Integrand size = 27, antiderivative size = 193 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {e g^2 p x^{-n}}{2 d n}+\frac {2 e f g p \log (x)}{d}-\frac {e^2 g^2 p \log (x)}{2 d^2}-\frac {2 e f g p \log \left (d+e x^n\right )}{d n}+\frac {e^2 g^2 p \log \left (d+e x^n\right )}{2 d^2 n}-\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]
-1/2*e*g^2*p/d/n/(x^n)+2*e*f*g*p*ln(x)/d-1/2*e^2*g^2*p*ln(x)/d^2-2*e*f*g*p *ln(d+e*x^n)/d/n+1/2*e^2*g^2*p*ln(d+e*x^n)/d^2/n-1/2*g^2*ln(c*(d+e*x^n)^p) /n/(x^(2*n))-2*f*g*ln(c*(d+e*x^n)^p)/n/(x^n)+f^2*ln(-e*x^n/d)*ln(c*(d+e*x^ n)^p)/n+f^2*p*polylog(2,1+e*x^n/d)/n
Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.83 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {-4 d e f g n p \log (x)+4 d e f g p \log \left (d+e x^n\right )+e g^2 p \left (d x^{-n}+e n \log (x)-e \log \left (d+e x^n\right )\right )+d^2 g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )+4 d^2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )-2 d^2 f^2 \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )\right )}{2 d^2 n} \]
-1/2*(-4*d*e*f*g*n*p*Log[x] + 4*d*e*f*g*p*Log[d + e*x^n] + e*g^2*p*(d/x^n + e*n*Log[x] - e*Log[d + e*x^n]) + (d^2*g^2*Log[c*(d + e*x^n)^p])/x^(2*n) + (4*d^2*f*g*Log[c*(d + e*x^n)^p])/x^n - 2*d^2*f^2*(Log[-((e*x^n)/d)]*Log[ c*(d + e*x^n)^p] + p*PolyLog[2, 1 + (e*x^n)/d]))/(d^2*n)
Time = 0.43 (sec) , antiderivative size = 180, 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.
| \(\displaystyle \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int x^{-2 n-1} \left (f x^n+g\right )^2 \log \left (c \left (d+e x^n\right )^p\right )dx\) |
| \(\Big \downarrow \) 2925 |
| \(\displaystyle \frac {\int x^{-3 n} \left (f x^n+g\right )^2 \log \left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \frac {\int \left (g^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{-3 n}+2 f g \log \left (c \left (e x^n+d\right )^p\right ) x^{-2 n}+f^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{-n}\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )-2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {1}{2} g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {e^2 g^2 p \log \left (x^n\right )}{2 d^2}+\frac {e^2 g^2 p \log \left (d+e x^n\right )}{2 d^2}+f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )+\frac {2 e f g p \log \left (x^n\right )}{d}-\frac {2 e f g p \log \left (d+e x^n\right )}{d}-\frac {e g^2 p x^{-n}}{2 d}}{n}\) |
(-1/2*(e*g^2*p)/(d*x^n) + (2*e*f*g*p*Log[x^n])/d - (e^2*g^2*p*Log[x^n])/(2 *d^2) - (2*e*f*g*p*Log[d + e*x^n])/d + (e^2*g^2*p*Log[d + e*x^n])/(2*d^2) - (g^2*Log[c*(d + e*x^n)^p])/(2*x^(2*n)) - (2*f*g*Log[c*(d + e*x^n)^p])/x^ n + f^2*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + f^2*p*PolyLog[2, 1 + (e*x ^n)/d])/n
3.4.68.3.1 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 = 6.57 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.73
| method | result | size |
| risch | \(\frac {\left (2 f^{2} \ln \left (x \right ) n \,x^{2 n}-4 f g \,x^{n}-g^{2}\right ) x^{-2 n} \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-2 f g \,x^{-n}+f^{2} \ln \left (x^{n}\right )-\frac {g^{2} x^{-2 n}}{2}\right )}{n}-\frac {2 e f g p \ln \left (d +e \,x^{n}\right )}{d n}+\frac {2 p e f g \ln \left (x^{n}\right )}{n d}+\frac {e^{2} g^{2} p \ln \left (d +e \,x^{n}\right )}{2 d^{2} n}-\frac {e \,g^{2} p \,x^{-n}}{2 d n}-\frac {p \,e^{2} g^{2} \ln \left (x^{n}\right )}{2 n \,d^{2}}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\) | \(334\) |
1/2*(2*f^2*ln(x)*n*(x^n)^2-4*f*g*x^n-g^2)/n/(x^n)^2*ln((d+e*x^n)^p)+(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))/n*(-2*f*g/(x^n)+f^2*ln(x^n)-1/ 2*g^2/(x^n)^2)-2*e*f*g*p*ln(d+e*x^n)/d/n+2*p*e/n*f*g/d*ln(x^n)+1/2*e^2*g^2 *p*ln(d+e*x^n)/d^2/n-1/2*e*g^2*p/d/n/(x^n)-1/2*p*e^2/n*g^2/d^2*ln(x^n)-p/n *f^2*dilog((d+e*x^n)/d)-p*f^2*ln(x)*ln((d+e*x^n)/d)
Time = 0.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.09 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {2 \, d^{2} f^{2} n p x^{2 \, n} \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + 2 \, d^{2} f^{2} p x^{2 \, n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + d^{2} g^{2} \log \left (c\right ) - {\left (2 \, d^{2} f^{2} n \log \left (c\right ) + {\left (4 \, d e f g - e^{2} g^{2}\right )} n p\right )} x^{2 \, n} \log \left (x\right ) + {\left (d e g^{2} p + 4 \, d^{2} f g \log \left (c\right )\right )} x^{n} + {\left (4 \, d^{2} f g p x^{n} + d^{2} g^{2} p - {\left (2 \, d^{2} f^{2} n p \log \left (x\right ) - {\left (4 \, d e f g - e^{2} g^{2}\right )} p\right )} x^{2 \, n}\right )} \log \left (e x^{n} + d\right )}{2 \, d^{2} n x^{2 \, n}} \]
-1/2*(2*d^2*f^2*n*p*x^(2*n)*log(x)*log((e*x^n + d)/d) + 2*d^2*f^2*p*x^(2*n )*dilog(-(e*x^n + d)/d + 1) + d^2*g^2*log(c) - (2*d^2*f^2*n*log(c) + (4*d* e*f*g - e^2*g^2)*n*p)*x^(2*n)*log(x) + (d*e*g^2*p + 4*d^2*f*g*log(c))*x^n + (4*d^2*f*g*p*x^n + d^2*g^2*p - (2*d^2*f^2*n*p*log(x) - (4*d*e*f*g - e^2* g^2)*p)*x^(2*n))*log(e*x^n + d))/(d^2*n*x^(2*n))
\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {x^{- 2 n} \left (f x^{n} + g\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]
\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
-1/2*(d*g^2*log(c) + (d*f^2*n^2*p*log(x)^2 - 2*d*f^2*n*log(c)*log(x))*x^(2 *n) + (e*g^2*p + 4*d*f*g*log(c))*x^n - (2*d*f^2*n*x^(2*n)*log(x) - 4*d*f*g *x^n - d*g^2)*log((e*x^n + d)^p))/(d*n*x^(2*n)) + integrate(1/2*(2*d^2*f^2 *n*p*log(x) + 4*d*e*f*g*p - e^2*g^2*p)/(d*e*x*x^n + d^2*x), x)
\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+\frac {g}{x^n}\right )}^2}{x} \,d x \]