3.892 \(\int \frac {1}{x^3 (a-b x^2+c x^4)} \, dx\)

Optimal. Leaf size=89 \[ \frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b-2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a-b x^2+c x^4\right )}{4 a^2}+\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \]

[Out]

-1/2/a/x^2+b*ln(x)/a^2-1/4*b*ln(c*x^4-b*x^2+a)/a^2+1/2*(-2*a*c+b^2)*arctanh((-2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/a
^2/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1114, 709, 800, 634, 618, 206, 628} \[ \frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b-2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a-b x^2+c x^4\right )}{4 a^2}+\frac {b \log (x)}{a^2}-\frac {1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2 + c*x^4)),x]

[Out]

-1/(2*a*x^2) + ((b^2 - 2*a*c)*ArcTanh[(b - 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^2*Sqrt[b^2 - 4*a*c]) + (b*Log[x])
/a^2 - (b*Log[a - b*x^2 + c*x^4])/(4*a^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 709

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && LtQ[m, -1]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1114

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 139, normalized size = 1.56 \[ \frac {\frac {\left (-b \sqrt {b^2-4 a c}-2 a c+b^2\right ) \log \left (-\sqrt {b^2-4 a c}-b+2 c x^2\right )}{\sqrt {b^2-4 a c}}-\frac {\left (b \sqrt {b^2-4 a c}-2 a c+b^2\right ) \log \left (\sqrt {b^2-4 a c}-b+2 c x^2\right )}{\sqrt {b^2-4 a c}}-\frac {2 a}{x^2}+4 b \log (x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a - b*x^2 + c*x^4)),x]

[Out]

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

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fricas [A]  time = 0.92, size = 298, normalized size = 3.35 \[ \left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{4} - 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} - b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} - b x^{2} + a}\right ) + {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} - b x^{2} + a\right ) - 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \relax (x) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {{\left (2 \, c x^{2} - b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} - b x^{2} + a\right ) - 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \relax (x) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4-b*x^2+a),x, algorithm="fricas")

[Out]

[-1/4*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*x^2*log((2*c^2*x^4 - 2*b*c*x^2 + b^2 - 2*a*c + (2*c*x^2 - b)*sqrt(b^2 -
4*a*c))/(c*x^4 - b*x^2 + a)) + (b^3 - 4*a*b*c)*x^2*log(c*x^4 - b*x^2 + a) - 4*(b^3 - 4*a*b*c)*x^2*log(x) + 2*
a*b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^2), -1/4*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*x^2*arctan(-(2*c*x^2 - b)
*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (b^3 - 4*a*b*c)*x^2*log(c*x^4 - b*x^2 + a) - 4*(b^3 - 4*a*b*c)*x^2*log(x)
+ 2*a*b^2 - 8*a^2*c)/((a^2*b^2 - 4*a^3*c)*x^2)]

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giac [A]  time = 0.59, size = 95, normalized size = 1.07 \[ -\frac {b \log \left (c x^{4} - b x^{2} + a\right )}{4 \, a^{2}} + \frac {b \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{2} - b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} - \frac {b x^{2} + a}{2 \, a^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4-b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 123, normalized size = 1.38 \[ -\frac {c \arctan \left (\frac {2 c \,x^{2}-b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, a}+\frac {b^{2} \arctan \left (\frac {2 c \,x^{2}-b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, a^{2}}+\frac {b \ln \relax (x )}{a^{2}}-\frac {b \ln \left (c \,x^{4}-b \,x^{2}+a \right )}{4 a^{2}}-\frac {1}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^4-b*x^2+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4-b*x^2+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 = 5.84, size = 2032, normalized size = 22.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a - b*x^2 + c*x^4)),x)

[Out]

(b*log(x))/a^2 - 1/(2*a*x^2) + (log(a - b*x^2 + c*x^4)*(2*b^3 - 8*a*b*c))/(2*(16*a^3*c - 4*a^2*b^2)) + (atan((
16*a^6*x^2*(((3*b^4 + a^2*c^2 - 9*a*b^2*c)*(c^5/a^3 + ((2*b^3 - 8*a*b*c)*((6*b*c^4)/a^2 + ((2*b^3 - 8*a*b*c)*(
(20*a^3*c^4 + 2*a^2*b^2*c^3)/a^3 + ((2*b^3 - 8*a*b*c)*(40*a^4*b*c^3 - 12*a^3*b^3*c^2))/(2*a^3*(16*a^3*c - 4*a^
2*b^2))))/(2*(16*a^3*c - 4*a^2*b^2))))/(2*(16*a^3*c - 4*a^2*b^2)) - ((((2*a*c - b^2)*((20*a^3*c^4 + 2*a^2*b^2*
c^3)/a^3 + ((2*b^3 - 8*a*b*c)*(40*a^4*b*c^3 - 12*a^3*b^3*c^2))/(2*a^3*(16*a^3*c - 4*a^2*b^2))))/(4*a^2*(4*a*c
- b^2)^(1/2)) + ((2*b^3 - 8*a*b*c)*(40*a^4*b*c^3 - 12*a^3*b^3*c^2)*(2*a*c - b^2))/(8*a^5*(4*a*c - b^2)^(1/2)*(
16*a^3*c - 4*a^2*b^2)))*(2*a*c - b^2))/(4*a^2*(4*a*c - b^2)^(1/2)) - ((2*b^3 - 8*a*b*c)*(40*a^4*b*c^3 - 12*a^3
*b^3*c^2)*(2*a*c - b^2)^2)/(32*a^7*(4*a*c - b^2)*(16*a^3*c - 4*a^2*b^2))))/(8*a^3*c^2*(a^2*c^2 - 6*b^4 + 24*a*
b^2*c)) + ((((2*b^3 - 8*a*b*c)*(((2*a*c - b^2)*((20*a^3*c^4 + 2*a^2*b^2*c^3)/a^3 + ((2*b^3 - 8*a*b*c)*(40*a^4*
b*c^3 - 12*a^3*b^3*c^2))/(2*a^3*(16*a^3*c - 4*a^2*b^2))))/(4*a^2*(4*a*c - b^2)^(1/2)) + ((2*b^3 - 8*a*b*c)*(40
*a^4*b*c^3 - 12*a^3*b^3*c^2)*(2*a*c - b^2))/(8*a^5*(4*a*c - b^2)^(1/2)*(16*a^3*c - 4*a^2*b^2))))/(2*(16*a^3*c
- 4*a^2*b^2)) - ((40*a^4*b*c^3 - 12*a^3*b^3*c^2)*(2*a*c - b^2)^3)/(64*a^9*(4*a*c - b^2)^(3/2)) + (((6*b*c^4)/a
^2 + ((2*b^3 - 8*a*b*c)*((20*a^3*c^4 + 2*a^2*b^2*c^3)/a^3 + ((2*b^3 - 8*a*b*c)*(40*a^4*b*c^3 - 12*a^3*b^3*c^2)
)/(2*a^3*(16*a^3*c - 4*a^2*b^2))))/(2*(16*a^3*c - 4*a^2*b^2)))*(2*a*c - b^2))/(4*a^2*(4*a*c - b^2)^(1/2)))*(3*
b^5 + 13*a^2*b*c^2 - 15*a*b^3*c))/(8*a^3*c^2*(4*a*c - b^2)^(1/2)*(a^2*c^2 - 6*b^4 + 24*a*b^2*c)))*(4*a*c - b^2
)^(3/2))/(4*a^2*c^4 + b^4*c^2 - 4*a*b^2*c^3) + (2*a^3*(4*a*c - b^2)*(3*b^5 + 13*a^2*b*c^2 - 15*a*b^3*c)*(((2*b
^3 - 8*a*b*c)*((((4*a^3*b*c^3 - 4*a^2*b^3*c^2)/a^3 + (2*a*b^2*c^2*(2*b^3 - 8*a*b*c))/(16*a^3*c - 4*a^2*b^2))*(
2*a*c - b^2))/(4*a^2*(4*a*c - b^2)^(1/2)) + (b^2*c^2*(2*b^3 - 8*a*b*c)*(2*a*c - b^2))/(2*a*(4*a*c - b^2)^(1/2)
*(16*a^3*c - 4*a^2*b^2))))/(2*(16*a^3*c - 4*a^2*b^2)) + ((2*a*c - b^2)*((a^2*c^4 - 4*a*b^2*c^3)/a^3 + ((2*b^3
- 8*a*b*c)*((4*a^3*b*c^3 - 4*a^2*b^3*c^2)/a^3 + (2*a*b^2*c^2*(2*b^3 - 8*a*b*c))/(16*a^3*c - 4*a^2*b^2)))/(2*(1
6*a^3*c - 4*a^2*b^2))))/(4*a^2*(4*a*c - b^2)^(1/2)) - (b^2*c^2*(2*a*c - b^2)^3)/(16*a^5*(4*a*c - b^2)^(3/2))))
/(c^2*(a^2*c^2 - 6*b^4 + 24*a*b^2*c)*(4*a^2*c^4 + b^4*c^2 - 4*a*b^2*c^3)) - (2*a^3*(4*a*c - b^2)^(3/2)*(3*b^4
+ a^2*c^2 - 9*a*b^2*c)*((b*c^4)/a^3 - ((2*b^3 - 8*a*b*c)*((a^2*c^4 - 4*a*b^2*c^3)/a^3 + ((2*b^3 - 8*a*b*c)*((4
*a^3*b*c^3 - 4*a^2*b^3*c^2)/a^3 + (2*a*b^2*c^2*(2*b^3 - 8*a*b*c))/(16*a^3*c - 4*a^2*b^2)))/(2*(16*a^3*c - 4*a^
2*b^2))))/(2*(16*a^3*c - 4*a^2*b^2)) + ((2*a*c - b^2)*((((4*a^3*b*c^3 - 4*a^2*b^3*c^2)/a^3 + (2*a*b^2*c^2*(2*b
^3 - 8*a*b*c))/(16*a^3*c - 4*a^2*b^2))*(2*a*c - b^2))/(4*a^2*(4*a*c - b^2)^(1/2)) + (b^2*c^2*(2*b^3 - 8*a*b*c)
*(2*a*c - b^2))/(2*a*(4*a*c - b^2)^(1/2)*(16*a^3*c - 4*a^2*b^2))))/(4*a^2*(4*a*c - b^2)^(1/2)) + (b^2*c^2*(2*b
^3 - 8*a*b*c)*(2*a*c - b^2)^2)/(8*a^3*(4*a*c - b^2)*(16*a^3*c - 4*a^2*b^2))))/(c^2*(a^2*c^2 - 6*b^4 + 24*a*b^2
*c)*(4*a^2*c^4 + b^4*c^2 - 4*a*b^2*c^3)))*(2*a*c - b^2))/(2*a^2*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 142.97, size = 350, normalized size = 3.93 \[ \left (- \frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- 8 a^{3} c \left (- \frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (- \frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) - 3 a b c + b^{3}}{2 a c^{2} - b^{2} c} \right )} + \left (- \frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- 8 a^{3} c \left (- \frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (- \frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) - 3 a b c + b^{3}}{2 a c^{2} - b^{2} c} \right )} - \frac {1}{2 a x^{2}} + \frac {b \log {\relax (x )}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(c*x**4-b*x**2+a),x)

[Out]

(-b/(4*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2)))*log(x**2 + (-8*a**3*c*(-b/(4*a**2)
- sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2))) + 2*a**2*b**2*(-b/(4*a**2) - sqrt(-4*a*c + b**2)
*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2))) - 3*a*b*c + b**3)/(2*a*c**2 - b**2*c)) + (-b/(4*a**2) + sqrt(-4*a*c +
b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2)))*log(x**2 + (-8*a**3*c*(-b/(4*a**2) + sqrt(-4*a*c + b**2)*(2*a*c
- b**2)/(4*a**2*(4*a*c - b**2))) + 2*a**2*b**2*(-b/(4*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a
*c - b**2))) - 3*a*b*c + b**3)/(2*a*c**2 - b**2*c)) - 1/(2*a*x**2) + b*log(x)/a**2

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