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

Optimal. Leaf size=343 \[ \frac{\left (-b^2 c \left (3 a d^2-c e^2\right )+4 a b c^2 d e+a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (-4 a^2 c^3 d e+8 a b^2 c^2 d e-b^3 c \left (5 a d^2-c e^2\right )+a b c^2 \left (5 a d^2-3 c e^2\right )-2 b^4 c d e+b^5 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{x (2 a d+b e)}{a^2 e^3}+\frac{d^5}{e^4 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac{d^4 \log (d+e x) \left (3 a d^2-e (4 b d-5 c e)\right )}{e^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{x^2}{2 a e^2} \]

[Out]

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

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Rubi [A]  time = 0.907376, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1569, 1628, 634, 618, 206, 628} \[ \frac{\left (-b^2 c \left (3 a d^2-c e^2\right )+4 a b c^2 d e+a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )^2}+\frac{\left (-4 a^2 c^3 d e+8 a b^2 c^2 d e-b^3 c \left (5 a d^2-c e^2\right )+a b c^2 \left (5 a d^2-3 c e^2\right )-2 b^4 c d e+b^5 d^2\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )^2}-\frac{x (2 a d+b e)}{a^2 e^3}+\frac{d^5}{e^4 (d+e x) \left (a d^2-e (b d-c e)\right )}+\frac{d^4 \log (d+e x) \left (3 a d^2-e (4 b d-5 c e)\right )}{e^4 \left (a d^2-e (b d-c e)\right )^2}+\frac{x^2}{2 a e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1569

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], 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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.379752, size = 338, normalized size = 0.99 \[ \frac{\left (b^2 c \left (c e^2-3 a d^2\right )+4 a b c^2 d e+a c^2 \left (a d^2-c e^2\right )-2 b^3 c d e+b^4 d^2\right ) \log (x (a x+b)+c)}{2 a^3 \left (a d^2+e (c e-b d)\right )^2}-\frac{\left (-4 a^2 c^3 d e+8 a b^2 c^2 d e+b^3 c \left (c e^2-5 a d^2\right )+a b c^2 \left (5 a d^2-3 c e^2\right )-2 b^4 c d e+b^5 d^2\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{a^3 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )^2}-\frac{x (2 a d+b e)}{a^2 e^3}+\frac{d^5}{e^4 (d+e x) \left (a d^2+e (c e-b d)\right )}+\frac{\log (d+e x) \left (3 a d^6+d^4 e (5 c e-4 b d)\right )}{e^4 \left (a d^2+e (c e-b d)\right )^2}+\frac{x^2}{2 a e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 943, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/e^4*d^6/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*a-4/e^3*d^5/(a*d^2-b*d*e+c*e^2)^2*ln(e*x+d)*b+5/e^2*d^4/(a*d^2-b*d*e
+c*e^2)^2*ln(e*x+d)*c+1/e^4*d^5/(a*d^2-b*d*e+c*e^2)/(e*x+d)+1/2*x^2/a/e^2-2/a/e^3*x*d-1/a^2/e^2*b*x+1/2/(a*d^2
-b*d*e+c*e^2)^2/a*ln(a*x^2+b*x+c)*c^2*d^2-3/2/(a*d^2-b*d*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*b^2*c*d^2+2/(a*d^2-b*d
*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*b*c^2*d*e-1/2/(a*d^2-b*d*e+c*e^2)^2/a^2*ln(a*x^2+b*x+c)*c^3*e^2+1/2/(a*d^2-b*d
*e+c*e^2)^2/a^3*ln(a*x^2+b*x+c)*b^4*d^2-1/(a*d^2-b*d*e+c*e^2)^2/a^3*ln(a*x^2+b*x+c)*b^3*c*d*e+1/2/(a*d^2-b*d*e
+c*e^2)^2/a^3*ln(a*x^2+b*x+c)*b^2*c^2*e^2-5/(a*d^2-b*d*e+c*e^2)^2/a/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-
b^2)^(1/2))*b*c^2*d^2+4/(a*d^2-b*d*e+c*e^2)^2/a/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*c^3*d*e+
5/(a*d^2-b*d*e+c*e^2)^2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^3*c*d^2-8/(a*d^2-b*d*e+c*e
^2)^2/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^2*c^2*d*e+3/(a*d^2-b*d*e+c*e^2)^2/a^2/(4*a*c
-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b*c^3*e^2-1/(a*d^2-b*d*e+c*e^2)^2/a^3/(4*a*c-b^2)^(1/2)*arctan
((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^5*d^2+2/(a*d^2-b*d*e+c*e^2)^2/a^3/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^
2)^(1/2))*b^4*c*d*e-1/(a*d^2-b*d*e+c*e^2)^2/a^3/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^3*c^2*
e^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(x^3/(a+c/x^2+b/x)/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.13396, size = 763, normalized size = 2.22 \begin{align*} \frac{d^{5} e^{4}}{{\left (a d^{2} e^{8} - b d e^{9} + c e^{10}\right )}{\left (x e + d\right )}} + \frac{{\left (b^{5} d^{2} e^{2} - 5 \, a b^{3} c d^{2} e^{2} + 5 \, a^{2} b c^{2} d^{2} e^{2} - 2 \, b^{4} c d e^{3} + 8 \, a b^{2} c^{2} d e^{3} - 4 \, a^{2} c^{3} d e^{3} + b^{3} c^{2} e^{4} - 3 \, a b c^{3} e^{4}\right )} \arctan \left (-\frac{{\left (2 \, a d - \frac{2 \, a d^{2}}{x e + d} - b e + \frac{2 \, b d e}{x e + d} - \frac{2 \, c e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-b^{2} + 4 \, a c}}\right ) e^{\left (-2\right )}}{{\left (a^{5} d^{4} - 2 \, a^{4} b d^{3} e + a^{3} b^{2} d^{2} e^{2} + 2 \, a^{4} c d^{2} e^{2} - 2 \, a^{3} b c d e^{3} + a^{3} c^{2} e^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (a^{2} - \frac{2 \,{\left (3 \, a^{2} d e + a b e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )}{\left (x e + d\right )}^{2} e^{\left (-4\right )}}{2 \, a^{3}} + \frac{{\left (b^{4} d^{2} - 3 \, a b^{2} c d^{2} + a^{2} c^{2} d^{2} - 2 \, b^{3} c d e + 4 \, a b c^{2} d e + b^{2} c^{2} e^{2} - a c^{3} e^{2}\right )} \log \left (-a + \frac{2 \, a d}{x e + d} - \frac{a d^{2}}{{\left (x e + d\right )}^{2}} - \frac{b e}{x e + d} + \frac{b d e}{{\left (x e + d\right )}^{2}} - \frac{c e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (a^{5} d^{4} - 2 \, a^{4} b d^{3} e + a^{3} b^{2} d^{2} e^{2} + 2 \, a^{4} c d^{2} e^{2} - 2 \, a^{3} b c d e^{3} + a^{3} c^{2} e^{4}\right )}} - \frac{{\left (3 \, a^{2} d^{2} + 2 \, a b d e + b^{2} e^{2} - a c e^{2}\right )} e^{\left (-4\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^5*e^4/((a*d^2*e^8 - b*d*e^9 + c*e^10)*(x*e + d)) + (b^5*d^2*e^2 - 5*a*b^3*c*d^2*e^2 + 5*a^2*b*c^2*d^2*e^2 -
2*b^4*c*d*e^3 + 8*a*b^2*c^2*d*e^3 - 4*a^2*c^3*d*e^3 + b^3*c^2*e^4 - 3*a*b*c^3*e^4)*arctan(-(2*a*d - 2*a*d^2/(x
*e + d) - b*e + 2*b*d*e/(x*e + d) - 2*c*e^2/(x*e + d))*e^(-1)/sqrt(-b^2 + 4*a*c))*e^(-2)/((a^5*d^4 - 2*a^4*b*d
^3*e + a^3*b^2*d^2*e^2 + 2*a^4*c*d^2*e^2 - 2*a^3*b*c*d*e^3 + a^3*c^2*e^4)*sqrt(-b^2 + 4*a*c)) + 1/2*(a^2 - 2*(
3*a^2*d*e + a*b*e^2)*e^(-1)/(x*e + d))*(x*e + d)^2*e^(-4)/a^3 + 1/2*(b^4*d^2 - 3*a*b^2*c*d^2 + a^2*c^2*d^2 - 2
*b^3*c*d*e + 4*a*b*c^2*d*e + b^2*c^2*e^2 - a*c^3*e^2)*log(-a + 2*a*d/(x*e + d) - a*d^2/(x*e + d)^2 - b*e/(x*e
+ d) + b*d*e/(x*e + d)^2 - c*e^2/(x*e + d)^2)/(a^5*d^4 - 2*a^4*b*d^3*e + a^3*b^2*d^2*e^2 + 2*a^4*c*d^2*e^2 - 2
*a^3*b*c*d*e^3 + a^3*c^2*e^4) - (3*a^2*d^2 + 2*a*b*d*e + b^2*e^2 - a*c*e^2)*e^(-4)*log(abs(x*e + d)*e^(-1)/(x*
e + d)^2)/a^3