Integrand size = 20, antiderivative size = 89 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=-\frac {1}{2 a x^2}-\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 a^2 \sqrt {b^2-4 a c}}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^2+c x^4\right )}{4 a^2} \] Output:
-1/2/a/x^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)-b*ln(x)/a^2+1/4*b*ln(c*x^4+b*x^2+a)/a^2
Time = 0.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.52 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\frac {-\frac {2 a}{x^2}-4 b \log (x)+\frac {\left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}+\frac {\left (-b^2+2 a c+b \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{\sqrt {b^2-4 a c}}}{4 a^2} \] Input:
Integrate[1/(x^2*(a*x + b*x^3 + c*x^5)),x]
Output:
((-2*a)/x^2 - 4*b*Log[x] + ((b^2 - 2*a*c + b*Sqrt[b^2 - 4*a*c])*Log[b - Sq rt[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)
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {9, 1434, 1145, 25, 1200, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^2+c x^4\right )}dx\) |
| \(\Big \downarrow \) 1434 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (c x^4+b x^2+a\right )}dx^2\) |
| \(\Big \downarrow \) 1145 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {c x^2+b}{x^2 \left (c x^4+b x^2+a\right )}dx^2}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {c x^2+b}{x^2 \left (c x^4+b x^2+a\right )}dx^2}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \left (\frac {b}{a x^2}+\frac {-b^2-c x^2 b+a c}{a \left (c x^4+b x^2+a\right )}\right )dx^2}{a}-\frac {1}{a x^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\left (b^2-2 a c\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^2+c x^4\right )}{2 a}+\frac {b \log \left (x^2\right )}{a}}{a}-\frac {1}{a x^2}\right )\) |
Input:
Int[1/(x^2*(a*x + b*x^3 + c*x^5)),x]
Output:
(-(1/(a*x^2)) - (((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/ (a*Sqrt[b^2 - 4*a*c]) + (b*Log[x^2])/a - (b*Log[a + b*x^2 + c*x^4])/(2*a)) /a)/2
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
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] + Simp [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}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(-\frac {-\frac {b \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2}+\frac {2 \left (a c -\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 a^{2}}-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}\) | \(85\) |
| risch | \(-\frac {1}{2 a \,x^{2}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{3} c -a^{2} b^{2}\right ) \textit {\_Z}^{2}+\left (-4 a b c +b^{3}\right ) \textit {\_Z} +c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (10 a^{3} c -3 a^{2} b^{2}\right ) \textit {\_R}^{2}-4 \textit {\_R} a b c +2 c^{2}\right ) x^{2}-a^{3} b \,\textit {\_R}^{2}+\left (a^{2} c -2 b^{2} a \right ) \textit {\_R} +2 b c \right )\right )}{2}\) | \(124\) |
Input:
int(1/x^2/(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)
Output:
-1/2/a^2*(-1/2*b*ln(c*x^4+b*x^2+a)+2*(a*c-1/2*b^2)/(4*a*c-b^2)^(1/2)*arcta n((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))-1/2/a/x^2-b*ln(x)/a^2
Time = 0.12 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.29 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} x^{2} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} + b x^{2} + a\right ) + 4 \, {\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \, {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}\right ] \] Input:
integrate(1/x^2/(c*x^5+b*x^3+a*x),x, algorithm="fricas")
Output:
[-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)]
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (82) = 164\).
Time = 108.92 (sec) , antiderivative size = 345, normalized size of antiderivative = 3.88 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\left (\frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- 8 a^{3} c \left (\frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (\frac {b}{4 a^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) + 3 a b c - b^{3}}{2 a c^{2} - b^{2} c} \right )} + \left (\frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- 8 a^{3} c \left (\frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (\frac {b}{4 a^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \cdot \left (2 a c - b^{2}\right )}{4 a^{2} \cdot \left (4 a c - b^{2}\right )}\right ) + 3 a b c - b^{3}}{2 a c^{2} - b^{2} c} \right )} - \frac {1}{2 a x^{2}} - \frac {b \log {\left (x \right )}}{a^{2}} \] Input:
integrate(1/x**2/(c*x**5+b*x**3+a*x),x)
Output:
(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
\[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\int { \frac {1}{{\left (c x^{5} + b x^{3} + a x\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(c*x^5+b*x^3+a*x),x, algorithm="maxima")
Output:
-b*log(x)/a^2 + integrate((b*c*x^3 + (b^2 - a*c)*x)/(c*x^4 + b*x^2 + a), x )/a^2 - 1/2/(a*x^2)
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\frac {b \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{2}} - \frac {b \log \left (x^{2}\right )}{2 \, a^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} a^{2}} + \frac {b x^{2} - a}{2 \, a^{2} x^{2}} \] Input:
integrate(1/x^2/(c*x^5+b*x^3+a*x),x, algorithm="giac")
Output:
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)
Time = 13.05 (sec) , antiderivative size = 2033, normalized size of antiderivative = 22.84 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\text {Too large to display} \] Input:
int(1/(x^2*(a*x + b*x^3 + c*x^5)),x)
Output:
(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...
Time = 0.22 (sec) , antiderivative size = 409, normalized size of antiderivative = 4.60 \[ \int \frac {1}{x^2 \left (a x+b x^3+c x^5\right )} \, dx=\frac {4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) a c \,x^{2}-2 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}-2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b^{2} x^{2}+4 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) a c \,x^{2}-2 \sqrt {2 \sqrt {c}\, \sqrt {a}+b}\, \sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, \mathit {atan} \left (\frac {\sqrt {2 \sqrt {c}\, \sqrt {a}-b}+2 \sqrt {c}\, x}{\sqrt {2 \sqrt {c}\, \sqrt {a}+b}}\right ) b^{2} x^{2}+4 \,\mathrm {log}\left (-\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a b c \,x^{2}-\mathrm {log}\left (-\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) b^{3} x^{2}+4 \,\mathrm {log}\left (\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) a b c \,x^{2}-\mathrm {log}\left (\sqrt {2 \sqrt {c}\, \sqrt {a}-b}\, x +\sqrt {a}+\sqrt {c}\, x^{2}\right ) b^{3} x^{2}-16 \,\mathrm {log}\left (x \right ) a b c \,x^{2}+4 \,\mathrm {log}\left (x \right ) b^{3} x^{2}-8 a^{2} c +2 a \,b^{2}}{4 a^{2} x^{2} \left (4 a c -b^{2}\right )} \] Input:
int(1/x^2/(c*x^5+b*x^3+a*x),x)
Output:
(4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sq rt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*x**2 - 2*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqr t(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b**2*x**2 + 4*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqr t(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*c*x**2 - 2 *sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt (c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*b**2*x**2 + 4 *log( - sqrt(2*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*a*b*c*x**2 - log( - sqrt(2*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*b**3*x** 2 + 4*log(sqrt(2*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*a*b*c*x* *2 - log(sqrt(2*sqrt(c)*sqrt(a) - b)*x + sqrt(a) + sqrt(c)*x**2)*b**3*x**2 - 16*log(x)*a*b*c*x**2 + 4*log(x)*b**3*x**2 - 8*a**2*c + 2*a*b**2)/(4*a** 2*x**2*(4*a*c - b**2))