\(\int \frac {(d+e x) (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [575]

Optimal result

Integrand size = 27, antiderivative size = 122 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {(e f+d g)^2}{8 d^2 e^3 (d-e x)^2}+\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d-e x)}-\frac {(e f-d g)^2}{8 d^3 e^3 (d+e x)}+\frac {(e f-d g) (3 e f+d g) \text {arctanh}\left (\frac {e x}{d}\right )}{8 d^4 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {813, 90, 214} \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right ) (d g+3 e f) (e f-d g)}{8 d^4 e^3}-\frac {(e f-d g)^2}{8 d^3 e^3 (d+e x)}+\frac {(d g+e f)^2}{8 d^2 e^3 (d-e x)^2}+\frac {e^2 f^2-d^2 g^2}{4 d^3 e^3 (d-e x)} \]

[In]

Int[((d + e*x)*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

(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)) - (e*f - d*g)^2/(8*d^3*e^3*(
d + e*x)) + ((e*f - d*g)*(3*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 813

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^m*(
f + g*x)^(p + 1)*(a/f + (c/g)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*f^2 + a*g^2, 0] && (Integer
Q[p] || (GtQ[a, 0] && GtQ[f, 0] && EqQ[p, -1]))

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\frac {2 d^2 (e f+d g)^2}{(d-e x)^2}+\frac {4 d e^2 f^2-4 d^3 g^2}{d-e x}-\frac {2 d (e f-d g)^2}{d+e x}+\left (-3 e^2 f^2+2 d e f g+d^2 g^2\right ) \log (d-e x)+\left (3 e^2 f^2-2 d e f g-d^2 g^2\right ) \log (d+e x)}{16 d^4 e^3} \]

[In]

Integrate[((d + e*x)*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

((2*d^2*(e*f + d*g)^2)/(d - e*x)^2 + (4*d*e^2*f^2 - 4*d^3*g^2)/(d - e*x) - (2*d*(e*f - d*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 = 0.44 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {-d^{2} g^{2}+e^{2} f^{2}}{4 d^{3} e^{3} \left (-e x +d \right )}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{8 e^{3} d^{2} \left (-e x +d \right )^{2}}+\frac {\left (d^{2} g^{2}+2 d e f g -3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}+\frac {\left (-d^{2} g^{2}-2 d e f g +3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{8 e^{3} d^{3} \left (e x +d \right )}\)	\(183\)
norman	\(\frac {\frac {-d^{2} g^{2} e +2 d f g \,e^{2}+f^{2} e^{3}}{4 e^{4}}+\frac {\left (d^{2} g^{2}+2 d e f g -3 e^{2} f^{2}\right ) x^{3}}{8 d^{3}}+\frac {g^{2} x^{2}}{2 e}+\frac {\left (d^{2} g^{2}+2 d e f g +5 e^{2} f^{2}\right ) x}{8 d \,e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}+\frac {\left (d^{2} g^{2}+2 d e f g -3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}-\frac {\left (d^{2} g^{2}+2 d e f g -3 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}\)	\(193\)
risch	\(\frac {\frac {\left (d^{2} g^{2}+2 d e f g -3 e^{2} f^{2}\right ) x^{2}}{8 d^{3} e}+\frac {\left (3 d^{2} g^{2}-2 d e f g +3 e^{2} f^{2}\right ) x}{8 e^{2} d^{2}}-\frac {d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{4 d \,e^{3}}}{\left (-e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )}-\frac {\ln \left (-e x -d \right ) g^{2}}{16 e^{3} d^{2}}-\frac {\ln \left (-e x -d \right ) f g}{8 e^{2} d^{3}}+\frac {3 \ln \left (-e x -d \right ) f^{2}}{16 e \,d^{4}}+\frac {\ln \left (e x -d \right ) g^{2}}{16 e^{3} d^{2}}+\frac {\ln \left (e x -d \right ) f g}{8 e^{2} d^{3}}-\frac {3 \ln \left (e x -d \right ) f^{2}}{16 e \,d^{4}}\)	\(236\)
parallelrisch	\(\frac {-2 \ln \left (e x +d \right ) d^{4} e f g +\ln \left (e x -d \right ) x^{3} d^{2} e^{3} g^{2}+2 \ln \left (e x +d \right ) x \,d^{3} e^{2} f g -2 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g -2 \ln \left (e x -d \right ) x^{2} d^{2} e^{3} f g +2 \ln \left (e x +d \right ) x^{2} d^{2} e^{3} f g -2 \ln \left (e x +d \right ) x^{3} d \,e^{4} f g +2 \ln \left (e x -d \right ) x^{3} d \,e^{4} f g -3 \ln \left (e x +d \right ) x \,d^{2} e^{3} f^{2}-4 g^{2} d^{5}+6 x \,d^{4} e \,g^{2}+6 x \,d^{2} e^{3} f^{2}+2 x^{2} d^{3} e^{2} g^{2}-6 x^{2} d \,e^{4} f^{2}-3 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}+8 f g \,d^{4} e +4 f^{2} d^{3} e^{2}-3 \ln \left (e x -d \right ) x^{3} e^{5} f^{2}+4 x^{2} d^{2} e^{3} f g +2 \ln \left (e x -d \right ) d^{4} e f g -4 x \,d^{3} e^{2} f g -\ln \left (e x +d \right ) d^{5} g^{2}-\ln \left (e x +d \right ) x^{3} d^{2} e^{3} g^{2}-\ln \left (e x -d \right ) x^{2} d^{3} e^{2} g^{2}+3 \ln \left (e x -d \right ) x^{2} d \,e^{4} f^{2}+\ln \left (e x +d \right ) x^{2} d^{3} e^{2} g^{2}-3 \ln \left (e x +d \right ) x^{2} d \,e^{4} f^{2}-\ln \left (e x -d \right ) x \,d^{4} e \,g^{2}+3 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}+\ln \left (e x +d \right ) x \,d^{4} e \,g^{2}+3 \ln \left (e x +d \right ) d^{3} e^{2} f^{2}+3 \ln \left (e x +d \right ) x^{3} e^{5} f^{2}+\ln \left (e x -d \right ) d^{5} g^{2}}{16 e^{3} d^{4} \left (e x -d \right ) \left (e^{2} x^{2}-d^{2}\right )}\)	\(567\)

[In]

int((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (116) = 232\).

Time = 0.29 (sec) , antiderivative size = 417, normalized size of antiderivative = 3.42 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {4 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g - 4 \, d^{5} g^{2} - 2 \, {\left (3 \, d e^{4} f^{2} - 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} + 2 \, {\left (3 \, d^{2} e^{3} f^{2} - 2 \, d^{3} e^{2} f g + 3 \, d^{4} e g^{2}\right )} x + {\left (3 \, d^{3} e^{2} f^{2} - 2 \, d^{4} e f g - d^{5} g^{2} + {\left (3 \, e^{5} f^{2} - 2 \, d e^{4} f g - d^{2} e^{3} g^{2}\right )} x^{3} - {\left (3 \, d e^{4} f^{2} - 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} - {\left (3 \, d^{2} e^{3} f^{2} - 2 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x\right )} \log \left (e x + d\right ) - {\left (3 \, d^{3} e^{2} f^{2} - 2 \, d^{4} e f g - d^{5} g^{2} + {\left (3 \, e^{5} f^{2} - 2 \, d e^{4} f g - d^{2} e^{3} g^{2}\right )} x^{3} - {\left (3 \, d e^{4} f^{2} - 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} - {\left (3 \, d^{2} e^{3} f^{2} - 2 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{16 \, {\left (d^{4} e^{6} x^{3} - d^{5} e^{5} x^{2} - d^{6} e^{4} x + d^{7} e^{3}\right )}} \]

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2)^3,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. 277 vs. \(2 (105) = 210\).

Time = 0.58 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.27 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {2 d^{4} g^{2} - 4 d^{3} e f g - 2 d^{2} e^{2} f^{2} + x^{2} \left (- d^{2} e^{2} g^{2} - 2 d e^{3} f g + 3 e^{4} f^{2}\right ) + x \left (- 3 d^{3} e g^{2} + 2 d^{2} e^{2} f g - 3 d e^{3} f^{2}\right )}{8 d^{6} e^{3} - 8 d^{5} e^{4} x - 8 d^{4} e^{5} x^{2} + 8 d^{3} e^{6} x^{3}} + \frac {\left (d g - e f\right ) \left (d g + 3 e f\right ) \log {\left (- \frac {d \left (d g - e f\right ) \left (d g + 3 e f\right )}{e \left (d^{2} g^{2} + 2 d e f g - 3 e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} - \frac {\left (d g - e f\right ) \left (d g + 3 e f\right ) \log {\left (\frac {d \left (d g - e f\right ) \left (d g + 3 e f\right )}{e \left (d^{2} g^{2} + 2 d e f g - 3 e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} \]

[In]

integrate((e*x+d)*(g*x+f)**2/(-e**2*x**2+d**2)**3,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 - e*f)*(d*g + 3*e*f)*log(-d*(d*g - e*f)*(d*g + 3*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 - e*f)*(d*g + 3*e*f)*log(d*(d*g - e*f)*(d*g + 3*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 = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.73 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {2 \, d^{2} e^{2} f^{2} + 4 \, d^{3} e f g - 2 \, d^{4} g^{2} - {\left (3 \, e^{4} f^{2} - 2 \, d e^{3} f g - d^{2} e^{2} g^{2}\right )} x^{2} + {\left (3 \, d e^{3} f^{2} - 2 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x}{8 \, {\left (d^{3} e^{6} x^{3} - d^{4} e^{5} x^{2} - d^{5} e^{4} x + d^{6} e^{3}\right )}} + \frac {{\left (3 \, e^{2} f^{2} - 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x + d\right )}{16 \, d^{4} e^{3}} - \frac {{\left (3 \, e^{2} f^{2} - 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x - d\right )}{16 \, d^{4} e^{3}} \]

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2)^3,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 = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.64 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {{\left (3 \, e^{2} f^{2} - 2 \, d e f g - d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{16 \, d^{4} e^{3}} - \frac {{\left (3 \, e^{2} f^{2} - 2 \, d e f g - d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{16 \, d^{4} e^{3}} + \frac {2 \, d^{3} e^{2} f^{2} + 4 \, d^{4} e f g - 2 \, d^{5} g^{2} - {\left (3 \, d e^{4} f^{2} - 2 \, d^{2} e^{3} f g - d^{3} e^{2} g^{2}\right )} x^{2} + {\left (3 \, d^{2} e^{3} f^{2} - 2 \, d^{3} e^{2} f g + 3 \, d^{4} e g^{2}\right )} x}{8 \, {\left (e x + d\right )} {\left (e x - d\right )}^{2} d^{4} e^{3}} \]

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2)^3,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)*(e*x - d)^2*d^4*e^3)

Mupad [B] (verification not implemented)

Time = 11.81 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.62 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\frac {-d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2}{4\,d\,e^3}+\frac {x\,\left (3\,d^2\,g^2-2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{8\,d^2\,e^2}+\frac {x^2\,\left (d^2\,g^2+2\,d\,e\,f\,g-3\,e^2\,f^2\right )}{8\,d^3\,e}}{d^3-d^2\,e\,x-d\,e^2\,x^2+e^3\,x^3}-\frac {\mathrm {atanh}\left (\frac {e\,x\,\left (d\,g-e\,f\right )\,\left (d\,g+3\,e\,f\right )}{d\,\left (d^2\,g^2+2\,d\,e\,f\,g-3\,e^2\,f^2\right )}\right )\,\left (d\,g-e\,f\right )\,\left (d\,g+3\,e\,f\right )}{8\,d^4\,e^3} \]

[In]

int(((f + g*x)^2*(d + e*x))/(d^2 - e^2*x^2)^3,x)

[Out]

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