\(\int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx\) [277]

Optimal result

Integrand size = 39, antiderivative size = 122 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {(8 c d-3 b e+2 c e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 c^2 e}+\frac {3 (2 c d-b e)^2 \arctan \left (\frac {\sqrt {c} (d+e x)}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{4 c^{5/2} e} \] Output:

-1/4*(2*c*e*x-3*b*e+8*c*d)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/c^2/e+3/ 
4*(-b*e+2*c*d)^2*arctan(c^(1/2)*(e*x+d)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^( 
1/2))/c^(5/2)/e
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {-\sqrt {c} (d+e x) \left (3 b^2 e^2+b c e (-11 d+e x)+c^2 \left (8 d^2-6 d e x-2 e^2 x^2\right )\right )-3 (-2 c d+b e)^2 \sqrt {d+e x} \sqrt {c d-b e-c e x} \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {c} \sqrt {d+e x}}\right )}{4 c^{5/2} e \sqrt {(d+e x) (-b e+c (d-e x))}} \] Input:

Integrate[(d + e*x)^2/Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2],x]
 

Output:

(-(Sqrt[c]*(d + e*x)*(3*b^2*e^2 + b*c*e*(-11*d + e*x) + c^2*(8*d^2 - 6*d*e 
*x - 2*e^2*x^2))) - 3*(-2*c*d + b*e)^2*Sqrt[d + e*x]*Sqrt[c*d - b*e - c*e* 
x]*ArcTan[Sqrt[c*d - b*e - c*e*x]/(Sqrt[c]*Sqrt[d + e*x])])/(4*c^(5/2)*e*S 
qrt[(d + e*x)*(-(b*e) + c*(d - e*x))])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1134, 1160, 1092, 217}

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.

\(\displaystyle \int \frac {(d+e x)^2}{\sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}} \, dx\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {3 (2 c d-b e) \int \frac {d+e x}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{4 c}-\frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c e}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {3 (2 c d-b e) \left (\frac {(2 c d-b e) \int \frac {1}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{2 c}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e}\right )}{4 c}-\frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c e}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {3 (2 c d-b e) \left (\frac {(2 c d-b e) \int \frac {1}{-\frac {(b+2 c x)^2 e^4}{-c x^2 e^2-b x e^2+d (c d-b e)}-4 c e^2}d\left (-\frac {e^2 (b+2 c x)}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}\right )}{c}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e}\right )}{4 c}-\frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c e}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 (2 c d-b e) \left (\frac {(2 c d-b e) \arctan \left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 c^{3/2} e}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{c e}\right )}{4 c}-\frac {(d+e x) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 c e}\)

Input:

Int[(d + e*x)^2/Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2],x]
 

Output:

-1/2*((d + e*x)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(c*e) + (3*(2*c 
*d - b*e)*(-(Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]/(c*e)) + ((2*c*d - 
b*e)*ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^ 
2*x^2])])/(2*c^(3/2)*e)))/(4*c)
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1134

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d + e*x)^ 
(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ 
c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 
*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(377\) vs. \(2(110)=220\).

Time = 1.74 (sec) , antiderivative size = 378, normalized size of antiderivative = 3.10

method	result	size
default	\(\frac {d^{2} \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}\right )}{\sqrt {c \,e^{2}}}+e^{2} \left (-\frac {x \sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}{2 c \,e^{2}}-\frac {3 b \left (-\frac {\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}{c \,e^{2}}-\frac {b \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}\right )}{2 c \sqrt {c \,e^{2}}}\right )}{4 c}+\frac {\left (-b d e +c \,d^{2}\right ) \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}\right )}{2 c \,e^{2} \sqrt {c \,e^{2}}}\right )+2 d e \left (-\frac {\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}{c \,e^{2}}-\frac {b \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {b}{2 c}\right )}{\sqrt {-x^{2} c \,e^{2}-x b \,e^{2}-b d e +c \,d^{2}}}\right )}{2 c \sqrt {c \,e^{2}}}\right )\)	\(378\)

Input:

int((e*x+d)^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x,method=_RETURNVERBO 
SE)
 

Output:

d^2/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d 
*e+c*d^2)^(1/2))+e^2*(-1/2*x/c/e^2*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)- 
3/4*b/c*(-1/c/e^2*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)-1/2*b/c/(c*e^2)^( 
1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/ 
2)))+1/2*(-b*d*e+c*d^2)/c/e^2/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2*b/ 
c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)))+2*d*e*(-1/c/e^2*(-c*e^2*x^2-b* 
e^2*x-b*d*e+c*d^2)^(1/2)-1/2*b/c/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+1/2 
*b/c)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.78 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\left [-\frac {3 \, {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} - 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {-c}\right ) + 4 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c^{2} e x + 8 \, c^{2} d - 3 \, b c e\right )}}{16 \, c^{3} e}, -\frac {3 \, {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c e x + b e\right )} \sqrt {c}}{2 \, {\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (2 \, c^{2} e x + 8 \, c^{2} d - 3 \, b c e\right )}}{8 \, c^{3} e}\right ] \] Input:

integrate((e*x+d)^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="f 
ricas")
 

Output:

[-1/16*(3*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sqrt(-c)*log(8*c^2*e^2*x^2 + 8 
*b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^2*e^2 - 4*sqrt(-c*e^2*x^2 - b*e^2*x 
 + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(-c)) + 4*sqrt(-c*e^2*x^2 - b*e^2*x 
+ c*d^2 - b*d*e)*(2*c^2*e*x + 8*c^2*d - 3*b*c*e))/(c^3*e), -1/8*(3*(4*c^2* 
d^2 - 4*b*c*d*e + b^2*e^2)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^2*x + 
c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^2*d^2 
+ b*c*d*e)) + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c^2*e*x + 8* 
c^2*d - 3*b*c*e))/(c^3*e)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (109) = 218\).

Time = 1.06 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.90 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\begin {cases} \left (- \frac {x}{2 c} - \frac {- \frac {3 b e^{2}}{4 c} + 2 d e}{c e^{2}}\right ) \sqrt {- b d e - b e^{2} x + c d^{2} - c e^{2} x^{2}} + \left (- \frac {b \left (- \frac {3 b e^{2}}{4 c} + 2 d e\right )}{2 c} + d^{2} + \frac {- b d e + c d^{2}}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (- b e^{2} - 2 c e^{2} x + 2 \sqrt {- c e^{2}} \sqrt {- b d e - b e^{2} x + c d^{2} - c e^{2} x^{2}} \right )}}{\sqrt {- c e^{2}}} & \text {for}\: \frac {b^{2} e^{2}}{4 c} - b d e + c d^{2} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {- c e^{2} \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c e^{2} \neq 0 \\- \frac {2 \left (\frac {c^{2} d^{4} \sqrt {- b d e - b e^{2} x + c d^{2}}}{b^{2} e^{2}} - \frac {2 c d^{2} \left (- b d e - b e^{2} x + c d^{2}\right )^{\frac {3}{2}}}{3 b^{2} e^{2}} + \frac {\left (- b d e - b e^{2} x + c d^{2}\right )^{\frac {5}{2}}}{5 b^{2} e^{2}}\right )}{b e^{2}} & \text {for}\: b e^{2} \neq 0 \\\frac {\begin {cases} d^{2} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{3}}{3 e} & \text {otherwise} \end {cases}}{\sqrt {- b d e + c d^{2}}} & \text {otherwise} \end {cases} \] Input:

integrate((e*x+d)**2/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2),x)
 

Output:

Piecewise(((-x/(2*c) - (-3*b*e**2/(4*c) + 2*d*e)/(c*e**2))*sqrt(-b*d*e - b 
*e**2*x + c*d**2 - c*e**2*x**2) + (-b*(-3*b*e**2/(4*c) + 2*d*e)/(2*c) + d* 
*2 + (-b*d*e + c*d**2)/(2*c))*Piecewise((log(-b*e**2 - 2*c*e**2*x + 2*sqrt 
(-c*e**2)*sqrt(-b*d*e - b*e**2*x + c*d**2 - c*e**2*x**2))/sqrt(-c*e**2), N 
e(b**2*e**2/(4*c) - b*d*e + c*d**2, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/s 
qrt(-c*e**2*(b/(2*c) + x)**2), True)), Ne(c*e**2, 0)), (-2*(c**2*d**4*sqrt 
(-b*d*e - b*e**2*x + c*d**2)/(b**2*e**2) - 2*c*d**2*(-b*d*e - b*e**2*x + c 
*d**2)**(3/2)/(3*b**2*e**2) + (-b*d*e - b*e**2*x + c*d**2)**(5/2)/(5*b**2* 
e**2))/(b*e**2), Ne(b*e**2, 0)), (Piecewise((d**2*x, Eq(e, 0)), ((d + e*x) 
**3/(3*e), True))/sqrt(-b*d*e + c*d**2), True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="m 
axima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e*(b*e-2*c*d)>0)', see `assume?` 
 for more
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (\frac {2 \, x}{c} + \frac {8 \, c d - 3 \, b e}{c^{2} e}\right )} - \frac {3 \, {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \log \left ({\left | -b e^{2} + 2 \, {\left (\sqrt {-c e^{2}} x - \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}\right )} \sqrt {-c} {\left | e \right |} \right |}\right )}{8 \, \sqrt {-c} c^{2} {\left | e \right |}} \] Input:

integrate((e*x+d)^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x, algorithm="g 
iac")
 

Output:

-1/4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*x/c + (8*c*d - 3*b*e)/( 
c^2*e)) - 3/8*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*log(abs(-b*e^2 + 2*(sqrt(- 
c*e^2)*x - sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))*sqrt(-c)*abs(e)))/( 
sqrt(-c)*c^2*abs(e))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \] Input:

int((d + e*x)^2/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2),x)
 

Output:

int((d + e*x)^2/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.58 \[ \int \frac {(d+e x)^2}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {i \left (3 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b^{3} e^{3}-18 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b^{2} c d \,e^{2}+36 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) b \,c^{2} d^{2} e -24 \sqrt {c}\, \mathit {asinh} \left (\frac {\sqrt {-c e x -b e +c d}\, i}{\sqrt {-b e +2 c d}}\right ) c^{3} d^{3}+3 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, b c e -8 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, c^{2} d -2 \sqrt {e x +d}\, \sqrt {b e -2 c d}\, \sqrt {-b e +2 c d}\, \sqrt {-c e x -b e +c d}\, c^{2} e x \right )}{4 c^{3} e \left (b e -2 c d \right )} \] Input:

int((e*x+d)^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2),x)
 

Output:

(i*(3*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b 
**3*e**3 - 18*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2 
*c*d))*b**2*c*d*e**2 + 36*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqr 
t( - b*e + 2*c*d))*b*c**2*d**2*e - 24*sqrt(c)*asinh((sqrt( - b*e + c*d - c 
*e*x)*i)/sqrt( - b*e + 2*c*d))*c**3*d**3 + 3*sqrt(d + e*x)*sqrt(b*e - 2*c* 
d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e + c*d - c*e*x)*b*c*e - 8*sqrt(d + e*x) 
*sqrt(b*e - 2*c*d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e + c*d - c*e*x)*c**2*d 
- 2*sqrt(d + e*x)*sqrt(b*e - 2*c*d)*sqrt( - b*e + 2*c*d)*sqrt( - b*e + c*d 
 - c*e*x)*c**2*e*x))/(4*c**3*e*(b*e - 2*c*d))