3.1511 \(\int \frac{(b+2 c x) (a+b x+c x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=219 \[ \frac{c x \left (-c e (15 b d-4 a e)+4 b^2 e^2+12 c^2 d^2\right )}{e^5}-\frac{2 \left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6 (d+e x)}-\frac{(2 c d-b e) \log (d+e x) \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )}{e^6}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{2 e^6 (d+e x)^2}-\frac{c^2 x^2 (6 c d-5 b e)}{2 e^4}+\frac{2 c^3 x^3}{3 e^3} \]

[Out]

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

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Rubi [A]  time = 0.249652, antiderivative size = 219, 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} \[ \frac{c x \left (-c e (15 b d-4 a e)+4 b^2 e^2+12 c^2 d^2\right )}{e^5}-\frac{2 \left (a e^2-b d e+c d^2\right ) \left (-c e (5 b d-a e)+b^2 e^2+5 c^2 d^2\right )}{e^6 (d+e x)}-\frac{(2 c d-b e) \log (d+e x) \left (-2 c e (5 b d-3 a e)+b^2 e^2+10 c^2 d^2\right )}{e^6}+\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^2}{2 e^6 (d+e x)^2}-\frac{c^2 x^2 (6 c d-5 b e)}{2 e^4}+\frac{2 c^3 x^3}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(12*c^2*d^2 + 4*b^2*e^2 - c*e*(15*b*d - 4*a*e))*x)/e^5 - (c^2*(6*c*d - 5*b*e)*x^2)/(2*e^4) + (2*c^3*x^3)/(3
*e^3) + ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^2)/(2*e^6*(d + e*x)^2) - (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)))/(e^6*(d + e*x)) - ((2*c*d - b*e)*(10*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 3*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)^3} \, dx &=\int \left (\frac{c \left (12 c^2 d^2+4 b^2 e^2-c e (15 b d-4 a e)\right )}{e^5}-\frac{c^2 (6 c d-5 b e) x}{e^4}+\frac{2 c^3 x^2}{e^3}+\frac{(-2 c d+b e) \left (c d^2-b d e+a e^2\right )^2}{e^5 (d+e x)^3}+\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)^2}+\frac{(2 c d-b e) \left (-10 c^2 d^2-b^2 e^2+2 c e (5 b d-3 a e)\right )}{e^5 (d+e x)}\right ) \, dx\\ &=\frac{c \left (12 c^2 d^2+4 b^2 e^2-c e (15 b d-4 a e)\right ) x}{e^5}-\frac{c^2 (6 c d-5 b e) x^2}{2 e^4}+\frac{2 c^3 x^3}{3 e^3}+\frac{(2 c d-b e) \left (c d^2-b d e+a e^2\right )^2}{2 e^6 (d+e x)^2}-\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 )}{e^6 (d+e x)}-\frac{(2 c d-b e) \left (10 c^2 d^2+b^2 e^2-2 c e (5 b d-3 a e)\right ) \log (d+e x)}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.0866134, size = 233, normalized size = 1.06 \[ \frac{-\frac{12 \left (c e^2 \left (a^2 e^2-6 a b d e+6 b^2 d^2\right )+b^2 e^3 (a e-b d)+2 c^2 d^2 e (3 a e-5 b d)+5 c^3 d^4\right )}{d+e x}+6 c e x \left (c e (4 a e-15 b d)+4 b^2 e^2+12 c^2 d^2\right )-6 (2 c d-b e) \log (d+e x) \left (2 c e (3 a e-5 b d)+b^2 e^2+10 c^2 d^2\right )+\frac{3 (2 c d-b e) \left (e (a e-b d)+c d^2\right )^2}{(d+e x)^2}-3 c^2 e^2 x^2 (6 c d-5 b e)+4 c^3 e^3 x^3}{6 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 471, normalized size = 2.2 \begin{align*} -15\,{\frac{bd{c}^{2}x}{{e}^{4}}}+{\frac{d{a}^{2}c}{{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{{b}^{2}da}{{e}^{2} \left ( ex+d \right ) ^{2}}}+2\,{\frac{a{c}^{2}{d}^{3}}{{e}^{4} \left ( ex+d \right ) ^{2}}}+2\,{\frac{{b}^{2}{d}^{3}c}{{e}^{4} \left ( ex+d \right ) ^{2}}}-{\frac{5\,b{d}^{4}{c}^{2}}{2\,{e}^{5} \left ( ex+d \right ) ^{2}}}-12\,{\frac{a{c}^{2}{d}^{2}}{{e}^{4} \left ( ex+d \right ) }}-12\,{\frac{{b}^{2}{d}^{2}c}{{e}^{4} \left ( ex+d \right ) }}+20\,{\frac{b{d}^{3}{c}^{2}}{{e}^{5} \left ( ex+d \right ) }}+6\,{\frac{\ln \left ( ex+d \right ) cab}{{e}^{3}}}-12\,{\frac{\ln \left ( ex+d \right ) a{c}^{2}d}{{e}^{4}}}-12\,{\frac{\ln \left ( ex+d \right ){b}^{2}cd}{{e}^{4}}}+30\,{\frac{\ln \left ( ex+d \right ) b{c}^{2}{d}^{2}}{{e}^{5}}}+{\frac{2\,{c}^{3}{x}^{3}}{3\,{e}^{3}}}+12\,{\frac{abcd}{{e}^{3} \left ( ex+d \right ) }}-3\,{\frac{{d}^{2}abc}{{e}^{3} \left ( ex+d \right ) ^{2}}}-3\,{\frac{{c}^{3}{x}^{2}d}{{e}^{4}}}+4\,{\frac{a{c}^{2}x}{{e}^{3}}}+4\,{\frac{{b}^{2}cx}{{e}^{3}}}+12\,{\frac{{c}^{3}{d}^{2}x}{{e}^{5}}}+{\frac{{c}^{3}{d}^{5}}{{e}^{6} \left ( ex+d \right ) ^{2}}}-20\,{\frac{\ln \left ( ex+d \right ){c}^{3}{d}^{3}}{{e}^{6}}}-2\,{\frac{c{a}^{2}}{{e}^{2} \left ( ex+d \right ) }}-2\,{\frac{{b}^{2}a}{{e}^{2} \left ( ex+d \right ) }}+2\,{\frac{{b}^{3}d}{{e}^{3} \left ( ex+d \right ) }}-10\,{\frac{{c}^{3}{d}^{4}}{{e}^{6} \left ( ex+d \right ) }}+{\frac{5\,{c}^{2}{x}^{2}b}{2\,{e}^{3}}}-{\frac{{a}^{2}b}{2\,e \left ( ex+d \right ) ^{2}}}-{\frac{{d}^{2}{b}^{3}}{2\,{e}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{\ln \left ( ex+d \right ){b}^{3}}{{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)^3,x)

[Out]

-15*c^2/e^4*b*d*x+1/e^2/(e*x+d)^2*d*a^2*c+1/e^2/(e*x+d)^2*d*a*b^2+2/e^4/(e*x+d)^2*a*c^2*d^3+2/e^4/(e*x+d)^2*b^
2*c*d^3-5/2/e^5/(e*x+d)^2*b*c^2*d^4-12/e^4/(e*x+d)*a*c^2*d^2-12/e^4/(e*x+d)*b^2*c*d^2+20/e^5/(e*x+d)*b*c^2*d^3
+6/e^3*ln(e*x+d)*c*a*b-12/e^4*ln(e*x+d)*a*c^2*d-12/e^4*ln(e*x+d)*b^2*c*d+30/e^5*ln(e*x+d)*b*c^2*d^2+2/3*c^3*x^
3/e^3+12/e^3/(e*x+d)*a*b*c*d-3/e^3/(e*x+d)^2*d^2*a*b*c-3*c^3/e^4*x^2*d+4*c^2/e^3*a*x+4*c/e^3*b^2*x+12*c^3/e^5*
d^2*x+1/e^6/(e*x+d)^2*c^3*d^5-20/e^6*ln(e*x+d)*c^3*d^3-2/e^2/(e*x+d)*c*a^2-2/e^2/(e*x+d)*a*b^2+2/e^3/(e*x+d)*b
^3*d-10/e^6/(e*x+d)*c^3*d^4+5/2*c^2/e^3*x^2*b-1/2/e/(e*x+d)^2*a^2*b-1/2/e^3/(e*x+d)^2*d^2*b^3+1/e^3*ln(e*x+d)*
b^3

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Maxima [A]  time = 1.05594, size = 428, normalized size = 1.95 \begin{align*} -\frac{18 \, c^{3} d^{5} - 35 \, b c^{2} d^{4} e + a^{2} b e^{5} + 20 \,{\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} - 3 \,{\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} + 2 \,{\left (a b^{2} + a^{2} c\right )} d e^{4} + 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}{2 \,{\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} + \frac{4 \, c^{3} e^{2} x^{3} - 3 \,{\left (6 \, c^{3} d e - 5 \, b c^{2} e^{2}\right )} x^{2} + 6 \,{\left (12 \, c^{3} d^{2} - 15 \, b c^{2} d e + 4 \,{\left (b^{2} c + a c^{2}\right )} e^{2}\right )} x}{6 \, e^{5}} - \frac{{\left (20 \, c^{3} d^{3} - 30 \, b c^{2} d^{2} e + 12 \,{\left (b^{2} c + a c^{2}\right )} d e^{2} -{\left (b^{3} + 6 \, a b c\right )} e^{3}\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)^3,x, algorithm="maxima")

[Out]

-1/2*(18*c^3*d^5 - 35*b*c^2*d^4*e + a^2*b*e^5 + 20*(b^2*c + a*c^2)*d^3*e^2 - 3*(b^3 + 6*a*b*c)*d^2*e^3 + 2*(a*
b^2 + a^2*c)*d*e^4 + 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)/(e^8*x^2 + 2*d*e^7*x + d^2*e^6) + 1/6*(4*c^3*e^2*x^3 - 3*(6*c^3*d*e - 5*b*c^2*e^2)*x^2
+ 6*(12*c^3*d^2 - 15*b*c^2*d*e + 4*(b^2*c + a*c^2)*e^2)*x)/e^5 - (20*c^3*d^3 - 30*b*c^2*d^2*e + 12*(b^2*c + a*
c^2)*d*e^2 - (b^3 + 6*a*b*c)*e^3)*log(e*x + d)/e^6

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Fricas [B]  time = 1.49383, size = 1021, normalized size = 4.66 \begin{align*} \frac{4 \, c^{3} e^{5} x^{5} - 54 \, c^{3} d^{5} + 105 \, b c^{2} d^{4} e - 3 \, a^{2} b e^{5} - 60 \,{\left (b^{2} c + a c^{2}\right )} d^{3} e^{2} + 9 \,{\left (b^{3} + 6 \, a b c\right )} d^{2} e^{3} - 6 \,{\left (a b^{2} + a^{2} c\right )} d e^{4} - 5 \,{\left (2 \, c^{3} d e^{4} - 3 \, b c^{2} e^{5}\right )} x^{4} + 4 \,{\left (10 \, c^{3} d^{2} e^{3} - 15 \, b c^{2} d e^{4} + 6 \,{\left (b^{2} c + a c^{2}\right )} e^{5}\right )} x^{3} + 3 \,{\left (42 \, c^{3} d^{3} e^{2} - 55 \, b c^{2} d^{2} e^{3} + 16 \,{\left (b^{2} c + a c^{2}\right )} d e^{4}\right )} x^{2} + 6 \,{\left (2 \, c^{3} d^{4} e + 5 \, b c^{2} d^{3} e^{2} - 8 \,{\left (b^{2} c + a c^{2}\right )} d^{2} e^{3} + 2 \,{\left (b^{3} + 6 \, a b c\right )} d e^{4} - 2 \,{\left (a b^{2} + a^{2} c\right )} e^{5}\right )} x - 6 \,{\left (20 \, c^{3} d^{5} - 30 \, b c^{2} d^{4} e + 12 \,{\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 (20 \, c^{3} d^{3} e^{2} - 30 \, b c^{2} d^{2} e^{3} + 12 \,{\left (b^{2} c + a c^{2}\right )} d e^{4} -{\left (b^{3} + 6 \, a b c\right )} e^{5}\right )} x^{2} + 2 \,{\left (20 \, c^{3} d^{4} e - 30 \, b c^{2} d^{3} e^{2} + 12 \,{\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\right )} \log \left (e x + d\right )}{6 \,{\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

1/6*(4*c^3*e^5*x^5 - 54*c^3*d^5 + 105*b*c^2*d^4*e - 3*a^2*b*e^5 - 60*(b^2*c + a*c^2)*d^3*e^2 + 9*(b^3 + 6*a*b*
c)*d^2*e^3 - 6*(a*b^2 + a^2*c)*d*e^4 - 5*(2*c^3*d*e^4 - 3*b*c^2*e^5)*x^4 + 4*(10*c^3*d^2*e^3 - 15*b*c^2*d*e^4
+ 6*(b^2*c + a*c^2)*e^5)*x^3 + 3*(42*c^3*d^3*e^2 - 55*b*c^2*d^2*e^3 + 16*(b^2*c + a*c^2)*d*e^4)*x^2 + 6*(2*c^3
*d^4*e + 5*b*c^2*d^3*e^2 - 8*(b^2*c + a*c^2)*d^2*e^3 + 2*(b^3 + 6*a*b*c)*d*e^4 - 2*(a*b^2 + a^2*c)*e^5)*x - 6*
(20*c^3*d^5 - 30*b*c^2*d^4*e + 12*(b^2*c + a*c^2)*d^3*e^2 - (b^3 + 6*a*b*c)*d^2*e^3 + (20*c^3*d^3*e^2 - 30*b*c
^2*d^2*e^3 + 12*(b^2*c + a*c^2)*d*e^4 - (b^3 + 6*a*b*c)*e^5)*x^2 + 2*(20*c^3*d^4*e - 30*b*c^2*d^3*e^2 + 12*(b^
2*c + a*c^2)*d^2*e^3 - (b^3 + 6*a*b*c)*d*e^4)*x)*log(e*x + d))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)

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Sympy [A]  time = 7.20122, size = 357, normalized size = 1.63 \begin{align*} \frac{2 c^{3} x^{3}}{3 e^{3}} - \frac{a^{2} b e^{5} + 2 a^{2} c d e^{4} + 2 a b^{2} d e^{4} - 18 a b c d^{2} e^{3} + 20 a c^{2} d^{3} e^{2} - 3 b^{3} d^{2} e^{3} + 20 b^{2} c d^{3} e^{2} - 35 b c^{2} d^{4} e + 18 c^{3} d^{5} + x \left (4 a^{2} c e^{5} + 4 a b^{2} e^{5} - 24 a b c d e^{4} + 24 a c^{2} d^{2} e^{3} - 4 b^{3} d e^{4} + 24 b^{2} c d^{2} e^{3} - 40 b c^{2} d^{3} e^{2} + 20 c^{3} d^{4} e\right )}{2 d^{2} e^{6} + 4 d e^{7} x + 2 e^{8} x^{2}} + \frac{x^{2} \left (5 b c^{2} e - 6 c^{3} d\right )}{2 e^{4}} + \frac{x \left (4 a c^{2} e^{2} + 4 b^{2} c e^{2} - 15 b c^{2} d e + 12 c^{3} d^{2}\right )}{e^{5}} + \frac{\left (b e - 2 c d\right ) \left (6 a c e^{2} + b^{2} e^{2} - 10 b c d e + 10 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)**3,x)

[Out]

2*c**3*x**3/(3*e**3) - (a**2*b*e**5 + 2*a**2*c*d*e**4 + 2*a*b**2*d*e**4 - 18*a*b*c*d**2*e**3 + 20*a*c**2*d**3*
e**2 - 3*b**3*d**2*e**3 + 20*b**2*c*d**3*e**2 - 35*b*c**2*d**4*e + 18*c**3*d**5 + x*(4*a**2*c*e**5 + 4*a*b**2*
e**5 - 24*a*b*c*d*e**4 + 24*a*c**2*d**2*e**3 - 4*b**3*d*e**4 + 24*b**2*c*d**2*e**3 - 40*b*c**2*d**3*e**2 + 20*
c**3*d**4*e))/(2*d**2*e**6 + 4*d*e**7*x + 2*e**8*x**2) + x**2*(5*b*c**2*e - 6*c**3*d)/(2*e**4) + x*(4*a*c**2*e
**2 + 4*b**2*c*e**2 - 15*b*c**2*d*e + 12*c**3*d**2)/e**5 + (b*e - 2*c*d)*(6*a*c*e**2 + b**2*e**2 - 10*b*c*d*e
+ 10*c**2*d**2)*log(d + e*x)/e**6

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Giac [A]  time = 1.15305, size = 428, normalized size = 1.95 \begin{align*} -{\left (20 \, c^{3} d^{3} - 30 \, b c^{2} d^{2} e + 12 \, b^{2} c d e^{2} + 12 \, a c^{2} d e^{2} - b^{3} e^{3} - 6 \, a b c e^{3}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{6} \,{\left (4 \, c^{3} x^{3} e^{6} - 18 \, c^{3} d x^{2} e^{5} + 72 \, c^{3} d^{2} x e^{4} + 15 \, b c^{2} x^{2} e^{6} - 90 \, b c^{2} d x e^{5} + 24 \, b^{2} c x e^{6} + 24 \, a c^{2} x e^{6}\right )} e^{\left (-9\right )} - \frac{{\left (18 \, c^{3} d^{5} - 35 \, b c^{2} d^{4} e + 20 \, b^{2} c d^{3} e^{2} + 20 \, a c^{2} d^{3} e^{2} - 3 \, b^{3} d^{2} e^{3} - 18 \, a b c d^{2} e^{3} + 2 \, a b^{2} d e^{4} + 2 \, a^{2} c d e^{4} + a^{2} b e^{5} + 4 \,{\left (5 \, c^{3} d^{4} e - 10 \, b c^{2} d^{3} e^{2} + 6 \, b^{2} c d^{2} e^{3} + 6 \, a c^{2} d^{2} e^{3} - b^{3} d e^{4} - 6 \, a b c d e^{4} + a b^{2} e^{5} + a^{2} c e^{5}\right )} x\right )} e^{\left (-6\right )}}{2 \,{\left (x e + d\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

-(20*c^3*d^3 - 30*b*c^2*d^2*e + 12*b^2*c*d*e^2 + 12*a*c^2*d*e^2 - b^3*e^3 - 6*a*b*c*e^3)*e^(-6)*log(abs(x*e +
d)) + 1/6*(4*c^3*x^3*e^6 - 18*c^3*d*x^2*e^5 + 72*c^3*d^2*x*e^4 + 15*b*c^2*x^2*e^6 - 90*b*c^2*d*x*e^5 + 24*b^2*
c*x*e^6 + 24*a*c^2*x*e^6)*e^(-9) - 1/2*(18*c^3*d^5 - 35*b*c^2*d^4*e + 20*b^2*c*d^3*e^2 + 20*a*c^2*d^3*e^2 - 3*
b^3*d^2*e^3 - 18*a*b*c*d^2*e^3 + 2*a*b^2*d*e^4 + 2*a^2*c*d*e^4 + a^2*b*e^5 + 4*(5*c^3*d^4*e - 10*b*c^2*d^3*e^2
+ 6*b^2*c*d^2*e^3 + 6*a*c^2*d^2*e^3 - b^3*d*e^4 - 6*a*b*c*d*e^4 + a*b^2*e^5 + a^2*c*e^5)*x)*e^(-6)/(x*e + d)^
2