Integrand size = 25, antiderivative size = 124 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {d g p x^n}{2 e n}-\frac {g p x^{2 n}}{4 n}-\frac {d^2 g p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]
1/2*d*g*p*x^n/e/n-1/4*g*p*x^(2*n)/n-1/2*d^2*g*p*ln(d+e*x^n)/e^2/n+1/2*g*x^ (2*n)*ln(c*(d+e*x^n)^p)/n+f*ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/n+f*p*polylog(2 ,1+e*x^n/d)/n
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {-e g p x^n \left (-2 d+e x^n\right )-2 d^2 g p \log \left (d+e x^n\right )+2 e^2 \left (g x^{2 n}+2 f \log \left (-\frac {e x^n}{d}\right )\right ) \log \left (c \left (d+e x^n\right )^p\right )+4 e^2 f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{4 e^2 n} \]
(-(e*g*p*x^n*(-2*d + e*x^n)) - 2*d^2*g*p*Log[d + e*x^n] + 2*e^2*(g*x^(2*n) + 2*f*Log[-((e*x^n)/d)])*Log[c*(d + e*x^n)^p] + 4*e^2*f*p*PolyLog[2, 1 + (e*x^n)/d])/(4*e^2*n)
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89, 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.
\(\displaystyle \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {\int x^{-n} \left (g x^{2 n}+f\right ) \log \left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \frac {\int \left (f \log \left (c \left (e x^n+d\right )^p\right ) x^{-n}+g \log \left (c \left (e x^n+d\right )^p\right ) x^n\right )dx^n}{n}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+\frac {1}{2} g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {d^2 g p \log \left (d+e x^n\right )}{2 e^2}+f p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )+\frac {d g p x^n}{2 e}-\frac {1}{4} g p x^{2 n}}{n}\) |
((d*g*p*x^n)/(2*e) - (g*p*x^(2*n))/4 - (d^2*g*p*Log[d + e*x^n])/(2*e^2) + (g*x^(2*n)*Log[c*(d + e*x^n)^p])/2 + f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n) ^p] + f*p*PolyLog[2, 1 + (e*x^n)/d])/n
3.4.61.3.1 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 = 2.67 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.01
method | result | size |
risch | \(\frac {\left (2 f \ln \left (x \right ) n +g \,x^{2 n}\right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (f \ln \left (x \right )+\frac {g \,x^{2 n}}{2 n}\right )-\frac {g p \,x^{2 n}}{4 n}+\frac {d g p \,x^{n}}{2 e n}-\frac {d^{2} g p \ln \left (d +e \,x^{n}\right )}{2 e^{2} n}-\frac {p f \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p f \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\) | \(249\) |
1/2*(2*f*ln(x)*n+g*(x^n)^2)/n*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))*(f*ln(x)+1/2*g*(x^n)^2/n)-1/4*p/n*g*(x^n)^2+1/2*d*g* p*x^n/e/n-1/2*d^2*g*p*ln(d+e*x^n)/e^2/n-p/n*f*dilog((d+e*x^n)/d)-p*f*ln(x) *ln((d+e*x^n)/d)
Time = 0.38 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {4 \, e^{2} f n p \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) - 4 \, e^{2} f n \log \left (c\right ) \log \left (x\right ) - 2 \, d e g p x^{n} + 4 \, e^{2} f p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + {\left (e^{2} g p - 2 \, e^{2} g \log \left (c\right )\right )} x^{2 \, n} - 2 \, {\left (2 \, e^{2} f n p \log \left (x\right ) + e^{2} g p x^{2 \, n} - d^{2} g p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]
-1/4*(4*e^2*f*n*p*log(x)*log((e*x^n + d)/d) - 4*e^2*f*n*log(c)*log(x) - 2* d*e*g*p*x^n + 4*e^2*f*p*dilog(-(e*x^n + d)/d + 1) + (e^2*g*p - 2*e^2*g*log (c))*x^(2*n) - 2*(2*e^2*f*n*p*log(x) + e^2*g*p*x^(2*n) - d^2*g*p)*log(e*x^ n + d))/(e^2*n)
\[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\left (f + g x^{2 n}\right ) \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]
\[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{2 \, n} + f\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
-1/4*(2*e^2*f*n^2*p*log(x)^2 - 2*d*e*g*p*x^n + (e^2*g*p - 2*e^2*g*log(c))* x^(2*n) - 2*(2*e^2*f*n*log(x) + e^2*g*x^(2*n))*log((e*x^n + d)^p) + 2*(d^2 *g*n*p - 2*e^2*f*n*log(c))*log(x))/(e^2*n) + integrate(1/2*(2*d*e^2*f*n*p* log(x) + d^3*g*p)/(e^3*x*x^n + d*e^2*x), x)
\[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{2 \, n} + f\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,\left (f+g\,x^{2\,n}\right )}{x} \,d x \]