3.1225 \(\int \frac{(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^4} \, dx\)

Optimal. Leaf size=145 \[ -\frac{5 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{7/2} d^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}+\frac{5 (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^3 d^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3} \]

[Out]

(5*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3*d^4) - (5*(a + b*x + c*x^2)^(3/2))/(24*c^2*d^4*(b + 2*c*x)) - (a
+ b*x + c*x^2)^(5/2)/(6*c*d^4*(b + 2*c*x)^3) - (5*(b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x +
c*x^2])])/(128*c^(7/2)*d^4)

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Rubi [A]  time = 0.0695159, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {684, 612, 621, 206} \[ -\frac{5 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{7/2} d^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}+\frac{5 (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^3 d^4}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^4,x]

[Out]

(5*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3*d^4) - (5*(a + b*x + c*x^2)^(3/2))/(24*c^2*d^4*(b + 2*c*x)) - (a
+ b*x + c*x^2)^(5/2)/(6*c*d^4*(b + 2*c*x)^3) - (5*(b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x +
c*x^2])])/(128*c^(7/2)*d^4)

Rule 684

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[(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] && NeQ[b^2 - 4*a*c, 0] && 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]

Rule 612

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] - Dist[(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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^4} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3}+\frac{5 \int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^2} \, dx}{12 c d^2}\\ &=-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3}+\frac{5 \int \sqrt{a+b x+c x^2} \, dx}{16 c^2 d^4}\\ &=\frac{5 (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^3 d^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3}-\frac{\left (5 \left (b^2-4 a c\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{128 c^3 d^4}\\ &=\frac{5 (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^3 d^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3}-\frac{\left (5 \left (b^2-4 a 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 )}{64 c^3 d^4}\\ &=\frac{5 (b+2 c x) \sqrt{a+b x+c x^2}}{64 c^3 d^4}-\frac{5 \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d^4 (b+2 c x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{6 c d^4 (b+2 c x)^3}-\frac{5 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{7/2} d^4}\\ \end{align*}

Mathematica [C]  time = 0.0474012, size = 97, normalized size = 0.67 \[ -\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{192 c^3 d^4 (b+2 c x)^3 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^4,x]

[Out]

-((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -3/2, -1/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(192
*c^3*d^4*(b + 2*c*x)^3*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.196, size = 1022, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^4,x)

[Out]

-1/12/d^4/c^3/(4*a*c-b^2)/(x+1/2*b/c)^3*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)-4/3/d^4/c/(4*a*c-b^2)^2/(x+1
/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(7/2)+4/3/d^4/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5
/2)*x+2/3/d^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)*b+5/3/d^4/(4*a*c-b^2)^2*((x+1/2*b/c)^2
*c+1/4*(4*a*c-b^2)/c)^(3/2)*x*a-5/12/d^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*x*b^2+5/6/d
^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b*a-5/24/d^4/c^2/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1
/4*(4*a*c-b^2)/c)^(3/2)*b^3+5/2/d^4/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*a^2-5/4/d^4/c/(4
*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*a*b^2+5/32/d^4/c^2/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*
(4*a*c-b^2)/c)^(1/2)*x*b^4+5/4/d^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b*a^2-5/8/d^4/c^2
/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b^3*a+5/64/d^4/c^3/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4
*(4*a*c-b^2)/c)^(1/2)*b^5+5/2/d^4/c^(1/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2
)/c)^(1/2))*a^3-15/8/d^4/c^(3/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2
))*b^2*a^2+15/32/d^4/c^(5/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b
^4*a-5/128/d^4/c^(7/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b^6

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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((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 14.8094, size = 1170, normalized size = 8.07 \begin{align*} \left [-\frac{15 \,{\left (b^{5} - 4 \, a b^{3} c + 8 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 12 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 6 \,{\left (b^{4} c - 4 \, a b^{2} 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 (48 \, c^{5} x^{4} + 96 \, b c^{4} x^{3} + 15 \, b^{4} c - 40 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 32 \,{\left (4 \, b^{2} c^{3} - 7 \, a c^{4}\right )} x^{2} + 16 \,{\left (5 \, b^{3} c^{2} - 14 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{768 \,{\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}}, \frac{15 \,{\left (b^{5} - 4 \, a b^{3} c + 8 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 12 \,{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 6 \,{\left (b^{4} c - 4 \, a b^{2} 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 (48 \, c^{5} x^{4} + 96 \, b c^{4} x^{3} + 15 \, b^{4} c - 40 \, a b^{2} c^{2} - 32 \, a^{2} c^{3} + 32 \,{\left (4 \, b^{2} c^{3} - 7 \, a c^{4}\right )} x^{2} + 16 \,{\left (5 \, b^{3} c^{2} - 14 \, a b c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{384 \,{\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^4,x, algorithm="fricas")

[Out]

[-1/768*(15*(b^5 - 4*a*b^3*c + 8*(b^2*c^3 - 4*a*c^4)*x^3 + 12*(b^3*c^2 - 4*a*b*c^3)*x^2 + 6*(b^4*c - 4*a*b^2*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*(48*c
^5*x^4 + 96*b*c^4*x^3 + 15*b^4*c - 40*a*b^2*c^2 - 32*a^2*c^3 + 32*(4*b^2*c^3 - 7*a*c^4)*x^2 + 16*(5*b^3*c^2 -
14*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2*c^5*d^4*x + b^3*c^4*d^4), 1/38
4*(15*(b^5 - 4*a*b^3*c + 8*(b^2*c^3 - 4*a*c^4)*x^3 + 12*(b^3*c^2 - 4*a*b*c^3)*x^2 + 6*(b^4*c - 4*a*b^2*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*(48*c^5*x^4 + 96*
b*c^4*x^3 + 15*b^4*c - 40*a*b^2*c^2 - 32*a^2*c^3 + 32*(4*b^2*c^3 - 7*a*c^4)*x^2 + 16*(5*b^3*c^2 - 14*a*b*c^3)*
x)*sqrt(c*x^2 + b*x + a))/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2*c^5*d^4*x + b^3*c^4*d^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{b^{2} x^{2} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{c^{2} x^{4} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{2 a b x \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{2 a c x^{2} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{2 b c x^{3} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**4,x)

[Out]

(Integral(a**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4),
x) + Integral(b**2*x**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c
**4*x**4), x) + Integral(c**2*x**4*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x
**3 + 16*c**4*x**4), x) + Integral(2*a*b*x*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*
b*c**3*x**3 + 16*c**4*x**4), x) + Integral(2*a*c*x**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2
*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x) + Integral(2*b*c*x**3*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 2
4*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x))/d**4

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Giac [B]  time = 1.92642, size = 841, normalized size = 5.8 \begin{align*} \frac{1}{64} \, \sqrt{c x^{2} + b x + a}{\left (\frac{2 \, x}{c^{2} d^{4}} + \frac{b}{c^{3} d^{4}}\right )} + \frac{5 \,{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{128 \, c^{\frac{7}{2}} d^{4}} + \frac{36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} b^{4} c^{\frac{5}{2}} - 288 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} a b^{2} c^{\frac{7}{2}} + 576 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} a^{2} c^{\frac{9}{2}} + 72 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} b^{5} c^{2} - 576 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a b^{3} c^{3} + 1152 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} a^{2} b c^{4} + 66 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} b^{6} c^{\frac{3}{2}} - 576 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a b^{4} c^{\frac{5}{2}} + 1440 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a^{2} b^{2} c^{\frac{7}{2}} - 768 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} a^{3} c^{\frac{9}{2}} + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} b^{7} c - 288 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a b^{5} c^{2} + 864 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{2} b^{3} c^{3} - 768 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} a^{3} b c^{4} + 7 \, b^{8} \sqrt{c} - 82 \, a b^{6} c^{\frac{3}{2}} + 348 \, a^{2} b^{4} c^{\frac{5}{2}} - 640 \, a^{3} b^{2} c^{\frac{7}{2}} + 448 \, a^{4} c^{\frac{9}{2}}}{192 \,{\left (2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} c^{\frac{3}{2}} + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} b c + b^{2} \sqrt{c} - 2 \, a c^{\frac{3}{2}}\right )}^{3} c^{\frac{5}{2}} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^4,x, algorithm="giac")

[Out]

1/64*sqrt(c*x^2 + b*x + a)*(2*x/(c^2*d^4) + b/(c^3*d^4)) + 5/128*(b^2 - 4*a*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))*sqrt(c) + b))/(c^(7/2)*d^4) + 1/192*(36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^4*c^(5/2) - 288
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a*b^2*c^(7/2) + 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^2*c^(9/2) +
72*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^5*c^2 - 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^3*c^3 + 1152
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*b*c^4 + 66*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^6*c^(3/2) - 576*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^4*c^(5/2) + 1440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*b^2*c^(7/
2) - 768*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^3*c^(9/2) + 30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^7*c - 28
8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^5*c^2 + 864*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*b^3*c^3 - 768*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b*c^4 + 7*b^8*sqrt(c) - 82*a*b^6*c^(3/2) + 348*a^2*b^4*c^(5/2) - 640*a^3
*b^2*c^(7/2) + 448*a^4*c^(9/2))/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(3/2) + 2*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))*b*c + b^2*sqrt(c) - 2*a*c^(3/2))^3*c^(5/2)*d^4)