3.1724 \(\int \frac{(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=260 \[ -\frac{5 e^3 \sqrt{d+e x}}{64 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(-5*e^3*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^2*Sqrt[d + e*x])/(32*b^3*(a +
b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(d + e*x)^(3/2))/(24*b^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) - (d + e*x)^(5/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^4*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(7/2)*(b*d - a*e)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.139203, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {646, 47, 51, 63, 208} \[ -\frac{5 e^3 \sqrt{d+e x}}{64 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^3*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^2*Sqrt[d + e*x])/(32*b^3*(a +
b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(d + e*x)^(3/2))/(24*b^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
) - (d + e*x)^(5/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^4*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(7/2)*(b*d - a*e)^(3/2)*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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{(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 b^2 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^4} \, dx}{8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{\left (a b+b^2 x\right )^3} \, dx}{16 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^2 \sqrt{d+e x}} \, dx}{64 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 e^4 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{128 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 e^3 \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 )}{64 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0402505, size = 67, normalized size = 0.26 \[ -\frac{2 e^4 (a+b x) (d+e x)^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{b (d+e x)}{b d-a e}\right )}{7 \sqrt{(a+b x)^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^4*(a + b*x)*(d + e*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, (b*(d + e*x))/(b*d - a*e)])/(7*(b*d - a*e)^5*
Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.275, size = 477, normalized size = 1.8 \begin{align*}{\frac{bx+a}{192\, \left ( ae-bd \right ){b}^{3}} \left ( 15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{4}{b}^{4}{e}^{4}+60\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{3}a{b}^{3}{e}^{4}+15\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{7/2}{b}^{3}+90\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}-73\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}a{b}^{2}e+73\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{3}d+60\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{3}b{e}^{4}-55\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{a}^{2}b{e}^{2}+110\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}de-55\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{3}{d}^{2}+15\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{4}{e}^{4}-15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{3}{e}^{3}+45\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}bd{e}^{2}-45\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}{d}^{2}e+15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(15*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^4*b^4*e^4+60*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2
))*x^3*a*b^3*e^4+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*b^3+90*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*a
^2*b^2*e^4-73*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^2*e+73*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^3*d+60*arctan(b
*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^3*b*e^4-55*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b*e^2+110*((a*e-b*d)*
b)^(1/2)*(e*x+d)^(3/2)*a*b^2*d*e-55*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^3*d^2+15*arctan(b*(e*x+d)^(1/2)/((a*e-
b*d)*b)^(1/2))*a^4*e^4-15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*e^3+45*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b
*d*e^2-45*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^2*d^2*e+15*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^3*d^3)*(b*x+a)/
((a*e-b*d)*b)^(1/2)/b^3/(a*e-b*d)/((b*x+a)^2)^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.73771, size = 1859, normalized size = 7.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*
log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) + 2*(48*b^5*d^4 - 56*a*b^4*d^3*e -
2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^5*d^2*e^2 - 191*a
*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55*a^3*b^2*e^4)*x)*
sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8*e^2)*x^4 + 4*(a*
b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^2 + 4*(a^3*b^7*d^
2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*x), -1/192*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e
^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (48*b^5*d^4 -
56*a*b^4*d^3*e - 2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^
5*d^2*e^2 - 191*a*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55
*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8
*e^2)*x^4 + 4*(a*b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^
2 + 4*(a^3*b^7*d^2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*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((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.23455, size = 477, normalized size = 1.83 \begin{align*} -\frac{5 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \,{\left (b^{4} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{3} e \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{15 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{3} e^{4} + 73 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{4} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{4} + 15 \, \sqrt{x e + d} b^{3} d^{3} e^{4} - 73 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{5} + 110 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{5} - 45 \, \sqrt{x e + d} a b^{2} d^{2} e^{5} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{6} + 45 \, \sqrt{x e + d} a^{2} b d e^{6} - 15 \, \sqrt{x e + d} a^{3} e^{7}}{192 \,{\left (b^{4} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{3} e \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 )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^3*e*sg
n((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/192*(15*(x*e + d)^(7/2)*b^3*e^4 + 73*(x*e + d)^(5/
2)*b^3*d*e^4 - 55*(x*e + d)^(3/2)*b^3*d^2*e^4 + 15*sqrt(x*e + d)*b^3*d^3*e^4 - 73*(x*e + d)^(5/2)*a*b^2*e^5 +
110*(x*e + d)^(3/2)*a*b^2*d*e^5 - 45*sqrt(x*e + d)*a*b^2*d^2*e^5 - 55*(x*e + d)^(3/2)*a^2*b*e^6 + 45*sqrt(x*e
+ d)*a^2*b*d*e^6 - 15*sqrt(x*e + d)*a^3*e^7)/((b^4*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^3*e*sgn((x*e + d
)*b*e - b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)^4)