3.2361 \(\int \frac{(a+b x+c x^2)^{5/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=331 \[ \frac{5 \sqrt{a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{8 e^5}-\frac{5 (2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c} e^6}+\frac{5 \sqrt{a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{8 e^6}+\frac{5 \left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.548146, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {732, 812, 814, 843, 621, 206, 724} \[ \frac{5 \sqrt{a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{8 e^5}-\frac{5 (2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c} e^6}+\frac{5 \sqrt{a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{8 e^6}+\frac{5 \left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

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

Rule 732

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[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 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{\left (a+b x+c x^2\right )^{5/2}}{(d+e x)^3} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{5 \int \frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^2} \, dx}{4 e}\\ &=\frac{5 (8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 \int \frac{\left (8 b c d-3 b^2 e-4 a c e+8 c (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx}{8 e^3}\\ &=\frac{5 \left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}+\frac{5 \int \frac{2 c \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-2 c (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{32 c e^5}\\ &=\frac{5 \left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{\left (5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 e^6}+\frac{\left (5 \left (2 c d (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+2 c e \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{32 c e^6}\\ &=\frac{5 \left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{\left (5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\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 )}{8 e^6}-\frac{\left (5 \left (2 c d (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+2 c e \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac{-b d+2 a e-(2 c d-b e) x}{\sqrt{a+b x+c x^2}}\right )}{16 c e^6}\\ &=\frac{5 \left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{8 e^5}+\frac{5 (8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)}-\frac{\left (a+b x+c x^2\right )^{5/2}}{2 e (d+e x)^2}-\frac{5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 \sqrt{c} e^6}+\frac{5 \sqrt{c d^2-b d e+a e^2} \left (16 c^2 d^2-16 b c d e+3 b^2 e^2+4 a c e^2\right ) \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x}{2 \sqrt{c d^2-b d e+a e^2} \sqrt{a+b x+c x^2}}\right )}{8 e^6}\\ \end{align*}

Mathematica [A]  time = 1.97118, size = 445, normalized size = 1.34 \[ \frac{\frac{5 \left (\frac{-2 c^2 e \sqrt{a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c e (a e-5 b d+b e x)+5 b^2 e^2+8 c^2 d (2 d-e x)\right )+2 c^2 \left (4 c e (a e-4 b d)+3 b^2 e^2+16 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+c^{3/2} (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \left (4 c e (3 a e-4 b d)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{4 c^2 e^5}-\frac{(a+x (b+c x))^{3/2} \left (c e (2 a e-11 b d+3 b e x)+3 b^2 e^2+2 c^2 d (4 d-3 e x)\right )}{3 e^2}\right )}{e (b d-a e)-c d^2}+\frac{5 (a+x (b+c x))^{5/2} (2 c d-b e)}{(d+e x) \left (e (a e-b d)+c d^2\right )}-\frac{2 (a+x (b+c x))^{5/2}}{(d+e x)^2}}{4 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

((-2*(a + x*(b + c*x))^(5/2))/(d + e*x)^2 + (5*(2*c*d - b*e)*(a + x*(b + c*x))^(5/2))/((c*d^2 + e*(-(b*d) + a*
e))*(d + e*x)) + (5*(-((a + x*(b + c*x))^(3/2)*(3*b^2*e^2 + 2*c^2*d*(4*d - 3*e*x) + c*e*(-11*b*d + 2*a*e + 3*b
*e*x)))/(3*e^2) + (-2*c^2*e*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b + c*x)]*(5*b^2*e^2 + 8*c^2*d*(2*d - e*x)
+ 4*c*e*(-5*b*d + a*e + b*e*x)) + c^(3/2)*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*(16*c^2*d^2 + b^2*e^2 + 4*c
*e*(-4*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*c^2*(16*c^2*d^2 + 3*b^2*e^2 +
4*c*e*(-4*b*d + a*e))*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^
2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(4*c^2*e^5)))/(-(c*d^2) + e*(b*d - a*e)))/(4*e)

________________________________________________________________________________________

Maple [B]  time = 0.237, size = 14002, normalized size = 42.3 \begin{align*} \text{output 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)/(e*x+d)^3,x)

[Out]

result too large to display

________________________________________________________________________________________

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)/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [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((c*x^2+b*x+a)^(5/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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((c*x**2+b*x+a)**(5/2)/(e*x+d)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.12558, size = 1968, normalized size = 5.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

5/4*(16*c^3*d^4 - 32*b*c^2*d^3*e + 19*b^2*c*d^2*e^2 + 20*a*c^2*d^2*e^2 - 3*b^3*d*e^3 - 20*a*b*c*d*e^3 + 3*a*b^
2*e^4 + 4*a^2*c*e^4)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))
*e^(-6)/sqrt(-c*d^2 + b*d*e - a*e^2) + 5/16*(32*c^3*d^3 - 48*b*c^2*d^2*e + 18*b^2*c*d*e^2 + 24*a*c^2*d*e^2 - b
^3*e^3 - 12*a*b*c*e^3)*e^(-6)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + 1/24*sqrt(
c*x^2 + b*x + a)*(2*(4*c^2*x*e^(-3) - (18*c^4*d*e^14 - 13*b*c^3*e^15)*e^(-18)/c^2)*x + (144*c^4*d^2*e^13 - 162
*b*c^3*d*e^14 + 33*b^2*c^2*e^15 + 56*a*c^3*e^15)*e^(-18)/c^2) + 1/4*(40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*
c^3*d^4*e + 72*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(7/2)*d^5 - 120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b
*c^(5/2)*d^4*e + 72*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c^3*d^5 - 80*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b
*c^2*d^3*e^2 - 124*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c^2*d^4*e - 104*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
*a*c^3*d^4*e + 18*b^2*c^(5/2)*d^5 + 51*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*c^(3/2)*d^3*e^2 + 36*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^2*a*c^(5/2)*d^3*e^2 - 27*b^3*c^(3/2)*d^4*e - 52*a*b*c^(5/2)*d^4*e + 49*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^3*b^2*c*d^2*e^3 + 44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c^2*d^2*e^3 + 59*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))*b^3*c*d^3*e^2 + 244*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c^2*d^3*e^2 - 3*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^2*b^3*sqrt(c)*d^2*e^3 + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b*c^(3/2)*d^
2*e^3 + 9*b^4*sqrt(c)*d^3*e^2 + 95*a*b^2*c^(3/2)*d^3*e^2 + 36*a^2*c^(5/2)*d^3*e^2 - 9*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^3*b^3*d*e^4 - 44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*c*d*e^4 - 7*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))*b^4*d^2*e^3 - 127*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^2*c*d^2*e^3 - 100*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*a^2*c^2*d^2*e^3 - 21*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^2*sqrt(c)*d*e^4 - 36*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))^2*a^2*c^(3/2)*d*e^4 - 34*a*b^3*sqrt(c)*d^2*e^3 - 104*a^2*b*c^(3/2)*d^2*e^3 + 9*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^3*a*b^2*e^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*c*e^5 + 14*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))*a*b^3*d*e^4 + 64*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*b*c*d*e^4 + 24*(sqrt(c)*x - sqrt
(c*x^2 + b*x + a))^2*a^2*b*sqrt(c)*e^5 + 41*a^2*b^2*sqrt(c)*d*e^4 + 36*a^3*c^(3/2)*d*e^4 - 7*(sqrt(c)*x - sqrt
(c*x^2 + b*x + a))*a^2*b^2*e^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*c*e^5 - 16*a^3*b*sqrt(c)*e^5)*e^(-6
)/((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c)*d + b*d - a*e)^2