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

Optimal. Leaf size=151 \[ \frac {e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^3}-\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {e^2 x (3 c d-b e)}{c^2}+\frac {e^3 x^2}{2 c} \]

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Rubi [A]  time = 0.17, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {701, 634, 618, 206, 628} \begin {gather*} \frac {e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^3}-\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {e^2 x (3 c d-b e)}{c^2}+\frac {e^3 x^2}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + b*x + c*x^2),x]

[Out]

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 148, normalized size = 0.98 \begin {gather*} \frac {e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log (a+x (b+c x))+\frac {2 (2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^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}}+2 c e^2 x (3 c d-b e)+c^2 e^3 x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 526, normalized size = 3.48 \begin {gather*} \left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 3 \, a b c\right )} e^{3}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, {\left (3 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} x + {\left (3 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 3 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} e^{3} x^{2} - 2 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 3 \, a b c\right )} e^{3}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (3 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d e^{2} - {\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} x + {\left (3 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 3 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((b^2*c^2 - 4*a*c^3)*e^3*x^2 + (2*c^3*d^3 - 3*b*c^2*d^2*e + 3*(b^2*c - 2*a*c^2)*d*e^2 - (b^3 - 3*a*b*c)*e
^3)*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*e^2 - (b^3*c - 4*a*b*c^2)*e^3)*x + (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 - 5*a*b^2*c + 4*a^2*c^2)*e^3)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4), 1/2*((b^2*c^2
- 4*a*c^3)*e^3*x^2 - 2*(2*c^3*d^3 - 3*b*c^2*d^2*e + 3*(b^2*c - 2*a*c^2)*d*e^2 - (b^3 - 3*a*b*c)*e^3)*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*e^2 - (b^3*c - 4*
a*b*c^2)*e^3)*x + (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 - 5*a*b^2*c + 4*a^2*c^2)*e
^3)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4)]

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giac [A]  time = 0.18, size = 161, normalized size = 1.07 \begin {gather*} \frac {c x^{2} e^{3} + 6 \, c d x e^{2} - 2 \, b x e^{3}}{2 \, c^{2}} + \frac {{\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3} - a c e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} + \frac {{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - 6 \, a c^{2} d e^{2} - b^{3} e^{3} + 3 \, a b 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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 366, normalized size = 2.42 \begin {gather*} \frac {3 a b \,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 {6 a 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 {b^{3} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {3 b^{2} d \,e^{2} \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^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {e^{3} x^{2}}{2 c}+\frac {2 d^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {a \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {b^{2} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}-\frac {3 b d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {b \,e^{3} x}{c^{2}}+\frac {3 d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {3 d \,e^{2} x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*x+d)^3/(c*x^2+b*x+a),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 = 0.24, size = 208, normalized size = 1.38 \begin {gather*} \frac {e^3\,x^2}{2\,c}-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (4\,a^2\,c^2\,e^3-5\,a\,b^2\,c\,e^3+12\,a\,b\,c^2\,d\,e^2-12\,a\,c^3\,d^2\,e+b^4\,e^3-3\,b^3\,c\,d\,e^2+3\,b^2\,c^2\,d^2\,e\right )}{2\,\left (4\,a\,c^4-b^2\,c^3\right )}-x\,\left (\frac {b\,e^3}{c^2}-\frac {3\,d\,e^2}{c}\right )-\frac {\mathrm {atan}\left (\frac {b+2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (b\,e-2\,c\,d\right )\,\left (b^2\,e^2-b\,c\,d\,e+c^2\,d^2-3\,a\,c\,e^2\right )}{c^3\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + b*x + c*x^2),x)

[Out]

(e^3*x^2)/(2*c) - (log(a + b*x + c*x^2)*(b^4*e^3 + 4*a^2*c^2*e^3 + 3*b^2*c^2*d^2*e - 5*a*b^2*c*e^3 - 12*a*c^3*
d^2*e - 3*b^3*c*d*e^2 + 12*a*b*c^2*d*e^2))/(2*(4*a*c^4 - b^2*c^3)) - x*((b*e^3)/c^2 - (3*d*e^2)/c) - (atan((b
+ 2*c*x)/(4*a*c - b^2)^(1/2))*(b*e - 2*c*d)*(b^2*e^2 + c^2*d^2 - 3*a*c*e^2 - b*c*d*e))/(c^3*(4*a*c - b^2)^(1/2
))

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sympy [B]  time = 4.31, size = 892, normalized size = 5.91 \begin {gather*} x \left (- \frac {b e^{3}}{c^{2}} + \frac {3 d e^{2}}{c}\right ) + \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {2 a^{2} c e^{3} - a b^{2} e^{3} + 3 a b c d e^{2} + 4 a c^{3} \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) - 6 a c^{2} d^{2} e - b^{2} c^{2} \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} - \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) + b c^{2} d^{3}}{3 a b c e^{3} - 6 a c^{2} d e^{2} - b^{3} e^{3} + 3 b^{2} c d e^{2} - 3 b c^{2} d^{2} e + 2 c^{3} d^{3}} \right )} + \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {2 a^{2} c e^{3} - a b^{2} e^{3} + 3 a b c d e^{2} + 4 a c^{3} \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) - 6 a c^{2} d^{2} e - b^{2} c^{2} \left (- \frac {e \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )}\right ) + b c^{2} d^{3}}{3 a b c e^{3} - 6 a c^{2} d e^{2} - b^{3} e^{3} + 3 b^{2} c d e^{2} - 3 b c^{2} d^{2} e + 2 c^{3} d^{3}} \right )} + \frac {e^{3} x^{2}}{2 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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