Integrand size = 26, antiderivative size = 204 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\frac {b^2 e n^2 r}{8 x^2}-\frac {b e n (2 a+b n) r}{8 x^2}-\frac {e \left (2 a^2+2 a b n+b^2 n^2\right ) r}{8 x^2}-\frac {b^2 e n r \log \left (c x^n\right )}{4 x^2}-\frac {b e (2 a+b n) r \log \left (c x^n\right )}{4 x^2}-\frac {b^2 e r \log ^2\left (c x^n\right )}{4 x^2}-\frac {b^2 n^2 \left (d+e \log \left (f x^r\right )\right )}{4 x^2}-\frac {b n \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 x^2} \]
-1/8*b^2*e*n^2*r/x^2-1/8*b*e*n*(b*n+2*a)*r/x^2-1/8*e*(b^2*n^2+2*a*b*n+2*a^ 2)*r/x^2-1/4*b^2*e*n*r*ln(c*x^n)/x^2-1/4*b*e*(b*n+2*a)*r*ln(c*x^n)/x^2-1/4 *b^2*e*r*ln(c*x^n)^2/x^2-1/4*b^2*n^2*(d+e*ln(f*x^r))/x^2-1/2*b*n*(a+b*ln(c *x^n))*(d+e*ln(f*x^r))/x^2-1/2*(a+b*ln(c*x^n))^2*(d+e*ln(f*x^r))/x^2
Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\frac {4 a^2 d+4 a b d n+2 b^2 d n^2+2 a^2 e r+4 a b e n r+3 b^2 e n^2 r+2 e \left (2 a^2+2 a b n+b^2 n^2\right ) \log \left (f x^r\right )+2 b^2 \log ^2\left (c x^n\right ) \left (2 d+e r+2 e \log \left (f x^r\right )\right )+4 b \log \left (c x^n\right ) \left (2 a d+b d n+a e r+b e n r+e (2 a+b n) \log \left (f x^r\right )\right )}{8 x^2} \]
-1/8*(4*a^2*d + 4*a*b*d*n + 2*b^2*d*n^2 + 2*a^2*e*r + 4*a*b*e*n*r + 3*b^2* e*n^2*r + 2*e*(2*a^2 + 2*a*b*n + b^2*n^2)*Log[f*x^r] + 2*b^2*Log[c*x^n]^2* (2*d + e*r + 2*e*Log[f*x^r]) + 4*b*Log[c*x^n]*(2*a*d + b*d*n + a*e*r + b*e *n*r + e*(2*a + b*n)*Log[f*x^r]))/x^2
Time = 0.46 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2813, 27, 2010, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2813 |
| \(\displaystyle -e r \int -\frac {2 a^2+2 b n a+b^2 n^2+2 b^2 \log ^2\left (c x^n\right )+2 b (2 a+b n) \log \left (c x^n\right )}{4 x^3}dx-\frac {b n \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {b^2 n^2 \left (d+e \log \left (f x^r\right )\right )}{4 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} e r \int \frac {2 a^2+2 b n a+b^2 n^2+2 b^2 \log ^2\left (c x^n\right )+2 b (2 a+b n) \log \left (c x^n\right )}{x^3}dx-\frac {b n \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {b^2 n^2 \left (d+e \log \left (f x^r\right )\right )}{4 x^2}\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {1}{4} e r \int \left (\frac {2 b^2 \log ^2\left (c x^n\right )}{x^3}+\frac {2 b (2 a+b n) \log \left (c x^n\right )}{x^3}+\frac {2 a^2+2 b n a+b^2 n^2}{x^3}\right )dx-\frac {b n \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {b^2 n^2 \left (d+e \log \left (f x^r\right )\right )}{4 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} e r \left (-\frac {2 a^2+2 a b n+b^2 n^2}{2 x^2}-\frac {b (2 a+b n) \log \left (c x^n\right )}{x^2}-\frac {b n (2 a+b n)}{2 x^2}-\frac {b^2 \log ^2\left (c x^n\right )}{x^2}-\frac {b^2 n \log \left (c x^n\right )}{x^2}-\frac {b^2 n^2}{2 x^2}\right )-\frac {b n \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{2 x^2}-\frac {b^2 n^2 \left (d+e \log \left (f x^r\right )\right )}{4 x^2}\) |
(e*r*(-1/2*(b^2*n^2)/x^2 - (b*n*(2*a + b*n))/(2*x^2) - (2*a^2 + 2*a*b*n + b^2*n^2)/(2*x^2) - (b^2*n*Log[c*x^n])/x^2 - (b*(2*a + b*n)*Log[c*x^n])/x^2 - (b^2*Log[c*x^n]^2)/x^2))/4 - (b^2*n^2*(d + e*Log[f*x^r]))/(4*x^2) - (b* n*(a + b*Log[c*x^n])*(d + e*Log[f*x^r]))/(2*x^2) - ((a + b*Log[c*x^n])^2*( d + e*Log[f*x^r]))/(2*x^2)
3.2.68.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ .)]*(e_.))*((g_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Simp[(d + e*Log[f*x^r]) u, x] - Simp[e*r Int[Simp lifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] && !(EqQ[p, 1] && EqQ[a, 0] && NeQ[d, 0])
Time = 8.05 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.31
| method | result | size |
| parallelrisch | \(-\frac {8 \ln \left (c \,x^{n}\right ) \ln \left (f \,x^{r}\right ) a b e \,n^{4}+4 \ln \left (c \,x^{n}\right ) a b e \,n^{4} r +2 b^{2} d \,n^{6}+4 a^{2} d \,n^{4}+3 b^{2} e \,n^{6} r +2 a^{2} e \,n^{4} r +4 a b d \,n^{5}+2 \ln \left (f \,x^{r}\right ) b^{2} e \,n^{6}+4 \ln \left (f \,x^{r}\right ) a^{2} e \,n^{4}+4 \ln \left (c \,x^{n}\right ) b^{2} d \,n^{5}+4 \ln \left (c \,x^{n}\right )^{2} b^{2} d \,n^{4}+4 a b e \,n^{5} r +4 e \,b^{2} \ln \left (f \,x^{r}\right ) \ln \left (c \,x^{n}\right )^{2} n^{4}+4 \ln \left (f \,x^{r}\right ) a b e \,n^{5}+4 \ln \left (c \,x^{n}\right ) \ln \left (f \,x^{r}\right ) b^{2} e \,n^{5}+4 \ln \left (c \,x^{n}\right ) b^{2} e \,n^{5} r +8 \ln \left (c \,x^{n}\right ) a b d \,n^{4}+2 \ln \left (c \,x^{n}\right )^{2} b^{2} e \,n^{4} r}{8 x^{2} n^{4}}\) | \(268\) |
| risch | \(\text {Expression too large to display}\) | \(8407\) |
-1/8/x^2*(8*ln(c*x^n)*ln(f*x^r)*a*b*e*n^4+4*ln(c*x^n)*a*b*e*n^4*r+2*b^2*d* n^6+4*a^2*d*n^4+3*b^2*e*n^6*r+2*a^2*e*n^4*r+4*a*b*d*n^5+2*ln(f*x^r)*b^2*e* n^6+4*ln(f*x^r)*a^2*e*n^4+4*ln(c*x^n)*b^2*d*n^5+4*ln(c*x^n)^2*b^2*d*n^4+4* a*b*e*n^5*r+4*e*b^2*ln(f*x^r)*ln(c*x^n)^2*n^4+4*ln(f*x^r)*a*b*e*n^5+4*ln(c *x^n)*ln(f*x^r)*b^2*e*n^5+4*ln(c*x^n)*b^2*e*n^5*r+8*ln(c*x^n)*a*b*d*n^4+2* ln(c*x^n)^2*b^2*e*n^4*r)/n^4
Time = 0.28 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\frac {4 \, b^{2} e n^{2} r \log \left (x\right )^{3} + 2 \, b^{2} d n^{2} + 4 \, a b d n + 4 \, a^{2} d + 2 \, {\left (b^{2} e r + 2 \, b^{2} d\right )} \log \left (c\right )^{2} + 2 \, {\left (4 \, b^{2} e n r \log \left (c\right ) + 2 \, b^{2} e n^{2} \log \left (f\right ) + 2 \, b^{2} d n^{2} + {\left (3 \, b^{2} e n^{2} + 4 \, a b e n\right )} r\right )} \log \left (x\right )^{2} + {\left (3 \, b^{2} e n^{2} + 4 \, a b e n + 2 \, a^{2} e\right )} r + 4 \, {\left (b^{2} d n + 2 \, a b d + {\left (b^{2} e n + a b e\right )} r\right )} \log \left (c\right ) + 2 \, {\left (b^{2} e n^{2} + 2 \, b^{2} e \log \left (c\right )^{2} + 2 \, a b e n + 2 \, a^{2} e + 2 \, {\left (b^{2} e n + 2 \, a b e\right )} \log \left (c\right )\right )} \log \left (f\right ) + 2 \, {\left (2 \, b^{2} e r \log \left (c\right )^{2} + 2 \, b^{2} d n^{2} + 4 \, a b d n + {\left (3 \, b^{2} e n^{2} + 4 \, a b e n + 2 \, a^{2} e\right )} r + 4 \, {\left (b^{2} d n + {\left (b^{2} e n + a b e\right )} r\right )} \log \left (c\right ) + 2 \, {\left (b^{2} e n^{2} + 2 \, b^{2} e n \log \left (c\right ) + 2 \, a b e n\right )} \log \left (f\right )\right )} \log \left (x\right )}{8 \, x^{2}} \]
-1/8*(4*b^2*e*n^2*r*log(x)^3 + 2*b^2*d*n^2 + 4*a*b*d*n + 4*a^2*d + 2*(b^2* e*r + 2*b^2*d)*log(c)^2 + 2*(4*b^2*e*n*r*log(c) + 2*b^2*e*n^2*log(f) + 2*b ^2*d*n^2 + (3*b^2*e*n^2 + 4*a*b*e*n)*r)*log(x)^2 + (3*b^2*e*n^2 + 4*a*b*e* n + 2*a^2*e)*r + 4*(b^2*d*n + 2*a*b*d + (b^2*e*n + a*b*e)*r)*log(c) + 2*(b ^2*e*n^2 + 2*b^2*e*log(c)^2 + 2*a*b*e*n + 2*a^2*e + 2*(b^2*e*n + 2*a*b*e)* log(c))*log(f) + 2*(2*b^2*e*r*log(c)^2 + 2*b^2*d*n^2 + 4*a*b*d*n + (3*b^2* e*n^2 + 4*a*b*e*n + 2*a^2*e)*r + 4*(b^2*d*n + (b^2*e*n + a*b*e)*r)*log(c) + 2*(b^2*e*n^2 + 2*b^2*e*n*log(c) + 2*a*b*e*n)*log(f))*log(x))/x^2
Time = 0.86 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=- \frac {a^{2} d}{2 x^{2}} - \frac {a^{2} e r}{4 x^{2}} - \frac {a^{2} e \log {\left (f x^{r} \right )}}{2 x^{2}} - \frac {a b d n}{2 x^{2}} - \frac {a b d \log {\left (c x^{n} \right )}}{x^{2}} - \frac {a b e n r}{2 x^{2}} - \frac {a b e n \log {\left (f x^{r} \right )}}{2 x^{2}} - \frac {a b e r \log {\left (c x^{n} \right )}}{2 x^{2}} - \frac {a b e \log {\left (c x^{n} \right )} \log {\left (f x^{r} \right )}}{x^{2}} - \frac {b^{2} d n^{2}}{4 x^{2}} - \frac {b^{2} d n \log {\left (c x^{n} \right )}}{2 x^{2}} - \frac {b^{2} d \log {\left (c x^{n} \right )}^{2}}{2 x^{2}} - \frac {3 b^{2} e n^{2} r}{8 x^{2}} - \frac {b^{2} e n^{2} \log {\left (f x^{r} \right )}}{4 x^{2}} - \frac {b^{2} e n r \log {\left (c x^{n} \right )}}{2 x^{2}} - \frac {b^{2} e n \log {\left (c x^{n} \right )} \log {\left (f x^{r} \right )}}{2 x^{2}} - \frac {b^{2} e r \log {\left (c x^{n} \right )}^{2}}{4 x^{2}} - \frac {b^{2} e \log {\left (c x^{n} \right )}^{2} \log {\left (f x^{r} \right )}}{2 x^{2}} \]
-a**2*d/(2*x**2) - a**2*e*r/(4*x**2) - a**2*e*log(f*x**r)/(2*x**2) - a*b*d *n/(2*x**2) - a*b*d*log(c*x**n)/x**2 - a*b*e*n*r/(2*x**2) - a*b*e*n*log(f* x**r)/(2*x**2) - a*b*e*r*log(c*x**n)/(2*x**2) - a*b*e*log(c*x**n)*log(f*x* *r)/x**2 - b**2*d*n**2/(4*x**2) - b**2*d*n*log(c*x**n)/(2*x**2) - b**2*d*l og(c*x**n)**2/(2*x**2) - 3*b**2*e*n**2*r/(8*x**2) - b**2*e*n**2*log(f*x**r )/(4*x**2) - b**2*e*n*r*log(c*x**n)/(2*x**2) - b**2*e*n*log(c*x**n)*log(f* x**r)/(2*x**2) - b**2*e*r*log(c*x**n)**2/(4*x**2) - b**2*e*log(c*x**n)**2* log(f*x**r)/(2*x**2)
Time = 0.20 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\frac {1}{4} \, b^{2} e {\left (\frac {r}{x^{2}} + \frac {2 \, \log \left (f x^{r}\right )}{x^{2}}\right )} \log \left (c x^{n}\right )^{2} - \frac {1}{2} \, a b e {\left (\frac {r}{x^{2}} + \frac {2 \, \log \left (f x^{r}\right )}{x^{2}}\right )} \log \left (c x^{n}\right ) - \frac {1}{8} \, b^{2} e {\left (\frac {{\left (2 \, r \log \left (x\right ) + 3 \, r + 2 \, \log \left (f\right )\right )} n^{2}}{x^{2}} + \frac {4 \, n {\left (r + \log \left (f\right ) + \log \left (x^{r}\right )\right )} \log \left (c x^{n}\right )}{x^{2}}\right )} - \frac {1}{4} \, b^{2} d {\left (\frac {n^{2}}{x^{2}} + \frac {2 \, n \log \left (c x^{n}\right )}{x^{2}}\right )} - \frac {a b e n {\left (r + \log \left (f\right ) + \log \left (x^{r}\right )\right )}}{2 \, x^{2}} - \frac {b^{2} d \log \left (c x^{n}\right )^{2}}{2 \, x^{2}} - \frac {a b d n}{2 \, x^{2}} - \frac {a^{2} e r}{4 \, x^{2}} - \frac {a b d \log \left (c x^{n}\right )}{x^{2}} - \frac {a^{2} e \log \left (f x^{r}\right )}{2 \, x^{2}} - \frac {a^{2} d}{2 \, x^{2}} \]
-1/4*b^2*e*(r/x^2 + 2*log(f*x^r)/x^2)*log(c*x^n)^2 - 1/2*a*b*e*(r/x^2 + 2* log(f*x^r)/x^2)*log(c*x^n) - 1/8*b^2*e*((2*r*log(x) + 3*r + 2*log(f))*n^2/ x^2 + 4*n*(r + log(f) + log(x^r))*log(c*x^n)/x^2) - 1/4*b^2*d*(n^2/x^2 + 2 *n*log(c*x^n)/x^2) - 1/2*a*b*e*n*(r + log(f) + log(x^r))/x^2 - 1/2*b^2*d*l og(c*x^n)^2/x^2 - 1/2*a*b*d*n/x^2 - 1/4*a^2*e*r/x^2 - a*b*d*log(c*x^n)/x^2 - 1/2*a^2*e*log(f*x^r)/x^2 - 1/2*a^2*d/x^2
Time = 0.30 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\frac {b^{2} e n^{2} r \log \left (x\right )^{3}}{2 \, x^{2}} - \frac {{\left (3 \, b^{2} e n^{2} r + 4 \, b^{2} e n r \log \left (c\right ) + 2 \, b^{2} e n^{2} \log \left (f\right ) + 2 \, b^{2} d n^{2} + 4 \, a b e n r\right )} \log \left (x\right )^{2}}{4 \, x^{2}} - \frac {{\left (3 \, b^{2} e n^{2} r + 4 \, b^{2} e n r \log \left (c\right ) + 2 \, b^{2} e r \log \left (c\right )^{2} + 2 \, b^{2} e n^{2} \log \left (f\right ) + 4 \, b^{2} e n \log \left (c\right ) \log \left (f\right ) + 2 \, b^{2} d n^{2} + 4 \, a b e n r + 4 \, b^{2} d n \log \left (c\right ) + 4 \, a b e r \log \left (c\right ) + 4 \, a b e n \log \left (f\right ) + 4 \, a b d n + 2 \, a^{2} e r\right )} \log \left (x\right )}{4 \, x^{2}} - \frac {3 \, b^{2} e n^{2} r + 4 \, b^{2} e n r \log \left (c\right ) + 2 \, b^{2} e r \log \left (c\right )^{2} + 2 \, b^{2} e n^{2} \log \left (f\right ) + 4 \, b^{2} e n \log \left (c\right ) \log \left (f\right ) + 4 \, b^{2} e \log \left (c\right )^{2} \log \left (f\right ) + 2 \, b^{2} d n^{2} + 4 \, a b e n r + 4 \, b^{2} d n \log \left (c\right ) + 4 \, a b e r \log \left (c\right ) + 4 \, b^{2} d \log \left (c\right )^{2} + 4 \, a b e n \log \left (f\right ) + 8 \, a b e \log \left (c\right ) \log \left (f\right ) + 4 \, a b d n + 2 \, a^{2} e r + 8 \, a b d \log \left (c\right ) + 4 \, a^{2} e \log \left (f\right ) + 4 \, a^{2} d}{8 \, x^{2}} \]
-1/2*b^2*e*n^2*r*log(x)^3/x^2 - 1/4*(3*b^2*e*n^2*r + 4*b^2*e*n*r*log(c) + 2*b^2*e*n^2*log(f) + 2*b^2*d*n^2 + 4*a*b*e*n*r)*log(x)^2/x^2 - 1/4*(3*b^2* e*n^2*r + 4*b^2*e*n*r*log(c) + 2*b^2*e*r*log(c)^2 + 2*b^2*e*n^2*log(f) + 4 *b^2*e*n*log(c)*log(f) + 2*b^2*d*n^2 + 4*a*b*e*n*r + 4*b^2*d*n*log(c) + 4* a*b*e*r*log(c) + 4*a*b*e*n*log(f) + 4*a*b*d*n + 2*a^2*e*r)*log(x)/x^2 - 1/ 8*(3*b^2*e*n^2*r + 4*b^2*e*n*r*log(c) + 2*b^2*e*r*log(c)^2 + 2*b^2*e*n^2*l og(f) + 4*b^2*e*n*log(c)*log(f) + 4*b^2*e*log(c)^2*log(f) + 2*b^2*d*n^2 + 4*a*b*e*n*r + 4*b^2*d*n*log(c) + 4*a*b*e*r*log(c) + 4*b^2*d*log(c)^2 + 4*a *b*e*n*log(f) + 8*a*b*e*log(c)*log(f) + 4*a*b*d*n + 2*a^2*e*r + 8*a*b*d*lo g(c) + 4*a^2*e*log(f) + 4*a^2*d)/x^2
Time = 0.80 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \left (d+e \log \left (f x^r\right )\right )}{x^3} \, dx=-\ln \left (f\,x^r\right )\,\left (\ln \left (c\,x^n\right )\,\left (\frac {a\,b\,e}{x^2}+\frac {b^2\,e\,n}{2\,x^2}\right )+\frac {a^2\,e}{2\,x^2}+\frac {b^2\,e\,n^2}{4\,x^2}+\frac {b^2\,e\,{\ln \left (c\,x^n\right )}^2}{2\,x^2}+\frac {a\,b\,e\,n}{2\,x^2}\right )-\frac {\frac {a^2\,d}{2}+\frac {b^2\,d\,n^2}{4}+\frac {a^2\,e\,r}{4}+\frac {3\,b^2\,e\,n^2\,r}{8}+\frac {a\,b\,d\,n}{2}+\frac {a\,b\,e\,n\,r}{2}}{x^2}-\frac {b^2\,{\ln \left (c\,x^n\right )}^2\,\left (2\,d+e\,r\right )}{4\,x^2}-\frac {b\,\ln \left (c\,x^n\right )\,\left (2\,a\,d+b\,d\,n+a\,e\,r+b\,e\,n\,r\right )}{2\,x^2} \]
- log(f*x^r)*(log(c*x^n)*((a*b*e)/x^2 + (b^2*e*n)/(2*x^2)) + (a^2*e)/(2*x^ 2) + (b^2*e*n^2)/(4*x^2) + (b^2*e*log(c*x^n)^2)/(2*x^2) + (a*b*e*n)/(2*x^2 )) - ((a^2*d)/2 + (b^2*d*n^2)/4 + (a^2*e*r)/4 + (3*b^2*e*n^2*r)/8 + (a*b*d *n)/2 + (a*b*e*n*r)/2)/x^2 - (b^2*log(c*x^n)^2*(2*d + e*r))/(4*x^2) - (b*l og(c*x^n)*(2*a*d + b*d*n + a*e*r + b*e*n*r))/(2*x^2)