\(\int \frac {x (d+e x^2+f x^4+g x^6)}{a+b x^2+c x^4} \, dx\) [125]
Optimal result
Integrand size = 33, antiderivative size = 149 \[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\frac {(c f-b g) x^2}{2 c^2}+\frac {g x^4}{4 c}-\frac {\left (2 c^3 d-c^2 (b e+2 a f)-b^3 g+b c (b f+3 a g)\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (c^2 e+b^2 g-c (b f+a g)\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}
\]
[Out]
1/2*(-b*g+c*f)*x^2/c^2+1/4*g*x^4/c+1/4*(c^2*e+b^2*g-c*(a*g+b*f))*ln(c*x^4+b*x^2+a)/c^3-1/2*(2*c^3*d-c^2*(2*a*f
+b*e)-b^3*g+b*c*(3*a*g+b*f))*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(1/2)
Rubi [A] (verified)
Time = 0.19 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1677, 1671, 648, 632, 212,
642} \[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=-\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right ) \left (-c^2 (2 a f+b e)+b c (3 a g+b f)+b^3 (-g)+2 c^3 d\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^2+c x^4\right ) \left (-c (a g+b f)+b^2 g+c^2 e\right )}{4 c^3}+\frac {x^2 (c f-b g)}{2 c^2}+\frac {g x^4}{4 c}
\]
[In]
Int[(x*(d + e*x^2 + f*x^4 + g*x^6))/(a + b*x^2 + c*x^4),x]
[Out]
((c*f - b*g)*x^2)/(2*c^2) + (g*x^4)/(4*c) - ((2*c^3*d - c^2*(b*e + 2*a*f) - b^3*g + b*c*(b*f + 3*a*g))*ArcTanh
[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*Sqrt[b^2 - 4*a*c]) + ((c^2*e + b^2*g - c*(b*f + a*g))*Log[a + b*x^2
+ c*x^4])/(4*c^3)
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 632
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 642
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 648
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]
Rule 1671
Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Rule 1677
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]
Rubi steps \begin{align*}
\text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {d+e x+f x^2+g x^3}{a+b x+c x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {c f-b g}{c^2}+\frac {g x}{c}+\frac {c^2 d-a c f+a b g+\left (c^2 e+b^2 g-c (b f+a g)\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {(c f-b g) x^2}{2 c^2}+\frac {g x^4}{4 c}+\frac {\text {Subst}\left (\int \frac {c^2 d-a c f+a b g+\left (c^2 e+b^2 g-c (b f+a g)\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2} \\ & = \frac {(c f-b g) x^2}{2 c^2}+\frac {g x^4}{4 c}+\frac {\left (c^2 e+b^2 g-c (b f+a g)\right ) \text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}+\frac {\left (2 c^3 d-c^2 (b e+2 a f)-b^3 g+b c (b f+3 a g)\right ) \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3} \\ & = \frac {(c f-b g) x^2}{2 c^2}+\frac {g x^4}{4 c}+\frac {\left (c^2 e+b^2 g-c (b f+a g)\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}-\frac {\left (2 c^3 d-c^2 (b e+2 a f)-b^3 g+b c (b f+3 a g)\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3} \\ & = \frac {(c f-b g) x^2}{2 c^2}+\frac {g x^4}{4 c}-\frac {\left (2 c^3 d-c^2 (b e+2 a f)-b^3 g+b c (b f+3 a g)\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (c^2 e+b^2 g-c (b f+a g)\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.07 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95
\[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\frac {2 c (c f-b g) x^2+c^2 g x^4+\frac {2 \left (2 c^3 d-c^2 (b e+2 a f)-b^3 g+b c (b f+3 a g)\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\left (c^2 e+b^2 g-c (b f+a g)\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}
\]
[In]
Integrate[(x*(d + e*x^2 + f*x^4 + g*x^6))/(a + b*x^2 + c*x^4),x]
[Out]
(2*c*(c*f - b*g)*x^2 + c^2*g*x^4 + (2*(2*c^3*d - c^2*(b*e + 2*a*f) - b^3*g + b*c*(b*f + 3*a*g))*ArcTan[(b + 2*
c*x^2)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (c^2*e + b^2*g - c*(b*f + a*g))*Log[a + b*x^2 + c*x^4])/(4*c^
3)
Maple [A] (verified)
Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01
| | |
| method | result | size |
| | |
| default |
\(-\frac {-\frac {1}{2} c g \,x^{4}+b g \,x^{2}-c f \,x^{2}}{2 c^{2}}+\frac {\frac {\left (-a c g +b^{2} g -f b c +e \,c^{2}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (a b g -a c f +c^{2} d -\frac {\left (-a c g +b^{2} g -f b c +e \,c^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 c^{2}}\) |
\(151\) |
| risch |
\(\text {Expression too large to display}\) |
\(3739\) |
| | |
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|
[In]
int(x*(g*x^6+f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)
[Out]
-1/2/c^2*(-1/2*c*g*x^4+b*g*x^2-c*f*x^2)+1/2/c^2*(1/2*(-a*c*g+b^2*g-b*c*f+c^2*e)/c*ln(c*x^4+b*x^2+a)+2*(a*b*g-a
*c*f+c^2*d-1/2*(-a*c*g+b^2*g-b*c*f+c^2*e)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))
Fricas [A] (verification not implemented)
none
Time = 0.34 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.26
\[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} g x^{4} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f - {\left (b^{3} c - 4 \, a b c^{2}\right )} g\right )} x^{2} + {\left (2 \, c^{3} d - b c^{2} e + {\left (b^{2} c - 2 \, a c^{2}\right )} f - {\left (b^{3} - 3 \, a b c\right )} g\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} f + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} g\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} g x^{4} + 2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} f - {\left (b^{3} c - 4 \, a b c^{2}\right )} g\right )} x^{2} - 2 \, {\left (2 \, c^{3} d - b c^{2} e + {\left (b^{2} c - 2 \, a c^{2}\right )} f - {\left (b^{3} - 3 \, a b c\right )} g\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} f + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} g\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ]
\]
[In]
integrate(x*(g*x^6+f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="fricas")
[Out]
[1/4*((b^2*c^2 - 4*a*c^3)*g*x^4 + 2*((b^2*c^2 - 4*a*c^3)*f - (b^3*c - 4*a*b*c^2)*g)*x^2 + (2*c^3*d - b*c^2*e +
(b^2*c - 2*a*c^2)*f - (b^3 - 3*a*b*c)*g)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^
2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + ((b^2*c^2 - 4*a*c^3)*e - (b^3*c - 4*a*b*c^2)*f + (b^4 - 5*a*b
^2*c + 4*a^2*c^2)*g)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4), 1/4*((b^2*c^2 - 4*a*c^3)*g*x^4 + 2*((b^2*c^2
- 4*a*c^3)*f - (b^3*c - 4*a*b*c^2)*g)*x^2 - 2*(2*c^3*d - b*c^2*e + (b^2*c - 2*a*c^2)*f - (b^3 - 3*a*b*c)*g)*s
qrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + ((b^2*c^2 - 4*a*c^3)*e - (b^3*c -
4*a*b*c^2)*f + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*g)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4)]
Sympy [F(-1)]
Timed out. \[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\text {Timed out}
\]
[In]
integrate(x*(g*x**6+f*x**4+e*x**2+d)/(c*x**4+b*x**2+a),x)
[Out]
Timed out
Maxima [F(-2)]
Exception generated. \[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(x*(g*x^6+f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta
Giac [A] (verification not implemented)
none
Time = 0.64 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97
\[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\frac {c g x^{4} + 2 \, c f x^{2} - 2 \, b g x^{2}}{4 \, c^{2}} + \frac {{\left (c^{2} e - b c f + b^{2} g - a c g\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} + \frac {{\left (2 \, c^{3} d - b c^{2} e + b^{2} c f - 2 \, a c^{2} f - b^{3} g + 3 \, a b c g\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{3}}
\]
[In]
integrate(x*(g*x^6+f*x^4+e*x^2+d)/(c*x^4+b*x^2+a),x, algorithm="giac")
[Out]
1/4*(c*g*x^4 + 2*c*f*x^2 - 2*b*g*x^2)/c^2 + 1/4*(c^2*e - b*c*f + b^2*g - a*c*g)*log(c*x^4 + b*x^2 + a)/c^3 + 1
/2*(2*c^3*d - b*c^2*e + b^2*c*f - 2*a*c^2*f - b^3*g + 3*a*b*c*g)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/(sqr
t(-b^2 + 4*a*c)*c^3)
Mupad [B] (verification not implemented)
Time = 9.00 (sec) , antiderivative size = 1834, normalized size of antiderivative = 12.31
\[
\int \frac {x \left (d+e x^2+f x^4+g x^6\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display}
\]
[In]
int((x*(d + e*x^2 + f*x^4 + g*x^6))/(a + b*x^2 + c*x^4),x)
[Out]
x^2*(f/(2*c) - (b*g)/(2*c^2)) + (g*x^4)/(4*c) - (log(a + b*x^2 + c*x^4)*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g -
8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a*b^2*c*g))/(2*(16*a*c^4 - 4*b^2*c^3)) + (atan((2*c^4*(4*a*c - b^2)*
(x^2*(((((4*c^6*d + 6*b^2*c^4*f - 6*b^3*c^3*g - 4*a*c^5*f - 6*b*c^5*e + 10*a*b*c^4*g)/c^4 - (4*b*c^2*(2*b^4*g
+ 2*b^2*c^2*e + 8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a*b^2*c*g))/(16*a*c^4 - 4*b^2*c^3))*(2*
c^3*d - b^3*g - 2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g))/(8*c^3*(4*a*c - b^2)^(1/2)) - (b*(2*c^3*d - b^3*g
- 2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g)*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*
a*b*c^2*f - 10*a*b^2*c*g))/(2*c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a + (b*((((4*c^6*d + 6*b^2*c^4*f
- 6*b^3*c^3*g - 4*a*c^5*f - 6*b*c^5*e + 10*a*b*c^4*g)/c^4 - (4*b*c^2*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g - 8*
a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a*b^2*c*g))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*
g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a*b^2*c*g))/(2*(16*a*c^4 - 4*b^2*c^3)) - (b^5*g^2 + b*c^4*e^2 + b
^3*c^2*f^2 - c^5*d*e + 2*a^2*b*c^2*g^2 + a*c^4*d*g + a*c^4*e*f + b*c^4*d*f - 2*b^4*c*f*g - a*b*c^3*f^2 - 3*a*b
^3*c*g^2 - b^2*c^3*d*g - 2*b^2*c^3*e*f - a^2*c^3*f*g + 2*b^3*c^2*e*g + 4*a*b^2*c^2*f*g - 3*a*b*c^3*e*g)/c^4 +
(b*(2*c^3*d - b^3*g - 2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g)^2)/(2*c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)
^(1/2))) + ((((8*a^2*c^4*g - 8*a*c^5*e + 8*a*b*c^4*f - 8*a*b^2*c^3*g)/c^4 - (8*a*c^2*(2*b^4*g + 2*b^2*c^2*e +
8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a*b^2*c*g))/(16*a*c^4 - 4*b^2*c^3))*(2*c^3*d - b^3*g -
2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g))/(8*c^3*(4*a*c - b^2)^(1/2)) - (a*(2*c^3*d - b^3*g - 2*a*c^2*f - b*
c^2*e + b^2*c*f + 3*a*b*c*g)*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*a
*b^2*c*g))/(c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a + (b*((((8*a^2*c^4*g - 8*a*c^5*e + 8*a*b*c^4*f -
8*a*b^2*c^3*g)/c^4 - (8*a*c^2*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f - 10*
a*b^2*c*g))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4*g + 2*b^2*c^2*e + 8*a^2*c^2*g - 8*a*c^3*e - 2*b^3*c*f + 8*a*b*c^2*f
- 10*a*b^2*c*g))/(2*(16*a*c^4 - 4*b^2*c^3)) - (a*c^4*e^2 + a*b^4*g^2 + a^3*c^2*g^2 + a*b^2*c^2*f^2 - 2*a^2*b^
2*c*g^2 - 2*a^2*c^3*e*g + 2*a*b^2*c^2*e*g + 2*a^2*b*c^2*f*g - 2*a*b*c^3*e*f - 2*a*b^3*c*f*g)/c^4 + (a*(2*c^3*d
- b^3*g - 2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g)^2)/(c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))))/(4*
c^6*d^2 + b^6*g^2 + 4*a^2*c^4*f^2 + b^2*c^4*e^2 + b^4*c^2*f^2 - 4*a*b^2*c^3*f^2 - 8*a*c^5*d*f - 4*b*c^5*d*e -
2*b^5*c*f*g + 9*a^2*b^2*c^2*g^2 - 6*a*b^4*c*g^2 + 4*b^2*c^4*d*f - 4*b^3*c^3*d*g - 2*b^3*c^3*e*f + 2*b^4*c^2*e*
g - 6*a*b^2*c^3*e*g + 10*a*b^3*c^2*f*g - 12*a^2*b*c^3*f*g + 12*a*b*c^4*d*g + 4*a*b*c^4*e*f))*(2*c^3*d - b^3*g
- 2*a*c^2*f - b*c^2*e + b^2*c*f + 3*a*b*c*g))/(2*c^3*(4*a*c - b^2)^(1/2))