3.1510 \(\int \frac{(b+2 c x) (a+b x+c x^2)^2}{(d+e x)^2} \, dx\)

Optimal. Leaf size=223 \[ \frac{c x^2 \left (-c e (5 b d-2 a e)+2 b^2 e^2+3 c^2 d^2\right )}{e^4}-\frac{x \left (-c^2 d e (15 b d-8 a e)+2 b c e^2 (4 b d-3 a e)-b^3 e^3+8 c^3 d^3\right )}{e^5}+\frac{2 \log (d+e x) \left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{e^6 (d+e x)}-\frac{c^2 x^3 (4 c d-5 b e)}{3 e^3}+\frac{c^3 x^4}{2 e^2} \]

[Out]

-(((8*c^3*d^3 - b^3*e^3 - c^2*d*e*(15*b*d - 8*a*e) + 2*b*c*e^2*(4*b*d - 3*a*e))*x)/e^5) + (c*(3*c^2*d^2 + 2*b^
2*e^2 - c*e*(5*b*d - 2*a*e))*x^2)/e^4 - (c^2*(4*c*d - 5*b*e)*x^3)/(3*e^3) + (c^3*x^4)/(2*e^2) + ((2*c*d - b*e)
*(c*d^2 - b*d*e + a*e^2)^2)/(e^6*(d + e*x)) + (2*(c*d^2 - b*d*e + a*e^2)*(5*c^2*d^2 + b^2*e^2 - c*e*(5*b*d - a
*e))*Log[d + e*x])/e^6

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Rubi [A]  time = 0.289959, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {771} \[ \frac{c x^2 \left (-c e (5 b d-2 a e)+2 b^2 e^2+3 c^2 d^2\right )}{e^4}-\frac{x \left (-c^2 d e (15 b d-8 a e)+2 b c e^2 (4 b d-3 a e)-b^3 e^3+8 c^3 d^3\right )}{e^5}+\frac{2 \log (d+e x) \left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{e^6 (d+e x)}-\frac{c^2 x^3 (4 c d-5 b e)}{3 e^3}+\frac{c^3 x^4}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((8*c^3*d^3 - b^3*e^3 - c^2*d*e*(15*b*d - 8*a*e) + 2*b*c*e^2*(4*b*d - 3*a*e))*x)/e^5) + (c*(3*c^2*d^2 + 2*b^
2*e^2 - c*e*(5*b*d - 2*a*e))*x^2)/e^4 - (c^2*(4*c*d - 5*b*e)*x^3)/(3*e^3) + (c^3*x^4)/(2*e^2) + ((2*c*d - b*e)
*(c*d^2 - b*d*e + a*e^2)^2)/(e^6*(d + e*x)) + (2*(c*d^2 - b*d*e + a*e^2)*(5*c^2*d^2 + b^2*e^2 - c*e*(5*b*d - a
*e))*Log[d + e*x])/e^6

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{(b+2 c x) \left (a+b x+c x^2\right )^2}{(d+e x)^2} \, dx &=\int \left (\frac{-8 c^3 d^3+b^3 e^3+c^2 d e (15 b d-8 a e)-2 b c e^2 (4 b d-3 a e)}{e^5}+\frac{2 c \left (3 c^2 d^2+2 b^2 e^2-c e (5 b d-2 a e)\right ) x}{e^4}-\frac{c^2 (4 c d-5 b e) x^2}{e^3}+\frac{2 c^3 x^3}{e^2}+\frac{(-2 c d+b e) \left (c d^2-b d e+a e^2\right )^2}{e^5 (d+e x)^2}+\frac{2 \left (c d^2-b d e+a e^2\right ) \left (5 c^2 d^2-5 b c d e+b^2 e^2+a c e^2\right )}{e^5 (d+e x)}\right ) \, dx\\ &=-\frac{\left (8 c^3 d^3-b^3 e^3-c^2 d e (15 b d-8 a e)+2 b c e^2 (4 b d-3 a e)\right ) x}{e^5}+\frac{c \left (3 c^2 d^2+2 b^2 e^2-c e (5 b d-2 a e)\right ) x^2}{e^4}-\frac{c^2 (4 c d-5 b e) x^3}{3 e^3}+\frac{c^3 x^4}{2 e^2}+\frac{(2 c d-b e) \left (c d^2-b d e+a e^2\right )^2}{e^6 (d+e x)}+\frac{2 \left (c d^2-b d e+a e^2\right ) \left (5 c^2 d^2+b^2 e^2-c e (5 b d-a e)\right ) \log (d+e x)}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.200251, size = 241, normalized size = 1.08 \[ \frac{12 \log (d+e x) \left (c e^2 \left (a^2 e^2-6 a b d e+6 b^2 d^2\right )+b^2 e^3 (a e-b d)+2 c^2 d^2 e (3 a e-5 b d)+5 c^3 d^4\right )+6 c e^2 x^2 \left (c e (2 a e-5 b d)+2 b^2 e^2+3 c^2 d^2\right )+6 e x \left (c^2 d e (15 b d-8 a e)+2 b c e^2 (3 a e-4 b d)+b^3 e^3-8 c^3 d^3\right )+\frac{6 (2 c d-b e) \left (e (a e-b d)+c d^2\right )^2}{d+e x}-2 c^2 e^3 x^3 (4 c d-5 b e)+3 c^3 e^4 x^4}{6 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*e*(-8*c^3*d^3 + b^3*e^3 + c^2*d*e*(15*b*d - 8*a*e) + 2*b*c*e^2*(-4*b*d + 3*a*e))*x + 6*c*e^2*(3*c^2*d^2 + 2
*b^2*e^2 + c*e*(-5*b*d + 2*a*e))*x^2 - 2*c^2*e^3*(4*c*d - 5*b*e)*x^3 + 3*c^3*e^4*x^4 + (6*(2*c*d - b*e)*(c*d^2
+ e*(-(b*d) + a*e))^2)/(d + e*x) + 12*(5*c^3*d^4 + b^2*e^3*(-(b*d) + a*e) + 2*c^2*d^2*e*(-5*b*d + 3*a*e) + c*
e^2*(6*b^2*d^2 - 6*a*b*d*e + a^2*e^2))*Log[d + e*x])/(6*e^6)

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Maple [B]  time = 0.012, size = 444, normalized size = 2. \begin{align*} 2\,{\frac{{b}^{2}da}{{e}^{2} \left ( ex+d \right ) }}+2\,{\frac{d{a}^{2}c}{{e}^{2} \left ( ex+d \right ) }}+{\frac{{c}^{3}{x}^{4}}{2\,{e}^{2}}}-12\,{\frac{\ln \left ( ex+d \right ) abcd}{{e}^{3}}}-6\,{\frac{abc{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}-8\,{\frac{{c}^{3}{d}^{3}x}{{e}^{5}}}+3\,{\frac{{x}^{2}{c}^{3}{d}^{2}}{{e}^{4}}}+2\,{\frac{{b}^{2}{x}^{2}c}{{e}^{2}}}+{\frac{5\,b{x}^{3}{c}^{2}}{3\,{e}^{2}}}-{\frac{4\,{x}^{3}{c}^{3}d}{3\,{e}^{3}}}+2\,{\frac{a{x}^{2}{c}^{2}}{{e}^{2}}}+2\,{\frac{{c}^{3}{d}^{5}}{{e}^{6} \left ( ex+d \right ) }}+10\,{\frac{\ln \left ( ex+d \right ){c}^{3}{d}^{4}}{{e}^{6}}}+2\,{\frac{\ln \left ( ex+d \right ) c{a}^{2}}{{e}^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) a{b}^{2}}{{e}^{2}}}-2\,{\frac{\ln \left ( ex+d \right ){b}^{3}d}{{e}^{3}}}-{\frac{{a}^{2}b}{e \left ( ex+d \right ) }}-{\frac{{b}^{3}{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}+{\frac{{b}^{3}x}{{e}^{2}}}+12\,{\frac{\ln \left ( ex+d \right ){b}^{2}c{d}^{2}}{{e}^{4}}}-20\,{\frac{\ln \left ( ex+d \right ) b{c}^{2}{d}^{3}}{{e}^{5}}}+6\,{\frac{abcx}{{e}^{2}}}-5\,{\frac{b{x}^{2}{c}^{2}d}{{e}^{3}}}+12\,{\frac{\ln \left ( ex+d \right ) a{c}^{2}{d}^{2}}{{e}^{4}}}-8\,{\frac{{b}^{2}cdx}{{e}^{3}}}+15\,{\frac{b{d}^{2}{c}^{2}x}{{e}^{4}}}+4\,{\frac{a{c}^{2}{d}^{3}}{{e}^{4} \left ( ex+d \right ) }}+4\,{\frac{{b}^{2}{d}^{3}c}{{e}^{4} \left ( ex+d \right ) }}-5\,{\frac{b{d}^{4}{c}^{2}}{{e}^{5} \left ( ex+d \right ) }}-8\,{\frac{a{c}^{2}dx}{{e}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e^2/(e*x+d)*d*a*b^2+2/e^2/(e*x+d)*d*a^2*c+1/2*c^3*x^4/e^2-12/e^3*ln(e*x+d)*a*b*c*d-6/e^3/(e*x+d)*a*b*c*d^2-8
/e^5*c^3*d^3*x+3/e^4*x^2*c^3*d^2+2/e^2*x^2*b^2*c+5/3/e^2*x^3*b*c^2-4/3/e^3*x^3*c^3*d+2/e^2*x^2*a*c^2+2/e^6/(e*
x+d)*c^3*d^5+10/e^6*ln(e*x+d)*c^3*d^4+2/e^2*ln(e*x+d)*c*a^2+2/e^2*ln(e*x+d)*a*b^2-2/e^3*ln(e*x+d)*b^3*d-1/e/(e
*x+d)*a^2*b-1/e^3/(e*x+d)*b^3*d^2+1/e^2*b^3*x+12/e^4*ln(e*x+d)*b^2*c*d^2-20/e^5*ln(e*x+d)*b*c^2*d^3+6/e^2*c*a*
b*x-5/e^3*x^2*b*c^2*d+12/e^4*ln(e*x+d)*a*c^2*d^2-8/e^3*b^2*c*d*x+15/e^4*b*d^2*c^2*x+4/e^4/(e*x+d)*a*c^2*d^3+4/
e^4/(e*x+d)*b^2*c*d^3-5/e^5/(e*x+d)*b*c^2*d^4-8/e^3*a*c^2*d*x

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Maxima [A]  time = 1.02214, size = 419, normalized size = 1.88 \begin{align*} \frac{2 \, c^{3} d^{5} - 5 \, b c^{2} d^{4} e - a^{2} b e^{5} + 4 \,{\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} -{\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} + 2 \,{\left (a b^{2} + a^{2} c\right )} d e^{4}}{e^{7} x + d e^{6}} + \frac{3 \, c^{3} e^{3} x^{4} - 2 \,{\left (4 \, c^{3} d e^{2} - 5 \, b c^{2} e^{3}\right )} x^{3} + 6 \,{\left (3 \, c^{3} d^{2} e - 5 \, b c^{2} d e^{2} + 2 \,{\left (b^{2} c + a c^{2}\right )} e^{3}\right )} x^{2} - 6 \,{\left (8 \, c^{3} d^{3} - 15 \, b c^{2} d^{2} e + 8 \,{\left (b^{2} c + a c^{2}\right )} d e^{2} -{\left (b^{3} + 6 \, a b c\right )} e^{3}\right )} x}{6 \, e^{5}} + \frac{2 \,{\left (5 \, c^{3} d^{4} - 10 \, b c^{2} d^{3} e + 6 \,{\left (b^{2} c + a c^{2}\right )} d^{2} e^{2} -{\left (b^{3} + 6 \, a b c\right )} d e^{3} +{\left (a b^{2} + a^{2} c\right )} e^{4}\right )} \log \left (e x + d\right )}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^3*d^5 - 5*b*c^2*d^4*e - a^2*b*e^5 + 4*(b^2*c + a*c^2)*d^3*e^2 - (b^3 + 6*a*b*c)*d^2*e^3 + 2*(a*b^2 + a^2*
c)*d*e^4)/(e^7*x + d*e^6) + 1/6*(3*c^3*e^3*x^4 - 2*(4*c^3*d*e^2 - 5*b*c^2*e^3)*x^3 + 6*(3*c^3*d^2*e - 5*b*c^2*
d*e^2 + 2*(b^2*c + a*c^2)*e^3)*x^2 - 6*(8*c^3*d^3 - 15*b*c^2*d^2*e + 8*(b^2*c + a*c^2)*d*e^2 - (b^3 + 6*a*b*c)
*e^3)*x)/e^5 + 2*(5*c^3*d^4 - 10*b*c^2*d^3*e + 6*(b^2*c + a*c^2)*d^2*e^2 - (b^3 + 6*a*b*c)*d*e^3 + (a*b^2 + a^
2*c)*e^4)*log(e*x + d)/e^6

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Fricas [B]  time = 1.49618, size = 921, normalized size = 4.13 \begin{align*} \frac{3 \, c^{3} e^{5} x^{5} + 12 \, c^{3} d^{5} - 30 \, b c^{2} d^{4} e - 6 \, a^{2} b e^{5} + 24 \,{\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} - 6 \,{\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} + 12 \,{\left (a b^{2} + a^{2} c\right )} d e^{4} - 5 \,{\left (c^{3} d e^{4} - 2 \, b c^{2} e^{5}\right )} x^{4} + 2 \,{\left (5 \, c^{3} d^{2} e^{3} - 10 \, b c^{2} d e^{4} + 6 \,{\left (b^{2} c + a c^{2}\right )} e^{5}\right )} x^{3} - 6 \,{\left (5 \, c^{3} d^{3} e^{2} - 10 \, b c^{2} d^{2} e^{3} + 6 \,{\left (b^{2} c + a c^{2}\right )} d e^{4} -{\left (b^{3} + 6 \, a b c\right )} e^{5}\right )} x^{2} - 6 \,{\left (8 \, c^{3} d^{4} e - 15 \, b c^{2} d^{3} e^{2} + 8 \,{\left (b^{2} c + a c^{2}\right )} d^{2} e^{3} -{\left (b^{3} + 6 \, a b c\right )} d e^{4}\right )} x + 12 \,{\left (5 \, c^{3} d^{5} - 10 \, b c^{2} d^{4} e + 6 \,{\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} -{\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} +{\left (a b^{2} + a^{2} c\right )} d e^{4} +{\left (5 \, c^{3} d^{4} e - 10 \, b c^{2} d^{3} e^{2} + 6 \,{\left (b^{2} c + a c^{2}\right )} d^{2} e^{3} -{\left (b^{3} + 6 \, a b c\right )} d e^{4} +{\left (a b^{2} + a^{2} c\right )} e^{5}\right )} x\right )} \log \left (e x + d\right )}{6 \,{\left (e^{7} x + d e^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*c^3*e^5*x^5 + 12*c^3*d^5 - 30*b*c^2*d^4*e - 6*a^2*b*e^5 + 24*(b^2*c + a*c^2)*d^3*e^2 - 6*(b^3 + 6*a*b*c
)*d^2*e^3 + 12*(a*b^2 + a^2*c)*d*e^4 - 5*(c^3*d*e^4 - 2*b*c^2*e^5)*x^4 + 2*(5*c^3*d^2*e^3 - 10*b*c^2*d*e^4 + 6
*(b^2*c + a*c^2)*e^5)*x^3 - 6*(5*c^3*d^3*e^2 - 10*b*c^2*d^2*e^3 + 6*(b^2*c + a*c^2)*d*e^4 - (b^3 + 6*a*b*c)*e^
5)*x^2 - 6*(8*c^3*d^4*e - 15*b*c^2*d^3*e^2 + 8*(b^2*c + a*c^2)*d^2*e^3 - (b^3 + 6*a*b*c)*d*e^4)*x + 12*(5*c^3*
d^5 - 10*b*c^2*d^4*e + 6*(b^2*c + a*c^2)*d^3*e^2 - (b^3 + 6*a*b*c)*d^2*e^3 + (a*b^2 + a^2*c)*d*e^4 + (5*c^3*d^
4*e - 10*b*c^2*d^3*e^2 + 6*(b^2*c + a*c^2)*d^2*e^3 - (b^3 + 6*a*b*c)*d*e^4 + (a*b^2 + a^2*c)*e^5)*x)*log(e*x +
d))/(e^7*x + d*e^6)

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Sympy [A]  time = 2.18455, size = 318, normalized size = 1.43 \begin{align*} \frac{c^{3} x^{4}}{2 e^{2}} - \frac{a^{2} b e^{5} - 2 a^{2} c d e^{4} - 2 a b^{2} d e^{4} + 6 a b c d^{2} e^{3} - 4 a c^{2} d^{3} e^{2} + b^{3} d^{2} e^{3} - 4 b^{2} c d^{3} e^{2} + 5 b c^{2} d^{4} e - 2 c^{3} d^{5}}{d e^{6} + e^{7} x} + \frac{x^{3} \left (5 b c^{2} e - 4 c^{3} d\right )}{3 e^{3}} + \frac{x^{2} \left (2 a c^{2} e^{2} + 2 b^{2} c e^{2} - 5 b c^{2} d e + 3 c^{3} d^{2}\right )}{e^{4}} + \frac{x \left (6 a b c e^{3} - 8 a c^{2} d e^{2} + b^{3} e^{3} - 8 b^{2} c d e^{2} + 15 b c^{2} d^{2} e - 8 c^{3} d^{3}\right )}{e^{5}} + \frac{2 \left (a e^{2} - b d e + c d^{2}\right ) \left (a c e^{2} + b^{2} e^{2} - 5 b c d e + 5 c^{2} d^{2}\right ) \log{\left (d + e x \right )}}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**3*x**4/(2*e**2) - (a**2*b*e**5 - 2*a**2*c*d*e**4 - 2*a*b**2*d*e**4 + 6*a*b*c*d**2*e**3 - 4*a*c**2*d**3*e**2
+ b**3*d**2*e**3 - 4*b**2*c*d**3*e**2 + 5*b*c**2*d**4*e - 2*c**3*d**5)/(d*e**6 + e**7*x) + x**3*(5*b*c**2*e -
4*c**3*d)/(3*e**3) + x**2*(2*a*c**2*e**2 + 2*b**2*c*e**2 - 5*b*c**2*d*e + 3*c**3*d**2)/e**4 + x*(6*a*b*c*e**3
- 8*a*c**2*d*e**2 + b**3*e**3 - 8*b**2*c*d*e**2 + 15*b*c**2*d**2*e - 8*c**3*d**3)/e**5 + 2*(a*e**2 - b*d*e +
c*d**2)*(a*c*e**2 + b**2*e**2 - 5*b*c*d*e + 5*c**2*d**2)*log(d + e*x)/e**6

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Giac [A]  time = 1.34768, size = 567, normalized size = 2.54 \begin{align*} \frac{1}{6} \,{\left (3 \, c^{3} - \frac{10 \,{\left (2 \, c^{3} d e - b c^{2} e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac{12 \,{\left (5 \, c^{3} d^{2} e^{2} - 5 \, b c^{2} d e^{3} + b^{2} c e^{4} + a c^{2} e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac{6 \,{\left (20 \, c^{3} d^{3} e^{3} - 30 \, b c^{2} d^{2} e^{4} + 12 \, b^{2} c d e^{5} + 12 \, a c^{2} d e^{5} - b^{3} e^{6} - 6 \, a b c e^{6}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}}\right )}{\left (x e + d\right )}^{4} e^{\left (-6\right )} - 2 \,{\left (5 \, c^{3} d^{4} - 10 \, b c^{2} d^{3} e + 6 \, b^{2} c d^{2} e^{2} + 6 \, a c^{2} d^{2} e^{2} - b^{3} d e^{3} - 6 \, a b c d e^{3} + a b^{2} e^{4} + a^{2} c e^{4}\right )} e^{\left (-6\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (\frac{2 \, c^{3} d^{5} e^{4}}{x e + d} - \frac{5 \, b c^{2} d^{4} e^{5}}{x e + d} + \frac{4 \, b^{2} c d^{3} e^{6}}{x e + d} + \frac{4 \, a c^{2} d^{3} e^{6}}{x e + d} - \frac{b^{3} d^{2} e^{7}}{x e + d} - \frac{6 \, a b c d^{2} e^{7}}{x e + d} + \frac{2 \, a b^{2} d e^{8}}{x e + d} + \frac{2 \, a^{2} c d e^{8}}{x e + d} - \frac{a^{2} b e^{9}}{x e + d}\right )} e^{\left (-10\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*c^3 - 10*(2*c^3*d*e - b*c^2*e^2)*e^(-1)/(x*e + d) + 12*(5*c^3*d^2*e^2 - 5*b*c^2*d*e^3 + b^2*c*e^4 + a*c
^2*e^4)*e^(-2)/(x*e + d)^2 - 6*(20*c^3*d^3*e^3 - 30*b*c^2*d^2*e^4 + 12*b^2*c*d*e^5 + 12*a*c^2*d*e^5 - b^3*e^6
- 6*a*b*c*e^6)*e^(-3)/(x*e + d)^3)*(x*e + d)^4*e^(-6) - 2*(5*c^3*d^4 - 10*b*c^2*d^3*e + 6*b^2*c*d^2*e^2 + 6*a*
c^2*d^2*e^2 - b^3*d*e^3 - 6*a*b*c*d*e^3 + a*b^2*e^4 + a^2*c*e^4)*e^(-6)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) +
(2*c^3*d^5*e^4/(x*e + d) - 5*b*c^2*d^4*e^5/(x*e + d) + 4*b^2*c*d^3*e^6/(x*e + d) + 4*a*c^2*d^3*e^6/(x*e + d)
- b^3*d^2*e^7/(x*e + d) - 6*a*b*c*d^2*e^7/(x*e + d) + 2*a*b^2*d*e^8/(x*e + d) + 2*a^2*c*d*e^8/(x*e + d) - a^2*
b*e^9/(x*e + d))*e^(-10)