Integrand size = 24, antiderivative size = 230 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=-\frac {7 b e m n}{144 d x^3}+\frac {3 b e^2 m n}{32 d^2 x^2}-\frac {5 b e^3 m n}{16 d^3 x}-\frac {b e^4 m n \log (x)}{16 d^4}-\frac {b e n \log \left (f x^m\right )}{12 d x^3}+\frac {b e^2 n \log \left (f x^m\right )}{8 d^2 x^2}-\frac {b e^3 n \log \left (f x^m\right )}{4 d^3 x}+\frac {b e^4 n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{4 d^4}+\frac {b e^4 m n \log (d+e x)}{16 d^4}-\frac {1}{16} \left (\frac {m}{x^4}+\frac {4 \log \left (f x^m\right )}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e^4 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{4 d^4} \]
-7/144*b*e*m*n/d/x^3+3/32*b*e^2*m*n/d^2/x^2-5/16*b*e^3*m*n/d^3/x-1/16*b*e^ 4*m*n*ln(x)/d^4-1/12*b*e*n*ln(f*x^m)/d/x^3+1/8*b*e^2*n*ln(f*x^m)/d^2/x^2-1 /4*b*e^3*n*ln(f*x^m)/d^3/x+1/4*b*e^4*n*ln(1+d/e/x)*ln(f*x^m)/d^4+1/16*b*e^ 4*m*n*ln(e*x+d)/d^4-1/16*(m/x^4+4*ln(f*x^m)/x^4)*(a+b*ln(c*(e*x+d)^n))-1/4 *b*e^4*m*n*polylog(2,-d/e/x)/d^4
Time = 0.12 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.19 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=-\frac {18 a d^4 m+14 b d^3 e m n x-27 b d^2 e^2 m n x^2+90 b d e^3 m n x^3-36 b e^4 m n x^4 \log ^2(x)+72 a d^4 \log \left (f x^m\right )+24 b d^3 e n x \log \left (f x^m\right )-36 b d^2 e^2 n x^2 \log \left (f x^m\right )+72 b d e^3 n x^3 \log \left (f x^m\right )-18 b e^4 m n x^4 \log (d+e x)-72 b e^4 n x^4 \log \left (f x^m\right ) \log (d+e x)+18 b d^4 m \log \left (c (d+e x)^n\right )+72 b d^4 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+18 b e^4 n x^4 \log (x) \left (m+4 \log \left (f x^m\right )+4 m \log (d+e x)-4 m \log \left (1+\frac {e x}{d}\right )\right )-72 b e^4 m n x^4 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{288 d^4 x^4} \]
-1/288*(18*a*d^4*m + 14*b*d^3*e*m*n*x - 27*b*d^2*e^2*m*n*x^2 + 90*b*d*e^3* m*n*x^3 - 36*b*e^4*m*n*x^4*Log[x]^2 + 72*a*d^4*Log[f*x^m] + 24*b*d^3*e*n*x *Log[f*x^m] - 36*b*d^2*e^2*n*x^2*Log[f*x^m] + 72*b*d*e^3*n*x^3*Log[f*x^m] - 18*b*e^4*m*n*x^4*Log[d + e*x] - 72*b*e^4*n*x^4*Log[f*x^m]*Log[d + e*x] + 18*b*d^4*m*Log[c*(d + e*x)^n] + 72*b*d^4*Log[f*x^m]*Log[c*(d + e*x)^n] + 18*b*e^4*n*x^4*Log[x]*(m + 4*Log[f*x^m] + 4*m*Log[d + e*x] - 4*m*Log[1 + ( e*x)/d]) - 72*b*e^4*m*n*x^4*PolyLog[2, -((e*x)/d)])/(d^4*x^4)
Time = 0.97 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {2873, 54, 2009, 2780, 2741, 2780, 2741, 2780, 2741, 2779, 2838}
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.
\(\displaystyle \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 2873 |
\(\displaystyle \frac {1}{4} b e n \int \frac {\log \left (f x^m\right )}{x^4 (d+e x)}dx+\frac {1}{16} b e m n \int \frac {1}{x^4 (d+e x)}dx-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{16} b e m n \int \left (\frac {e^4}{d^4 (d+e x)}-\frac {e^3}{d^4 x}+\frac {e^2}{d^3 x^2}-\frac {e}{d^2 x^3}+\frac {1}{d x^4}\right )dx+\frac {1}{4} b e n \int \frac {\log \left (f x^m\right )}{x^4 (d+e x)}dx-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} b e n \int \frac {\log \left (f x^m\right )}{x^4 (d+e x)}dx-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {\int \frac {\log \left (f x^m\right )}{x^4}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^3 (d+e x)}dx}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^3 (d+e x)}dx}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {\int \frac {\log \left (f x^m\right )}{x^3}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx}{d}\right )}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx}{d}\right )}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2780 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {\int \frac {\log \left (f x^m\right )}{x^2}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )}{d}\right )}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )}{d}\right )}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )}{d}\right )}{d}\right )-\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\frac {1}{16} \left (\frac {4 \log \left (f x^m\right )}{x^4}+\frac {m}{x^4}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{16} b e m n \left (-\frac {e^3 \log (x)}{d^4}+\frac {e^3 \log (d+e x)}{d^4}-\frac {e^2}{d^3 x}+\frac {e}{2 d^2 x^2}-\frac {1}{3 d x^3}\right )+\frac {1}{4} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{3 x^3}-\frac {m}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{2 x^2}-\frac {m}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )}{d}\right )}{d}\right )\) |
(b*e*m*n*(-1/3*1/(d*x^3) + e/(2*d^2*x^2) - e^2/(d^3*x) - (e^3*Log[x])/d^4 + (e^3*Log[d + e*x])/d^4))/16 - ((m/x^4 + (4*Log[f*x^m])/x^4)*(a + b*Log[c *(d + e*x)^n]))/16 + (b*e*n*((-1/9*m/x^3 - Log[f*x^m]/(3*x^3))/d - (e*((-1 /4*m/x^2 - Log[f*x^m]/(2*x^2))/d - (e*((-(m/x) - Log[f*x^m]/x)/d - (e*(-(( Log[1 + d/(e*x)]*Log[f*x^m])/d) + (m*PolyLog[2, -(d/(e*x))])/d))/d))/d))/d ))/4
3.4.66.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_ .))*((g_.)*(x_))^(q_.), x_Symbol] :> Simp[(-(g*(q + 1))^(-1))*(m*((g*x)^(q + 1)/(q + 1)) - (g*x)^(q + 1)*Log[f*x^m])*(a + b*Log[c*(d + e*x)^n]), x] + (-Simp[b*e*(n/(g*(q + 1))) Int[(g*x)^(q + 1)*(Log[f*x^m]/(d + e*x)), x], x] + Simp[b*e*m*(n/(g*(q + 1)^2)) Int[(g*x)^(q + 1)/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && NeQ[q, -1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 32.96 (sec) , antiderivative size = 1237, normalized size of antiderivative = 5.38
-1/24*I*e*n*b/d/x^3*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/8*I*e^4*n*b/d^4*ln(e* x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/8*I*e^4*n*b/d^4*ln(e*x+d)*Pi*csgn(I*x^ m)*csgn(I*f*x^m)^2-1/8*I*e^3*n*b/d^3/x*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/8*I* e^3*n*b/d^3/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/16*I*e^2*n*b/d^2/x^2*Pi*csg n(I*f)*csgn(I*f*x^m)^2+1/16*I*e^2*n*b/d^2/x^2*Pi*csgn(I*x^m)*csgn(I*f*x^m) ^2-1/8*I*e^4*n*b/d^4*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/8*I*e^4*n*b/d^4* ln(x)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/24*I*e*n*b/d/x^3*Pi*csgn(I*f)*csgn( I*f*x^m)^2+1/24*I*e*n*b/d/x^3*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/24* I*e*n*b/d/x^3*Pi*csgn(I*f*x^m)^3+1/8*I*e^3*n*b/d^3/x*Pi*csgn(I*f*x^m)^3-1/ 16*I*e^2*n*b/d^2/x^2*Pi*csgn(I*f*x^m)^3+1/8*I*e^4*n*b/d^4*ln(x)*Pi*csgn(I* f*x^m)^3-1/8*I*e^4*n*b/d^4*ln(e*x+d)*Pi*csgn(I*f*x^m)^3-1/16*I*e^2*n*b/d^2 /x^2*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*I*e^4*n*b/d^4*ln(x)*Pi*csg n(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/8*I*e^4*n*b/d^4*ln(e*x+d)*Pi*csgn(I*f)* csgn(I*x^m)*csgn(I*f*x^m)+1/8*I*e^3*n*b/d^3/x*Pi*csgn(I*f)*csgn(I*x^m)*csg n(I*f*x^m)-1/4*e^3*n*b/d^3/x*ln(f)-1/12*e*n*b/d/x^3*ln(f)+1/8*e^2*n*b/d^2/ x^2*ln(f)-1/4*e^4*n*b*m/d^4*ln(e*x+d)*ln(-e*x/d)+(-1/4*b/x^4*ln(x^m)-1/16* (-2*I*Pi*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+2*I*Pi*b*csgn(I*f)*csgn(I*f *x^m)^2+2*I*Pi*b*csgn(I*x^m)*csgn(I*f*x^m)^2-2*I*Pi*b*csgn(I*f*x^m)^3+4*b* ln(f)+b*m)/x^4)*ln((e*x+d)^n)+(-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csg n(I*c*(e*x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/4*I*b*Pi*...
\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.10 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=\frac {1}{288} \, {\left (\frac {72 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{4} n}{d^{4}} + \frac {18 \, b e^{4} n \log \left (e x + d\right )}{d^{4}} - \frac {72 \, b e^{4} n x^{4} \log \left (e x + d\right ) \log \left (x\right ) - 36 \, b e^{4} n x^{4} \log \left (x\right )^{2} + 18 \, b e^{4} n x^{4} \log \left (x\right ) + 90 \, b d e^{3} n x^{3} - 27 \, b d^{2} e^{2} n x^{2} + 14 \, b d^{3} e n x + 18 \, b d^{4} \log \left ({\left (e x + d\right )}^{n}\right ) + 18 \, b d^{4} \log \left (c\right ) + 18 \, a d^{4}}{d^{4} x^{4}}\right )} m + \frac {1}{24} \, {\left (b e n {\left (\frac {6 \, e^{3} \log \left (e x + d\right )}{d^{4}} - \frac {6 \, e^{3} \log \left (x\right )}{d^{4}} - \frac {6 \, e^{2} x^{2} - 3 \, d e x + 2 \, d^{2}}{d^{3} x^{3}}\right )} - \frac {6 \, b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{4}} - \frac {6 \, a}{x^{4}}\right )} \log \left (f x^{m}\right ) \]
1/288*(72*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*b*e^4*n/d^4 + 18*b*e^4*n *log(e*x + d)/d^4 - (72*b*e^4*n*x^4*log(e*x + d)*log(x) - 36*b*e^4*n*x^4*l og(x)^2 + 18*b*e^4*n*x^4*log(x) + 90*b*d*e^3*n*x^3 - 27*b*d^2*e^2*n*x^2 + 14*b*d^3*e*n*x + 18*b*d^4*log((e*x + d)^n) + 18*b*d^4*log(c) + 18*a*d^4)/( d^4*x^4))*m + 1/24*(b*e*n*(6*e^3*log(e*x + d)/d^4 - 6*e^3*log(x)/d^4 - (6* e^2*x^2 - 3*d*e*x + 2*d^2)/(d^3*x^3)) - 6*b*log((e*x + d)^n*c)/x^4 - 6*a/x ^4)*log(f*x^m)
\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^5} \, dx=\int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^5} \,d x \]