Integrand size = 24, antiderivative size = 156 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {3 b e m n}{4 d x}-\frac {b e^2 m n \log (x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}+\frac {b e^2 n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{2 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {1}{4} \left (\frac {m}{x^2}+\frac {2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e^2 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{2 d^2} \]
-3/4*b*e*m*n/d/x-1/4*b*e^2*m*n*ln(x)/d^2-1/2*b*e*n*ln(f*x^m)/d/x+1/2*b*e^2 *n*ln(1+d/e/x)*ln(f*x^m)/d^2+1/4*b*e^2*m*n*ln(e*x+d)/d^2-1/4*(m/x^2+2*ln(f *x^m)/x^2)*(a+b*ln(c*(e*x+d)^n))-1/2*b*e^2*m*n*polylog(2,-d/e/x)/d^2
Time = 0.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {a d^2 m+3 b d e m n x-b e^2 m n x^2 \log ^2(x)+2 a d^2 \log \left (f x^m\right )+2 b d e n x \log \left (f x^m\right )-b e^2 m n x^2 \log (d+e x)-2 b e^2 n x^2 \log \left (f x^m\right ) \log (d+e x)+b d^2 m \log \left (c (d+e x)^n\right )+2 b d^2 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b e^2 n x^2 \log (x) \left (m+2 \log \left (f x^m\right )+2 m \log (d+e x)-2 m \log \left (1+\frac {e x}{d}\right )\right )-2 b e^2 m n x^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^2 x^2} \]
-1/4*(a*d^2*m + 3*b*d*e*m*n*x - b*e^2*m*n*x^2*Log[x]^2 + 2*a*d^2*Log[f*x^m ] + 2*b*d*e*n*x*Log[f*x^m] - b*e^2*m*n*x^2*Log[d + e*x] - 2*b*e^2*n*x^2*Lo g[f*x^m]*Log[d + e*x] + b*d^2*m*Log[c*(d + e*x)^n] + 2*b*d^2*Log[f*x^m]*Lo g[c*(d + e*x)^n] + b*e^2*n*x^2*Log[x]*(m + 2*Log[f*x^m] + 2*m*Log[d + e*x] - 2*m*Log[1 + (e*x)/d]) - 2*b*e^2*m*n*x^2*PolyLog[2, -((e*x)/d)])/(d^2*x^ 2)
Time = 0.52 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2873, 54, 2009, 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.
| \(\displaystyle \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2873 |
| \(\displaystyle \frac {1}{2} b e n \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx+\frac {1}{4} b e m n \int \frac {1}{x^2 (d+e x)}dx-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {1}{4} b e m n \int \left (\frac {e^2}{d^2 (d+e x)}-\frac {e}{d^2 x}+\frac {1}{d x^2}\right )dx+\frac {1}{2} b e n \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} b e n \int \frac {\log \left (f x^m\right )}{x^2 (d+e x)}dx-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b e m n \left (-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}-\frac {1}{d x}\right )\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {1}{2} b e n \left (\frac {\int \frac {\log \left (f x^m\right )}{x^2}dx}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b e m n \left (-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}-\frac {1}{d x}\right )\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {1}{2} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \int \frac {\log \left (f x^m\right )}{x (d+e x)}dx}{d}\right )-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b e m n \left (-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}-\frac {1}{d x}\right )\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {1}{2} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b e m n \left (-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}-\frac {1}{d x}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle -\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b e m n \left (-\frac {e \log (x)}{d^2}+\frac {e \log (d+e x)}{d^2}-\frac {1}{d x}\right )+\frac {1}{2} b e n \left (\frac {-\frac {\log \left (f x^m\right )}{x}-\frac {m}{x}}{d}-\frac {e \left (\frac {m \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{d}\right )}{d}\right )\) |
(b*e*m*n*(-(1/(d*x)) - (e*Log[x])/d^2 + (e*Log[d + e*x])/d^2))/4 - ((m/x^2 + (2*Log[f*x^m])/x^2)*(a + b*Log[c*(d + e*x)^n]))/4 + (b*e*n*((-(m/x) - L og[f*x^m]/x)/d - (e*(-((Log[1 + d/(e*x)]*Log[f*x^m])/d) + (m*PolyLog[2, -( d/(e*x))])/d))/d))/2
3.4.64.3.1 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 = 5.39 (sec) , antiderivative size = 901, normalized size of antiderivative = 5.78
-1/2*e^2*n*b*m/d^2*ln(e*x+d)*ln(-e*x/d)+1/2*e^2*n*b*ln(x^m)/d^2*ln(e*x+d)- 1/2*e^2*n*b*ln(x^m)/d^2*ln(x)-1/2*e^2*n*b*m/d^2*dilog(-e*x/d)+1/4*e^2*n*b* m/d^2*ln(x)^2+1/2*e^2*n*b/d^2*ln(e*x+d)*ln(f)-1/2*e^2*n*b/d^2*ln(x)*ln(f)+ 1/4*I*e*n*b/d/x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I*e^2*n*b/d^2*l n(x)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi *csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*csgn(I *f*x^m)^3+1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*f*x^m)^3+1/4*I*e*n*b/d/x*Pi*cs gn(I*f*x^m)^3-1/2*e*n*b*ln(x^m)/d/x-1/4*I*e*n*b/d/x*Pi*csgn(I*f)*csgn(I*f* x^m)^2-1/4*I*e*n*b/d/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/4*I*e^2*n*b/d^2*ln (e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/4*I*e^2*n*b/d^2*ln(e*x+d)*Pi*csgn(I *x^m)*csgn(I*f*x^m)^2-1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2 -1/4*I*e^2*n*b/d^2*ln(x)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+(-1/4*I*b*Pi*csgn( I*c)*csgn(I*(e*x+d)^n)*csgn(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*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b*Pi* csgn(I*c*(e*x+d)^n)^3+1/2*b*ln(c)+1/2*a)*(-ln(x^m)/x^2-1/2*m/x^2-1/2*(-I*P i*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*csgn(I*f)*csgn(I*f*x^m)^2+I*Pi* csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi*csgn(I*f*x^m)^3+2*ln(f))/x^2)+(-1/2*b/x^2 *ln(x^m)-1/4*(-I*Pi*b*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*b*csgn(I*f) *csgn(I*f*x^m)^2+I*Pi*b*csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi*b*csgn(I*f*x^m)^3 +2*b*ln(f)+b*m)/x^2)*ln((e*x+d)^n)-1/2*e*n*b/d/x*ln(f)-3/4*b*e*m*n/d/x-...
\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{2} n}{d^{2}} + \frac {b e^{2} n \log \left (e x + d\right )}{d^{2}} - \frac {2 \, b e^{2} n x^{2} \log \left (e x + d\right ) \log \left (x\right ) - b e^{2} n x^{2} \log \left (x\right )^{2} + b e^{2} n x^{2} \log \left (x\right ) + 3 \, b d e n x + b d^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b d^{2} \log \left (c\right ) + a d^{2}}{d^{2} x^{2}}\right )} m + \frac {1}{2} \, {\left (b e n {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{2}} - \frac {a}{x^{2}}\right )} \log \left (f x^{m}\right ) \]
1/4*(2*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*b*e^2*n/d^2 + b*e^2*n*log(e *x + d)/d^2 - (2*b*e^2*n*x^2*log(e*x + d)*log(x) - b*e^2*n*x^2*log(x)^2 + b*e^2*n*x^2*log(x) + 3*b*d*e*n*x + b*d^2*log((e*x + d)^n) + b*d^2*log(c) + a*d^2)/(d^2*x^2))*m + 1/2*(b*e*n*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/( d*x)) - b*log((e*x + d)^n*c)/x^2 - a/x^2)*log(f*x^m)
\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^3} \,d x \]