Integrand size = 27, antiderivative size = 257 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {e g^2 p x^{-3 n}}{12 d n}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}-\frac {e f g p x^{-n}}{d n}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}-\frac {e^2 f g p \log (x)}{d^2}-\frac {e^4 g^2 p \log (x)}{4 d^4}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]
-1/12*e*g^2*p/d/n/(x^(3*n))+1/8*e^2*g^2*p/d^2/n/(x^(2*n))-e*f*g*p/d/n/(x^n )-1/4*e^3*g^2*p/d^3/n/(x^n)-e^2*f*g*p*ln(x)/d^2-1/4*e^4*g^2*p*ln(x)/d^4+e^ 2*f*g*p*ln(d+e*x^n)/d^2/n+1/4*e^4*g^2*p*ln(d+e*x^n)/d^4/n-1/4*g^2*ln(c*(d+ e*x^n)^p)/n/(x^(4*n))-f*g*ln(c*(d+e*x^n)^p)/n/(x^(2*n))+f^2*ln(-e*x^n/d)*l n(c*(d+e*x^n)^p)/n+f^2*p*polylog(2,1+e*x^n/d)/n
Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.73 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {\frac {24 e f g p \left (d x^{-n}+e n \log (x)-e \log \left (d+e x^n\right )\right )}{d^2}+\frac {e g^2 p \left (d x^{-3 n} \left (2 d^2-3 d e x^n+6 e^2 x^{2 n}\right )+6 e^3 n \log (x)-6 e^3 \log \left (d+e x^n\right )\right )}{d^4}+6 g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )+24 f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )-24 f^2 \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )\right )}{24 n} \]
-1/24*((24*e*f*g*p*(d/x^n + e*n*Log[x] - e*Log[d + e*x^n]))/d^2 + (e*g^2*p *((d*(2*d^2 - 3*d*e*x^n + 6*e^2*x^(2*n)))/x^(3*n) + 6*e^3*n*Log[x] - 6*e^3 *Log[d + e*x^n]))/d^4 + (6*g^2*Log[c*(d + e*x^n)^p])/x^(4*n) + (24*f*g*Log [c*(d + e*x^n)^p])/x^(2*n) - 24*f^2*(Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p ] + p*PolyLog[2, 1 + (e*x^n)/d]))/n
Time = 0.48 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.91, 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.
| \(\displaystyle \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int x^{-4 n-1} \left (f x^{2 n}+g\right )^2 \log \left (c \left (d+e x^n\right )^p\right )dx\) |
| \(\Big \downarrow \) 2925 |
| \(\displaystyle \frac {\int x^{-5 n} \left (f x^{2 n}+g\right )^2 \log \left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \frac {\int \left (g^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{-5 n}+2 f g \log \left (c \left (e x^n+d\right )^p\right ) x^{-3 n}+f^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{-n}\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )-f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {1}{4} g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {e^4 g^2 p \log \left (x^n\right )}{4 d^4}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4}-\frac {e^3 g^2 p x^{-n}}{4 d^3}-\frac {e^2 f g p \log \left (x^n\right )}{d^2}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2}+f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )-\frac {e f g p x^{-n}}{d}-\frac {e g^2 p x^{-3 n}}{12 d}}{n}\) |
(-1/12*(e*g^2*p)/(d*x^(3*n)) + (e^2*g^2*p)/(8*d^2*x^(2*n)) - (e*f*g*p)/(d* x^n) - (e^3*g^2*p)/(4*d^3*x^n) - (e^2*f*g*p*Log[x^n])/d^2 - (e^4*g^2*p*Log [x^n])/(4*d^4) + (e^2*f*g*p*Log[d + e*x^n])/d^2 + (e^4*g^2*p*Log[d + e*x^n ])/(4*d^4) - (g^2*Log[c*(d + e*x^n)^p])/(4*x^(4*n)) - (f*g*Log[c*(d + e*x^ n)^p])/x^(2*n) + f^2*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + f^2*p*PolyLo g[2, 1 + (e*x^n)/d])/n
3.4.69.3.1 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 = 6.13 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.54
| method | result | size |
| risch | \(\frac {\left (4 f^{2} \ln \left (x \right ) n \,x^{4 n}-4 f g \,x^{2 n}-g^{2}\right ) x^{-4 n} \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{4 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (f^{2} \ln \left (x^{n}\right )-f g \,x^{-2 n}-\frac {g^{2} x^{-4 n}}{4}\right )}{n}+\frac {e^{4} g^{2} p \ln \left (d +e \,x^{n}\right )}{4 d^{4} n}-\frac {p e \,g^{2} x^{-3 n}}{12 n d}-\frac {e^{3} g^{2} p \,x^{-n}}{4 d^{3} n}+\frac {p \,e^{2} g^{2} x^{-2 n}}{8 n \,d^{2}}-\frac {p \,e^{4} g^{2} \ln \left (x^{n}\right )}{4 n \,d^{4}}+\frac {e^{2} f g p \ln \left (d +e \,x^{n}\right )}{d^{2} n}-\frac {e f g p \,x^{-n}}{d n}-\frac {p \,e^{2} f g \ln \left (x^{n}\right )}{n \,d^{2}}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\) | \(396\) |
1/4*(4*f^2*ln(x)*n*(x^n)^4-4*f*g*(x^n)^2-g^2)/n/(x^n)^4*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*(f^2*ln(x^n)-f*g/(x^n)^2 -1/4*g^2/(x^n)^4)+1/4*e^4*g^2*p*ln(d+e*x^n)/d^4/n-1/12*p*e/n*g^2/d/(x^n)^3 -1/4*e^3*g^2*p/d^3/n/(x^n)+1/8*p*e^2/n*g^2/d^2/(x^n)^2-1/4*p*e^4/n*g^2/d^4 *ln(x^n)+e^2*f*g*p*ln(d+e*x^n)/d^2/n-e*f*g*p/d/n/(x^n)-p*e^2/n*f*g/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 = 0.38 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.03 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {24 \, d^{4} f^{2} n p x^{4 \, n} \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + 24 \, d^{4} f^{2} p x^{4 \, n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + 2 \, d^{3} e g^{2} p x^{n} + 6 \, d^{4} g^{2} \log \left (c\right ) + 6 \, {\left (4 \, d^{3} e f g + d e^{3} g^{2}\right )} p x^{3 \, n} - 6 \, {\left (4 \, d^{4} f^{2} n \log \left (c\right ) - {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} n p\right )} x^{4 \, n} \log \left (x\right ) - 3 \, {\left (d^{2} e^{2} g^{2} p - 8 \, d^{4} f g \log \left (c\right )\right )} x^{2 \, n} + 6 \, {\left (4 \, d^{4} f g p x^{2 \, n} + d^{4} g^{2} p - {\left (4 \, d^{4} f^{2} n p \log \left (x\right ) + {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} p\right )} x^{4 \, n}\right )} \log \left (e x^{n} + d\right )}{24 \, d^{4} n x^{4 \, n}} \]
-1/24*(24*d^4*f^2*n*p*x^(4*n)*log(x)*log((e*x^n + d)/d) + 24*d^4*f^2*p*x^( 4*n)*dilog(-(e*x^n + d)/d + 1) + 2*d^3*e*g^2*p*x^n + 6*d^4*g^2*log(c) + 6* (4*d^3*e*f*g + d*e^3*g^2)*p*x^(3*n) - 6*(4*d^4*f^2*n*log(c) - (4*d^2*e^2*f *g + e^4*g^2)*n*p)*x^(4*n)*log(x) - 3*(d^2*e^2*g^2*p - 8*d^4*f*g*log(c))*x ^(2*n) + 6*(4*d^4*f*g*p*x^(2*n) + d^4*g^2*p - (4*d^4*f^2*n*p*log(x) + (4*d ^2*e^2*f*g + e^4*g^2)*p)*x^(4*n))*log(e*x^n + d))/(d^4*n*x^(4*n))
\[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {x^{- 4 n} \left (f x^{2 n} + g\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]
\[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
-1/24*(2*d^2*e*g^2*p*x^n + 6*d^3*g^2*log(c) + 12*(d^3*f^2*n^2*p*log(x)^2 - 2*d^3*f^2*n*log(c)*log(x))*x^(4*n) + 6*(4*d^2*e*f*g*p + e^3*g^2*p)*x^(3*n ) - 3*(d*e^2*g^2*p - 8*d^3*f*g*log(c))*x^(2*n) - 6*(4*d^3*f^2*n*x^(4*n)*lo g(x) - 4*d^3*f*g*x^(2*n) - d^3*g^2)*log((e*x^n + d)^p))/(d^3*n*x^(4*n)) + integrate(1/4*(4*d^4*f^2*n*p*log(x) - 4*d^2*e^2*f*g*p - e^4*g^2*p)/(d^3*e* x*x^n + d^4*x), x)
\[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+\frac {g}{x^{2\,n}}\right )}^2}{x} \,d x \]