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

Optimal. Leaf size=126 \[ -\frac {3 e \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {3 e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {(d+e x)^3}{a+b x+c x^2}+\frac {3 e^3 x}{c} \]

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Rubi [A]  time = 0.14, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {768, 701, 634, 618, 206, 628} \begin {gather*} -\frac {3 e \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {3 e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {(d+e x)^3}{a+b x+c x^2}+\frac {3 e^3 x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*e^3*x)/c - (d + e*x)^3/(a + b*x + c*x^2) - (3*e*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*ArcTanh[(b + 2*c*
x)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]) + (3*e^2*(2*c*d - b*e)*Log[a + b*x + c*x^2])/(2*c^2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^3}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {(d+e x)^3}{a+b x+c x^2}+(3 e) \int \frac {(d+e x)^2}{a+b x+c x^2} \, dx\\ &=-\frac {(d+e x)^3}{a+b x+c x^2}+(3 e) \int \left (\frac {e^2}{c}+\frac {c d^2-a e^2+e (2 c d-b e) x}{c \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {3 e^3 x}{c}-\frac {(d+e x)^3}{a+b x+c x^2}+\frac {(3 e) \int \frac {c d^2-a e^2+e (2 c d-b e) x}{a+b x+c x^2} \, dx}{c}\\ &=\frac {3 e^3 x}{c}-\frac {(d+e x)^3}{a+b x+c x^2}+\frac {\left (3 e^2 (2 c d-b e)\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^2}+\frac {\left (3 e \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^2}\\ &=\frac {3 e^3 x}{c}-\frac {(d+e x)^3}{a+b x+c x^2}+\frac {3 e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}-\frac {\left (3 e \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^2}\\ &=\frac {3 e^3 x}{c}-\frac {(d+e x)^3}{a+b x+c x^2}-\frac {3 e \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}+\frac {3 e^2 (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 161, normalized size = 1.28 \begin {gather*} \frac {\frac {6 e \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {2 \left (-c e^2 (3 a d+a e x+3 b d x)+b e^3 (a+b x)+c^2 d^2 (d+3 e x)\right )}{a+x (b+c x)}-3 e^2 (b e-2 c d) \log (a+x (b+c x))+4 c e^3 x}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 1050, normalized size = 8.33

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*(b^2*c^2 - 4*a*c^3)*e^3*x^3 + 4*(b^3*c - 4*a*b*c^2)*e^3*x^2 - 2*(b^2*c^2 - 4*a*c^3)*d^3 + 6*(a*b^2*c -
4*a^2*c^2)*d*e^2 - 2*(a*b^3 - 4*a^2*b*c)*e^3 - 3*(2*a*c^2*d^2*e - 2*a*b*c*d*e^2 + (a*b^2 - 2*a^2*c)*e^3 + (2*
c^3*d^2*e - 2*b*c^2*d*e^2 + (b^2*c - 2*a*c^2)*e^3)*x^2 + (2*b*c^2*d^2*e - 2*b^2*c*d*e^2 + (b^3 - 2*a*b*c)*e^3)
*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a
)) - 2*(3*(b^2*c^2 - 4*a*c^3)*d^2*e - 3*(b^3*c - 4*a*b*c^2)*d*e^2 + (b^4 - 7*a*b^2*c + 12*a^2*c^2)*e^3)*x + 3*
(2*(a*b^2*c - 4*a^2*c^2)*d*e^2 - (a*b^3 - 4*a^2*b*c)*e^3 + (2*(b^2*c^2 - 4*a*c^3)*d*e^2 - (b^3*c - 4*a*b*c^2)*
e^3)*x^2 + (2*(b^3*c - 4*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c)*e^3)*x)*log(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c
^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x), 1/2*(4*(b^2*c^2 - 4*a*c^3)*e^3*x^3 + 4*(b^3*c - 4*a*b
*c^2)*e^3*x^2 - 2*(b^2*c^2 - 4*a*c^3)*d^3 + 6*(a*b^2*c - 4*a^2*c^2)*d*e^2 - 2*(a*b^3 - 4*a^2*b*c)*e^3 - 6*(2*a
*c^2*d^2*e - 2*a*b*c*d*e^2 + (a*b^2 - 2*a^2*c)*e^3 + (2*c^3*d^2*e - 2*b*c^2*d*e^2 + (b^2*c - 2*a*c^2)*e^3)*x^2
+ (2*b*c^2*d^2*e - 2*b^2*c*d*e^2 + (b^3 - 2*a*b*c)*e^3)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c
*x + b)/(b^2 - 4*a*c)) - 2*(3*(b^2*c^2 - 4*a*c^3)*d^2*e - 3*(b^3*c - 4*a*b*c^2)*d*e^2 + (b^4 - 7*a*b^2*c + 12*
a^2*c^2)*e^3)*x + 3*(2*(a*b^2*c - 4*a^2*c^2)*d*e^2 - (a*b^3 - 4*a^2*b*c)*e^3 + (2*(b^2*c^2 - 4*a*c^3)*d*e^2 -
(b^3*c - 4*a*b*c^2)*e^3)*x^2 + (2*(b^3*c - 4*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c)*e^3)*x)*log(c*x^2 + b*x + a))/
(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x)]

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giac [A]  time = 0.16, size = 173, normalized size = 1.37 \begin {gather*} \frac {2 \, x e^{3}}{c} + \frac {3 \, {\left (2 \, c d e^{2} - b e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{2}} + \frac {3 \, {\left (2 \, c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3} - 2 \, a c e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{2}} - \frac {c^{2} d^{3} - 3 \, a c d e^{2} + a b e^{3} + {\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3} - a c e^{3}\right )} x}{{\left (c x^{2} + b x + a\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*e^3/c + 3/2*(2*c*d*e^2 - b*e^3)*log(c*x^2 + b*x + a)/c^2 + 3*(2*c^2*d^2*e - 2*b*c*d*e^2 + b^2*e^3 - 2*a*c*
e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^2) - (c^2*d^3 - 3*a*c*d*e^2 + a*b*e^3 + (3*c
^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3 - a*c*e^3)*x)/((c*x^2 + b*x + a)*c^2)

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maple [B]  time = 0.06, size = 363, normalized size = 2.88 \begin {gather*} \frac {a \,e^{3} x}{\left (c \,x^{2}+b x +a \right ) c}-\frac {6 a \,e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {b^{2} e^{3} x}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {3 b^{2} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}+\frac {3 b d \,e^{2} x}{\left (c \,x^{2}+b x +a \right ) c}-\frac {6 b d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {3 d^{2} e x}{c \,x^{2}+b x +a}+\frac {6 d^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {a b \,e^{3}}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {3 a d \,e^{2}}{\left (c \,x^{2}+b x +a \right ) c}-\frac {3 b \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {3 d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {2 e^{3} x}{c}-\frac {d^{3}}{c \,x^{2}+b x +a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.95, size = 226, normalized size = 1.79 \begin {gather*} \frac {2\,e^3\,x}{c}-\frac {\frac {c^2\,d^3-3\,a\,c\,d\,e^2+a\,b\,e^3}{c}+\frac {x\,\left (b^2\,e^3-3\,b\,c\,d\,e^2+3\,c^2\,d^2\,e-a\,c\,e^3\right )}{c}}{c^2\,x^2+b\,c\,x+a\,c}+\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (3\,b^3\,e^3-6\,d\,b^2\,c\,e^2-12\,a\,b\,c\,e^3+24\,a\,d\,c^2\,e^2\right )}{2\,\left (4\,a\,c^3-b^2\,c^2\right )}+\frac {3\,e\,\mathrm {atan}\left (\frac {b+2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (b^2\,e^2-2\,b\,c\,d\,e+2\,c^2\,d^2-2\,a\,c\,e^2\right )}{c^2\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*e^3*x)/c - ((c^2*d^3 + a*b*e^3 - 3*a*c*d*e^2)/c + (x*(b^2*e^3 + 3*c^2*d^2*e - a*c*e^3 - 3*b*c*d*e^2))/c)/(a
*c + c^2*x^2 + b*c*x) + (log(a + b*x + c*x^2)*(3*b^3*e^3 - 12*a*b*c*e^3 + 24*a*c^2*d*e^2 - 6*b^2*c*d*e^2))/(2*
(4*a*c^3 - b^2*c^2)) + (3*e*atan((b + 2*c*x)/(4*a*c - b^2)^(1/2))*(b^2*e^2 + 2*c^2*d^2 - 2*a*c*e^2 - 2*b*c*d*e
))/(c^2*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 11.14, size = 733, normalized size = 5.82 \begin {gather*} \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 3 a b e^{3} - 4 a c^{2} \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 12 a c d e^{2} + b^{2} c \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} - \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - 3 b c d^{2} e}{6 a c e^{3} - 3 b^{2} e^{3} + 6 b c d e^{2} - 6 c^{2} d^{2} e} \right )} + \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} + \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {- 3 a b e^{3} - 4 a c^{2} \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} + \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) + 12 a c d e^{2} + b^{2} c \left (- \frac {3 e^{2} \left (b e - 2 c d\right )}{2 c^{2}} + \frac {3 e \sqrt {- 4 a c + b^{2}} \left (2 a c e^{2} - b^{2} e^{2} + 2 b c d e - 2 c^{2} d^{2}\right )}{2 c^{2} \left (4 a c - b^{2}\right )}\right ) - 3 b c d^{2} e}{6 a c e^{3} - 3 b^{2} e^{3} + 6 b c d e^{2} - 6 c^{2} d^{2} e} \right )} + \frac {- a b e^{3} + 3 a c d e^{2} - c^{2} d^{3} + x \left (a c e^{3} - b^{2} e^{3} + 3 b c d e^{2} - 3 c^{2} d^{2} e\right )}{a c^{2} + b c^{2} x + c^{3} x^{2}} + \frac {2 e^{3} x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*e**2*(b*e - 2*c*d)/(2*c**2) - 3*e*sqrt(-4*a*c + b**2)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(
2*c**2*(4*a*c - b**2)))*log(x + (-3*a*b*e**3 - 4*a*c**2*(-3*e**2*(b*e - 2*c*d)/(2*c**2) - 3*e*sqrt(-4*a*c + b*
*2)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2*(4*a*c - b**2))) + 12*a*c*d*e**2 + b**2*c*(-3*e
**2*(b*e - 2*c*d)/(2*c**2) - 3*e*sqrt(-4*a*c + b**2)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**
2*(4*a*c - b**2))) - 3*b*c*d**2*e)/(6*a*c*e**3 - 3*b**2*e**3 + 6*b*c*d*e**2 - 6*c**2*d**2*e)) + (-3*e**2*(b*e
- 2*c*d)/(2*c**2) + 3*e*sqrt(-4*a*c + b**2)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2*(4*a*c
- b**2)))*log(x + (-3*a*b*e**3 - 4*a*c**2*(-3*e**2*(b*e - 2*c*d)/(2*c**2) + 3*e*sqrt(-4*a*c + b**2)*(2*a*c*e**
2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2*(4*a*c - b**2))) + 12*a*c*d*e**2 + b**2*c*(-3*e**2*(b*e - 2*c
*d)/(2*c**2) + 3*e*sqrt(-4*a*c + b**2)*(2*a*c*e**2 - b**2*e**2 + 2*b*c*d*e - 2*c**2*d**2)/(2*c**2*(4*a*c - b**
2))) - 3*b*c*d**2*e)/(6*a*c*e**3 - 3*b**2*e**3 + 6*b*c*d*e**2 - 6*c**2*d**2*e)) + (-a*b*e**3 + 3*a*c*d*e**2 -
c**2*d**3 + x*(a*c*e**3 - b**2*e**3 + 3*b*c*d*e**2 - 3*c**2*d**2*e))/(a*c**2 + b*c**2*x + c**3*x**2) + 2*e**3*
x/c

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