\(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^2} \, dx\) [168]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2} \] Output:

3/16*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2/d^2-1/2*(c*x^2+b*x+a)^(3/2)/c/d^2/( 
2*c*x+b)-3/32*(-4*a*c+b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/ 
2))/c^(5/2)/d^2
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(113)=226\).

Time = 1.52 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.16 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (3 b^2+4 b c x+4 c \left (-2 a+c x^2\right )\right )}{b+2 c x}+\frac {2 \left (b^2-4 a c\right )^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}-3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )-\frac {2 \left (-b^2+4 a c\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b}}{16 c^{5/2} d^2} \] Input:

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

Output:

((Sqrt[c]*Sqrt[a + x*(b + c*x)]*(3*b^2 + 4*b*c*x + 4*c*(-2*a + c*x^2)))/(b 
 + 2*c*x) + (2*(b^2 - 4*a*c)^(3/2)*ArcTan[(Sqrt[c]*Sqrt[b^2 - 4*a*c]*x)/(S 
qrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/b - 3*(b^2 - 4*a*c)*ArcTan 
h[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])] - (2*(-b^2 + 4*a*c)^(3/2 
)*ArcTanh[(Sqrt[c]*Sqrt[-b^2 + 4*a*c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + 
 x*(b + c*x)])])/b)/(16*c^(5/2)*d^2)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1108, 1087, 1092, 219}

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.

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx\)

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {3 \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}\)

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{4 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}\)

Input:

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

Output:

-1/2*(a + b*x + c*x^2)^(3/2)/(c*d^2*(b + 2*c*x)) + (3*(((b + 2*c*x)*Sqrt[a 
 + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr 
t[a + b*x + c*x^2])])/(8*c^(3/2))))/(4*c*d^2)
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

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 1108

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[b*(p/(d*e*(m + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x 
], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 
3, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0] 
) && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.58

method	result	size
risch	\(\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{16 c^{2} d^{2}}+\frac {-\frac {\left (32 a^{2} c^{2}-16 c a \,b^{2}+2 b^{4}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}{c \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}-\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+12 a \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{2} d^{2}}\)	\(179\)
default	\(\frac {-\frac {4 c \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {16 c^{2} \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}{2}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\sqrt {c}\, \left (x +\frac {b}{2 c}\right )+\sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{4 a c -b^{2}}}{4 d^{2} c^{2}}\)	\(238\)

Input:

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^2,x,method=_RETURNVERBOSE)
 

Output:

1/16*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2/d^2+1/32/c^2*(-(32*a^2*c^2-16*a*b^2 
*c+2*b^4)/c/(4*a*c-b^2)/(x+1/2*b/c)*(c*(x+1/2*b/c)^2+1/4*(4*a*c-b^2)/c)^(1 
/2)-3*b^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+12*a*c^(1/2) 
*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))/d^2
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.49 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{64 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}}, \frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{32 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/64*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*sqrt(c)*log(-8*c^2*x^2 - 
 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4* 
(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b*x + a))/(2*c^4* 
d^2*x + b*c^3*d^2), 1/32*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*sqrt(- 
c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x 
+ a*c)) + 2*(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b*x + 
 a))/(2*c^4*d^2*x + b*c^3*d^2)]
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \] Input:

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

Output:

(Integral(a*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + In 
tegral(b*x*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x) + Int 
egral(c*x**2*sqrt(a + b*x + c*x**2)/(b**2 + 4*b*c*x + 4*c**2*x**2), x))/d* 
*2
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

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(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (95) = 190\).

Time = 0.53 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.89 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {1}{32} \, d^{2} {\left (\frac {3 \, {\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2} d^{4} {\left | c \right |}} + \frac {2 \, {\left (\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )}}{c^{3} d^{4} {\left | c \right |}} + \frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{{\left (\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}}\right )} c^{2} d^{4} {\left | c \right |}}\right )} {\left | c \right |} \] Input:

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

Output:

1/32*d^2*(3*(b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d 
*x + b*d))*sgn(c)*sgn(d))*arctan(sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c 
^2*d^2/(2*c*d*x + b*d)^2 + c)/sqrt(-c))/(sqrt(-c)*c^2*d^4*abs(c)) + 2*(sqr 
t(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^2*sg 
n(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 
 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn( 
d))/(c^3*d^4*abs(c)) + (sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2 
*c*d*x + b*d)^2 + c)*b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*sqrt(-b^ 
2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*sgn(1/( 
2*c*d*x + b*d))*sgn(c)*sgn(d))/((b^2*c*d^2/(2*c*d*x + b*d)^2 - 4*a*c^2*d^2 
/(2*c*d*x + b*d)^2)*c^2*d^4*abs(c)))*abs(c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.62 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx=\frac {-64 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2}+24 \sqrt {c \,x^{2}+b x +a}\, b^{2} c +32 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} x +32 \sqrt {c \,x^{2}+b x +a}\, c^{3} x^{2}+48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a b c +96 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,c^{2} x -12 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{3}-24 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2} c x -36 \sqrt {c}\, a b c -72 \sqrt {c}\, a \,c^{2} x +9 \sqrt {c}\, b^{3}+18 \sqrt {c}\, b^{2} c x}{128 c^{3} d^{2} \left (2 c x +b \right )} \] Input:

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

Output:

( - 64*sqrt(a + b*x + c*x**2)*a*c**2 + 24*sqrt(a + b*x + c*x**2)*b**2*c + 
32*sqrt(a + b*x + c*x**2)*b*c**2*x + 32*sqrt(a + b*x + c*x**2)*c**3*x**2 + 
 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c 
- b**2))*a*b*c + 96*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2* 
c*x)/sqrt(4*a*c - b**2))*a*c**2*x - 12*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x 
 + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**3 - 24*sqrt(c)*log((2*sqrt( 
c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c*x - 36*s 
qrt(c)*a*b*c - 72*sqrt(c)*a*c**2*x + 9*sqrt(c)*b**3 + 18*sqrt(c)*b**2*c*x) 
/(128*c**3*d**2*(b + 2*c*x))