Integrand size = 25, antiderivative size = 121 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (f+g x^n\right )}{e f-d g}\right )}{f n}-\frac {p \operatorname {PolyLog}\left (2,-\frac {g \left (d+e x^n\right )}{e f-d g}\right )}{f n}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{f n} \]
ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/f/n-ln(c*(d+e*x^n)^p)*ln(e*(f+g*x^n)/(-d*g+ e*f))/f/n-p*polylog(2,-g*(d+e*x^n)/(-d*g+e*f))/f/n+p*polylog(2,1+e*x^n/d)/ f/n
Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\frac {\log \left (c \left (d+e x^n\right )^p\right ) \left (\log \left (-\frac {e x^n}{d}\right )-\log \left (\frac {e \left (f+g x^n\right )}{e f-d g}\right )\right )-p \operatorname {PolyLog}\left (2,\frac {g \left (d+e x^n\right )}{-e f+d g}\right )+p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{f n} \]
(Log[c*(d + e*x^n)^p]*(Log[-((e*x^n)/d)] - Log[(e*(f + g*x^n))/(e*f - d*g) ]) - p*PolyLog[2, (g*(d + e*x^n))/(-(e*f) + d*g)] + p*PolyLog[2, 1 + (e*x^ n)/d])/(f*n)
Time = 0.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2925, 2863, 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.
\(\displaystyle \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {\int \frac {x^{-n} \log \left (c \left (e x^n+d\right )^p\right )}{g x^n+f}dx^n}{n}\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \frac {\int \left (\frac {x^{-n} \log \left (c \left (e x^n+d\right )^p\right )}{f}-\frac {g \log \left (c \left (e x^n+d\right )^p\right )}{f \left (g x^n+f\right )}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (f+g x^n\right )}{e f-d g}\right )}{f}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f}-\frac {p \operatorname {PolyLog}\left (2,-\frac {g \left (e x^n+d\right )}{e f-d g}\right )}{f}+\frac {p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{f}}{n}\) |
((Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/f - (Log[c*(d + e*x^n)^p]*Log[(e *(f + g*x^n))/(e*f - d*g)])/f - (p*PolyLog[2, -((g*(d + e*x^n))/(e*f - d*g ))])/f + (p*PolyLog[2, 1 + (e*x^n)/d])/f)/n
3.4.71.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.72 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.67
method | result | size |
risch | \(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (x^{n}\right )}{n f}-\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (f +g \,x^{n}\right )}{n f}-\frac {p \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n f}-\frac {p \ln \left (x^{n}\right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )}{n f}+\frac {p \operatorname {dilog}\left (\frac {\left (f +g \,x^{n}\right ) e +d g -e f}{d g -e f}\right )}{n f}+\frac {p \ln \left (f +g \,x^{n}\right ) \ln \left (\frac {\left (f +g \,x^{n}\right ) e +d g -e f}{d g -e f}\right )}{n f}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\ln \left (x^{n}\right )}{n f}-\frac {\ln \left (f +g \,x^{n}\right )}{n f}\right )\) | \(323\) |
1/n*ln((d+e*x^n)^p)/f*ln(x^n)-1/n*ln((d+e*x^n)^p)/f*ln(f+g*x^n)-1/n*p/f*di log((d+e*x^n)/d)-1/n*p/f*ln(x^n)*ln((d+e*x^n)/d)+1/n*p/f*dilog(((f+g*x^n)* e+d*g-e*f)/(d*g-e*f))+1/n*p/f*ln(f+g*x^n)*ln(((f+g*x^n)*e+d*g-e*f)/(d*g-e* f))+(1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)^2-1/2*I*Pi*csgn(I* (d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*csgn(I*c)-1/2*I*Pi*csgn(I*c*(d+e*x^n)^p )^3+1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)+ln(c))*(1/n/f*ln(x^n)-1/n/f *ln(f+g*x^n))
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (g x^{n} + f\right )} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=-e n p {\left (\frac {\log \left (x^{n}\right ) \log \left (\frac {e x^{n}}{d} + 1\right ) + {\rm Li}_2\left (-\frac {e x^{n}}{d}\right )}{e f n^{2}} - \frac {\log \left (g x^{n} + f\right ) \log \left (-\frac {e g x^{n} + e f}{e f - d g} + 1\right ) + {\rm Li}_2\left (\frac {e g x^{n} + e f}{e f - d g}\right )}{e f n^{2}}\right )} - {\left (\frac {\log \left (g x^{n} + f\right )}{f n} - \frac {\log \left (x^{n}\right )}{f n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \]
-e*n*p*((log(x^n)*log(e*x^n/d + 1) + dilog(-e*x^n/d))/(e*f*n^2) - (log(g*x ^n + f)*log(-(e*g*x^n + e*f)/(e*f - d*g) + 1) + dilog((e*g*x^n + e*f)/(e*f - d*g)))/(e*f*n^2)) - (log(g*x^n + f)/(f*n) - log(x^n)/(f*n))*log((e*x^n + d)^p*c)
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (g x^{n} + f\right )} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^n\right )} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{x\,\left (f+g\,x^n\right )} \,d x \]