3.2136 \(\int \frac{a+b x}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac{5 b e (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}-\frac{5 e (a+b x)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}+\frac{5 b^{3/2} e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]

[Out]

-(1/((b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (5*e*(a + b*x))/(3*(b*d - a*e)^2*(d + e*x)^
(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*b*e*(a + b*x))/((b*d - a*e)^3*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) + (5*b^(3/2)*e*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/((b*d - a*e)^(7/2)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.110092, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 21, 51, 63, 208} \[ -\frac{5 b e (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}-\frac{5 e (a+b x)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{1}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}+\frac{5 b^{3/2} e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/((b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (5*e*(a + b*x))/(3*(b*d - a*e)^2*(d + e*x)^
(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*b*e*(a + b*x))/((b*d - a*e)^3*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) + (5*b^(3/2)*e*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/((b*d - a*e)^(7/2)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{a+b x}{\left (a b+b^2 x\right )^3 (d+e x)^{5/2}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{(b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 e \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{2 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{(b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 e \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{(b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b e (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 b e \left (a b+b^2 x\right )\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{(b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b e (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 b \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{(b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b e (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b^{3/2} e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0263483, size = 66, normalized size = 0.31 \[ -\frac{2 e (a+b x) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{3 \sqrt{(a+b x)^2} (d+e x)^{3/2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(a + b*x)*Hypergeometric2F1[-3/2, 2, -1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^2*Sqrt[(a
+ b*x)^2]*(d + e*x)^(3/2))

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Maple [A]  time = 0.018, size = 242, normalized size = 1.1 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{3\, \left ( ae-bd \right ) ^{3}} \left ( 15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}x{b}^{3}e+15\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{3/2}a{b}^{2}e+15\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}{e}^{2}+10\,\sqrt{ \left ( ae-bd \right ) b}xab{e}^{2}+20\,\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}de-2\,\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{e}^{2}+14\,\sqrt{ \left ( ae-bd \right ) b}abde+3\,\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*x*b^3*e+15*arctan((e*x+d)^(1/2)*b/((a*e-b*d)
*b)^(1/2))*(e*x+d)^(3/2)*a*b^2*e+15*((a*e-b*d)*b)^(1/2)*x^2*b^2*e^2+10*((a*e-b*d)*b)^(1/2)*x*a*b*e^2+20*((a*e-
b*d)*b)^(1/2)*x*b^2*d*e-2*((a*e-b*d)*b)^(1/2)*a^2*e^2+14*((a*e-b*d)*b)^(1/2)*a*b*d*e+3*((a*e-b*d)*b)^(1/2)*b^2
*d^2)*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/(e*x+d)^(3/2)/(a*e-b*d)^3/((b*x+a)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.0975, size = 1592, normalized size = 7.51 \begin{align*} \left [-\frac{15 \,{\left (b^{2} e^{3} x^{3} + a b d^{2} e +{\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} +{\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt{\frac{b}{b d - a e}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \,{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{\frac{b}{b d - a e}}}{b x + a}\right ) + 2 \,{\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \,{\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt{e x + d}}{6 \,{\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} +{\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} +{\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} +{\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}, \frac{15 \,{\left (b^{2} e^{3} x^{3} + a b d^{2} e +{\left (2 \, b^{2} d e^{2} + a b e^{3}\right )} x^{2} +{\left (b^{2} d^{2} e + 2 \, a b d e^{2}\right )} x\right )} \sqrt{-\frac{b}{b d - a e}} \arctan \left (-\frac{{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{-\frac{b}{b d - a e}}}{b e x + b d}\right ) -{\left (15 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 14 \, a b d e - 2 \, a^{2} e^{2} + 10 \,{\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} +{\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} +{\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} +{\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2 + (b^2*d^2*e + 2*a*b*d*e^2)*x)*sqrt(b/(b*d -
a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(15*b^2*e^2*x
^2 + 3*b^2*d^2 + 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^2*b^2*d^
4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b
^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e - 3*a^2*b
^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x), 1/3*(15*(b^2*e^3*x^3 + a*b*d^2*e + (2*b^2*d*e^2 + a*b*e^3)*x^2
+ (b^2*d^2*e + 2*a*b*d*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b
*e*x + b*d)) - (15*b^2*e^2*x^2 + 3*b^2*d^2 + 14*a*b*d*e - 2*a^2*e^2 + 10*(2*b^2*d*e + a*b*e^2)*x)*sqrt(e*x + d
))/(a*b^3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d
*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4
*d^5 - a*b^3*d^4*e - 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*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)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.24274, size = 667, normalized size = 3.15 \begin{align*} -\frac{5 \, b^{2} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} - \frac{\sqrt{x e + d} b^{2} e^{2}}{{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}} - \frac{2 \,{\left (6 \,{\left (x e + d\right )} b e^{2} + b d e^{2} - a e^{3}\right )}}{3 \,{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left (x e + d\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^3*d^3*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b
^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^4*sgn
((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - sqrt(x*e + d)*b^2*e^2/((b^3*d^3*e*sgn((x*e + d)*b*e -
b*d*e + a*e^2) - 3*a*b^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^3*sgn((x*e + d)*b*e - b*d*e
+ a*e^2) - a^3*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)) - 2/3*(6*(x*e + d)*b*e^2 +
b*d*e^2 - a*e^3)/((b^3*d^3*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d^2*e^2*sgn((x*e + d)*b*e - b*d*e +
a*e^2) + 3*a^2*b*d*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^4*sgn((x*e + d)*b*e - b*d*e + a*e^2))*(x*e +
d)^(3/2))