### 3.286 $$\int \frac{d+e x+f x^2+g x^3}{x^4 \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=186 $\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right ) \left (8 a^2 (c e-2 a g)-6 a b^2 e-4 a b (3 c d-2 a f)+5 b^3 d\right )}{16 a^{7/2}}-\frac{\sqrt{a+b x+c x^2} \left (24 a^2 f-18 a b e-16 a c d+15 b^2 d\right )}{24 a^3 x}+\frac{\sqrt{a+b x+c x^2} (5 b d-6 a e)}{12 a^2 x^2}-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}$

[Out]

-(d*Sqrt[a + b*x + c*x^2])/(3*a*x^3) + ((5*b*d - 6*a*e)*Sqrt[a + b*x + c*x^2])/(12*a^2*x^2) - ((15*b^2*d - 16*
a*c*d - 18*a*b*e + 24*a^2*f)*Sqrt[a + b*x + c*x^2])/(24*a^3*x) + ((5*b^3*d - 6*a*b^2*e - 4*a*b*(3*c*d - 2*a*f)
+ 8*a^2*(c*e - 2*a*g))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(7/2))

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Rubi [A]  time = 0.319922, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.121, Rules used = {1650, 806, 724, 206} $\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right ) \left (8 a^2 (c e-2 a g)-6 a b^2 e-4 a b (3 c d-2 a f)+5 b^3 d\right )}{16 a^{7/2}}-\frac{\sqrt{a+b x+c x^2} \left (24 a^2 f-18 a b e-16 a c d+15 b^2 d\right )}{24 a^3 x}+\frac{\sqrt{a+b x+c x^2} (5 b d-6 a e)}{12 a^2 x^2}-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x + f*x^2 + g*x^3)/(x^4*Sqrt[a + b*x + c*x^2]),x]

[Out]

-(d*Sqrt[a + b*x + c*x^2])/(3*a*x^3) + ((5*b*d - 6*a*e)*Sqrt[a + b*x + c*x^2])/(12*a^2*x^2) - ((15*b^2*d - 16*
a*c*d - 18*a*b*e + 24*a^2*f)*Sqrt[a + b*x + c*x^2])/(24*a^3*x) + ((5*b^3*d - 6*a*b^2*e - 4*a*b*(3*c*d - 2*a*f)
+ 8*a^2*(c*e - 2*a*g))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(7/2))

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 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]

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])

Rubi steps

\begin{align*} \int \frac{d+e x+f x^2+g x^3}{x^4 \sqrt{a+b x+c x^2}} \, dx &=-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}-\frac{\int \frac{\frac{1}{2} (5 b d-6 a e)+(2 c d-3 a f) x-3 a g x^2}{x^3 \sqrt{a+b x+c x^2}} \, dx}{3 a}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}+\frac{(5 b d-6 a e) \sqrt{a+b x+c x^2}}{12 a^2 x^2}+\frac{\int \frac{\frac{1}{4} \left (15 b^2 d-16 a c d-18 a b e+24 a^2 f\right )+\frac{1}{2} \left (5 b c d-6 a c e+12 a^2 g\right ) x}{x^2 \sqrt{a+b x+c x^2}} \, dx}{6 a^2}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}+\frac{(5 b d-6 a e) \sqrt{a+b x+c x^2}}{12 a^2 x^2}-\frac{\left (15 b^2 d-16 a c d-18 a b e+24 a^2 f\right ) \sqrt{a+b x+c x^2}}{24 a^3 x}-\frac{\left (5 b^3 d-6 a b^2 e-4 a b (3 c d-2 a f)+8 a^2 (c e-2 a g)\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{16 a^3}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}+\frac{(5 b d-6 a e) \sqrt{a+b x+c x^2}}{12 a^2 x^2}-\frac{\left (15 b^2 d-16 a c d-18 a b e+24 a^2 f\right ) \sqrt{a+b x+c x^2}}{24 a^3 x}+\frac{\left (5 b^3 d-6 a b^2 e-4 a b (3 c d-2 a f)+8 a^2 (c e-2 a g)\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 )}{8 a^3}\\ &=-\frac{d \sqrt{a+b x+c x^2}}{3 a x^3}+\frac{(5 b d-6 a e) \sqrt{a+b x+c x^2}}{12 a^2 x^2}-\frac{\left (15 b^2 d-16 a c d-18 a b e+24 a^2 f\right ) \sqrt{a+b x+c x^2}}{24 a^3 x}+\frac{\left (5 b^3 d-6 a b^2 e-4 a b (3 c d-2 a f)+8 a^2 (c e-2 a g)\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.313395, size = 150, normalized size = 0.81 $\frac{\tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right ) \left (8 a^2 (c e-2 a g)-6 a b^2 e+4 a b (2 a f-3 c d)+5 b^3 d\right )}{16 a^{7/2}}-\frac{\sqrt{a+x (b+c x)} \left (4 a^2 (2 d+3 x (e+2 f x))-2 a x (5 b d+9 b e x+8 c d x)+15 b^2 d x^2\right )}{24 a^3 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3)/(x^4*Sqrt[a + b*x + c*x^2]),x]

[Out]

-(Sqrt[a + x*(b + c*x)]*(15*b^2*d*x^2 - 2*a*x*(5*b*d + 8*c*d*x + 9*b*e*x) + 4*a^2*(2*d + 3*x*(e + 2*f*x))))/(2
4*a^3*x^3) + ((5*b^3*d - 6*a*b^2*e + 4*a*b*(-3*c*d + 2*a*f) + 8*a^2*(c*e - 2*a*g))*ArcTanh[(2*a + b*x)/(2*Sqrt
[a]*Sqrt[a + x*(b + c*x)])])/(16*a^(7/2))

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Maple [B]  time = 0.057, size = 375, normalized size = 2. \begin{align*} -{g\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{d}{3\,a{x}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,bd}{12\,{a}^{2}{x}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,{b}^{2}d}{8\,x{a}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{3}d}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}-{\frac{3\,bcd}{4}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{2\,cd}{3\,{a}^{2}x}\sqrt{c{x}^{2}+bx+a}}-{\frac{e}{2\,a{x}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,be}{4\,{a}^{2}x}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{b}^{2}e}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{ce}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{f}{ax}\sqrt{c{x}^{2}+bx+a}}+{\frac{bf}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^3+f*x^2+e*x+d)/x^4/(c*x^2+b*x+a)^(1/2),x)

[Out]

-g/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-1/3*d*(c*x^2+b*x+a)^(1/2)/a/x^3+5/12*d*b/a^2/x^2*(c*x
^2+b*x+a)^(1/2)-5/8*d*b^2/a^3/x*(c*x^2+b*x+a)^(1/2)+5/16*d*b^3/a^(7/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/
2))/x)-3/4*d*b/a^(5/2)*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+2/3*d*c/a^2/x*(c*x^2+b*x+a)^(1/2)-1/2*e
/a/x^2*(c*x^2+b*x+a)^(1/2)+3/4*e*b/a^2/x*(c*x^2+b*x+a)^(1/2)-3/8*e*b^2/a^(5/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*
x+a)^(1/2))/x)+1/2*e*c/a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-f/a/x*(c*x^2+b*x+a)^(1/2)+1/2*f*b
/a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)

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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((g*x^3+f*x^2+e*x+d)/x^4/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 21.0011, size = 848, normalized size = 4.56 \begin{align*} \left [-\frac{3 \,{\left (8 \, a^{2} b f - 16 \, a^{3} g +{\left (5 \, b^{3} - 12 \, a b c\right )} d - 2 \,{\left (3 \, a b^{2} - 4 \, a^{2} c\right )} e\right )} \sqrt{a} x^{3} \log \left (-\frac{8 \, a b x +{\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{2}}\right ) + 4 \,{\left (8 \, a^{3} d -{\left (18 \, a^{2} b e - 24 \, a^{3} f -{\left (15 \, a b^{2} - 16 \, a^{2} c\right )} d\right )} x^{2} - 2 \,{\left (5 \, a^{2} b d - 6 \, a^{3} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{96 \, a^{4} x^{3}}, -\frac{3 \,{\left (8 \, a^{2} b f - 16 \, a^{3} g +{\left (5 \, b^{3} - 12 \, a b c\right )} d - 2 \,{\left (3 \, a b^{2} - 4 \, a^{2} c\right )} e\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{2} + a b x + a^{2}\right )}}\right ) + 2 \,{\left (8 \, a^{3} d -{\left (18 \, a^{2} b e - 24 \, a^{3} f -{\left (15 \, a b^{2} - 16 \, a^{2} c\right )} d\right )} x^{2} - 2 \,{\left (5 \, a^{2} b d - 6 \, a^{3} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{48 \, a^{4} x^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/x^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(8*a^2*b*f - 16*a^3*g + (5*b^3 - 12*a*b*c)*d - 2*(3*a*b^2 - 4*a^2*c)*e)*sqrt(a)*x^3*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*(8*a^3*d - (18*a^2*b*e - 24*
a^3*f - (15*a*b^2 - 16*a^2*c)*d)*x^2 - 2*(5*a^2*b*d - 6*a^3*e)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^3), -1/48*(3*(
8*a^2*b*f - 16*a^3*g + (5*b^3 - 12*a*b*c)*d - 2*(3*a*b^2 - 4*a^2*c)*e)*sqrt(-a)*x^3*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)) + 2*(8*a^3*d - (18*a^2*b*e - 24*a^3*f - (15*a*b^2 - 16*a^
2*c)*d)*x^2 - 2*(5*a^2*b*d - 6*a^3*e)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x + f x^{2} + g x^{3}}{x^{4} \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**3+f*x**2+e*x+d)/x**4/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x + f*x**2 + g*x**3)/(x**4*sqrt(a + b*x + c*x**2)), x)

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Giac [B]  time = 1.18562, size = 930, normalized size = 5. \begin{align*} -\frac{{\left (5 \, b^{3} d - 12 \, a b c d + 8 \, a^{2} b f - 16 \, a^{3} g - 6 \, a b^{2} e + 8 \, a^{2} c e\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{3}} + \frac{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} b^{3} d - 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} a b c d + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} a^{2} b f - 18 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} a b^{2} e + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} a^{2} c e + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} a^{3} \sqrt{c} f - 40 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a b^{3} d + 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a^{2} b c d - 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a^{3} b f + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a^{2} b^{2} e + 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a^{3} c^{\frac{3}{2}} d - 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a^{4} \sqrt{c} f + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a^{3} b \sqrt{c} e + 33 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{2} b^{3} d + 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{3} b c d + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{4} b f - 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{3} b^{2} e - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{4} c e + 48 \, a^{3} b^{2} \sqrt{c} d - 32 \, a^{4} c^{\frac{3}{2}} d + 48 \, a^{5} \sqrt{c} f - 48 \, a^{4} b \sqrt{c} e}{24 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )}^{3} a^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/x^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/8*(5*b^3*d - 12*a*b*c*d + 8*a^2*b*f - 16*a^3*g - 6*a*b^2*e + 8*a^2*c*e)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))/sqrt(-a))/(sqrt(-a)*a^3) + 1/24*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^3*d - 36*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))^5*a*b*c*d + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*b*f - 18*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^5*a*b^2*e + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*c*e + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^4*a^3*sqrt(c)*f - 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^3*d + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)^3*a^2*b*c*d - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^3*b*f + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^
2*b^2*e + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^3*c^(3/2)*d - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^
4*sqrt(c)*f + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^3*b*sqrt(c)*e + 33*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)*a^2*b^3*d + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b*c*d + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^4*b*
f - 30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b^2*e - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^4*c*e + 48*a^3
*b^2*sqrt(c)*d - 32*a^4*c^(3/2)*d + 48*a^5*sqrt(c)*f - 48*a^4*b*sqrt(c)*e)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^2 - a)^3*a^3)