### 3.1717 $$\int \frac{1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=223 $\frac{15 e^2 (a+b x)}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}-\frac{15 \sqrt{b} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac{5 e}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}$

[Out]

(5*e)/(4*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x]
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*e^2*(a + b*x))/(4*(b*d - a*e)^3*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]) - (15*Sqrt[b]*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(7/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.106826, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {646, 51, 63, 208} $\frac{15 e^2 (a+b x)}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}-\frac{15 \sqrt{b} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac{5 e}{4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*e)/(4*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x]
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*e^2*(a + b*x))/(4*(b*d - a*e)^3*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]) - (15*Sqrt[b]*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(7/2)*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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{1}{(d+e x)^{3/2} \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{1}{\left (a b+b^2 x\right )^3 (d+e x)^{3/2}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{1}{2 (b d-a e) (a+b x) \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}{\left (a b+b^2 x\right )^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e}{4 (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e}{4 (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 b e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e}{4 (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (15 b e \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{5 e}{4 (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{15 \sqrt{b} e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0215676, size = 65, normalized size = 0.29 $\frac{2 e^2 (a+b x) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{\sqrt{(a+b x)^2} \sqrt{d+e x} (b d-a e)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.283, size = 285, normalized size = 1.3 \begin{align*} -{\frac{bx+a}{4\, \left ( ae-bd \right ) ^{3}} \left ( 15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{x}^{2}{b}^{3}{e}^{2}+30\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}xa{b}^{2}{e}^{2}+15\,\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}{e}^{2}+15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{a}^{2}b{e}^{2}+25\,\sqrt{ \left ( ae-bd \right ) b}xab{e}^{2}+5\,\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}de+8\,\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{e}^{2}+9\,\sqrt{ \left ( ae-bd \right ) b}abde-2\,\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ex+d}}}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/4*(15*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^(1/2)*x^2*b^3*e^2+30*arctan(b*(e*x+d)^(1/2)/((a*e
-b*d)*b)^(1/2))*(e*x+d)^(1/2)*x*a*b^2*e^2+15*((a*e-b*d)*b)^(1/2)*x^2*b^2*e^2+15*arctan(b*(e*x+d)^(1/2)/((a*e-b
*d)*b)^(1/2))*(e*x+d)^(1/2)*a^2*b*e^2+25*((a*e-b*d)*b)^(1/2)*x*a*b*e^2+5*((a*e-b*d)*b)^(1/2)*x*b^2*d*e+8*((a*e
-b*d)*b)^(1/2)*a^2*e^2+9*((a*e-b*d)*b)^(1/2)*a*b*d*e-2*((a*e-b*d)*b)^(1/2)*b^2*d^2)*(b*x+a)/((a*e-b*d)*b)^(1/2
)/(e*x+d)^(1/2)/(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{1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(e*x + d)^(3/2)), x)

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Fricas [B]  time = 1.63644, size = 1582, normalized size = 7.09 \begin{align*} \left [-\frac{15 \,{\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} +{\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} +{\left (2 \, a b d e^{2} + a^{2} e^{3}\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} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \,{\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} +{\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} +{\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} +{\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}, -\frac{15 \,{\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} +{\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} +{\left (2 \, a b d e^{2} + a^{2} e^{3}\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} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \,{\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} +{\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} +{\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} +{\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 + (2*a*b*d*e^2 + a^2*e^3)*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
- 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2*b^3*d^4 - 3*a^3*b^2*d^3*
e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^2*e^4)*x^3 + (b^5*d^4
- a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^4 - 5*a^2*b^3*d^3*e + 3*a
^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x), -1/4*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 +
(2*a*b*d*e^2 + a^2*e^3)*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 - 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2
*b^3*d^4 - 3*a^3*b^2*d^3*e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^
3*b^2*e^4)*x^3 + (b^5*d^4 - a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^
4 - 5*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right )^{\frac{3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/((d + e*x)**(3/2)*((a + b*x)**2)**(3/2)), x)

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Giac [B]  time = 1.25694, size = 674, normalized size = 3.02 \begin{align*} \frac{15 \, b \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{3} d^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \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{2 \, e^{2}}{{\left (b^{3} d^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{x e + d}} + \frac{7 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} e^{2} - 9 \, \sqrt{x e + d} b^{2} d e^{2} + 9 \, \sqrt{x e + d} a b e^{3}}{4 \,{\left (b^{3} d^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \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 )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2
*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*
e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) + 2*e^2/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*
b^2*d^2*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + 3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn(
(x*e + d)*b*e - b*d*e + a*e^2))*sqrt(x*e + d)) + 1/4*(7*(x*e + d)^(3/2)*b^2*e^2 - 9*sqrt(x*e + d)*b^2*d*e^2 +
9*sqrt(x*e + d)*a*b*e^3)/((b^3*d^3*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 3*a*b^2*d^2*e*sgn((x*e + d)*b*e - b*d*
e + a*e^2) + 3*a^2*b*d*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a^3*e^3*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((
x*e + d)*b - b*d + a*e)^2)