3.2323 \(\int \frac{(A+B x) \left (a+b x+c x^2\right )^2}{d+e x} \, dx\)

Optimal. Leaf size=257 \[ -\frac{x^2 \left (B (c d-b e) \left (c d^2-e (b d-2 a e)\right )-A e \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{2 e^4}-\frac{x^3 \left (A c e (c d-2 b e)-B \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{3 e^3}-\frac{x \left (A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )-B \left (c d^2-e (b d-a e)\right )^2\right )}{e^5}-\frac{(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )^2}{e^6}-\frac{c x^4 (-A c e-2 b B e+B c d)}{4 e^2}+\frac{B c^2 x^5}{5 e} \]

[Out]

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

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Rubi [A]  time = 1.10751, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04 \[ -\frac{x^2 \left (B (c d-b e) \left (c d^2-e (b d-2 a e)\right )-A e \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{2 e^4}-\frac{x^3 \left (A c e (c d-2 b e)-B \left (-2 c e (b d-a e)+b^2 e^2+c^2 d^2\right )\right )}{3 e^3}-\frac{x \left (A e (c d-b e) \left (c d^2-e (b d-2 a e)\right )-B \left (c d^2-e (b d-a e)\right )^2\right )}{e^5}-\frac{(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )^2}{e^6}-\frac{c x^4 (-A c e-2 b B e+B c d)}{4 e^2}+\frac{B c^2 x^5}{5 e} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)*(c*x**2+b*x+a)**2/(e*x+d),x)

[Out]

Timed out

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Mathematica [A]  time = 0.546442, size = 298, normalized size = 1.16 \[ \frac{e x \left (B \left (10 e^2 \left (6 a^2 e^2+6 a b e (e x-2 d)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+10 c e \left (2 a e \left (6 d^2-3 d e x+2 e^2 x^2\right )+b \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )+c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+5 A e \left (4 c e \left (3 a e (e x-2 d)+b \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+6 b e^2 (4 a e-2 b d+b e x)+c^2 \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )-60 (B d-A e) \log (d+e x) \left (e (a e-b d)+c d^2\right )^2}{60 e^6} \]

Antiderivative was successfully verified.

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

[Out]

(e*x*(5*A*e*(6*b*e^2*(-2*b*d + 4*a*e + b*e*x) + c^2*(-12*d^3 + 6*d^2*e*x - 4*d*e
^2*x^2 + 3*e^3*x^3) + 4*c*e*(3*a*e*(-2*d + e*x) + b*(6*d^2 - 3*d*e*x + 2*e^2*x^2
))) + B*(c^2*(60*d^4 - 30*d^3*e*x + 20*d^2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4)
+ 10*e^2*(6*a^2*e^2 + 6*a*b*e*(-2*d + e*x) + b^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2))
+ 10*c*e*(2*a*e*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + b*(-12*d^3 + 6*d^2*e*x - 4*d*e^2
*x^2 + 3*e^3*x^3)))) - 60*(B*d - A*e)*(c*d^2 + e*(-(b*d) + a*e))^2*Log[d + e*x])
/(60*e^6)

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Maple [B]  time = 0.009, size = 558, normalized size = 2.2 \[ -{\frac{2\,B{x}^{3}bcd}{3\,{e}^{2}}}-{\frac{Ab{x}^{2}cd}{{e}^{2}}}-{\frac{aBc{x}^{2}d}{{e}^{2}}}-2\,{\frac{aAcdx}{{e}^{2}}}+2\,{\frac{A{d}^{2}bcx}{{e}^{3}}}-2\,{\frac{abBdx}{{e}^{2}}}+2\,{\frac{aBc{d}^{2}x}{{e}^{3}}}-2\,{\frac{Bbc{d}^{3}x}{{e}^{4}}}+{\frac{B{x}^{2}bc{d}^{2}}{{e}^{3}}}-2\,{\frac{\ln \left ( ex+d \right ) Aabd}{{e}^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) Aac{d}^{2}}{{e}^{3}}}-2\,{\frac{\ln \left ( ex+d \right ) Abc{d}^{3}}{{e}^{4}}}+2\,{\frac{\ln \left ( ex+d \right ) Bab{d}^{2}}{{e}^{3}}}-2\,{\frac{\ln \left ( ex+d \right ) Bac{d}^{3}}{{e}^{4}}}+2\,{\frac{\ln \left ( ex+d \right ) Bbc{d}^{4}}{{e}^{5}}}+{\frac{B{c}^{2}{x}^{5}}{5\,e}}+{\frac{{a}^{2}Bx}{e}}+{\frac{B{x}^{3}{b}^{2}}{3\,e}}+{\frac{A{b}^{2}{x}^{2}}{2\,e}}+{\frac{A{c}^{2}{x}^{4}}{4\,e}}+{\frac{\ln \left ( ex+d \right ) A{a}^{2}}{e}}+{\frac{B{c}^{2}{x}^{3}{d}^{2}}{3\,{e}^{3}}}+{\frac{2\,aBc{x}^{3}}{3\,e}}+{\frac{B{c}^{2}{d}^{4}x}{{e}^{5}}}-{\frac{B{c}^{2}{x}^{2}{d}^{3}}{2\,{e}^{4}}}+{\frac{aAc{x}^{2}}{e}}+{\frac{B{x}^{2}ab}{e}}-{\frac{A{c}^{2}{x}^{3}d}{3\,{e}^{2}}}+{\frac{B{x}^{4}bc}{2\,e}}-{\frac{B{c}^{2}{x}^{4}d}{4\,{e}^{2}}}+{\frac{2\,Ab{x}^{3}c}{3\,e}}-{\frac{A{b}^{2}dx}{{e}^{2}}}-{\frac{A{d}^{3}{c}^{2}x}{{e}^{4}}}+{\frac{{b}^{2}B{d}^{2}x}{{e}^{3}}}+2\,{\frac{aAbx}{e}}+{\frac{\ln \left ( ex+d \right ) A{b}^{2}{d}^{2}}{{e}^{3}}}+{\frac{A{c}^{2}{x}^{2}{d}^{2}}{2\,{e}^{3}}}-{\frac{{b}^{2}B{x}^{2}d}{2\,{e}^{2}}}-{\frac{\ln \left ( ex+d \right ) B{b}^{2}{d}^{3}}{{e}^{4}}}-{\frac{\ln \left ( ex+d \right ) B{c}^{2}{d}^{5}}{{e}^{6}}}+{\frac{\ln \left ( ex+d \right ) A{c}^{2}{d}^{4}}{{e}^{5}}}-{\frac{\ln \left ( ex+d \right ) B{a}^{2}d}{{e}^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((B*x+A)*(c*x^2+b*x+a)^2/(e*x+d),x)

[Out]

-2/3/e^2*B*x^3*b*c*d-1/e^2*A*x^2*b*c*d-1/e^2*B*x^2*a*c*d-2/e^2*A*a*c*d*x+2/e^3*A
*b*c*d^2*x-2/e^2*B*a*b*d*x+2/e^3*B*a*c*d^2*x-2/e^4*B*b*c*d^3*x+1/e^3*B*x^2*b*c*d
^2-2/e^2*ln(e*x+d)*A*a*b*d+2/e^3*ln(e*x+d)*A*a*c*d^2-2/e^4*ln(e*x+d)*A*b*c*d^3+2
/e^3*ln(e*x+d)*B*a*b*d^2-2/e^4*ln(e*x+d)*B*a*c*d^3+2/e^5*ln(e*x+d)*B*b*c*d^4+1/5
*B*c^2*x^5/e+1/e*B*a^2*x+1/3/e*B*x^3*b^2+1/2/e*A*x^2*b^2+1/4/e*A*x^4*c^2+1/e*ln(
e*x+d)*A*a^2+1/3/e^3*B*x^3*c^2*d^2+2/3/e*B*x^3*a*c+1/e^5*B*c^2*d^4*x-1/2/e^4*B*x
^2*c^2*d^3+1/e*A*x^2*a*c+1/e*B*x^2*a*b-1/3/e^2*A*x^3*c^2*d+1/2/e*B*x^4*b*c-1/4/e
^2*B*x^4*c^2*d+2/3/e*A*x^3*b*c-1/e^2*A*b^2*d*x-1/e^4*A*c^2*d^3*x+1/e^3*B*b^2*d^2
*x+2/e*A*a*b*x+1/e^3*ln(e*x+d)*A*b^2*d^2+1/2/e^3*A*x^2*c^2*d^2-1/2/e^2*B*x^2*b^2
*d-1/e^4*ln(e*x+d)*B*b^2*d^3-1/e^6*ln(e*x+d)*B*c^2*d^5+1/e^5*ln(e*x+d)*A*c^2*d^4
-1/e^2*ln(e*x+d)*B*a^2*d

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Maxima [A]  time = 0.703455, size = 510, normalized size = 1.98 \[ \frac{12 \, B c^{2} e^{4} x^{5} - 15 \,{\left (B c^{2} d e^{3} -{\left (2 \, B b c + A c^{2}\right )} e^{4}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{2} -{\left (2 \, B b c + A c^{2}\right )} d e^{3} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} e^{4}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e -{\left (2 \, B b c + A c^{2}\right )} d^{2} e^{2} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d e^{3} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{4}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} -{\left (2 \, B b c + A c^{2}\right )} d^{3} e +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{2} e^{2} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{3} +{\left (B a^{2} + 2 \, A a b\right )} e^{4}\right )} x}{60 \, e^{5}} - \frac{{\left (B c^{2} d^{5} - A a^{2} e^{5} -{\left (2 \, B b c + A c^{2}\right )} d^{4} e +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{3} e^{2} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} +{\left (B a^{2} + 2 \, A a b\right )} d e^{4}\right )} \log \left (e x + d\right )}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/(e*x + d),x, algorithm="maxima")

[Out]

1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - (2*B*b*c + A*c^2)*e^4)*x^4 + 20*(B*c^
2*d^2*e^2 - (2*B*b*c + A*c^2)*d*e^3 + (B*b^2 + 2*(B*a + A*b)*c)*e^4)*x^3 - 30*(B
*c^2*d^3*e - (2*B*b*c + A*c^2)*d^2*e^2 + (B*b^2 + 2*(B*a + A*b)*c)*d*e^3 - (2*B*
a*b + A*b^2 + 2*A*a*c)*e^4)*x^2 + 60*(B*c^2*d^4 - (2*B*b*c + A*c^2)*d^3*e + (B*b
^2 + 2*(B*a + A*b)*c)*d^2*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^3 + (B*a^2 + 2*A
*a*b)*e^4)*x)/e^5 - (B*c^2*d^5 - A*a^2*e^5 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 +
2*(B*a + A*b)*c)*d^3*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d^2*e^3 + (B*a^2 + 2*A*a*
b)*d*e^4)*log(e*x + d)/e^6

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Fricas [A]  time = 0.264005, size = 512, normalized size = 1.99 \[ \frac{12 \, B c^{2} e^{5} x^{5} - 15 \,{\left (B c^{2} d e^{4} -{\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{3} -{\left (2 \, B b c + A c^{2}\right )} d e^{4} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} e^{5}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e^{2} -{\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d e^{4} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{5}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} e -{\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{2} e^{3} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{4} +{\left (B a^{2} + 2 \, A a b\right )} e^{5}\right )} x - 60 \,{\left (B c^{2} d^{5} - A a^{2} e^{5} -{\left (2 \, B b c + A c^{2}\right )} d^{4} e +{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} d^{3} e^{2} -{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} +{\left (B a^{2} + 2 \, A a b\right )} d e^{4}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/(e*x + d),x, algorithm="fricas")

[Out]

1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 20*(B*c^
2*d^2*e^3 - (2*B*b*c + A*c^2)*d*e^4 + (B*b^2 + 2*(B*a + A*b)*c)*e^5)*x^3 - 30*(B
*c^2*d^3*e^2 - (2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*(B*a + A*b)*c)*d*e^4 - (2*
B*a*b + A*b^2 + 2*A*a*c)*e^5)*x^2 + 60*(B*c^2*d^4*e - (2*B*b*c + A*c^2)*d^3*e^2
+ (B*b^2 + 2*(B*a + A*b)*c)*d^2*e^3 - (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^4 + (B*a^2
+ 2*A*a*b)*e^5)*x - 60*(B*c^2*d^5 - A*a^2*e^5 - (2*B*b*c + A*c^2)*d^4*e + (B*b^
2 + 2*(B*a + A*b)*c)*d^3*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d^2*e^3 + (B*a^2 + 2*
A*a*b)*d*e^4)*log(e*x + d))/e^6

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Sympy [A]  time = 5.58348, size = 369, normalized size = 1.44 \[ \frac{B c^{2} x^{5}}{5 e} + \frac{x^{4} \left (A c^{2} e + 2 B b c e - B c^{2} d\right )}{4 e^{2}} + \frac{x^{3} \left (2 A b c e^{2} - A c^{2} d e + 2 B a c e^{2} + B b^{2} e^{2} - 2 B b c d e + B c^{2} d^{2}\right )}{3 e^{3}} + \frac{x^{2} \left (2 A a c e^{3} + A b^{2} e^{3} - 2 A b c d e^{2} + A c^{2} d^{2} e + 2 B a b e^{3} - 2 B a c d e^{2} - B b^{2} d e^{2} + 2 B b c d^{2} e - B c^{2} d^{3}\right )}{2 e^{4}} + \frac{x \left (2 A a b e^{4} - 2 A a c d e^{3} - A b^{2} d e^{3} + 2 A b c d^{2} e^{2} - A c^{2} d^{3} e + B a^{2} e^{4} - 2 B a b d e^{3} + 2 B a c d^{2} e^{2} + B b^{2} d^{2} e^{2} - 2 B b c d^{3} e + B c^{2} d^{4}\right )}{e^{5}} - \frac{\left (- A e + B d\right ) \left (a e^{2} - b d e + c d^{2}\right )^{2} \log{\left (d + e x \right )}}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x+A)*(c*x**2+b*x+a)**2/(e*x+d),x)

[Out]

B*c**2*x**5/(5*e) + x**4*(A*c**2*e + 2*B*b*c*e - B*c**2*d)/(4*e**2) + x**3*(2*A*
b*c*e**2 - A*c**2*d*e + 2*B*a*c*e**2 + B*b**2*e**2 - 2*B*b*c*d*e + B*c**2*d**2)/
(3*e**3) + x**2*(2*A*a*c*e**3 + A*b**2*e**3 - 2*A*b*c*d*e**2 + A*c**2*d**2*e + 2
*B*a*b*e**3 - 2*B*a*c*d*e**2 - B*b**2*d*e**2 + 2*B*b*c*d**2*e - B*c**2*d**3)/(2*
e**4) + x*(2*A*a*b*e**4 - 2*A*a*c*d*e**3 - A*b**2*d*e**3 + 2*A*b*c*d**2*e**2 - A
*c**2*d**3*e + B*a**2*e**4 - 2*B*a*b*d*e**3 + 2*B*a*c*d**2*e**2 + B*b**2*d**2*e*
*2 - 2*B*b*c*d**3*e + B*c**2*d**4)/e**5 - (-A*e + B*d)*(a*e**2 - b*d*e + c*d**2)
**2*log(d + e*x)/e**6

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GIAC/XCAS [A]  time = 0.257391, size = 625, normalized size = 2.43 \[ -{\left (B c^{2} d^{5} - 2 \, B b c d^{4} e - A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 2 \, B a c d^{3} e^{2} + 2 \, A b c d^{3} e^{2} - 2 \, B a b d^{2} e^{3} - A b^{2} d^{2} e^{3} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + 2 \, A a b d e^{4} - A a^{2} e^{5}\right )} e^{\left (-6\right )}{\rm ln}\left ({\left | x e + d \right |}\right ) + \frac{1}{60} \,{\left (12 \, B c^{2} x^{5} e^{4} - 15 \, B c^{2} d x^{4} e^{3} + 20 \, B c^{2} d^{2} x^{3} e^{2} - 30 \, B c^{2} d^{3} x^{2} e + 60 \, B c^{2} d^{4} x + 30 \, B b c x^{4} e^{4} + 15 \, A c^{2} x^{4} e^{4} - 40 \, B b c d x^{3} e^{3} - 20 \, A c^{2} d x^{3} e^{3} + 60 \, B b c d^{2} x^{2} e^{2} + 30 \, A c^{2} d^{2} x^{2} e^{2} - 120 \, B b c d^{3} x e - 60 \, A c^{2} d^{3} x e + 20 \, B b^{2} x^{3} e^{4} + 40 \, B a c x^{3} e^{4} + 40 \, A b c x^{3} e^{4} - 30 \, B b^{2} d x^{2} e^{3} - 60 \, B a c d x^{2} e^{3} - 60 \, A b c d x^{2} e^{3} + 60 \, B b^{2} d^{2} x e^{2} + 120 \, B a c d^{2} x e^{2} + 120 \, A b c d^{2} x e^{2} + 60 \, B a b x^{2} e^{4} + 30 \, A b^{2} x^{2} e^{4} + 60 \, A a c x^{2} e^{4} - 120 \, B a b d x e^{3} - 60 \, A b^{2} d x e^{3} - 120 \, A a c d x e^{3} + 60 \, B a^{2} x e^{4} + 120 \, A a b x e^{4}\right )} e^{\left (-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/(e*x + d),x, algorithm="giac")

[Out]

-(B*c^2*d^5 - 2*B*b*c*d^4*e - A*c^2*d^4*e + B*b^2*d^3*e^2 + 2*B*a*c*d^3*e^2 + 2*
A*b*c*d^3*e^2 - 2*B*a*b*d^2*e^3 - A*b^2*d^2*e^3 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4
+ 2*A*a*b*d*e^4 - A*a^2*e^5)*e^(-6)*ln(abs(x*e + d)) + 1/60*(12*B*c^2*x^5*e^4 -
15*B*c^2*d*x^4*e^3 + 20*B*c^2*d^2*x^3*e^2 - 30*B*c^2*d^3*x^2*e + 60*B*c^2*d^4*x
+ 30*B*b*c*x^4*e^4 + 15*A*c^2*x^4*e^4 - 40*B*b*c*d*x^3*e^3 - 20*A*c^2*d*x^3*e^3
+ 60*B*b*c*d^2*x^2*e^2 + 30*A*c^2*d^2*x^2*e^2 - 120*B*b*c*d^3*x*e - 60*A*c^2*d^3
*x*e + 20*B*b^2*x^3*e^4 + 40*B*a*c*x^3*e^4 + 40*A*b*c*x^3*e^4 - 30*B*b^2*d*x^2*e
^3 - 60*B*a*c*d*x^2*e^3 - 60*A*b*c*d*x^2*e^3 + 60*B*b^2*d^2*x*e^2 + 120*B*a*c*d^
2*x*e^2 + 120*A*b*c*d^2*x*e^2 + 60*B*a*b*x^2*e^4 + 30*A*b^2*x^2*e^4 + 60*A*a*c*x
^2*e^4 - 120*B*a*b*d*x*e^3 - 60*A*b^2*d*x*e^3 - 120*A*a*c*d*x*e^3 + 60*B*a^2*x*e
^4 + 120*A*a*b*x*e^4)*e^(-5)