Integrand size = 27, antiderivative size = 254 \[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {d f g p x^n}{e n}+\frac {d^3 g^2 p x^n}{4 e^3 n}-\frac {f g p x^{2 n}}{2 n}-\frac {d^2 g^2 p x^{2 n}}{8 e^2 n}+\frac {d g^2 p x^{3 n}}{12 e n}-\frac {g^2 p x^{4 n}}{16 n}-\frac {d^2 f g p \log \left (d+e x^n\right )}{e^2 n}-\frac {d^4 g^2 p \log \left (d+e x^n\right )}{4 e^4 n}+\frac {f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]
d*f*g*p*x^n/e/n+1/4*d^3*g^2*p*x^n/e^3/n-1/2*f*g*p*x^(2*n)/n-1/8*d^2*g^2*p* x^(2*n)/e^2/n+1/12*d*g^2*p*x^(3*n)/e/n-1/16*g^2*p*x^(4*n)/n-d^2*f*g*p*ln(d +e*x^n)/e^2/n-1/4*d^4*g^2*p*ln(d+e*x^n)/e^4/n+f*g*x^(2*n)*ln(c*(d+e*x^n)^p )/n+1/4*g^2*x^(4*n)*ln(c*(d+e*x^n)^p)/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 = 0.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.67 \[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {-e g p x^n \left (-12 d^3 g+6 d^2 e g x^n+3 e^3 x^n \left (8 f+g x^{2 n}\right )-4 d e^2 \left (12 f+g x^{2 n}\right )\right )-12 d^2 g \left (4 e^2 f+d^2 g\right ) p \log \left (d+e x^n\right )+12 e^4 \left (g x^{2 n} \left (4 f+g x^{2 n}\right )+4 f^2 \log \left (-\frac {e x^n}{d}\right )\right ) \log \left (c \left (d+e x^n\right )^p\right )+48 e^4 f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{48 e^4 n} \]
(-(e*g*p*x^n*(-12*d^3*g + 6*d^2*e*g*x^n + 3*e^3*x^n*(8*f + g*x^(2*n)) - 4* d*e^2*(12*f + g*x^(2*n)))) - 12*d^2*g*(4*e^2*f + d^2*g)*p*Log[d + e*x^n] + 12*e^4*(g*x^(2*n)*(4*f + g*x^(2*n)) + 4*f^2*Log[-((e*x^n)/d)])*Log[c*(d + e*x^n)^p] + 48*e^4*f^2*p*PolyLog[2, 1 + (e*x^n)/d])/(48*e^4*n)
Time = 0.43 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
| \(\displaystyle \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx\) |
| \(\Big \downarrow \) 2925 |
| \(\displaystyle \frac {\int x^{-n} \left (g x^{2 n}+f\right )^2 \log \left (c \left (e x^n+d\right )^p\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \frac {\int \left (f^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{-n}+2 f g \log \left (c \left (e x^n+d\right )^p\right ) x^n+g^2 \log \left (c \left (e x^n+d\right )^p\right ) x^{3 n}\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )+\frac {1}{4} g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {d^4 g^2 p \log \left (d+e x^n\right )}{4 e^4}+\frac {d^3 g^2 p x^n}{4 e^3}-\frac {d^2 f g p \log \left (d+e x^n\right )}{e^2}-\frac {d^2 g^2 p x^{2 n}}{8 e^2}+f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )+\frac {d f g p x^n}{e}+\frac {d g^2 p x^{3 n}}{12 e}-\frac {1}{2} f g p x^{2 n}-\frac {1}{16} g^2 p x^{4 n}}{n}\) |
((d*f*g*p*x^n)/e + (d^3*g^2*p*x^n)/(4*e^3) - (f*g*p*x^(2*n))/2 - (d^2*g^2* p*x^(2*n))/(8*e^2) + (d*g^2*p*x^(3*n))/(12*e) - (g^2*p*x^(4*n))/16 - (d^2* f*g*p*Log[d + e*x^n])/e^2 - (d^4*g^2*p*Log[d + e*x^n])/(4*e^4) + f*g*x^(2* n)*Log[c*(d + e*x^n)^p] + (g^2*x^(4*n)*Log[c*(d + e*x^n)^p])/4 + 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
3.4.66.3.1 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 = 5.57 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.48
| method | result | size |
| risch | \(\frac {\left (g^{2} x^{4 n}+4 f^{2} \ln \left (x \right ) n +4 f g \,x^{2 n}\right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{4 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {g^{2} x^{4 n}}{4}+f g \,x^{2 n}+f^{2} \ln \left (x^{n}\right )\right )}{n}-\frac {g^{2} p \,x^{4 n}}{16 n}+\frac {d \,g^{2} p \,x^{3 n}}{12 e n}-\frac {d^{2} g^{2} p \,x^{2 n}}{8 e^{2} n}+\frac {d^{3} g^{2} p \,x^{n}}{4 e^{3} n}-\frac {d^{4} g^{2} p \ln \left (d +e \,x^{n}\right )}{4 e^{4} n}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )-\frac {f g p \,x^{2 n}}{2 n}+\frac {d f g p \,x^{n}}{e n}-\frac {d^{2} f g p \ln \left (d +e \,x^{n}\right )}{e^{2} n}\) | \(375\) |
1/4*(g^2*(x^n)^4+4*f^2*ln(x)*n+4*f*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)*c sgn(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*c sgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)+ln(c))/n*(1/4*g^2*(x^n)^4+f*g*(x^n)^2+f^2 *ln(x^n))-1/16*p/n*g^2*(x^n)^4+1/12*p/e/n*g^2*d*(x^n)^3-1/8*p/e^2/n*g^2*d^ 2*(x^n)^2+1/4*d^3*g^2*p*x^n/e^3/n-1/4*d^4*g^2*p*ln(d+e*x^n)/e^4/n-p/n*f^2* dilog((d+e*x^n)/d)-p*f^2*ln(x)*ln((d+e*x^n)/d)-1/2*p/n*f*g*(x^n)^2+d*f*g*p *x^n/e/n-d^2*f*g*p*ln(d+e*x^n)/e^2/n
Time = 0.35 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.95 \[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {48 \, e^{4} f^{2} n p \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) - 48 \, e^{4} f^{2} n \log \left (c\right ) \log \left (x\right ) - 4 \, d e^{3} g^{2} p x^{3 \, n} + 48 \, e^{4} f^{2} p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) - 12 \, {\left (4 \, d e^{3} f g + d^{3} e g^{2}\right )} p x^{n} + 3 \, {\left (e^{4} g^{2} p - 4 \, e^{4} g^{2} \log \left (c\right )\right )} x^{4 \, n} - 6 \, {\left (8 \, e^{4} f g \log \left (c\right ) - {\left (4 \, e^{4} f g + d^{2} e^{2} g^{2}\right )} p\right )} x^{2 \, n} - 12 \, {\left (4 \, e^{4} f^{2} n p \log \left (x\right ) + e^{4} g^{2} p x^{4 \, n} + 4 \, e^{4} f g p x^{2 \, n} - {\left (4 \, d^{2} e^{2} f g + d^{4} g^{2}\right )} p\right )} \log \left (e x^{n} + d\right )}{48 \, e^{4} n} \]
-1/48*(48*e^4*f^2*n*p*log(x)*log((e*x^n + d)/d) - 48*e^4*f^2*n*log(c)*log( x) - 4*d*e^3*g^2*p*x^(3*n) + 48*e^4*f^2*p*dilog(-(e*x^n + d)/d + 1) - 12*( 4*d*e^3*f*g + d^3*e*g^2)*p*x^n + 3*(e^4*g^2*p - 4*e^4*g^2*log(c))*x^(4*n) - 6*(8*e^4*f*g*log(c) - (4*e^4*f*g + d^2*e^2*g^2)*p)*x^(2*n) - 12*(4*e^4*f ^2*n*p*log(x) + e^4*g^2*p*x^(4*n) + 4*e^4*f*g*p*x^(2*n) - (4*d^2*e^2*f*g + d^4*g^2)*p)*log(e*x^n + d))/(e^4*n)
\[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\left (f + g x^{2 n}\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]
\[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{2 \, n} + f\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
-1/48*(24*e^4*f^2*n^2*p*log(x)^2 - 4*d*e^3*g^2*p*x^(3*n) + 3*(e^4*g^2*p - 4*e^4*g^2*log(c))*x^(4*n) + 6*(4*e^4*f*g*p + d^2*e^2*g^2*p - 8*e^4*f*g*log (c))*x^(2*n) - 12*(4*d*e^3*f*g*p + d^3*e*g^2*p)*x^n - 12*(4*e^4*f^2*n*log( x) + e^4*g^2*x^(4*n) + 4*e^4*f*g*x^(2*n))*log((e*x^n + d)^p) + 12*(4*d^2*e ^2*f*g*n*p + d^4*g^2*n*p - 4*e^4*f^2*n*log(c))*log(x))/(e^4*n) + integrate (1/4*(4*d*e^4*f^2*n*p*log(x) + 4*d^3*e^2*f*g*p + d^5*g^2*p)/(e^5*x*x^n + d *e^4*x), x)
\[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (g x^{2 \, n} + f\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+g\,x^{2\,n}\right )}^2}{x} \,d x \]