### 3.1670 $$\int \frac{(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx$$

Optimal. Leaf size=208 $-\frac{3 e^4 \sqrt{d+e x}}{128 b^2 (a+b x) (b d-a e)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (a+b x)^2 (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (a+b x)^3 (b d-a e)}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}}-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}$

[Out]

(-3*e*Sqrt[d + e*x])/(40*b^2*(a + b*x)^4) - (e^2*Sqrt[d + e*x])/(80*b^2*(b*d - a*e)*(a + b*x)^3) + (e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)^2) - (3*e^4*Sqrt[d + e*x])/(128*b^2*(b*d - a*e)^3*(a + b*x)) - (d + e
*x)^(3/2)/(5*b*(a + b*x)^5) + (3*e^5*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(5/2)*(b*d - a*e
)^(7/2))

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Rubi [A]  time = 0.123246, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {27, 47, 51, 63, 208} $-\frac{3 e^4 \sqrt{d+e x}}{128 b^2 (a+b x) (b d-a e)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (a+b x)^2 (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (a+b x)^3 (b d-a e)}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}}-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*e*Sqrt[d + e*x])/(40*b^2*(a + b*x)^4) - (e^2*Sqrt[d + e*x])/(80*b^2*(b*d - a*e)*(a + b*x)^3) + (e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)^2) - (3*e^4*Sqrt[d + e*x])/(128*b^2*(b*d - a*e)^3*(a + b*x)) - (d + e
*x)^(3/2)/(5*b*(a + b*x)^5) + (3*e^5*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(5/2)*(b*d - a*e
)^(7/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{(d+e x)^{3/2}}{(a+b x)^6} \, dx\\ &=-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac{(3 e) \int \frac{\sqrt{d+e x}}{(a+b x)^5} \, dx}{10 b}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac{\left (3 e^2\right ) \int \frac{1}{(a+b x)^4 \sqrt{d+e x}} \, dx}{80 b^2}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (b d-a e) (a+b x)^3}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac{e^3 \int \frac{1}{(a+b x)^3 \sqrt{d+e x}} \, dx}{32 b^2 (b d-a e)}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac{\left (3 e^4\right ) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{128 b^2 (b d-a e)^2}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac{3 e^4 \sqrt{d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac{\left (3 e^5\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{256 b^2 (b d-a e)^3}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac{3 e^4 \sqrt{d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac{\left (3 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{128 b^2 (b d-a e)^3}\\ &=-\frac{3 e \sqrt{d+e x}}{40 b^2 (a+b x)^4}-\frac{e^2 \sqrt{d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac{e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac{3 e^4 \sqrt{d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac{(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac{3 e^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0181231, size = 52, normalized size = 0.25 $\frac{2 e^5 (d+e x)^{5/2} \, _2F_1\left (\frac{5}{2},6;\frac{7}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^6}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*e^5*(d + e*x)^(5/2)*Hypergeometric2F1[5/2, 6, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^6)

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Maple [A]  time = 0.232, size = 300, normalized size = 1.4 \begin{align*}{\frac{3\,{e}^{5}{b}^{2}}{128\, \left ( bxe+ae \right ) ^{5} \left ({a}^{3}{e}^{3}-3\,d{e}^{2}{a}^{2}b+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) } \left ( ex+d \right ) ^{{\frac{9}{2}}}}+{\frac{7\,{e}^{5}b}{64\, \left ( bxe+ae \right ) ^{5} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{7}{2}}}}+{\frac{{e}^{5}}{5\, \left ( bxe+ae \right ) ^{5} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{7\,{e}^{5}}{64\, \left ( bxe+ae \right ) ^{5}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{6}a}{128\, \left ( bxe+ae \right ) ^{5}{b}^{2}}\sqrt{ex+d}}+{\frac{3\,{e}^{5}d}{128\, \left ( bxe+ae \right ) ^{5}b}\sqrt{ex+d}}+{\frac{3\,{e}^{5}}{ \left ( 128\,{a}^{3}{e}^{3}-384\,d{e}^{2}{a}^{2}b+384\,a{b}^{2}{d}^{2}e-128\,{b}^{3}{d}^{3} \right ){b}^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*e^5/(b*e*x+a*e)^5/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*b^2*(e*x+d)^(9/2)+7/64*e^5/(b*e*x+a*e)^5
*b/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(7/2)+1/5*e^5/(b*e*x+a*e)^5/(a*e-b*d)*(e*x+d)^(5/2)-7/64*e^5/(b*e*x+a*e
)^5/b*(e*x+d)^(3/2)-3/128*e^6/(b*e*x+a*e)^5/b^2*(e*x+d)^(1/2)*a+3/128*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(1/2)*d+3/12
8*e^5/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/b^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*
b)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86769, size = 3065, normalized size = 14.74 \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)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/1280*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^
5)*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*(128*b^6
*d^5 - 464*a*b^5*d^4*e + 584*a^2*b^4*d^3*e^2 - 258*a^3*b^3*d^2*e^3 - 5*a^4*b^2*d*e^4 + 15*a^5*b*e^5 + 15*(b^6*
d*e^4 - a*b^5*e^5)*x^4 - 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(4*b^6*d^3*e^2 - 27*a*b^5*d^
2*e^3 + 87*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(88*b^6*d^4*e - 344*a*b^5*d^3*e^2 + 489*a^2*b^4*d^2*e^3 - 2
68*a^3*b^3*d*e^4 + 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^4 - 4*a^6*b^6*d^3*e + 6*a^7*b^5*d^2*e^2 - 4*a^
8*b^4*d*e^3 + a^9*b^3*e^4 + (b^12*d^4 - 4*a*b^11*d^3*e + 6*a^2*b^10*d^2*e^2 - 4*a^3*b^9*d*e^3 + a^4*b^8*e^4)*x
^5 + 5*(a*b^11*d^4 - 4*a^2*b^10*d^3*e + 6*a^3*b^9*d^2*e^2 - 4*a^4*b^8*d*e^3 + a^5*b^7*e^4)*x^4 + 10*(a^2*b^10*
d^4 - 4*a^3*b^9*d^3*e + 6*a^4*b^8*d^2*e^2 - 4*a^5*b^7*d*e^3 + a^6*b^6*e^4)*x^3 + 10*(a^3*b^9*d^4 - 4*a^4*b^8*d
^3*e + 6*a^5*b^7*d^2*e^2 - 4*a^6*b^6*d*e^3 + a^7*b^5*e^4)*x^2 + 5*(a^4*b^8*d^4 - 4*a^5*b^7*d^3*e + 6*a^6*b^6*d
^2*e^2 - 4*a^7*b^5*d*e^3 + a^8*b^4*e^4)*x), -1/640*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 1
0*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b
*e*x + b*d)) + (128*b^6*d^5 - 464*a*b^5*d^4*e + 584*a^2*b^4*d^3*e^2 - 258*a^3*b^3*d^2*e^3 - 5*a^4*b^2*d*e^4 +
15*a^5*b*e^5 + 15*(b^6*d*e^4 - a*b^5*e^5)*x^4 - 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(4*b^
6*d^3*e^2 - 27*a*b^5*d^2*e^3 + 87*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(88*b^6*d^4*e - 344*a*b^5*d^3*e^2 +
489*a^2*b^4*d^2*e^3 - 268*a^3*b^3*d*e^4 + 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^4 - 4*a^6*b^6*d^3*e + 6
*a^7*b^5*d^2*e^2 - 4*a^8*b^4*d*e^3 + a^9*b^3*e^4 + (b^12*d^4 - 4*a*b^11*d^3*e + 6*a^2*b^10*d^2*e^2 - 4*a^3*b^9
*d*e^3 + a^4*b^8*e^4)*x^5 + 5*(a*b^11*d^4 - 4*a^2*b^10*d^3*e + 6*a^3*b^9*d^2*e^2 - 4*a^4*b^8*d*e^3 + a^5*b^7*e
^4)*x^4 + 10*(a^2*b^10*d^4 - 4*a^3*b^9*d^3*e + 6*a^4*b^8*d^2*e^2 - 4*a^5*b^7*d*e^3 + a^6*b^6*e^4)*x^3 + 10*(a^
3*b^9*d^4 - 4*a^4*b^8*d^3*e + 6*a^5*b^7*d^2*e^2 - 4*a^6*b^6*d*e^3 + a^7*b^5*e^4)*x^2 + 5*(a^4*b^8*d^4 - 4*a^5*
b^7*d^3*e + 6*a^6*b^6*d^2*e^2 - 4*a^7*b^5*d*e^3 + a^8*b^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((e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.24847, size = 556, normalized size = 2.67 \begin{align*} -\frac{3 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{5}}{128 \,{\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} \sqrt{-b^{2} d + a b e}} - \frac{15 \,{\left (x e + d\right )}^{\frac{9}{2}} b^{4} e^{5} - 70 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{4} d e^{5} + 128 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} d^{2} e^{5} + 70 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d^{3} e^{5} - 15 \, \sqrt{x e + d} b^{4} d^{4} e^{5} + 70 \,{\left (x e + d\right )}^{\frac{7}{2}} a b^{3} e^{6} - 256 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{3} d e^{6} - 210 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} d^{2} e^{6} + 60 \, \sqrt{x e + d} a b^{3} d^{3} e^{6} + 128 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} b^{2} e^{7} + 210 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{2} d e^{7} - 90 \, \sqrt{x e + d} a^{2} b^{2} d^{2} e^{7} - 70 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b e^{8} + 60 \, \sqrt{x e + d} a^{3} b d e^{8} - 15 \, \sqrt{x e + d} a^{4} e^{9}}{640 \,{\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^5/((b^5*d^3 - 3*a*b^4*d^2*e + 3*a^2*b^3*d*e^2 - a^3*b^2*
e^3)*sqrt(-b^2*d + a*b*e)) - 1/640*(15*(x*e + d)^(9/2)*b^4*e^5 - 70*(x*e + d)^(7/2)*b^4*d*e^5 + 128*(x*e + d)^
(5/2)*b^4*d^2*e^5 + 70*(x*e + d)^(3/2)*b^4*d^3*e^5 - 15*sqrt(x*e + d)*b^4*d^4*e^5 + 70*(x*e + d)^(7/2)*a*b^3*e
^6 - 256*(x*e + d)^(5/2)*a*b^3*d*e^6 - 210*(x*e + d)^(3/2)*a*b^3*d^2*e^6 + 60*sqrt(x*e + d)*a*b^3*d^3*e^6 + 12
8*(x*e + d)^(5/2)*a^2*b^2*e^7 + 210*(x*e + d)^(3/2)*a^2*b^2*d*e^7 - 90*sqrt(x*e + d)*a^2*b^2*d^2*e^7 - 70*(x*e
+ d)^(3/2)*a^3*b*e^8 + 60*sqrt(x*e + d)*a^3*b*d*e^8 - 15*sqrt(x*e + d)*a^4*e^9)/((b^5*d^3 - 3*a*b^4*d^2*e + 3
*a^2*b^3*d*e^2 - a^3*b^2*e^3)*((x*e + d)*b - b*d + a*e)^5)