3.929 \(\int \frac{(A+B x) (a+b x+c x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=179 \[ -\frac{3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{a}}+\frac{3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c}}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac{3 \sqrt{a+b x+c x^2} (2 a B-x (2 A c+b B)+A b)}{4 x} \]

[Out]

(-3*(A*b + 2*a*B - (b*B + 2*A*c)*x)*Sqrt[a + b*x + c*x^2])/(4*x) - ((A - B*x)*(a + b*x + c*x^2)^(3/2))/(2*x^2)
- (3*(4*a*b*B + A*(b^2 + 4*a*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[a]) + (3*(b^
2*B + 4*A*b*c + 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c])

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Rubi [A]  time = 0.166527, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {812, 843, 621, 206, 724} \[ -\frac{3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{a}}+\frac{3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c}}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac{3 \sqrt{a+b x+c x^2} (2 a B-x (2 A c+b B)+A b)}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(A*b + 2*a*B - (b*B + 2*A*c)*x)*Sqrt[a + b*x + c*x^2])/(4*x) - ((A - B*x)*(a + b*x + c*x^2)^(3/2))/(2*x^2)
- (3*(4*a*b*B + A*(b^2 + 4*a*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[a]) + (3*(b^
2*B + 4*A*b*c + 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c])

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^3} \, dx &=-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac{3}{8} \int \frac{(-2 (A b+2 a B)-2 (b B+2 A c) x) \sqrt{a+b x+c x^2}}{x^2} \, dx\\ &=-\frac{3 (A b+2 a B-(b B+2 A c) x) \sqrt{a+b x+c x^2}}{4 x}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}+\frac{3}{16} \int \frac{2 \left (4 a b B+A \left (b^2+4 a c\right )\right )+2 \left (b^2 B+4 A b c+4 a B c\right ) x}{x \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{3 (A b+2 a B-(b B+2 A c) x) \sqrt{a+b x+c x^2}}{4 x}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}+\frac{1}{8} \left (3 \left (b^2 B+4 A b c+4 a B c\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx+\frac{1}{8} \left (3 \left (4 a b B+A \left (b^2+4 a c\right )\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{3 (A b+2 a B-(b B+2 A c) x) \sqrt{a+b x+c x^2}}{4 x}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}+\frac{1}{4} \left (3 \left (b^2 B+4 A b c+4 a B c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )-\frac{1}{4} \left (3 \left (4 a b B+A \left (b^2+4 a c\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{2 a+b x}{\sqrt{a+b x+c x^2}}\right )\\ &=-\frac{3 (A b+2 a B-(b B+2 A c) x) \sqrt{a+b x+c x^2}}{4 x}-\frac{(A-B x) \left (a+b x+c x^2\right )^{3/2}}{2 x^2}-\frac{3 \left (4 a b B+A \left (b^2+4 a c\right )\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{a}}+\frac{3 \left (b^2 B+4 A b c+4 a B c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.49054, size = 162, normalized size = 0.91 \[ \frac{1}{8} \left (-\frac{3 \left (A \left (4 a c+b^2\right )+4 a b B\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{\sqrt{a}}+\frac{3 \left (4 a B c+4 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{\sqrt{c}}+\frac{2 \sqrt{a+x (b+c x)} (x (A (4 c x-5 b)+B x (5 b+2 c x))-2 a (A+2 B x))}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a + x*(b + c*x)]*(-2*a*(A + 2*B*x) + x*(B*x*(5*b + 2*c*x) + A*(-5*b + 4*c*x))))/x^2 - (3*(4*a*b*B + A
*(b^2 + 4*a*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/Sqrt[a] + (3*(b^2*B + 4*A*b*c + 4*a*B*
c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c])/8

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Maple [B]  time = 0.009, size = 463, normalized size = 2.6 \begin{align*} -{\frac{A}{2\,a{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}-{\frac{Ab}{4\,{a}^{2}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{b}^{2}}{4\,{a}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,A{b}^{2}}{4\,a}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,A{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{Abcx}{4\,{a}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Abcx}{4\,a}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,Ab}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) }+{\frac{Ac}{2\,a} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Ac}{2}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,Ac}{2}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) }-{\frac{B}{ax} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{bB}{a} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{2}B}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{9\,bB}{4}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,bB}{2}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) }+{\frac{Bcx}{a} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Bcx}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,aB}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*A/a/x^2*(c*x^2+b*x+a)^(5/2)-1/4*A/a^2*b/x*(c*x^2+b*x+a)^(5/2)+1/4*A/a^2*b^2*(c*x^2+b*x+a)^(3/2)+3/4*A/a*b
^2*(c*x^2+b*x+a)^(1/2)-3/8*A/a^(1/2)*b^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/4*A/a^2*b*c*(c*x^2+b*
x+a)^(3/2)*x+3/4*A/a*b*c*(c*x^2+b*x+a)^(1/2)*x+3/2*A*b*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2
*A/a*c*(c*x^2+b*x+a)^(3/2)+3/2*A*c*(c*x^2+b*x+a)^(1/2)-3/2*A*a^(1/2)*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/
2))/x)-B/a/x*(c*x^2+b*x+a)^(5/2)+B/a*b*(c*x^2+b*x+a)^(3/2)+3/8*B*b^2/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))+9/4*B*b*(c*x^2+b*x+a)^(1/2)-3/2*B*a^(1/2)*b*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+B/a*c*(c*
x^2+b*x+a)^(3/2)*x+3/2*B*c*(c*x^2+b*x+a)^(1/2)*x+3/2*B*a*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.89007, size = 2138, normalized size = 11.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*c)*sqrt(c)*x^2*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) + 3*(4*A*a*c^2 + (4*B*a*b + A*b^2)*c)*sqrt(a)*x^2*log(-(8*a*b*x + (b^2 + 4*a*c)*x
^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 4*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A
*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2), -1/16*(6*(B*a*b^2 + 4*(B*a^2 + A*a*
b)*c)*sqrt(-c)*x^2*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)) - 3*(4*A*a*c
^2 + (4*B*a*b + A*b^2)*c)*sqrt(a)*x^2*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*
sqrt(a) + 8*a^2)/x^2) - 4*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*
sqrt(c*x^2 + b*x + a))/(a*c*x^2), 1/16*(6*(4*A*a*c^2 + (4*B*a*b + A*b^2)*c)*sqrt(-a)*x^2*arctan(1/2*sqrt(c*x^2
+ b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + 3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*c)*sqrt(c)*x^2*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*(2*B*a*c^2*x^3 - 2*A*a^2
*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2), 1/8*(3*(4*A*a*c^
2 + (4*B*a*b + A*b^2)*c)*sqrt(-a)*x^2*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x +
a^2)) - 3*(B*a*b^2 + 4*(B*a^2 + A*a*b)*c)*sqrt(-c)*x^2*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*(2*B*a*c^2*x^3 - 2*A*a^2*c - (4*B*a^2 + 5*A*a*b)*c*x + (5*B*a*b*c + 4*A*a*c^2)*x^
2)*sqrt(c*x^2 + b*x + a))/(a*c*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac{3}{2}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(3/2)/x**3, x)

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Giac [B]  time = 1.37759, size = 556, normalized size = 3.11 \begin{align*} \frac{1}{4} \,{\left (2 \, B c x + \frac{5 \, B b c + 4 \, A c^{2}}{c}\right )} \sqrt{c x^{2} + b x + a} + \frac{3 \,{\left (4 \, B a b + A b^{2} + 4 \, A a c\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a}} - \frac{3 \,{\left (B b^{2} + 4 \, B a c + 4 \, A b c\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{8 \, \sqrt{c}} + \frac{4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} B a b + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A b^{2} + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a c + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} B a^{2} \sqrt{c} + 16 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} A a b \sqrt{c} - 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{2} b - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a b^{2} + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{2} c - 8 \, B a^{3} \sqrt{c} - 8 \, A a^{2} b \sqrt{c}}{4 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*B*c*x + (5*B*b*c + 4*A*c^2)/c)*sqrt(c*x^2 + b*x + a) + 3/4*(4*B*a*b + A*b^2 + 4*A*a*c)*arctan(-(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) - 3/8*(B*b^2 + 4*B*a*c + 4*A*b*c)*log(abs(2*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + 1/4*(4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a*b + 5*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))^3*A*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*c + 8*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^2*B*a^2*sqrt(c) + 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a*b*sqrt(c) - 4*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*B*a^2*b - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a
^2*c - 8*B*a^3*sqrt(c) - 8*A*a^2*b*sqrt(c))/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^2