3.827 \(\int \frac {(d+e x)^2 (a+b x+c x^2)}{(f+g x)^{3/2}} \, dx\)

Optimal. Leaf size=210 \[ -\frac {2 (f+g x)^{3/2} \left (e g (-a e g-2 b d g+3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{3 g^5}-\frac {2 (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}{g^5 \sqrt {f+g x}}-\frac {2 \sqrt {f+g x} (e f-d g) (2 c f (2 e f-d g)-g (-2 a e g-b d g+3 b e f))}{g^5}-\frac {2 e (f+g x)^{5/2} (-b e g-2 c d g+4 c e f)}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \]

[Out]

-2/3*(e*g*(-a*e*g-2*b*d*g+3*b*e*f)-c*(d^2*g^2-6*d*e*f*g+6*e^2*f^2))*(g*x+f)^(3/2)/g^5-2/5*e*(-b*e*g-2*c*d*g+4*
c*e*f)*(g*x+f)^(5/2)/g^5+2/7*c*e^2*(g*x+f)^(7/2)/g^5-2*(-d*g+e*f)^2*(a*g^2-b*f*g+c*f^2)/g^5/(g*x+f)^(1/2)-2*(-
d*g+e*f)*(2*c*f*(-d*g+2*e*f)-g*(-2*a*e*g-b*d*g+3*b*e*f))*(g*x+f)^(1/2)/g^5

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Rubi [A]  time = 0.29, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {897, 1261} \[ -\frac {2 (f+g x)^{3/2} \left (e g (-a e g-2 b d g+3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{3 g^5}-\frac {2 (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}{g^5 \sqrt {f+g x}}-\frac {2 \sqrt {f+g x} (e f-d g) (2 c f (2 e f-d g)-g (-2 a e g-b d g+3 b e f))}{g^5}-\frac {2 e (f+g x)^{5/2} (-b e g-2 c d g+4 c e f)}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e*f - d*g)^2*(c*f^2 - b*f*g + a*g^2))/(g^5*Sqrt[f + g*x]) - (2*(e*f - d*g)*(2*c*f*(2*e*f - d*g) - g*(3*b*
e*f - b*d*g - 2*a*e*g))*Sqrt[f + g*x])/g^5 - (2*(e*g*(3*b*e*f - 2*b*d*g - a*e*g) - c*(6*e^2*f^2 - 6*d*e*f*g +
d^2*g^2))*(f + g*x)^(3/2))/(3*g^5) - (2*e*(4*c*e*f - 2*c*d*g - b*e*g)*(f + g*x)^(5/2))/(5*g^5) + (2*c*e^2*(f +
g*x)^(7/2))/(7*g^5)

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2 \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2 \left (\frac {c f^2-b f g+a g^2}{g^2}-\frac {(2 c f-b g) x^2}{g^2}+\frac {c x^4}{g^2}\right )}{x^2} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (\frac {(e f-d g) (-2 c f (2 e f-d g)+g (3 b e f-b d g-2 a e g))}{g^4}+\frac {(-e f+d g)^2 \left (c f^2-b f g+a g^2\right )}{g^4 x^2}+\frac {\left (-e g (3 b e f-2 b d g-a e g)+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) x^2}{g^4}+\frac {e (-4 c e f+2 c d g+b e g) x^4}{g^4}+\frac {c e^2 x^6}{g^4}\right ) \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {2 (e f-d g)^2 \left (c f^2-b f g+a g^2\right )}{g^5 \sqrt {f+g x}}-\frac {2 (e f-d g) (2 c f (2 e f-d g)-g (3 b e f-b d g-2 a e g)) \sqrt {f+g x}}{g^5}-\frac {2 \left (e g (3 b e f-2 b d g-a e g)-c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{3/2}}{3 g^5}-\frac {2 e (4 c e f-2 c d g-b e g) (f+g x)^{5/2}}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 184, normalized size = 0.88 \[ \frac {2 \left (-35 (f+g x)^2 \left (-e g (a e g+2 b d g-3 b e f)-c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )-105 (e f-d g)^2 \left (g (a g-b f)+c f^2\right )-105 (f+g x) (e f-d g) (g (2 a e g+b d g-3 b e f)+2 c f (2 e f-d g))-21 e (f+g x)^3 (-b e g-2 c d g+4 c e f)+15 c e^2 (f+g x)^4\right )}{105 g^5 \sqrt {f+g x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-105*(e*f - d*g)^2*(c*f^2 + g*(-(b*f) + a*g)) - 105*(e*f - d*g)*(2*c*f*(2*e*f - d*g) + g*(-3*b*e*f + b*d*g
+ 2*a*e*g))*(f + g*x) - 35*(-(e*g*(-3*b*e*f + 2*b*d*g + a*e*g)) - c*(6*e^2*f^2 - 6*d*e*f*g + d^2*g^2))*(f + g
*x)^2 - 21*e*(4*c*e*f - 2*c*d*g - b*e*g)*(f + g*x)^3 + 15*c*e^2*(f + g*x)^4))/(105*g^5*Sqrt[f + g*x])

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fricas [A]  time = 0.99, size = 269, normalized size = 1.28 \[ \frac {2 \, {\left (15 \, c e^{2} g^{4} x^{4} - 384 \, c e^{2} f^{4} - 105 \, a d^{2} g^{4} + 336 \, {\left (2 \, c d e + b e^{2}\right )} f^{3} g - 280 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f^{2} g^{2} + 210 \, {\left (b d^{2} + 2 \, a d e\right )} f g^{3} - 3 \, {\left (8 \, c e^{2} f g^{3} - 7 \, {\left (2 \, c d e + b e^{2}\right )} g^{4}\right )} x^{3} + {\left (48 \, c e^{2} f^{2} g^{2} - 42 \, {\left (2 \, c d e + b e^{2}\right )} f g^{3} + 35 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} g^{4}\right )} x^{2} - {\left (192 \, c e^{2} f^{3} g - 168 \, {\left (2 \, c d e + b e^{2}\right )} f^{2} g^{2} + 140 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f g^{3} - 105 \, {\left (b d^{2} + 2 \, a d e\right )} g^{4}\right )} x\right )} \sqrt {g x + f}}{105 \, {\left (g^{6} x + f g^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="fricas")

[Out]

2/105*(15*c*e^2*g^4*x^4 - 384*c*e^2*f^4 - 105*a*d^2*g^4 + 336*(2*c*d*e + b*e^2)*f^3*g - 280*(c*d^2 + 2*b*d*e +
a*e^2)*f^2*g^2 + 210*(b*d^2 + 2*a*d*e)*f*g^3 - 3*(8*c*e^2*f*g^3 - 7*(2*c*d*e + b*e^2)*g^4)*x^3 + (48*c*e^2*f^
2*g^2 - 42*(2*c*d*e + b*e^2)*f*g^3 + 35*(c*d^2 + 2*b*d*e + a*e^2)*g^4)*x^2 - (192*c*e^2*f^3*g - 168*(2*c*d*e +
b*e^2)*f^2*g^2 + 140*(c*d^2 + 2*b*d*e + a*e^2)*f*g^3 - 105*(b*d^2 + 2*a*d*e)*g^4)*x)*sqrt(g*x + f)/(g^6*x + f
*g^5)

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giac [B]  time = 0.23, size = 404, normalized size = 1.92 \[ -\frac {2 \, {\left (c d^{2} f^{2} g^{2} - b d^{2} f g^{3} + a d^{2} g^{4} - 2 \, c d f^{3} g e + 2 \, b d f^{2} g^{2} e - 2 \, a d f g^{3} e + c f^{4} e^{2} - b f^{3} g e^{2} + a f^{2} g^{2} e^{2}\right )}}{\sqrt {g x + f} g^{5}} + \frac {2 \, {\left (35 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{32} - 210 \, \sqrt {g x + f} c d^{2} f g^{32} + 105 \, \sqrt {g x + f} b d^{2} g^{33} + 42 \, {\left (g x + f\right )}^{\frac {5}{2}} c d g^{31} e - 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c d f g^{31} e + 630 \, \sqrt {g x + f} c d f^{2} g^{31} e + 70 \, {\left (g x + f\right )}^{\frac {3}{2}} b d g^{32} e - 420 \, \sqrt {g x + f} b d f g^{32} e + 210 \, \sqrt {g x + f} a d g^{33} e + 15 \, {\left (g x + f\right )}^{\frac {7}{2}} c g^{30} e^{2} - 84 \, {\left (g x + f\right )}^{\frac {5}{2}} c f g^{30} e^{2} + 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c f^{2} g^{30} e^{2} - 420 \, \sqrt {g x + f} c f^{3} g^{30} e^{2} + 21 \, {\left (g x + f\right )}^{\frac {5}{2}} b g^{31} e^{2} - 105 \, {\left (g x + f\right )}^{\frac {3}{2}} b f g^{31} e^{2} + 315 \, \sqrt {g x + f} b f^{2} g^{31} e^{2} + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{32} e^{2} - 210 \, \sqrt {g x + f} a f g^{32} e^{2}\right )}}{105 \, g^{35}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="giac")

[Out]

-2*(c*d^2*f^2*g^2 - b*d^2*f*g^3 + a*d^2*g^4 - 2*c*d*f^3*g*e + 2*b*d*f^2*g^2*e - 2*a*d*f*g^3*e + c*f^4*e^2 - b*
f^3*g*e^2 + a*f^2*g^2*e^2)/(sqrt(g*x + f)*g^5) + 2/105*(35*(g*x + f)^(3/2)*c*d^2*g^32 - 210*sqrt(g*x + f)*c*d^
2*f*g^32 + 105*sqrt(g*x + f)*b*d^2*g^33 + 42*(g*x + f)^(5/2)*c*d*g^31*e - 210*(g*x + f)^(3/2)*c*d*f*g^31*e + 6
30*sqrt(g*x + f)*c*d*f^2*g^31*e + 70*(g*x + f)^(3/2)*b*d*g^32*e - 420*sqrt(g*x + f)*b*d*f*g^32*e + 210*sqrt(g*
x + f)*a*d*g^33*e + 15*(g*x + f)^(7/2)*c*g^30*e^2 - 84*(g*x + f)^(5/2)*c*f*g^30*e^2 + 210*(g*x + f)^(3/2)*c*f^
2*g^30*e^2 - 420*sqrt(g*x + f)*c*f^3*g^30*e^2 + 21*(g*x + f)^(5/2)*b*g^31*e^2 - 105*(g*x + f)^(3/2)*b*f*g^31*e
^2 + 315*sqrt(g*x + f)*b*f^2*g^31*e^2 + 35*(g*x + f)^(3/2)*a*g^32*e^2 - 210*sqrt(g*x + f)*a*f*g^32*e^2)/g^35

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maple [A]  time = 0.01, size = 315, normalized size = 1.50 \[ -\frac {2 \left (-15 e^{2} c \,x^{4} g^{4}-21 b \,e^{2} g^{4} x^{3}-42 c d e \,g^{4} x^{3}+24 c \,e^{2} f \,g^{3} x^{3}-35 a \,e^{2} g^{4} x^{2}-70 b d e \,g^{4} x^{2}+42 b \,e^{2} f \,g^{3} x^{2}-35 c \,d^{2} g^{4} x^{2}+84 c d e f \,g^{3} x^{2}-48 c \,e^{2} f^{2} g^{2} x^{2}-210 a d e \,g^{4} x +140 a \,e^{2} f \,g^{3} x -105 b \,d^{2} g^{4} x +280 b d e f \,g^{3} x -168 b \,e^{2} f^{2} g^{2} x +140 c \,d^{2} f \,g^{3} x -336 c d e \,f^{2} g^{2} x +192 c \,e^{2} f^{3} g x +105 a \,d^{2} g^{4}-420 a d e f \,g^{3}+280 a \,e^{2} f^{2} g^{2}-210 b \,d^{2} f \,g^{3}+560 b d e \,f^{2} g^{2}-336 b \,e^{2} f^{3} g +280 c \,d^{2} f^{2} g^{2}-672 c d e \,f^{3} g +384 c \,e^{2} f^{4}\right )}{105 \sqrt {g x +f}\, g^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(c*x^2+b*x+a)/(g*x+f)^(3/2),x)

[Out]

-2/105/(g*x+f)^(1/2)*(-15*c*e^2*g^4*x^4-21*b*e^2*g^4*x^3-42*c*d*e*g^4*x^3+24*c*e^2*f*g^3*x^3-35*a*e^2*g^4*x^2-
70*b*d*e*g^4*x^2+42*b*e^2*f*g^3*x^2-35*c*d^2*g^4*x^2+84*c*d*e*f*g^3*x^2-48*c*e^2*f^2*g^2*x^2-210*a*d*e*g^4*x+1
40*a*e^2*f*g^3*x-105*b*d^2*g^4*x+280*b*d*e*f*g^3*x-168*b*e^2*f^2*g^2*x+140*c*d^2*f*g^3*x-336*c*d*e*f^2*g^2*x+1
92*c*e^2*f^3*g*x+105*a*d^2*g^4-420*a*d*e*f*g^3+280*a*e^2*f^2*g^2-210*b*d^2*f*g^3+560*b*d*e*f^2*g^2-336*b*e^2*f
^3*g+280*c*d^2*f^2*g^2-672*c*d*e*f^3*g+384*c*e^2*f^4)/g^5

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maxima [A]  time = 0.45, size = 269, normalized size = 1.28 \[ \frac {2 \, {\left (\frac {15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e^{2} - 21 \, {\left (4 \, c e^{2} f - {\left (2 \, c d e + b e^{2}\right )} g\right )} {\left (g x + f\right )}^{\frac {5}{2}} + 35 \, {\left (6 \, c e^{2} f^{2} - 3 \, {\left (2 \, c d e + b e^{2}\right )} f g + {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} - 105 \, {\left (4 \, c e^{2} f^{3} - 3 \, {\left (2 \, c d e + b e^{2}\right )} f^{2} g + 2 \, {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f g^{2} - {\left (b d^{2} + 2 \, a d e\right )} g^{3}\right )} \sqrt {g x + f}}{g^{4}} - \frac {105 \, {\left (c e^{2} f^{4} + a d^{2} g^{4} - {\left (2 \, c d e + b e^{2}\right )} f^{3} g + {\left (c d^{2} + 2 \, b d e + a e^{2}\right )} f^{2} g^{2} - {\left (b d^{2} + 2 \, a d e\right )} f g^{3}\right )}}{\sqrt {g x + f} g^{4}}\right )}}{105 \, g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="maxima")

[Out]

2/105*((15*(g*x + f)^(7/2)*c*e^2 - 21*(4*c*e^2*f - (2*c*d*e + b*e^2)*g)*(g*x + f)^(5/2) + 35*(6*c*e^2*f^2 - 3*
(2*c*d*e + b*e^2)*f*g + (c*d^2 + 2*b*d*e + a*e^2)*g^2)*(g*x + f)^(3/2) - 105*(4*c*e^2*f^3 - 3*(2*c*d*e + b*e^2
)*f^2*g + 2*(c*d^2 + 2*b*d*e + a*e^2)*f*g^2 - (b*d^2 + 2*a*d*e)*g^3)*sqrt(g*x + f))/g^4 - 105*(c*e^2*f^4 + a*d
^2*g^4 - (2*c*d*e + b*e^2)*f^3*g + (c*d^2 + 2*b*d*e + a*e^2)*f^2*g^2 - (b*d^2 + 2*a*d*e)*f*g^3)/(sqrt(g*x + f)
*g^4))/g

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mupad [B]  time = 3.13, size = 270, normalized size = 1.29 \[ \frac {{\left (f+g\,x\right )}^{5/2}\,\left (2\,b\,e^2\,g-8\,c\,e^2\,f+4\,c\,d\,e\,g\right )}{5\,g^5}-\frac {2\,c\,d^2\,f^2\,g^2-2\,b\,d^2\,f\,g^3+2\,a\,d^2\,g^4-4\,c\,d\,e\,f^3\,g+4\,b\,d\,e\,f^2\,g^2-4\,a\,d\,e\,f\,g^3+2\,c\,e^2\,f^4-2\,b\,e^2\,f^3\,g+2\,a\,e^2\,f^2\,g^2}{g^5\,\sqrt {f+g\,x}}+\frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,c\,d^2\,g^2-12\,c\,d\,e\,f\,g+4\,b\,d\,e\,g^2+12\,c\,e^2\,f^2-6\,b\,e^2\,f\,g+2\,a\,e^2\,g^2\right )}{3\,g^5}+\frac {2\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )\,\left (2\,a\,e\,g^2+b\,d\,g^2+4\,c\,e\,f^2-3\,b\,e\,f\,g-2\,c\,d\,f\,g\right )}{g^5}+\frac {2\,c\,e^2\,{\left (f+g\,x\right )}^{7/2}}{7\,g^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x)^2*(a + b*x + c*x^2))/(f + g*x)^(3/2),x)

[Out]

((f + g*x)^(5/2)*(2*b*e^2*g - 8*c*e^2*f + 4*c*d*e*g))/(5*g^5) - (2*a*d^2*g^4 + 2*c*e^2*f^4 + 2*a*e^2*f^2*g^2 +
2*c*d^2*f^2*g^2 - 2*b*d^2*f*g^3 - 2*b*e^2*f^3*g + 4*b*d*e*f^2*g^2 - 4*a*d*e*f*g^3 - 4*c*d*e*f^3*g)/(g^5*(f +
g*x)^(1/2)) + ((f + g*x)^(3/2)*(2*a*e^2*g^2 + 2*c*d^2*g^2 + 12*c*e^2*f^2 + 4*b*d*e*g^2 - 6*b*e^2*f*g - 12*c*d*
e*f*g))/(3*g^5) + (2*(f + g*x)^(1/2)*(d*g - e*f)*(2*a*e*g^2 + b*d*g^2 + 4*c*e*f^2 - 3*b*e*f*g - 2*c*d*f*g))/g^
5 + (2*c*e^2*(f + g*x)^(7/2))/(7*g^5)

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sympy [A]  time = 79.49, size = 272, normalized size = 1.30 \[ \frac {2 c e^{2} \left (f + g x\right )^{\frac {7}{2}}}{7 g^{5}} + \frac {\left (f + g x\right )^{\frac {5}{2}} \left (2 b e^{2} g + 4 c d e g - 8 c e^{2} f\right )}{5 g^{5}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (2 a e^{2} g^{2} + 4 b d e g^{2} - 6 b e^{2} f g + 2 c d^{2} g^{2} - 12 c d e f g + 12 c e^{2} f^{2}\right )}{3 g^{5}} + \frac {\sqrt {f + g x} \left (4 a d e g^{3} - 4 a e^{2} f g^{2} + 2 b d^{2} g^{3} - 8 b d e f g^{2} + 6 b e^{2} f^{2} g - 4 c d^{2} f g^{2} + 12 c d e f^{2} g - 8 c e^{2} f^{3}\right )}{g^{5}} - \frac {2 \left (d g - e f\right )^{2} \left (a g^{2} - b f g + c f^{2}\right )}{g^{5} \sqrt {f + g x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(c*x**2+b*x+a)/(g*x+f)**(3/2),x)

[Out]

2*c*e**2*(f + g*x)**(7/2)/(7*g**5) + (f + g*x)**(5/2)*(2*b*e**2*g + 4*c*d*e*g - 8*c*e**2*f)/(5*g**5) + (f + g*
x)**(3/2)*(2*a*e**2*g**2 + 4*b*d*e*g**2 - 6*b*e**2*f*g + 2*c*d**2*g**2 - 12*c*d*e*f*g + 12*c*e**2*f**2)/(3*g**
5) + sqrt(f + g*x)*(4*a*d*e*g**3 - 4*a*e**2*f*g**2 + 2*b*d**2*g**3 - 8*b*d*e*f*g**2 + 6*b*e**2*f**2*g - 4*c*d*
*2*f*g**2 + 12*c*d*e*f**2*g - 8*c*e**2*f**3)/g**5 - 2*(d*g - e*f)**2*(a*g**2 - b*f*g + c*f**2)/(g**5*sqrt(f +
g*x))

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