x])/(16*d^4*e^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\left (d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}+\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{8 e^{3} d^{3} \left (-e x +d \right )}-\frac {-d^{2} g^{2}+e^{2} f^{2}}{4 e^{3} d^{3} \left (e x +d \right )}+\frac {\left (-d^{2} g^{2}+2 d e f g +3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{8 d^{2} e^{3} \left (e x +d \right )^{2}}\)	\(180\)
norman	\(\frac {-\frac {-d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{4 d \,e^{3}}-\frac {\left (-3 d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) x}{8 e^{2} d^{2}}+\frac {\left (-d^{2} g^{2}+2 d e f g +3 e^{2} f^{2}\right ) x^{2}}{8 d^{3} e}}{\left (-e x +d \right ) \left (e x +d \right )^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}-\frac {\left (d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}\)	\(188\)
risch	\(\frac {-\frac {\left (d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) x^{2}}{8 e \,d^{3}}+\frac {\left (3 d^{2} g^{2}+2 d e f g +3 e^{2} f^{2}\right ) x}{8 d^{2} e^{2}}+\frac {d^{2} g^{2}+2 d e f g -e^{2} f^{2}}{4 d \,e^{3}}}{\left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )}+\frac {\ln \left (e x -d \right ) g^{2}}{16 e^{3} d^{2}}-\frac {\ln \left (e x -d \right ) f g}{8 e^{2} d^{3}}-\frac {3 \ln \left (e x -d \right ) f^{2}}{16 e \,d^{4}}-\frac {\ln \left (-e x -d \right ) g^{2}}{16 e^{3} d^{2}}+\frac {\ln \left (-e x -d \right ) f g}{8 e^{2} d^{3}}+\frac {3 \ln \left (-e x -d \right ) f^{2}}{16 e \,d^{4}}\)	\(235\)
parallelrisch	\(\frac {-2 \ln \left (e x +d \right ) d^{4} e f g +\ln \left (e x -d \right ) x^{3} d^{2} e^{3} g^{2}-2 \ln \left (e x +d \right ) x \,d^{3} e^{2} f g +2 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g -2 \ln \left (e x -d \right ) x^{2} d^{2} e^{3} f g +2 \ln \left (e x +d \right ) x^{2} d^{2} e^{3} f g +2 \ln \left (e x +d \right ) x^{3} d \,e^{4} f g -2 \ln \left (e x -d \right ) x^{3} d \,e^{4} f g -3 \ln \left (e x +d \right ) x \,d^{2} e^{3} f^{2}-4 g^{2} d^{5}-6 x \,d^{4} e \,g^{2}-6 x \,d^{2} e^{3} f^{2}+2 x^{2} d^{3} e^{2} g^{2}-6 x^{2} d \,e^{4} f^{2}+3 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}-8 f g \,d^{4} e +4 f^{2} d^{3} e^{2}-3 \ln \left (e x -d \right ) x^{3} e^{5} f^{2}-4 x^{2} d^{2} e^{3} f g +2 \ln \left (e x -d \right ) d^{4} e f g -4 x \,d^{3} e^{2} f g +\ln \left (e x +d \right ) d^{5} g^{2}-\ln \left (e x +d \right ) x^{3} d^{2} e^{3} g^{2}+\ln \left (e x -d \right ) x^{2} d^{3} e^{2} g^{2}-3 \ln \left (e x -d \right ) x^{2} d \,e^{4} f^{2}-\ln \left (e x +d \right ) x^{2} d^{3} e^{2} g^{2}+3 \ln \left (e x +d \right ) x^{2} d \,e^{4} f^{2}-\ln \left (e x -d \right ) x \,d^{4} e \,g^{2}+3 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}+\ln \left (e x +d \right ) x \,d^{4} e \,g^{2}-3 \ln \left (e x +d \right ) d^{3} e^{2} f^{2}+3 \ln \left (e x +d \right ) x^{3} e^{5} f^{2}-\ln \left (e x -d \right ) d^{5} g^{2}}{16 d^{4} e^{3} \left (e^{2} x^{2}-d^{2}\right ) \left (e x +d \right )}\)	\(565\)

[In]

int((g*x+f)^2/(e*x+d)/(-e^2*x^2+d^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/16/e^3*(d^2*g^2-2*d*e*f*g-3*e^2*f^2)/d^4*ln(-e*x+d)+1/8*(d^2*g^2+2*d*e*f*g+e^2*f^2)/e^3/d^3/(-e*x+d)-1/4*(-d
^2*g^2+e^2*f^2)/e^3/d^3/(e*x+d)+1/16*(-d^2*g^2+2*d*e*f*g+3*e^2*f^2)/e^3/d^4*ln(e*x+d)-1/8*(d^2*g^2-2*d*e*f*g+e
^2*f^2)/d^2/e^3/(e*x+d)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (114) = 228\).

Time = 0.35 (sec) , antiderivative size = 417, normalized size of antiderivative = 3.45 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 \, d^{3} e^{2} f^{2} - 8 \, d^{4} e f g - 4 \, d^{5} g^{2} - 2 \, {\left (3 \, d e^{4} f^{2} + 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (3 \, d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g + 3 \, d^{4} e g^{2}\right )} x - {\left (3 \, d^{3} e^{2} f^{2} + 2 \, d^{4} e f g - d^{5} g^{2} - {\left (3 \, e^{5} f^{2} + 2 \, d e^{4} f g - d^{2} e^{3} g^{2}\right )} x^{3} - {\left (3 \, d e^{4} f^{2} + 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} + {\left (3 \, d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + {\left (3 \, d^{3} e^{2} f^{2} + 2 \, d^{4} e f g - d^{5} g^{2} - {\left (3 \, e^{5} f^{2} + 2 \, d e^{4} f g - d^{2} e^{3} g^{2}\right )} x^{3} - {\left (3 \, d e^{4} f^{2} + 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} + {\left (3 \, d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{16 \, {\left (d^{4} e^{6} x^{3} + d^{5} e^{5} x^{2} - d^{6} e^{4} x - d^{7} e^{3}\right )}} \]

[In]

integrate((g*x+f)^2/(e*x+d)/(-e^2*x^2+d^2)^2,x, algorithm="fricas")

[Out]

1/16*(4*d^3*e^2*f^2 - 8*d^4*e*f*g - 4*d^5*g^2 - 2*(3*d*e^4*f^2 + 2*d^2*e^3*f*g - d^3*e^2*g^2)*x^2 - 2*(3*d^2*e
^3*f^2 + 2*d^3*e^2*f*g + 3*d^4*e*g^2)*x - (3*d^3*e^2*f^2 + 2*d^4*e*f*g - d^5*g^2 - (3*e^5*f^2 + 2*d*e^4*f*g -
d^2*e^3*g^2)*x^3 - (3*d*e^4*f^2 + 2*d^2*e^3*f*g - d^3*e^2*g^2)*x^2 + (3*d^2*e^3*f^2 + 2*d^3*e^2*f*g - d^4*e*g^
2)*x)*log(e*x + d) + (3*d^3*e^2*f^2 + 2*d^4*e*f*g - d^5*g^2 - (3*e^5*f^2 + 2*d*e^4*f*g - d^2*e^3*g^2)*x^3 - (3
*d*e^4*f^2 + 2*d^2*e^3*f*g - d^3*e^2*g^2)*x^2 + (3*d^2*e^3*f^2 + 2*d^3*e^2*f*g - d^4*e*g^2)*x)*log(e*x - d))/(
d^4*e^6*x^3 + d^5*e^5*x^2 - d^6*e^4*x - d^7*e^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (105) = 210\).

Time = 0.58 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.31 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {- 2 d^{4} g^{2} - 4 d^{3} e f g + 2 d^{2} e^{2} f^{2} + x^{2} \left (d^{2} e^{2} g^{2} - 2 d e^{3} f g - 3 e^{4} f^{2}\right ) + x \left (- 3 d^{3} e g^{2} - 2 d^{2} e^{2} f g - 3 d e^{3} f^{2}\right )}{- 8 d^{6} e^{3} - 8 d^{5} e^{4} x + 8 d^{4} e^{5} x^{2} + 8 d^{3} e^{6} x^{3}} + \frac {\left (d g - 3 e f\right ) \left (d g + e f\right ) \log {\left (- \frac {d \left (d g - 3 e f\right ) \left (d g + e f\right )}{e \left (d^{2} g^{2} - 2 d e f g - 3 e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} - \frac {\left (d g - 3 e f\right ) \left (d g + e f\right ) \log {\left (\frac {d \left (d g - 3 e f\right ) \left (d g + e f\right )}{e \left (d^{2} g^{2} - 2 d e f g - 3 e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} \]

[In]

integrate((g*x+f)**2/(e*x+d)/(-e**2*x**2+d**2)**2,x)

[Out]

(-2*d**4*g**2 - 4*d**3*e*f*g + 2*d**2*e**2*f**2 + x**2*(d**2*e**2*g**2 - 2*d*e**3*f*g - 3*e**4*f**2) + x*(-3*d
**3*e*g**2 - 2*d**2*e**2*f*g - 3*d*e**3*f**2))/(-8*d**6*e**3 - 8*d**5*e**4*x + 8*d**4*e**5*x**2 + 8*d**3*e**6*
x**3) + (d*g - 3*e*f)*(d*g + e*f)*log(-d*(d*g - 3*e*f)*(d*g + e*f)/(e*(d**2*g**2 - 2*d*e*f*g - 3*e**2*f**2)) +
x)/(16*d**4*e**3) - (d*g - 3*e*f)*(d*g + e*f)*log(d*(d*g - 3*e*f)*(d*g + e*f)/(e*(d**2*g**2 - 2*d*e*f*g - 3*e
**2*f**2)) + x)/(16*d**4*e**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.75 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {2 \, d^{2} e^{2} f^{2} - 4 \, d^{3} e f g - 2 \, d^{4} g^{2} - {\left (3 \, e^{4} f^{2} + 2 \, d e^{3} f g - d^{2} e^{2} g^{2}\right )} x^{2} - {\left (3 \, d e^{3} f^{2} + 2 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x}{8 \, {\left (d^{3} e^{6} x^{3} + d^{4} e^{5} x^{2} - d^{5} e^{4} x - d^{6} e^{3}\right )}} + \frac {{\left (3 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x + d\right )}{16 \, d^{4} e^{3}} - \frac {{\left (3 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x - d\right )}{16 \, d^{4} e^{3}} \]

[In]

integrate((g*x+f)^2/(e*x+d)/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

1/8*(2*d^2*e^2*f^2 - 4*d^3*e*f*g - 2*d^4*g^2 - (3*e^4*f^2 + 2*d*e^3*f*g - d^2*e^2*g^2)*x^2 - (3*d*e^3*f^2 + 2*
d^2*e^2*f*g + 3*d^3*e*g^2)*x)/(d^3*e^6*x^3 + d^4*e^5*x^2 - d^5*e^4*x - d^6*e^3) + 1/16*(3*e^2*f^2 + 2*d*e*f*g
- d^2*g^2)*log(e*x + d)/(d^4*e^3) - 1/16*(3*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*log(e*x - d)/(d^4*e^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.66 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {{\left (3 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{16 \, d^{4} e^{3}} - \frac {{\left (3 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{16 \, d^{4} e^{3}} + \frac {2 \, d^{3} e^{2} f^{2} - 4 \, d^{4} e f g - 2 \, d^{5} g^{2} - {\left (3 \, d e^{4} f^{2} + 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g + 3 \, d^{4} e g^{2}\right )} x}{8 \, {\left (e x + d\right )}^{2} {\left (e x - d\right )} d^{4} e^{3}} \]

[In]

integrate((g*x+f)^2/(e*x+d)/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

1/16*(3*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*log(abs(e*x + d))/(d^4*e^3) - 1/16*(3*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*lo
g(abs(e*x - d))/(d^4*e^3) + 1/8*(2*d^3*e^2*f^2 - 4*d^4*e*f*g - 2*d^5*g^2 - (3*d*e^4*f^2 + 2*d^2*e^3*f*g - d^3*
e^2*g^2)*x^2 - (3*d^2*e^3*f^2 + 2*d^3*e^2*f*g + 3*d^4*e*g^2)*x)/((e*x + d)^2*(e*x - d)*d^4*e^3)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.64 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^2} \, dx=\frac {\frac {d^2\,g^2+2\,d\,e\,f\,g-e^2\,f^2}{4\,d\,e^3}+\frac {x\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{8\,d^2\,e^2}+\frac {x^2\,\left (-d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{8\,d^3\,e}}{d^3+d^2\,e\,x-d\,e^2\,x^2-e^3\,x^3}+\frac {\mathrm {atanh}\left (\frac {e\,x\,\left (d\,g+e\,f\right )\,\left (d\,g-3\,e\,f\right )}{d\,\left (-d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}\right )\,\left (d\,g+e\,f\right )\,\left (d\,g-3\,e\,f\right )}{8\,d^4\,e^3} \]

[In]

int((f + g*x)^2/((d^2 - e^2*x^2)^2*(d + e*x)),x)

[Out]

((d^2*g^2 - e^2*f^2 + 2*d*e*f*g)/(4*d*e^3) + (x*(3*d^2*g^2 + 3*e^2*f^2 + 2*d*e*f*g))/(8*d^2*e^2) + (x^2*(3*e^2
*f^2 - d^2*g^2 + 2*d*e*f*g))/(8*d^3*e))/(d^3 - e^3*x^3 - d*e^2*x^2 + d^2*e*x) + (atanh((e*x*(d*g + e*f)*(d*g -
3*e*f))/(d*(3*e^2*f^2 - d^2*g^2 + 2*d*e*f*g)))*(d*g + e*f)*(d*g - 3*e*f))/(8*d^4*e^3)