3.5.28 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^2 x^5} \, dx\) [428]

Optimal. Leaf size=108 \[ \frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x+c x^2\right )}{2 a^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1368, 754, 814, 648, 632, 212, 642} \begin {gather*} \frac {b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}-\frac {\log \left (a+b x+c x^2\right )}{2 a^2}+\frac {\log (x)}{a^2}+\frac {-2 a c+b^2+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^2*x^5),x]

[Out]

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

Rule 212

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

Rule 632

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 642

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 648

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 754

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(b
*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e +
a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), 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[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

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 1368

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^2 x^5} \, dx &=\int \frac {1}{x \left (a+b x+c x^2\right )^2} \, dx\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {-b^2+4 a c-b c x}{x \left (a+b x+c x^2\right )} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {-b^2+4 a c}{a x}+\frac {b \left (b^2-5 a c\right )+c \left (b^2-4 a c\right ) x}{a \left (a+b x+c x^2\right )}\right ) \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\log (x)}{a^2}-\frac {\int \frac {b \left (b^2-5 a c\right )+c \left (b^2-4 a c\right ) x}{a+b x+c x^2} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\log (x)}{a^2}-\frac {\int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 a^2}-\frac {\left (b \left (b^2-6 a c\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x+c x^2\right )}{2 a^2}+\frac {\left (b \left (b^2-6 a c\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \left (b^2-4 a c\right )^{3/2}}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x+c x^2\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 107, normalized size = 0.99 \begin {gather*} \frac {\frac {2 a \left (b^2-2 a c+b c x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {2 b \left (b^2-6 a c\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+2 \log (x)-\log (a+x (b+c x))}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^2*x^5),x]

[Out]

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

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Maple [A]
time = 0.06, size = 177, normalized size = 1.64

method	result	size
default	\(-\frac {\frac {\frac {a b c x}{4 a c -b^{2}}-\frac {a \left (2 a c -b^{2}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (4 c^{2} a -b^{2} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (5 a b c -b^{3}-\frac {\left (4 c^{2} a -b^{2} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{2}}+\frac {\ln \left (x \right )}{a^{2}}\)	\(177\)
risch	\(\text {Expression too large to display}\)	\(2292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/a^2*((a*b*c/(4*a*c-b^2)*x-a*(2*a*c-b^2)/(4*a*c-b^2))/(c*x^2+b*x+a)+1/(4*a*c-b^2)*(1/2*(4*a*c^2-b^2*c)/c*ln(
c*x^2+b*x+a)+2*(5*a*b*c-b^3-1/2*(4*a*c^2-b^2*c)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))+l
n(x)/a^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(1/(c+a/x^2+b/x)^2/x^5,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 deta

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (102) = 204\).
time = 0.44, size = 781, normalized size = 7.23 \begin {gather*} \left [\frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (a b^{3} - 6 \, a^{2} b c + {\left (b^{3} c - 6 \, a b c^{2}\right )} x^{2} + {\left (b^{4} - 6 \, a b^{2} c\right )} x\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 (a b^{3} c - 4 \, a^{2} b c^{2}\right )} x - {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}, \frac {2 \, a b^{4} - 12 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + 2 \, {\left (a b^{3} - 6 \, a^{2} b c + {\left (b^{3} c - 6 \, a b c^{2}\right )} x^{2} + {\left (b^{4} - 6 \, a b^{2} c\right )} x\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 (a b^{3} c - 4 \, a^{2} b c^{2}\right )} x - {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{4} - 8 \, a^{4} b^{2} c + 16 \, a^{5} c^{2} + {\left (a^{2} b^{4} c - 8 \, a^{3} b^{2} c^{2} + 16 \, a^{4} c^{3}\right )} x^{2} + {\left (a^{2} b^{5} - 8 \, a^{3} b^{3} c + 16 \, a^{4} b c^{2}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^2/x^5,x, algorithm="fricas")

[Out]

[1/2*(2*a*b^4 - 12*a^2*b^2*c + 16*a^3*c^2 + (a*b^3 - 6*a^2*b*c + (b^3*c - 6*a*b*c^2)*x^2 + (b^4 - 6*a*b^2*c)*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*(a*b^3*c - 4*a^2*b*c^2)*x - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 +
(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*log(c*x^2 + b*x + a) + 2*(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*
b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*log(x))/(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 +
(a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^2 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x), 1/2*(2*a*b^4 - 12*a^
2*b^2*c + 16*a^3*c^2 + 2*(a*b^3 - 6*a^2*b*c + (b^3*c - 6*a*b*c^2)*x^2 + (b^4 - 6*a*b^2*c)*x)*sqrt(-b^2 + 4*a*c
)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(a*b^3*c - 4*a^2*b*c^2)*x - (a*b^4 - 8*a^2*b^2*c +
16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x)*log(c*x^2 + b*x + a
) + 2*(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b
*c^2)*x)*log(x))/(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2 + (a^2*b^4*c - 8*a^3*b^2*c^2 + 16*a^4*c^3)*x^2 + (a^2*b^5
- 8*a^3*b^3*c + 16*a^4*b*c^2)*x)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**2/x**5,x)

[Out]

Timed out

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Giac [A]
time = 5.71, size = 126, normalized size = 1.17 \begin {gather*} -\frac {{\left (b^{3} - 6 \, a b c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {\log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2}} + \frac {a b c x + a b^{2} - 2 \, a^{2} c}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^2/x^5,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.10, size = 620, normalized size = 5.74 \begin {gather*} \frac {\ln \left (x\right )}{a^2}+\frac {\frac {2\,a\,c-b^2}{a\,\left (4\,a\,c-b^2\right )}-\frac {b\,c\,x}{a\,\left (4\,a\,c-b^2\right )}}{c\,x^2+b\,x+a}+\frac {\ln \left (2\,a\,b^6+2\,b^7\,x-96\,a^4\,c^3+2\,a\,b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-23\,a^2\,b^4\,c+2\,b^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+84\,a^3\,b^2\,c^2+94\,a^2\,b^3\,c^2\,x+12\,a^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-24\,a\,b^5\,c\,x-9\,a^2\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-120\,a^3\,b\,c^3\,x-12\,a\,b^2\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (b^6-64\,a^3\,c^3+b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c-6\,a\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^2\,{\left (4\,a\,c-b^2\right )}^3}+\frac {\ln \left (96\,a^4\,c^3-2\,b^7\,x-2\,a\,b^6+2\,a\,b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+23\,a^2\,b^4\,c+2\,b^4\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-84\,a^3\,b^2\,c^2-94\,a^2\,b^3\,c^2\,x+12\,a^2\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+24\,a\,b^5\,c\,x-9\,a^2\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+120\,a^3\,b\,c^3\,x-12\,a\,b^2\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (b^6-64\,a^3\,c^3-b^3\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+6\,a\,b\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )}{2\,a^2\,{\left (4\,a\,c-b^2\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(c + a/x^2 + b/x)^2),x)

[Out]

log(x)/a^2 + ((2*a*c - b^2)/(a*(4*a*c - b^2)) - (b*c*x)/(a*(4*a*c - b^2)))/(a + b*x + c*x^2) + (log(2*a*b^6 +
2*b^7*x - 96*a^4*c^3 + 2*a*b^3*(-(4*a*c - b^2)^3)^(1/2) - 23*a^2*b^4*c + 2*b^4*x*(-(4*a*c - b^2)^3)^(1/2) + 84
*a^3*b^2*c^2 + 94*a^2*b^3*c^2*x + 12*a^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2) - 24*a*b^5*c*x - 9*a^2*b*c*(-(4*a*c -
b^2)^3)^(1/2) - 120*a^3*b*c^3*x - 12*a*b^2*c*x*(-(4*a*c - b^2)^3)^(1/2))*(b^6 - 64*a^3*c^3 + b^3*(-(4*a*c - b^
2)^3)^(1/2) + 48*a^2*b^2*c^2 - 12*a*b^4*c - 6*a*b*c*(-(4*a*c - b^2)^3)^(1/2)))/(2*a^2*(4*a*c - b^2)^3) + (log(
96*a^4*c^3 - 2*b^7*x - 2*a*b^6 + 2*a*b^3*(-(4*a*c - b^2)^3)^(1/2) + 23*a^2*b^4*c + 2*b^4*x*(-(4*a*c - b^2)^3)^
(1/2) - 84*a^3*b^2*c^2 - 94*a^2*b^3*c^2*x + 12*a^2*c^2*x*(-(4*a*c - b^2)^3)^(1/2) + 24*a*b^5*c*x - 9*a^2*b*c*(
-(4*a*c - b^2)^3)^(1/2) + 120*a^3*b*c^3*x - 12*a*b^2*c*x*(-(4*a*c - b^2)^3)^(1/2))*(b^6 - 64*a^3*c^3 - b^3*(-(
4*a*c - b^2)^3)^(1/2) + 48*a^2*b^2*c^2 - 12*a*b^4*c + 6*a*b*c*(-(4*a*c - b^2)^3)^(1/2)))/(2*a^2*(4*a*c - b^2)^
3)

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