3.10.92 \(\int \frac {(A+B x) (d+e x)^4}{b x+c x^2} \, dx\)

Optimal. Leaf size=207 \[ \frac {e^2 x^2 \left (A c e (4 c d-b e)+B \left (b^2 e^2-4 b c d e+6 c^2 d^2\right )\right )}{2 c^3}+\frac {e x \left (A c e \left (b^2 e^2-4 b c d e+6 c^2 d^2\right )+B \left (-b^3 e^3+4 b^2 c d e^2-6 b c^2 d^2 e+4 c^3 d^3\right )\right )}{c^4}+\frac {(b B-A c) (c d-b e)^4 \log (b+c x)}{b c^5}+\frac {e^3 x^3 (A c e-b B e+4 B c d)}{3 c^2}+\frac {A d^4 \log (x)}{b}+\frac {B e^4 x^4}{4 c} \]

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Rubi [A]  time = 0.26, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {771} \begin {gather*} \frac {e^2 x^2 \left (A c e (4 c d-b e)+B \left (b^2 e^2-4 b c d e+6 c^2 d^2\right )\right )}{2 c^3}+\frac {e x \left (A c e \left (b^2 e^2-4 b c d e+6 c^2 d^2\right )+B \left (4 b^2 c d e^2-b^3 e^3-6 b c^2 d^2 e+4 c^3 d^3\right )\right )}{c^4}+\frac {e^3 x^3 (A c e-b B e+4 B c d)}{3 c^2}+\frac {(b B-A c) (c d-b e)^4 \log (b+c x)}{b c^5}+\frac {A d^4 \log (x)}{b}+\frac {B e^4 x^4}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^4)/(b*x + c*x^2),x]

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 187, normalized size = 0.90 \begin {gather*} \frac {e x \left (2 A c e \left (6 b^2 e^2-3 b c e (8 d+e x)+2 c^2 \left (18 d^2+6 d e x+e^2 x^2\right )\right )+B \left (-12 b^3 e^3+6 b^2 c e^2 (8 d+e x)-4 b c^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+c^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )\right )}{12 c^4}+\frac {(b B-A c) (c d-b e)^4 \log (b+c x)}{b c^5}+\frac {A d^4 \log (x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^4)/(b*x + c*x^2),x]

[Out]

(e*x*(2*A*c*e*(6*b^2*e^2 - 3*b*c*e*(8*d + e*x) + 2*c^2*(18*d^2 + 6*d*e*x + e^2*x^2)) + B*(-12*b^3*e^3 + 6*b^2*
c*e^2*(8*d + e*x) - 4*b*c^2*e*(18*d^2 + 6*d*e*x + e^2*x^2) + c^3*(48*d^3 + 36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x
^3))))/(12*c^4) + (A*d^4*Log[x])/b + ((b*B - A*c)*(c*d - b*e)^4*Log[b + c*x])/(b*c^5)

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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)^4}{b x+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(b*x + c*x^2),x]

[Out]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(b*x + c*x^2), x]

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fricas [A]  time = 0.42, size = 331, normalized size = 1.60 \begin {gather*} \frac {3 \, B b c^{4} e^{4} x^{4} + 12 \, A c^{5} d^{4} \log \relax (x) + 4 \, {\left (4 \, B b c^{4} d e^{3} - {\left (B b^{2} c^{3} - A b c^{4}\right )} e^{4}\right )} x^{3} + 6 \, {\left (6 \, B b c^{4} d^{2} e^{2} - 4 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d e^{3} + {\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} e^{4}\right )} x^{2} + 12 \, {\left (4 \, B b c^{4} d^{3} e - 6 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e^{2} + 4 \, {\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} d e^{3} - {\left (B b^{4} c - A b^{3} c^{2}\right )} e^{4}\right )} x + 12 \, {\left ({\left (B b c^{4} - A c^{5}\right )} d^{4} - 4 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{3} e + 6 \, {\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} d^{2} e^{2} - 4 \, {\left (B b^{4} c - A b^{3} c^{2}\right )} d e^{3} + {\left (B b^{5} - A b^{4} c\right )} e^{4}\right )} \log \left (c x + b\right )}{12 \, b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 330, normalized size = 1.59 \begin {gather*} \frac {A d^{4} \log \left ({\left | x \right |}\right )}{b} + \frac {3 \, B c^{3} x^{4} e^{4} + 16 \, B c^{3} d x^{3} e^{3} + 36 \, B c^{3} d^{2} x^{2} e^{2} + 48 \, B c^{3} d^{3} x e - 4 \, B b c^{2} x^{3} e^{4} + 4 \, A c^{3} x^{3} e^{4} - 24 \, B b c^{2} d x^{2} e^{3} + 24 \, A c^{3} d x^{2} e^{3} - 72 \, B b c^{2} d^{2} x e^{2} + 72 \, A c^{3} d^{2} x e^{2} + 6 \, B b^{2} c x^{2} e^{4} - 6 \, A b c^{2} x^{2} e^{4} + 48 \, B b^{2} c d x e^{3} - 48 \, A b c^{2} d x e^{3} - 12 \, B b^{3} x e^{4} + 12 \, A b^{2} c x e^{4}}{12 \, c^{4}} + \frac {{\left (B b c^{4} d^{4} - A c^{5} d^{4} - 4 \, B b^{2} c^{3} d^{3} e + 4 \, A b c^{4} d^{3} e + 6 \, B b^{3} c^{2} d^{2} e^{2} - 6 \, A b^{2} c^{3} d^{2} e^{2} - 4 \, B b^{4} c d e^{3} + 4 \, A b^{3} c^{2} d e^{3} + B b^{5} e^{4} - A b^{4} c e^{4}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+b*x),x, algorithm="giac")

[Out]

A*d^4*log(abs(x))/b + 1/12*(3*B*c^3*x^4*e^4 + 16*B*c^3*d*x^3*e^3 + 36*B*c^3*d^2*x^2*e^2 + 48*B*c^3*d^3*x*e - 4
*B*b*c^2*x^3*e^4 + 4*A*c^3*x^3*e^4 - 24*B*b*c^2*d*x^2*e^3 + 24*A*c^3*d*x^2*e^3 - 72*B*b*c^2*d^2*x*e^2 + 72*A*c
^3*d^2*x*e^2 + 6*B*b^2*c*x^2*e^4 - 6*A*b*c^2*x^2*e^4 + 48*B*b^2*c*d*x*e^3 - 48*A*b*c^2*d*x*e^3 - 12*B*b^3*x*e^
4 + 12*A*b^2*c*x*e^4)/c^4 + (B*b*c^4*d^4 - A*c^5*d^4 - 4*B*b^2*c^3*d^3*e + 4*A*b*c^4*d^3*e + 6*B*b^3*c^2*d^2*e
^2 - 6*A*b^2*c^3*d^2*e^2 - 4*B*b^4*c*d*e^3 + 4*A*b^3*c^2*d*e^3 + B*b^5*e^4 - A*b^4*c*e^4)*log(abs(c*x + b))/(b
*c^5)

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maple [A]  time = 0.05, size = 396, normalized size = 1.91 \begin {gather*} \frac {B \,e^{4} x^{4}}{4 c}+\frac {A \,e^{4} x^{3}}{3 c}-\frac {B b \,e^{4} x^{3}}{3 c^{2}}+\frac {4 B d \,e^{3} x^{3}}{3 c}-\frac {A b \,e^{4} x^{2}}{2 c^{2}}+\frac {2 A d \,e^{3} x^{2}}{c}+\frac {B \,b^{2} e^{4} x^{2}}{2 c^{3}}-\frac {2 B b d \,e^{3} x^{2}}{c^{2}}+\frac {3 B \,d^{2} e^{2} x^{2}}{c}-\frac {A \,b^{3} e^{4} \ln \left (c x +b \right )}{c^{4}}+\frac {4 A \,b^{2} d \,e^{3} \ln \left (c x +b \right )}{c^{3}}+\frac {A \,b^{2} e^{4} x}{c^{3}}-\frac {6 A b \,d^{2} e^{2} \ln \left (c x +b \right )}{c^{2}}-\frac {4 A b d \,e^{3} x}{c^{2}}+\frac {A \,d^{4} \ln \relax (x )}{b}-\frac {A \,d^{4} \ln \left (c x +b \right )}{b}+\frac {4 A \,d^{3} e \ln \left (c x +b \right )}{c}+\frac {6 A \,d^{2} e^{2} x}{c}+\frac {B \,b^{4} e^{4} \ln \left (c x +b \right )}{c^{5}}-\frac {4 B \,b^{3} d \,e^{3} \ln \left (c x +b \right )}{c^{4}}-\frac {B \,b^{3} e^{4} x}{c^{4}}+\frac {6 B \,b^{2} d^{2} e^{2} \ln \left (c x +b \right )}{c^{3}}+\frac {4 B \,b^{2} d \,e^{3} x}{c^{3}}-\frac {4 B b \,d^{3} e \ln \left (c x +b \right )}{c^{2}}-\frac {6 B b \,d^{2} e^{2} x}{c^{2}}+\frac {B \,d^{4} \ln \left (c x +b \right )}{c}+\frac {4 B \,d^{3} e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^4/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 0.50, size = 309, normalized size = 1.49 \begin {gather*} \frac {A d^{4} \log \relax (x)}{b} + \frac {3 \, B c^{3} e^{4} x^{4} + 4 \, {\left (4 \, B c^{3} d e^{3} - {\left (B b c^{2} - A c^{3}\right )} e^{4}\right )} x^{3} + 6 \, {\left (6 \, B c^{3} d^{2} e^{2} - 4 \, {\left (B b c^{2} - A c^{3}\right )} d e^{3} + {\left (B b^{2} c - A b c^{2}\right )} e^{4}\right )} x^{2} + 12 \, {\left (4 \, B c^{3} d^{3} e - 6 \, {\left (B b c^{2} - A c^{3}\right )} d^{2} e^{2} + 4 \, {\left (B b^{2} c - A b c^{2}\right )} d e^{3} - {\left (B b^{3} - A b^{2} c\right )} e^{4}\right )} x}{12 \, c^{4}} + \frac {{\left ({\left (B b c^{4} - A c^{5}\right )} d^{4} - 4 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{3} e + 6 \, {\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} d^{2} e^{2} - 4 \, {\left (B b^{4} c - A b^{3} c^{2}\right )} d e^{3} + {\left (B b^{5} - A b^{4} c\right )} e^{4}\right )} \log \left (c x + b\right )}{b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+b*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.67, size = 322, normalized size = 1.56 \begin {gather*} x\,\left (\frac {b\,\left (\frac {b\,\left (\frac {A\,e^4+4\,B\,d\,e^3}{c}-\frac {B\,b\,e^4}{c^2}\right )}{c}-\frac {2\,d\,e^2\,\left (2\,A\,e+3\,B\,d\right )}{c}\right )}{c}+\frac {2\,d^2\,e\,\left (3\,A\,e+2\,B\,d\right )}{c}\right )-x^2\,\left (\frac {b\,\left (\frac {A\,e^4+4\,B\,d\,e^3}{c}-\frac {B\,b\,e^4}{c^2}\right )}{2\,c}-\frac {d\,e^2\,\left (2\,A\,e+3\,B\,d\right )}{c}\right )+x^3\,\left (\frac {A\,e^4+4\,B\,d\,e^3}{3\,c}-\frac {B\,b\,e^4}{3\,c^2}\right )-\ln \left (b+c\,x\right )\,\left (\frac {A\,d^4}{b}-\frac {c^4\,\left (B\,b\,d^4+4\,A\,b\,e\,d^3\right )-c\,\left (A\,b^4\,e^4+4\,B\,d\,b^4\,e^3\right )-c^3\,\left (4\,B\,b^2\,d^3\,e+6\,A\,b^2\,d^2\,e^2\right )+c^2\,\left (6\,B\,b^3\,d^2\,e^2+4\,A\,b^3\,d\,e^3\right )+B\,b^5\,e^4}{b\,c^5}\right )+\frac {A\,d^4\,\ln \relax (x)}{b}+\frac {B\,e^4\,x^4}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^4)/(b*x + c*x^2),x)

[Out]

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

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sympy [A]  time = 7.18, size = 396, normalized size = 1.91 \begin {gather*} \frac {A d^{4} \log {\relax (x )}}{b} + \frac {B e^{4} x^{4}}{4 c} + x^{3} \left (\frac {A e^{4}}{3 c} - \frac {B b e^{4}}{3 c^{2}} + \frac {4 B d e^{3}}{3 c}\right ) + x^{2} \left (- \frac {A b e^{4}}{2 c^{2}} + \frac {2 A d e^{3}}{c} + \frac {B b^{2} e^{4}}{2 c^{3}} - \frac {2 B b d e^{3}}{c^{2}} + \frac {3 B d^{2} e^{2}}{c}\right ) + x \left (\frac {A b^{2} e^{4}}{c^{3}} - \frac {4 A b d e^{3}}{c^{2}} + \frac {6 A d^{2} e^{2}}{c} - \frac {B b^{3} e^{4}}{c^{4}} + \frac {4 B b^{2} d e^{3}}{c^{3}} - \frac {6 B b d^{2} e^{2}}{c^{2}} + \frac {4 B d^{3} e}{c}\right ) + \frac {\left (- A c + B b\right ) \left (b e - c d\right )^{4} \log {\left (x + \frac {- A b c^{4} d^{4} + \frac {b \left (- A c + B b\right ) \left (b e - c d\right )^{4}}{c}}{- A b^{4} c e^{4} + 4 A b^{3} c^{2} d e^{3} - 6 A b^{2} c^{3} d^{2} e^{2} + 4 A b c^{4} d^{3} e - 2 A c^{5} d^{4} + B b^{5} e^{4} - 4 B b^{4} c d e^{3} + 6 B b^{3} c^{2} d^{2} e^{2} - 4 B b^{2} c^{3} d^{3} e + B b c^{4} d^{4}} \right )}}{b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**4/(c*x**2+b*x),x)

[Out]

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

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