3.1197 \(\int \frac{A+B x}{(d+e x)^3 \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{\left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 d^{5/2} (c d-b e)^{5/2}}-\frac{\sqrt{b x+c x^2} (3 A e (2 c d-b e)-B d (b e+2 c d))}{4 d^2 (d+e x) (c d-b e)^2}+\frac{\sqrt{b x+c x^2} (B d-A e)}{2 d (d+e x)^2 (c d-b e)} \]

[Out]

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

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Rubi [A]  time = 0.263283, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {834, 806, 724, 206} \[ \frac{\left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{8 d^{5/2} (c d-b e)^{5/2}}-\frac{\sqrt{b x+c x^2} (3 A e (2 c d-b e)-B d (b e+2 c d))}{4 d^2 (d+e x) (c d-b e)^2}+\frac{\sqrt{b x+c x^2} (B d-A e)}{2 d (d+e x)^2 (c d-b e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(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*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

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{A+B x}{(d+e x)^3 \sqrt{b x+c x^2}} \, dx &=\frac{(B d-A e) \sqrt{b x+c x^2}}{2 d (c d-b e) (d+e x)^2}-\frac{\int \frac{\frac{1}{2} (b B d-4 A c d+3 A b e)-c (B d-A e) x}{(d+e x)^2 \sqrt{b x+c x^2}} \, dx}{2 d (c d-b e)}\\ &=\frac{(B d-A e) \sqrt{b x+c x^2}}{2 d (c d-b e) (d+e x)^2}-\frac{(3 A e (2 c d-b e)-B d (2 c d+b e)) \sqrt{b x+c x^2}}{4 d^2 (c d-b e)^2 (d+e x)}+\frac{\left (8 A c^2 d^2-4 b c d (B d+2 A e)+b^2 e (B d+3 A e)\right ) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{8 d^2 (c d-b e)^2}\\ &=\frac{(B d-A e) \sqrt{b x+c x^2}}{2 d (c d-b e) (d+e x)^2}-\frac{(3 A e (2 c d-b e)-B d (2 c d+b e)) \sqrt{b x+c x^2}}{4 d^2 (c d-b e)^2 (d+e x)}-\frac{\left (8 A c^2 d^2-4 b c d (B d+2 A e)+b^2 e (B d+3 A e)\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac{-b d-(2 c d-b e) x}{\sqrt{b x+c x^2}}\right )}{4 d^2 (c d-b e)^2}\\ &=\frac{(B d-A e) \sqrt{b x+c x^2}}{2 d (c d-b e) (d+e x)^2}-\frac{(3 A e (2 c d-b e)-B d (2 c d+b e)) \sqrt{b x+c x^2}}{4 d^2 (c d-b e)^2 (d+e x)}+\frac{\left (8 A c^2 d^2-4 b c d (B d+2 A e)+b^2 e (B d+3 A e)\right ) \tanh ^{-1}\left (\frac{b d+(2 c d-b e) x}{2 \sqrt{d} \sqrt{c d-b e} \sqrt{b x+c x^2}}\right )}{8 d^{5/2} (c d-b e)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.484227, size = 217, normalized size = 1. \[ \frac{\sqrt{x} \left (\frac{\sqrt{b+c x} \left (b^2 e (3 A e+B d)-4 b c d (2 A e+B d)+8 A c^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{2 d^{3/2} (b e-c d)^{3/2}}-\frac{\sqrt{x} (b+c x) (3 A e (b e-2 c d)+B d (b e+2 c d))}{2 d (d+e x) (c d-b e)}+\frac{\sqrt{x} (b+c x) (A e-B d)}{(d+e x)^2}\right )}{2 d \sqrt{x (b+c x)} (b e-c d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(((-(B*d) + A*e)*Sqrt[x]*(b + c*x))/(d + e*x)^2 - ((3*A*e*(-2*c*d + b*e) + B*d*(2*c*d + b*e))*Sqrt[x]
*(b + c*x))/(2*d*(c*d - b*e)*(d + e*x)) + ((8*A*c^2*d^2 - 4*b*c*d*(B*d + 2*A*e) + b^2*e*(B*d + 3*A*e))*Sqrt[b
+ c*x]*ArcTan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(2*d^(3/2)*(-(c*d) + b*e)^(3/2))))/(2*d*(
-(c*d) + b*e)*Sqrt[x*(b + c*x)])

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Maple [B]  time = 0.013, size = 1821, normalized size = 8.4 \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)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x)

[Out]

1/2/e/d/(b*e-c*d)/(x+d/e)^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*A-1/2/e^2/(b*e-c*d)/(x+d
/e)^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*B+3/4*e/d^2/(b*e-c*d)^2/(x+d/e)*((x+d/e)^2*c+(
b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*b*A-3/4/d/(b*e-c*d)^2/(x+d/e)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d
*(b*e-c*d)/e^2)^(1/2)*b*B-3/2/d/(b*e-c*d)^2/(x+d/e)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*
c*A+3/2/e/(b*e-c*d)^2/(x+d/e)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)*c*B-3/8*e/d^2/(b*e-c*d
)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^
2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*b^2*A+3/8/d/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln
((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*
e-c*d)/e^2)^(1/2))/(x+d/e))*b^2*B+3/2/d/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d
)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*b*c
*A-3/2/e/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2
)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*b*c*B-3/2/e/(b*e-c*d)^2/(-d*(b*e-c
*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d
)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*c^2*A+3/2/e^2/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-
c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^
(1/2))/(x+d/e))*c^2*B*d-1/2/e*c/d/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d
/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*A+3/2/e^2*c
/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((
x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d/e))*B+B/e/d/(b*e-c*d)/(x+d/e)*((x+d/e)^2*c+(b*e-
2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2)-1/2*B/e/d/(b*e-c*d)/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b
*e-2*c*d)/e*(x+d/e)+2*(-d*(b*e-c*d)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)-d*(b*e-c*d)/e^2)^(1/2))/(x+d
/e))*b

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.81227, size = 1955, normalized size = 9.05 \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)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/8*((3*A*b^2*d^2*e^2 - 4*(B*b*c - 2*A*c^2)*d^4 + (B*b^2 - 8*A*b*c)*d^3*e + (3*A*b^2*e^4 - 4*(B*b*c - 2*A*c^2
)*d^2*e^2 + (B*b^2 - 8*A*b*c)*d*e^3)*x^2 + 2*(3*A*b^2*d*e^3 - 4*(B*b*c - 2*A*c^2)*d^3*e + (B*b^2 - 8*A*b*c)*d^
2*e^2)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d))
+ 2*(4*B*c^2*d^5 - 5*A*b^2*d^2*e^3 - (5*B*b*c + 8*A*c^2)*d^4*e + (B*b^2 + 13*A*b*c)*d^3*e^2 + (2*B*c^2*d^4*e
- 3*A*b^2*d*e^4 - (B*b*c + 6*A*c^2)*d^3*e^2 - (B*b^2 - 9*A*b*c)*d^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^3*d^8 - 3*b*
c^2*d^7*e + 3*b^2*c*d^6*e^2 - b^3*d^5*e^3 + (c^3*d^6*e^2 - 3*b*c^2*d^5*e^3 + 3*b^2*c*d^4*e^4 - b^3*d^3*e^5)*x^
2 + 2*(c^3*d^7*e - 3*b*c^2*d^6*e^2 + 3*b^2*c*d^5*e^3 - b^3*d^4*e^4)*x), 1/4*((3*A*b^2*d^2*e^2 - 4*(B*b*c - 2*A
*c^2)*d^4 + (B*b^2 - 8*A*b*c)*d^3*e + (3*A*b^2*e^4 - 4*(B*b*c - 2*A*c^2)*d^2*e^2 + (B*b^2 - 8*A*b*c)*d*e^3)*x^
2 + 2*(3*A*b^2*d*e^3 - 4*(B*b*c - 2*A*c^2)*d^3*e + (B*b^2 - 8*A*b*c)*d^2*e^2)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-
sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) + (4*B*c^2*d^5 - 5*A*b^2*d^2*e^3 - (5*B*b*c + 8*A*c^2)
*d^4*e + (B*b^2 + 13*A*b*c)*d^3*e^2 + (2*B*c^2*d^4*e - 3*A*b^2*d*e^4 - (B*b*c + 6*A*c^2)*d^3*e^2 - (B*b^2 - 9*
A*b*c)*d^2*e^3)*x)*sqrt(c*x^2 + b*x))/(c^3*d^8 - 3*b*c^2*d^7*e + 3*b^2*c*d^6*e^2 - b^3*d^5*e^3 + (c^3*d^6*e^2
- 3*b*c^2*d^5*e^3 + 3*b^2*c*d^4*e^4 - b^3*d^3*e^5)*x^2 + 2*(c^3*d^7*e - 3*b*c^2*d^6*e^2 + 3*b^2*c*d^5*e^3 - b^
3*d^4*e^4)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**3/(c*x**2+b*x)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.45274, size = 1038, normalized size = 4.81 \begin{align*} -\frac{{\left (4 \, B b c d^{2} - 8 \, A c^{2} d^{2} - B b^{2} d e + 8 \, A b c d e - 3 \, A b^{2} e^{2}\right )} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} + b d e}}\right )}{4 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \sqrt{-c d^{2} + b d e}} + \frac{8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B c^{\frac{5}{2}} d^{4} - 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b c^{\frac{3}{2}} d^{3} e - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A c^{\frac{5}{2}} d^{3} e + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} B b c^{2} d^{4} + 4 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b c d^{2} e^{2} - 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A c^{2} d^{2} e^{2} - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b c^{2} d^{3} e + 2 \, B b^{2} c^{\frac{3}{2}} d^{4} + 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{2} \sqrt{c} d^{2} e^{2} + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b c^{\frac{3}{2}} d^{2} e^{2} + B b^{3} \sqrt{c} d^{3} e - 6 \, A b^{2} c^{\frac{3}{2}} d^{3} e -{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{2} d e^{3} + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b c d e^{3} +{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} B b^{3} d^{2} e^{2} + 20 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{2} c d^{2} e^{2} - 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{2} \sqrt{c} d e^{3} + 3 \, A b^{3} \sqrt{c} d^{2} e^{2} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{2} e^{4} - 5 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{3} d e^{3}}{4 \,{\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )}{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} d + b d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*B*b*c*d^2 - 8*A*c^2*d^2 - B*b^2*d*e + 8*A*b*c*d*e - 3*A*b^2*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*
x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*sqrt(-c*d^2 + b*d*e)) + 1/4*(8
*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*c^(5/2)*d^4 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b*c^(3/2)*d^3*e - 24*
(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*c^(5/2)*d^3*e + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b*c^2*d^4 + 4*(sqrt(c)
*x - sqrt(c*x^2 + b*x))^3*B*b*c*d^2*e^2 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*c^2*d^2*e^2 - 24*(sqrt(c)*x -
sqrt(c*x^2 + b*x))*A*b*c^2*d^3*e + 2*B*b^2*c^(3/2)*d^4 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2*sqrt(c)*d^2
*e^2 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b*c^(3/2)*d^2*e^2 + B*b^3*sqrt(c)*d^3*e - 6*A*b^2*c^(3/2)*d^3*e
- (sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^2*d*e^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b*c*d*e^3 + (sqrt(c)*x
- sqrt(c*x^2 + b*x))*B*b^3*d^2*e^2 + 20*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*c*d^2*e^2 - 9*(sqrt(c)*x - sqrt
(c*x^2 + b*x))^2*A*b^2*sqrt(c)*d*e^3 + 3*A*b^3*sqrt(c)*d^2*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^2*e^4
- 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*d*e^3)/((c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*((sqrt(c)*x - sqr
t(c*x^2 + b*x))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c)*d + b*d)^2)