3.15.82 \(\int \frac {(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=144 \[ \frac {e^2 \log (a+b x) (-4 a B e+A b e+3 b B d)}{b^5}-\frac {3 e (b d-a e) (-2 a B e+A b e+b B d)}{b^5 (a+b x)}-\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{2 b^5 (a+b x)^2}-\frac {(A b-a B) (b d-a e)^3}{3 b^5 (a+b x)^3}+\frac {B e^3 x}{b^4} \]

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Rubi [A]  time = 0.15, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 77} \begin {gather*} \frac {e^2 \log (a+b x) (-4 a B e+A b e+3 b B d)}{b^5}-\frac {3 e (b d-a e) (-2 a B e+A b e+b B d)}{b^5 (a+b x)}-\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{2 b^5 (a+b x)^2}-\frac {(A b-a B) (b d-a e)^3}{3 b^5 (a+b x)^3}+\frac {B e^3 x}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 217, normalized size = 1.51 \begin {gather*} -\frac {A b (b d-a e) \left (11 a^2 e^2+a b e (5 d+27 e x)+b^2 \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )+B \left (26 a^4 e^3+3 a^3 b e^2 (18 e x-11 d)+3 a^2 b^2 e \left (2 d^2-27 d e x+6 e^2 x^2\right )+a b^3 \left (d^3+18 d^2 e x-54 d e^2 x^2-18 e^3 x^3\right )+3 b^4 x \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )-6 e^2 (a+b x)^3 \log (a+b x) (-4 a B e+A b e+3 b B d)}{6 b^5 (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(A*b*(b*d - a*e)*(11*a^2*e^2 + a*b*e*(5*d + 27*e*x) + b^2*(2*d^2 + 9*d*e*x + 18*e^2*x^2)) + B*(26*a^4*e^3
+ 3*a^3*b*e^2*(-11*d + 18*e*x) + 3*a^2*b^2*e*(2*d^2 - 27*d*e*x + 6*e^2*x^2) + a*b^3*(d^3 + 18*d^2*e*x - 54*d*
e^2*x^2 - 18*e^3*x^3) + 3*b^4*x*(d^3 + 6*d^2*e*x - 2*e^3*x^3)) - 6*e^2*(3*b*B*d + A*b*e - 4*a*B*e)*(a + b*x)^3
*Log[a + b*x])/(b^5*(a + b*x)^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 427, normalized size = 2.97 \begin {gather*} \frac {6 \, B b^{4} e^{3} x^{4} + 18 \, B a b^{3} e^{3} x^{3} - {\left (B a b^{3} + 2 \, A b^{4}\right )} d^{3} - 3 \, {\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} d^{2} e + 3 \, {\left (11 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} d e^{2} - {\left (26 \, B a^{4} - 11 \, A a^{3} b\right )} e^{3} - 18 \, {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} - 3 \, {\left (B b^{4} d^{3} + 3 \, {\left (2 \, B a b^{3} + A b^{4}\right )} d^{2} e - 3 \, {\left (9 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} d e^{2} + 9 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (3 \, B a^{3} b d e^{2} - {\left (4 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (3 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (3 \, B a b^{3} d e^{2} - {\left (4 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{2} b^{2} d e^{2} - {\left (4 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*B*b^4*e^3*x^4 + 18*B*a*b^3*e^3*x^3 - (B*a*b^3 + 2*A*b^4)*d^3 - 3*(2*B*a^2*b^2 + A*a*b^3)*d^2*e + 3*(11*
B*a^3*b - 2*A*a^2*b^2)*d*e^2 - (26*B*a^4 - 11*A*a^3*b)*e^3 - 18*(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (B*
a^2*b^2 - A*a*b^3)*e^3)*x^2 - 3*(B*b^4*d^3 + 3*(2*B*a*b^3 + A*b^4)*d^2*e - 3*(9*B*a^2*b^2 - 2*A*a*b^3)*d*e^2 +
9*(2*B*a^3*b - A*a^2*b^2)*e^3)*x + 6*(3*B*a^3*b*d*e^2 - (4*B*a^4 - A*a^3*b)*e^3 + (3*B*b^4*d*e^2 - (4*B*a*b^3
- A*b^4)*e^3)*x^3 + 3*(3*B*a*b^3*d*e^2 - (4*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3*(3*B*a^2*b^2*d*e^2 - (4*B*a^3*b
- A*a^2*b^2)*e^3)*x)*log(b*x + a))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)

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giac [A]  time = 0.19, size = 266, normalized size = 1.85 \begin {gather*} \frac {B x e^{3}}{b^{4}} + \frac {{\left (3 \, B b d e^{2} - 4 \, B a e^{3} + A b e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac {B a b^{3} d^{3} + 2 \, A b^{4} d^{3} + 6 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e - 33 \, B a^{3} b d e^{2} + 6 \, A a^{2} b^{2} d e^{2} + 26 \, B a^{4} e^{3} - 11 \, A a^{3} b e^{3} + 18 \, {\left (B b^{4} d^{2} e - 3 \, B a b^{3} d e^{2} + A b^{4} d e^{2} + 2 \, B a^{2} b^{2} e^{3} - A a b^{3} e^{3}\right )} x^{2} + 3 \, {\left (B b^{4} d^{3} + 6 \, B a b^{3} d^{2} e + 3 \, A b^{4} d^{2} e - 27 \, B a^{2} b^{2} d e^{2} + 6 \, A a b^{3} d e^{2} + 20 \, B a^{3} b e^{3} - 9 \, A a^{2} b^{2} e^{3}\right )} x}{6 \, {\left (b x + a\right )}^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x*e^3/b^4 + (3*B*b*d*e^2 - 4*B*a*e^3 + A*b*e^3)*log(abs(b*x + a))/b^5 - 1/6*(B*a*b^3*d^3 + 2*A*b^4*d^3 + 6*B
*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e - 33*B*a^3*b*d*e^2 + 6*A*a^2*b^2*d*e^2 + 26*B*a^4*e^3 - 11*A*a^3*b*e^3 + 18*(
B*b^4*d^2*e - 3*B*a*b^3*d*e^2 + A*b^4*d*e^2 + 2*B*a^2*b^2*e^3 - A*a*b^3*e^3)*x^2 + 3*(B*b^4*d^3 + 6*B*a*b^3*d^
2*e + 3*A*b^4*d^2*e - 27*B*a^2*b^2*d*e^2 + 6*A*a*b^3*d*e^2 + 20*B*a^3*b*e^3 - 9*A*a^2*b^2*e^3)*x)/((b*x + a)^3
*b^5)

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maple [B]  time = 0.06, size = 419, normalized size = 2.91 \begin {gather*} \frac {A \,a^{3} e^{3}}{3 \left (b x +a \right )^{3} b^{4}}-\frac {A \,a^{2} d \,e^{2}}{\left (b x +a \right )^{3} b^{3}}+\frac {A a \,d^{2} e}{\left (b x +a \right )^{3} b^{2}}-\frac {A \,d^{3}}{3 \left (b x +a \right )^{3} b}-\frac {B \,a^{4} e^{3}}{3 \left (b x +a \right )^{3} b^{5}}+\frac {B \,a^{3} d \,e^{2}}{\left (b x +a \right )^{3} b^{4}}-\frac {B \,a^{2} d^{2} e}{\left (b x +a \right )^{3} b^{3}}+\frac {B a \,d^{3}}{3 \left (b x +a \right )^{3} b^{2}}-\frac {3 A \,a^{2} e^{3}}{2 \left (b x +a \right )^{2} b^{4}}+\frac {3 A a d \,e^{2}}{\left (b x +a \right )^{2} b^{3}}-\frac {3 A \,d^{2} e}{2 \left (b x +a \right )^{2} b^{2}}+\frac {2 B \,a^{3} e^{3}}{\left (b x +a \right )^{2} b^{5}}-\frac {9 B \,a^{2} d \,e^{2}}{2 \left (b x +a \right )^{2} b^{4}}+\frac {3 B a \,d^{2} e}{\left (b x +a \right )^{2} b^{3}}-\frac {B \,d^{3}}{2 \left (b x +a \right )^{2} b^{2}}+\frac {3 A a \,e^{3}}{\left (b x +a \right ) b^{4}}-\frac {3 A d \,e^{2}}{\left (b x +a \right ) b^{3}}+\frac {A \,e^{3} \ln \left (b x +a \right )}{b^{4}}-\frac {6 B \,a^{2} e^{3}}{\left (b x +a \right ) b^{5}}+\frac {9 B a d \,e^{2}}{\left (b x +a \right ) b^{4}}-\frac {4 B a \,e^{3} \ln \left (b x +a \right )}{b^{5}}-\frac {3 B \,d^{2} e}{\left (b x +a \right ) b^{3}}+\frac {3 B d \,e^{2} \ln \left (b x +a \right )}{b^{4}}+\frac {B \,e^{3} x}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e^3*x/b^4+1/3/b^4/(b*x+a)^3*A*a^3*e^3-1/b^3/(b*x+a)^3*A*a^2*d*e^2+1/b^2/(b*x+a)^3*A*a*d^2*e-1/3/b/(b*x+a)^3*
A*d^3-1/3/b^5/(b*x+a)^3*B*a^4*e^3+1/b^4/(b*x+a)^3*a^3*B*d*e^2-1/b^3/(b*x+a)^3*B*a^2*d^2*e+1/3/b^2/(b*x+a)^3*B*
a*d^3-3/2/b^4/(b*x+a)^2*A*a^2*e^3+3/b^3/(b*x+a)^2*A*a*d*e^2-3/2/b^2/(b*x+a)^2*A*d^2*e+2/b^5/(b*x+a)^2*B*a^3*e^
3-9/2/b^4/(b*x+a)^2*B*a^2*d*e^2+3/b^3/(b*x+a)^2*B*a*d^2*e-1/2/b^2/(b*x+a)^2*B*d^3+1/b^4*e^3*ln(b*x+a)*A-4/b^5*
e^3*ln(b*x+a)*a*B+3/b^4*e^2*ln(b*x+a)*B*d+3*e^3/b^4/(b*x+a)*A*a-3*e^2/b^3/(b*x+a)*A*d-6*e^3/b^5/(b*x+a)*B*a^2+
9*e^2/b^4/(b*x+a)*B*d*a-3*e/b^3/(b*x+a)*B*d^2

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maxima [B]  time = 0.64, size = 292, normalized size = 2.03 \begin {gather*} \frac {B e^{3} x}{b^{4}} - \frac {{\left (B a b^{3} + 2 \, A b^{4}\right )} d^{3} + 3 \, {\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} d^{2} e - 3 \, {\left (11 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} d e^{2} + {\left (26 \, B a^{4} - 11 \, A a^{3} b\right )} e^{3} + 18 \, {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (B b^{4} d^{3} + 3 \, {\left (2 \, B a b^{3} + A b^{4}\right )} d^{2} e - 3 \, {\left (9 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} d e^{2} + {\left (20 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} e^{3}\right )} x}{6 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} + \frac {{\left (3 \, B b d e^{2} - {\left (4 \, B a - A b\right )} e^{3}\right )} \log \left (b x + a\right )}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e^3*x/b^4 - 1/6*((B*a*b^3 + 2*A*b^4)*d^3 + 3*(2*B*a^2*b^2 + A*a*b^3)*d^2*e - 3*(11*B*a^3*b - 2*A*a^2*b^2)*d*
e^2 + (26*B*a^4 - 11*A*a^3*b)*e^3 + 18*(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)
*x^2 + 3*(B*b^4*d^3 + 3*(2*B*a*b^3 + A*b^4)*d^2*e - 3*(9*B*a^2*b^2 - 2*A*a*b^3)*d*e^2 + (20*B*a^3*b - 9*A*a^2*
b^2)*e^3)*x)/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5) + (3*B*b*d*e^2 - (4*B*a - A*b)*e^3)*log(b*x + a)/
b^5

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mupad [B]  time = 2.34, size = 301, normalized size = 2.09 \begin {gather*} \frac {\ln \left (a+b\,x\right )\,\left (A\,b\,e^3-4\,B\,a\,e^3+3\,B\,b\,d\,e^2\right )}{b^5}-\frac {\frac {26\,B\,a^4\,e^3-33\,B\,a^3\,b\,d\,e^2-11\,A\,a^3\,b\,e^3+6\,B\,a^2\,b^2\,d^2\,e+6\,A\,a^2\,b^2\,d\,e^2+B\,a\,b^3\,d^3+3\,A\,a\,b^3\,d^2\,e+2\,A\,b^4\,d^3}{6\,b}+x\,\left (10\,B\,a^3\,e^3-\frac {27\,B\,a^2\,b\,d\,e^2}{2}-\frac {9\,A\,a^2\,b\,e^3}{2}+3\,B\,a\,b^2\,d^2\,e+3\,A\,a\,b^2\,d\,e^2+\frac {B\,b^3\,d^3}{2}+\frac {3\,A\,b^3\,d^2\,e}{2}\right )+x^2\,\left (6\,B\,a^2\,b\,e^3-9\,B\,a\,b^2\,d\,e^2-3\,A\,a\,b^2\,e^3+3\,B\,b^3\,d^2\,e+3\,A\,b^3\,d\,e^2\right )}{a^3\,b^4+3\,a^2\,b^5\,x+3\,a\,b^6\,x^2+b^7\,x^3}+\frac {B\,e^3\,x}{b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a + b*x)*(A*b*e^3 - 4*B*a*e^3 + 3*B*b*d*e^2))/b^5 - ((2*A*b^4*d^3 + 26*B*a^4*e^3 - 11*A*a^3*b*e^3 + B*a*b
^3*d^3 + 6*A*a^2*b^2*d*e^2 + 6*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e - 33*B*a^3*b*d*e^2)/(6*b) + x*(10*B*a^3*e^3 +
(B*b^3*d^3)/2 - (9*A*a^2*b*e^3)/2 + (3*A*b^3*d^2*e)/2 + 3*A*a*b^2*d*e^2 + 3*B*a*b^2*d^2*e - (27*B*a^2*b*d*e^2
)/2) + x^2*(6*B*a^2*b*e^3 - 3*A*a*b^2*e^3 + 3*A*b^3*d*e^2 + 3*B*b^3*d^2*e - 9*B*a*b^2*d*e^2))/(a^3*b^4 + b^7*x
^3 + 3*a^2*b^5*x + 3*a*b^6*x^2) + (B*e^3*x)/b^4

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sympy [B]  time = 12.68, size = 337, normalized size = 2.34 \begin {gather*} \frac {B e^{3} x}{b^{4}} + \frac {11 A a^{3} b e^{3} - 6 A a^{2} b^{2} d e^{2} - 3 A a b^{3} d^{2} e - 2 A b^{4} d^{3} - 26 B a^{4} e^{3} + 33 B a^{3} b d e^{2} - 6 B a^{2} b^{2} d^{2} e - B a b^{3} d^{3} + x^{2} \left (18 A a b^{3} e^{3} - 18 A b^{4} d e^{2} - 36 B a^{2} b^{2} e^{3} + 54 B a b^{3} d e^{2} - 18 B b^{4} d^{2} e\right ) + x \left (27 A a^{2} b^{2} e^{3} - 18 A a b^{3} d e^{2} - 9 A b^{4} d^{2} e - 60 B a^{3} b e^{3} + 81 B a^{2} b^{2} d e^{2} - 18 B a b^{3} d^{2} e - 3 B b^{4} d^{3}\right )}{6 a^{3} b^{5} + 18 a^{2} b^{6} x + 18 a b^{7} x^{2} + 6 b^{8} x^{3}} - \frac {e^{2} \left (- A b e + 4 B a e - 3 B b d\right ) \log {\left (a + b x \right )}}{b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

B*e**3*x/b**4 + (11*A*a**3*b*e**3 - 6*A*a**2*b**2*d*e**2 - 3*A*a*b**3*d**2*e - 2*A*b**4*d**3 - 26*B*a**4*e**3
+ 33*B*a**3*b*d*e**2 - 6*B*a**2*b**2*d**2*e - B*a*b**3*d**3 + x**2*(18*A*a*b**3*e**3 - 18*A*b**4*d*e**2 - 36*B
*a**2*b**2*e**3 + 54*B*a*b**3*d*e**2 - 18*B*b**4*d**2*e) + x*(27*A*a**2*b**2*e**3 - 18*A*a*b**3*d*e**2 - 9*A*b
**4*d**2*e - 60*B*a**3*b*e**3 + 81*B*a**2*b**2*d*e**2 - 18*B*a*b**3*d**2*e - 3*B*b**4*d**3))/(6*a**3*b**5 + 18
*a**2*b**6*x + 18*a*b**7*x**2 + 6*b**8*x**3) - e**2*(-A*b*e + 4*B*a*e - 3*B*b*d)*log(a + b*x)/b**5

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