3.932 \(\int \frac{(A+B x) (a+b x+c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=170 \[ -\frac{3 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \sqrt{a+b x+c x^2}}{128 a^3 x^2}+\frac{3 \left (b^2-4 a c\right )^2 (A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{256 a^{7/2}}+\frac{(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5} \]

[Out]

(-3*(A*b - 2*a*B)*(b^2 - 4*a*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(128*a^3*x^2) + ((A*b - 2*a*B)*(2*a + b*x)*
(a + b*x + c*x^2)^(3/2))/(16*a^2*x^4) - (A*(a + b*x + c*x^2)^(5/2))/(5*a*x^5) + (3*(A*b - 2*a*B)*(b^2 - 4*a*c)
^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(256*a^(7/2))

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Rubi [A]  time = 0.0984788, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {806, 720, 724, 206} \[ -\frac{3 \left (b^2-4 a c\right ) (2 a+b x) (A b-2 a B) \sqrt{a+b x+c x^2}}{128 a^3 x^2}+\frac{3 \left (b^2-4 a c\right )^2 (A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{256 a^{7/2}}+\frac{(2 a+b x) (A b-2 a B) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6,x]

[Out]

(-3*(A*b - 2*a*B)*(b^2 - 4*a*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(128*a^3*x^2) + ((A*b - 2*a*B)*(2*a + b*x)*
(a + b*x + c*x^2)^(3/2))/(16*a^2*x^4) - (A*(a + b*x + c*x^2)^(5/2))/(5*a*x^5) + (3*(A*b - 2*a*B)*(b^2 - 4*a*c)
^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(256*a^(7/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), 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] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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{(A+B x) \left (a+b x+c x^2\right )^{3/2}}{x^6} \, dx &=-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}-\frac{(A b-2 a B) \int \frac{\left (a+b x+c x^2\right )^{3/2}}{x^5} \, dx}{2 a}\\ &=\frac{(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac{\left (3 (A b-2 a B) \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx}{32 a^2}\\ &=-\frac{3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{128 a^3 x^2}+\frac{(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}-\frac{\left (3 (A b-2 a B) \left (b^2-4 a c\right )^2\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{256 a^3}\\ &=-\frac{3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{128 a^3 x^2}+\frac{(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac{\left (3 (A b-2 a B) \left (b^2-4 a c\right )^2\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 )}{128 a^3}\\ &=-\frac{3 (A b-2 a B) \left (b^2-4 a c\right ) (2 a+b x) \sqrt{a+b x+c x^2}}{128 a^3 x^2}+\frac{(A b-2 a B) (2 a+b x) \left (a+b x+c x^2\right )^{3/2}}{16 a^2 x^4}-\frac{A \left (a+b x+c x^2\right )^{5/2}}{5 a x^5}+\frac{3 (A b-2 a B) \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{256 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.280297, size = 157, normalized size = 0.92 \[ \frac{(A b-2 a B) \left (16 a^{3/2} (2 a+b x) (a+x (b+c x))^{3/2}-3 x^2 \left (b^2-4 a c\right ) \left (2 \sqrt{a} (2 a+b x) \sqrt{a+x (b+c x)}-x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )\right )\right )}{256 a^{7/2} x^4}-\frac{A (a+x (b+c x))^{5/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^(3/2))/x^6,x]

[Out]

-(A*(a + x*(b + c*x))^(5/2))/(5*a*x^5) + ((A*b - 2*a*B)*(16*a^(3/2)*(2*a + b*x)*(a + x*(b + c*x))^(3/2) - 3*(b
^2 - 4*a*c)*x^2*(2*Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*x^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a
]*Sqrt[a + x*(b + c*x)])])))/(256*a^(7/2)*x^4)

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Maple [B]  time = 0.017, size = 978, normalized size = 5.8 \begin{align*} \text{result too large to display} \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^6,x)

[Out]

-1/5*A*(c*x^2+b*x+a)^(5/2)/a/x^5+3/32*A/a^3*b^2*c^2*(c*x^2+b*x+a)^(1/2)*x+1/16*A/a^3*b*c/x^2*(c*x^2+b*x+a)^(5/
2)+3/32*A/a^4*b^2*c^2*(c*x^2+b*x+a)^(3/2)*x-3/32*A/a^4*b^2*c/x*(c*x^2+b*x+a)^(5/2)-1/128*A/a^5*b^4*c*(c*x^2+b*
x+a)^(3/2)*x-3/128*A/a^4*b^4*c*(c*x^2+b*x+a)^(1/2)*x-3/16*B/a^3*b*c^2*(c*x^2+b*x+a)^(3/2)*x+1/64*B/a^4*b^3*c*(
c*x^2+b*x+a)^(3/2)*x+3/64*B/a^3*b^3*c*(c*x^2+b*x+a)^(1/2)*x+3/16*B/a^3*b*c/x*(c*x^2+b*x+a)^(5/2)-3/16*B/a^2*b*
c^2*(c*x^2+b*x+a)^(1/2)*x-3/8*B/a^(1/2)*c^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-3/128*B/a^(5/2)*b^4*
ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/8*B/a^2*c^2*(c*x^2+b*x+a)^(3/2)+3/8*B/a*c^2*(c*x^2+b*x+a)^(1/2
)+1/64*B/a^4*b^4*(c*x^2+b*x+a)^(3/2)+3/64*B/a^3*b^4*(c*x^2+b*x+a)^(1/2)-1/4*B/a/x^4*(c*x^2+b*x+a)^(5/2)+3/256*
A/a^(7/2)*b^5*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-1/128*A/a^5*b^5*(c*x^2+b*x+a)^(3/2)-3/128*A/a^4*b^
5*(c*x^2+b*x+a)^(1/2)+1/8*B/a^2*b/x^3*(c*x^2+b*x+a)^(5/2)-1/32*B/a^3*b^2/x^2*(c*x^2+b*x+a)^(5/2)-1/64*B/a^4*b^
3/x*(c*x^2+b*x+a)^(5/2)-5/32*B/a^3*b^2*c*(c*x^2+b*x+a)^(3/2)-9/32*B/a^2*b^2*c*(c*x^2+b*x+a)^(1/2)+1/8*A/a^2*b/
x^4*(c*x^2+b*x+a)^(5/2)+3/16*A/a^(3/2)*b*c^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-1/16*A/a^3*b^2/x^3*
(c*x^2+b*x+a)^(5/2)+1/64*A/a^4*b^3/x^2*(c*x^2+b*x+a)^(5/2)+1/128*A/a^5*b^4/x*(c*x^2+b*x+a)^(5/2)+5/64*A/a^4*b^
3*c*(c*x^2+b*x+a)^(3/2)+9/64*A/a^3*b^3*c*(c*x^2+b*x+a)^(1/2)-1/16*A/a^3*b*c^2*(c*x^2+b*x+a)^(3/2)-3/16*A/a^2*b
*c^2*(c*x^2+b*x+a)^(1/2)-3/32*A/a^(5/2)*b^3*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+3/16*B/a^(3/2)*b^2
*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-1/8*B/a^2*c/x^2*(c*x^2+b*x+a)^(5/2)

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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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 12.6327, size = 1258, normalized size = 7.4 \begin{align*} \left [\frac{15 \,{\left (2 \, B a b^{4} - A b^{5} + 16 \,{\left (2 \, B a^{3} - A a^{2} b\right )} c^{2} - 8 \,{\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} c\right )} \sqrt{a} x^{5} \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 (128 \, A a^{5} -{\left (30 \, B a^{2} b^{3} - 15 \, A a b^{4} - 128 \, A a^{3} c^{2} - 100 \,{\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} c\right )} x^{4} + 2 \,{\left (10 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} + 4 \,{\left (50 \, B a^{4} + 7 \, A a^{3} b\right )} c\right )} x^{3} + 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2} + 32 \, A a^{4} c\right )} x^{2} + 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{2560 \, a^{4} x^{5}}, \frac{15 \,{\left (2 \, B a b^{4} - A b^{5} + 16 \,{\left (2 \, B a^{3} - A a^{2} b\right )} c^{2} - 8 \,{\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} c\right )} \sqrt{-a} x^{5} \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 (128 \, A a^{5} -{\left (30 \, B a^{2} b^{3} - 15 \, A a b^{4} - 128 \, A a^{3} c^{2} - 100 \,{\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} c\right )} x^{4} + 2 \,{\left (10 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} + 4 \,{\left (50 \, B a^{4} + 7 \, A a^{3} b\right )} c\right )} x^{3} + 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2} + 32 \, A a^{4} c\right )} x^{2} + 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{1280 \, a^{4} x^{5}}\right ] \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^6,x, algorithm="fricas")

[Out]

[1/2560*(15*(2*B*a*b^4 - A*b^5 + 16*(2*B*a^3 - A*a^2*b)*c^2 - 8*(2*B*a^2*b^2 - A*a*b^3)*c)*sqrt(a)*x^5*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*(128*A*a^5 - (30*B*
a^2*b^3 - 15*A*a*b^4 - 128*A*a^3*c^2 - 100*(2*B*a^3*b - A*a^2*b^2)*c)*x^4 + 2*(10*B*a^3*b^2 - 5*A*a^2*b^3 + 4*
(50*B*a^4 + 7*A*a^3*b)*c)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2 + 32*A*a^4*c)*x^2 + 16*(10*B*a^5 + 11*A*a^4*b)*x)*sq
rt(c*x^2 + b*x + a))/(a^4*x^5), 1/1280*(15*(2*B*a*b^4 - A*b^5 + 16*(2*B*a^3 - A*a^2*b)*c^2 - 8*(2*B*a^2*b^2 -
A*a*b^3)*c)*sqrt(-a)*x^5*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*(1
28*A*a^5 - (30*B*a^2*b^3 - 15*A*a*b^4 - 128*A*a^3*c^2 - 100*(2*B*a^3*b - A*a^2*b^2)*c)*x^4 + 2*(10*B*a^3*b^2 -
5*A*a^2*b^3 + 4*(50*B*a^4 + 7*A*a^3*b)*c)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2 + 32*A*a^4*c)*x^2 + 16*(10*B*a^5 +
11*A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^5)]

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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^{6}}\, 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**6,x)

[Out]

Integral((A + B*x)*(a + b*x + c*x**2)**(3/2)/x**6, x)

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Giac [B]  time = 1.32266, size = 1832, normalized size = 10.78 \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^6,x, algorithm="giac")

[Out]

3/128*(2*B*a*b^4 - A*b^5 - 16*B*a^2*b^2*c + 8*A*a*b^3*c + 32*B*a^3*c^2 - 16*A*a^2*b*c^2)*arctan(-(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^3) - 1/640*(30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a*b^4 - 15
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*b^5 - 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^2*b^2*c + 120*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a*b^3*c - 800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^3*c^2 - 240*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^2*b*c^2 - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*B*a^3*b*c^(3/2) - 12
80*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^8*A*a^3*c^(5/2) - 140*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b^4 +
70*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a*b^5 - 1440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^3*b^2*c - 5
60*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*b^3*c + 320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^4*c^2 - 2
720*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b*c^2 - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^3*b^3*s
qrt(c) + 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^4*b*c^(3/2) - 5120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^6*A*a^3*b^2*c^(3/2) - 128*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^2*b^5 - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^5*A*a^3*b^3*c - 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^4*b*c^2 + 1280*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^4*B*a^4*b^3*sqrt(c) - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^3*b^4*sqrt(c) - 2560*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^4*B*a^5*b*c^(3/2) - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^4*b^2*c^(3/2) -
2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^5*c^(5/2) + 140*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b^
4 - 70*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^3*b^5 + 1440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^5*b^2*
c - 2000*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*b^3*c - 320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^6*c
^2 - 2400*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^5*b*c^2 + 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^6
*b*c^(3/2) - 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^5*b^2*c^(3/2) - 30*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*B*a^5*b^4 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^4*b^5 + 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*
a^6*b^2*c - 120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^5*b^3*c + 800*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^
7*c^2 - 1040*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^6*b*c^2 - 256*A*a^7*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 +
b*x + a))^2 - a)^5*a^3)