Integrand size = 27, antiderivative size = 156 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\frac {e g p \log \left (d+e x^n\right )}{f^2 (d f-e g) n}+\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{f^2 n \left (g+f x^n\right )}-\frac {e g p \log \left (g+f x^n\right )}{f^2 (d f-e g) n}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (g+f x^n\right )}{d f-e g}\right )}{f^2 n}+\frac {p \operatorname {PolyLog}\left (2,\frac {f \left (d+e x^n\right )}{d f-e g}\right )}{f^2 n} \]
e*g*p*ln(d+e*x^n)/f^2/(d*f-e*g)/n+g*ln(c*(d+e*x^n)^p)/f^2/n/(g+f*x^n)-e*g* p*ln(g+f*x^n)/f^2/(d*f-e*g)/n+ln(c*(d+e*x^n)^p)*ln(-e*(g+f*x^n)/(d*f-e*g)) /f^2/n+p*polylog(2,f*(d+e*x^n)/(d*f-e*g))/f^2/n
Leaf count is larger than twice the leaf count of optimal. \(433\) vs. \(2(156)=312\).
Time = 1.05 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.78 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\frac {g p \log \left (f-f x^{-n}\right )+f p x^n \log \left (f-f x^{-n}\right )-g n p \log (x) \log \left (f-f x^{-n}\right )-f n p x^n \log (x) \log \left (f-f x^{-n}\right )-p \log \left (e+d x^{-n}\right ) \left (-f x^n+\left (g+f x^n\right ) \log \left (f-f x^{-n}\right )\right )-f x^n \log \left (c \left (d+e x^n\right )^p\right )+g \log \left (f-f x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )+f x^n \log \left (f-f x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )+g n p \log (x) \log \left (1+\frac {f x^n}{g}\right )+f n p x^n \log (x) \log \left (1+\frac {f x^n}{g}\right )+p \left (g+f x^n\right ) \operatorname {PolyLog}\left (2,-\frac {f x^n}{g}\right )}{f^2 n \left (g+f x^n\right )}-\frac {p \left (-\frac {d f \log \left (e+d x^{-n}\right )}{d f-e g}+\frac {f x^n \log \left (e+d x^{-n}\right )}{g+f x^n}+\log \left (-\frac {d x^{-n}}{e}\right ) \log \left (e+d x^{-n}\right )+\frac {d f \log \left (f+g x^{-n}\right )}{d f-e g}-\log \left (e+d x^{-n}\right ) \log \left (\frac {d \left (f+g x^{-n}\right )}{d f-e g}\right )-\operatorname {PolyLog}\left (2,-\frac {g \left (e+d x^{-n}\right )}{d f-e g}\right )+\operatorname {PolyLog}\left (2,1+\frac {d x^{-n}}{e}\right )\right )}{f^2 n} \]
(g*p*Log[f - f/x^n] + f*p*x^n*Log[f - f/x^n] - g*n*p*Log[x]*Log[f - f/x^n] - f*n*p*x^n*Log[x]*Log[f - f/x^n] - p*Log[e + d/x^n]*(-(f*x^n) + (g + f*x ^n)*Log[f - f/x^n]) - f*x^n*Log[c*(d + e*x^n)^p] + g*Log[f - f/x^n]*Log[c* (d + e*x^n)^p] + f*x^n*Log[f - f/x^n]*Log[c*(d + e*x^n)^p] + g*n*p*Log[x]* Log[1 + (f*x^n)/g] + f*n*p*x^n*Log[x]*Log[1 + (f*x^n)/g] + p*(g + f*x^n)*P olyLog[2, -((f*x^n)/g)])/(f^2*n*(g + f*x^n)) - (p*(-((d*f*Log[e + d/x^n])/ (d*f - e*g)) + (f*x^n*Log[e + d/x^n])/(g + f*x^n) + Log[-(d/(e*x^n))]*Log[ e + d/x^n] + (d*f*Log[f + g/x^n])/(d*f - e*g) - Log[e + d/x^n]*Log[(d*(f + g/x^n))/(d*f - e*g)] - PolyLog[2, -((g*(e + d/x^n))/(d*f - e*g))] + PolyL og[2, 1 + d/(e*x^n)]))/(f^2*n)
Time = 0.43 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2005, 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.
\(\displaystyle \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \int \frac {x^{2 n-1} \log \left (c \left (d+e x^n\right )^p\right )}{\left (f x^n+g\right )^2}dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {\int \frac {x^n \log \left (c \left (e x^n+d\right )^p\right )}{\left (f x^n+g\right )^2}dx^n}{n}\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \frac {\int \left (\frac {\log \left (c \left (e x^n+d\right )^p\right )}{f \left (f x^n+g\right )}-\frac {g \log \left (c \left (e x^n+d\right )^p\right )}{f \left (f x^n+g\right )^2}\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{f^2 \left (f x^n+g\right )}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (f x^n+g\right )}{d f-e g}\right )}{f^2}+\frac {p \operatorname {PolyLog}\left (2,\frac {f \left (e x^n+d\right )}{d f-e g}\right )}{f^2}+\frac {e g p \log \left (d+e x^n\right )}{f^2 (d f-e g)}-\frac {e g p \log \left (f x^n+g\right )}{f^2 (d f-e g)}}{n}\) |
((e*g*p*Log[d + e*x^n])/(f^2*(d*f - e*g)) + (g*Log[c*(d + e*x^n)^p])/(f^2* (g + f*x^n)) - (e*g*p*Log[g + f*x^n])/(f^2*(d*f - e*g)) + (Log[c*(d + e*x^ n)^p]*Log[-((e*(g + f*x^n))/(d*f - e*g))])/f^2 + (p*PolyLog[2, (f*(d + e*x ^n))/(d*f - e*g)])/f^2)/n
3.4.76.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
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 = 7.88 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.28
method | result | size |
risch | \(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (g +f \,x^{n}\right )}{n \,f^{2}}+\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) g}{n \,f^{2} \left (g +f \,x^{n}\right )}-\frac {p \operatorname {dilog}\left (\frac {\left (g +f \,x^{n}\right ) e +d f -e g}{d f -e g}\right )}{n \,f^{2}}-\frac {p \ln \left (g +f \,x^{n}\right ) \ln \left (\frac {\left (g +f \,x^{n}\right ) e +d f -e g}{d f -e g}\right )}{n \,f^{2}}+\frac {p e g \ln \left (\left (g +f \,x^{n}\right ) e +d f -e g \right )}{n \,f^{2} \left (d f -e g \right )}-\frac {e g p \ln \left (g +f \,x^{n}\right )}{f^{2} \left (d f -e g \right ) n}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\ln \left (g +f \,x^{n}\right )}{n \,f^{2}}+\frac {g}{n \,f^{2} \left (g +f \,x^{n}\right )}\right )\) | \(356\) |
1/n*ln((d+e*x^n)^p)/f^2*ln(g+f*x^n)+1/n*ln((d+e*x^n)^p)*g/f^2/(g+f*x^n)-1/ n*p/f^2*dilog(((g+f*x^n)*e+d*f-e*g)/(d*f-e*g))-1/n*p/f^2*ln(g+f*x^n)*ln((( g+f*x^n)*e+d*f-e*g)/(d*f-e*g))+1/n*p*e/f^2*g/(d*f-e*g)*ln((g+f*x^n)*e+d*f- e*g)-e*g*p*ln(g+f*x^n)/f^2/(d*f-e*g)/n+(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*cs gn(I*c)+ln(c))*(1/n/f^2*ln(g+f*x^n)+1/n*g/f^2/(g+f*x^n))
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (f + \frac {g}{x^{n}}\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.34 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=e n p {\left (\frac {d \log \left (\frac {e x^{n} + d}{e}\right )}{d e f^{2} n^{2} - e^{2} f g n^{2}} - \frac {g \log \left (\frac {f x^{n} + g}{f}\right )}{d f^{3} n^{2} - e f^{2} g n^{2}} - \frac {\log \left (f x^{n} + g\right ) \log \left (\frac {e f x^{n} + e g}{d f - e g} + 1\right ) + {\rm Li}_2\left (-\frac {e f x^{n} + e g}{d f - e g}\right )}{e f^{2} n^{2}}\right )} - {\left (\frac {1}{f^{2} n + \frac {f g n}{x^{n}}} - \frac {\log \left (f + \frac {g}{x^{n}}\right )}{f^{2} n} + \frac {\log \left (\frac {1}{x^{n}}\right )}{f^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right ) \]
e*n*p*(d*log((e*x^n + d)/e)/(d*e*f^2*n^2 - e^2*f*g*n^2) - g*log((f*x^n + g )/f)/(d*f^3*n^2 - e*f^2*g*n^2) - (log(f*x^n + g)*log((e*f*x^n + e*g)/(d*f - e*g) + 1) + dilog(-(e*f*x^n + e*g)/(d*f - e*g)))/(e*f^2*n^2)) - (1/(f^2* n + f*g*n/x^n) - log(f + g/x^n)/(f^2*n) + log(1/(x^n))/(f^2*n))*log((e*x^n + d)^p*c)
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (f + \frac {g}{x^{n}}\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-n}\right )^2} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{x\,{\left (f+\frac {g}{x^n}\right )}^2} \,d x \]