Integrand size = 16, antiderivative size = 138 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=-\frac {1}{a^2 x}-\frac {b \left (b^2-3 a c\right )+c \left (b^2-2 a c\right ) x}{a^2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 \left (b^4-6 a b^2 c+6 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac {2 b \log (x)}{a^3}+\frac {b \log \left (a+b x+c x^2\right )}{a^3} \] Output:
-1/a^2/x-(b*(-3*a*c+b^2)+c*(-2*a*c+b^2)*x)/a^2/(-4*a*c+b^2)/(c*x^2+b*x+a)- 2*(6*a^2*c^2-6*a*b^2*c+b^4)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/a^3/(-4* a*c+b^2)^(3/2)-2*b*ln(x)/a^3+b*ln(c*x^2+b*x+a)/a^3
Time = 0.18 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=-\frac {\frac {a}{x}+\frac {a \left (b^3-3 a b c+b^2 c x-2 a c^2 x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}+\frac {2 \left (b^4-6 a b^2 c+6 a^2 c^2\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+2 b \log (x)-b \log (a+x (b+c x))}{a^3} \] Input:
Integrate[1/(x^2*(a + b*x + c*x^2)^2),x]
Output:
-((a/x + (a*(b^3 - 3*a*b*c + b^2*c*x - 2*a*c^2*x))/((b^2 - 4*a*c)*(a + x*( b + c*x))) + (2*(b^4 - 6*a*b^2*c + 6*a^2*c^2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2) + 2*b*Log[x] - b*Log[a + x*(b + c*x)])/a^ 3)
Time = 0.66 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1165, 27, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1165 |
\(\displaystyle \frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int -\frac {2 \left (b^2+c x b-3 a c\right )}{x^2 \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {b^2+c x b-3 a c}{x^2 \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {2 \int \left (\frac {b^2-3 a c}{a x^2}+\frac {4 a b c-b^3}{a^2 x}+\frac {b^4-5 a c b^2+c \left (b^2-4 a c\right ) x b+3 a^2 c^2}{a^2 \left (c x^2+b x+a\right )}\right )dx}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {\left (6 a^2 c^2-6 a b^2 c+b^4\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}+\frac {b \left (b^2-4 a c\right ) \log \left (a+b x+c x^2\right )}{2 a^2}-\frac {b \log (x) \left (b^2-4 a c\right )}{a^2}-\frac {b^2-3 a c}{a x}\right )}{a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
Input:
Int[1/(x^2*(a + b*x + c*x^2)^2),x]
Output:
(b^2 - 2*a*c + b*c*x)/(a*(b^2 - 4*a*c)*x*(a + b*x + c*x^2)) + (2*(-((b^2 - 3*a*c)/(a*x)) - ((b^4 - 6*a*b^2*c + 6*a^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b ^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]) - (b*(b^2 - 4*a*c)*Log[x])/a^2 + (b* (b^2 - 4*a*c)*Log[a + b*x + c*x^2])/(2*a^2)))/(a*(b^2 - 4*a*c))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> 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] + Simp[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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
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]
Time = 0.71 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.49
method | result | size |
default | \(-\frac {1}{a^{2} x}-\frac {2 b \ln \left (x \right )}{a^{3}}-\frac {\frac {\frac {a c \left (2 a c -b^{2}\right ) x}{4 a c -b^{2}}+\frac {a b \left (3 a c -b^{2}\right )}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (-4 a b \,c^{2}+b^{3} c \right ) \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {4 \left (3 a^{2} c^{2}-5 c a \,b^{2}+b^{4}-\frac {\left (-4 a b \,c^{2}+b^{3} c \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{a^{3}}\) | \(205\) |
risch | \(\text {Expression too large to display}\) | \(3181\) |
Input:
int(1/x^2/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/a^2/x-2*b*ln(x)/a^3-1/a^3*((a*c*(2*a*c-b^2)/(4*a*c-b^2)*x+a*b*(3*a*c-b^ 2)/(4*a*c-b^2))/(c*x^2+b*x+a)+2/(4*a*c-b^2)*(1/2*(-4*a*b*c^2+b^3*c)/c*ln(c *x^2+b*x+a)+2*(3*a^2*c^2-5*c*a*b^2+b^4-1/2*(-4*a*b*c^2+b^3*c)*b/c)/(4*a*c- b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (134) = 268\).
Time = 0.22 (sec) , antiderivative size = 975, normalized size of antiderivative = 7.07 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
[-(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^ 3*c^3)*x^2 + ((b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*x^3 + (b^5 - 6*a*b^3*c + 6 *a^2*b*c^2)*x^2 + (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*x)*sqrt(b^2 - 4*a*c)*l og((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^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*x - ((b^5*c - 8* a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + ( a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(c*x^2 + b*x + a) + 2*((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(x))/((a^3*b^4*c - 8*a^4*b^2*c ^2 + 16*a^5*c^3)*x^3 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2 + (a^4*b ^4 - 8*a^5*b^2*c + 16*a^6*c^2)*x), -(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*x^2 + 2*((b^4*c - 6*a*b^2*c^2 + 6 *a^2*c^3)*x^3 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*x^2 + (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*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^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*x - ((b^5*c - 8* a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + ( a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(c*x^2 + b*x + a) + 2*((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(x))/((a^3*b^4*c - 8*a^4*b^2*c ^2 + 16*a^5*c^3)*x^3 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2 + (a^...
Timed out. \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/x**2/(c*x**2+b*x+a)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=\frac {2 \, {\left (b^{4} - 6 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, b^{2} c x^{2} - 6 \, a c^{2} x^{2} + 2 \, b^{3} x - 7 \, a b c x + a b^{2} - 4 \, a^{2} c}{{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} {\left (c x^{3} + b x^{2} + a x\right )}} + \frac {b \log \left (c x^{2} + b x + a\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
2*(b^4 - 6*a*b^2*c + 6*a^2*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((a ^3*b^2 - 4*a^4*c)*sqrt(-b^2 + 4*a*c)) - (2*b^2*c*x^2 - 6*a*c^2*x^2 + 2*b^3 *x - 7*a*b*c*x + a*b^2 - 4*a^2*c)/((a^2*b^2 - 4*a^3*c)*(c*x^3 + b*x^2 + a* x)) + b*log(c*x^2 + b*x + a)/a^3 - 2*b*log(abs(x))/a^3
Time = 9.98 (sec) , antiderivative size = 775, normalized size of antiderivative = 5.62 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx=\ln \left (2\,a\,b^7+2\,b^8\,x+2\,a\,b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-23\,a^2\,b^5\,c-108\,a^4\,b\,c^3+24\,a^4\,c^4\,x+2\,b^5\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+87\,a^3\,b^3\,c^2+3\,a^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-9\,a^2\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+97\,a^2\,b^4\,c^2\,x-138\,a^3\,b^2\,c^3\,x-24\,a\,b^6\,c\,x-12\,a\,b^3\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+15\,a^2\,b\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+6\,a^2\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-6\,a\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^6\,c^3+48\,a^5\,b^2\,c^2-12\,a^4\,b^4\,c+a^3\,b^6}+\frac {b}{a^3}\right )-\frac {\frac {1}{a}-\frac {x\,\left (2\,b^3-7\,a\,b\,c\right )}{a^2\,\left (4\,a\,c-b^2\right )}+\frac {2\,c\,x^2\,\left (3\,a\,c-b^2\right )}{a^2\,\left (4\,a\,c-b^2\right )}}{c\,x^3+b\,x^2+a\,x}-\ln \left (2\,a\,b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-2\,b^8\,x-2\,a\,b^7+23\,a^2\,b^5\,c+108\,a^4\,b\,c^3-24\,a^4\,c^4\,x+2\,b^5\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-87\,a^3\,b^3\,c^2+3\,a^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-9\,a^2\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-97\,a^2\,b^4\,c^2\,x+138\,a^3\,b^2\,c^3\,x+24\,a\,b^6\,c\,x-12\,a\,b^3\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+15\,a^2\,b\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+6\,a^2\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-6\,a\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^6\,c^3+48\,a^5\,b^2\,c^2-12\,a^4\,b^4\,c+a^3\,b^6}-\frac {b}{a^3}\right )-\frac {2\,b\,\ln \left (x\right )}{a^3} \] Input:
int(1/(x^2*(a + b*x + c*x^2)^2),x)
Output:
log(2*a*b^7 + 2*b^8*x + 2*a*b^4*(-(4*a*c - b^2)^3)^(1/2) - 23*a^2*b^5*c - 108*a^4*b*c^3 + 24*a^4*c^4*x + 2*b^5*x*(-(4*a*c - b^2)^3)^(1/2) + 87*a^3*b ^3*c^2 + 3*a^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 9*a^2*b^2*c*(-(4*a*c - b^2)^ 3)^(1/2) + 97*a^2*b^4*c^2*x - 138*a^3*b^2*c^3*x - 24*a*b^6*c*x - 12*a*b^3* c*x*(-(4*a*c - b^2)^3)^(1/2) + 15*a^2*b*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*(( b^4*(-(4*a*c - b^2)^3)^(1/2) + 6*a^2*c^2*(-(4*a*c - b^2)^3)^(1/2) - 6*a*b^ 2*c*(-(4*a*c - b^2)^3)^(1/2))/(a^3*b^6 - 64*a^6*c^3 - 12*a^4*b^4*c + 48*a^ 5*b^2*c^2) + b/a^3) - (1/a - (x*(2*b^3 - 7*a*b*c))/(a^2*(4*a*c - b^2)) + ( 2*c*x^2*(3*a*c - b^2))/(a^2*(4*a*c - b^2)))/(a*x + b*x^2 + c*x^3) - log(2* a*b^4*(-(4*a*c - b^2)^3)^(1/2) - 2*b^8*x - 2*a*b^7 + 23*a^2*b^5*c + 108*a^ 4*b*c^3 - 24*a^4*c^4*x + 2*b^5*x*(-(4*a*c - b^2)^3)^(1/2) - 87*a^3*b^3*c^2 + 3*a^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 9*a^2*b^2*c*(-(4*a*c - b^2)^3)^(1/ 2) - 97*a^2*b^4*c^2*x + 138*a^3*b^2*c^3*x + 24*a*b^6*c*x - 12*a*b^3*c*x*(- (4*a*c - b^2)^3)^(1/2) + 15*a^2*b*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*((b^4*(- (4*a*c - b^2)^3)^(1/2) + 6*a^2*c^2*(-(4*a*c - b^2)^3)^(1/2) - 6*a*b^2*c*(- (4*a*c - b^2)^3)^(1/2))/(a^3*b^6 - 64*a^6*c^3 - 12*a^4*b^4*c + 48*a^5*b^2* c^2) - b/a^3) - (2*b*log(x))/a^3
Time = 0.22 (sec) , antiderivative size = 877, normalized size of antiderivative = 6.36 \[ \int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/x^2/(c*x^2+b*x+a)^2,x)
Output:
( - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b*c**2 *x + 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3* c*x - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**2 *c**2*x**2 - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a* *2*b*c**3*x**3 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2)) *a*b**5*x + 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b **4*c*x**2 + 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a* b**3*c**2*x**3 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2)) *b**6*x**2 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b** 5*c*x**3 + 16*log(a + b*x + c*x**2)*a**3*b**2*c**2*x - 8*log(a + b*x + c*x **2)*a**2*b**4*c*x + 16*log(a + b*x + c*x**2)*a**2*b**3*c**2*x**2 + 16*log (a + b*x + c*x**2)*a**2*b**2*c**3*x**3 + log(a + b*x + c*x**2)*a*b**6*x - 8*log(a + b*x + c*x**2)*a*b**5*c*x**2 - 8*log(a + b*x + c*x**2)*a*b**4*c** 2*x**3 + log(a + b*x + c*x**2)*b**7*x**2 + log(a + b*x + c*x**2)*b**6*c*x* *3 - 32*log(x)*a**3*b**2*c**2*x + 16*log(x)*a**2*b**4*c*x - 32*log(x)*a**2 *b**3*c**2*x**2 - 32*log(x)*a**2*b**2*c**3*x**3 - 2*log(x)*a*b**6*x + 16*l og(x)*a*b**5*c*x**2 + 16*log(x)*a*b**4*c**2*x**3 - 2*log(x)*b**7*x**2 - 2* log(x)*b**6*c*x**3 - 16*a**4*b*c**2 + 24*a**4*c**3*x + 8*a**3*b**3*c - 42* a**3*b**2*c**2*x + 24*a**3*c**4*x**3 - a**2*b**5 + 17*a**2*b**4*c*x - 14*a **2*b**2*c**3*x**3 - 2*a*b**6*x + 2*a*b**4*c**2*x**3)/(a**3*b*x*(16*a**...