Integrand size = 30, antiderivative size = 255 \[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {2 b k n (g x)^m}{g m^2}-\frac {k (g x)^m \left (a+b \log \left (c x^n\right )\right )}{g m}-\frac {b e k n x^{-m} (g x)^m \log \left (e+f x^m\right )}{f g m^2}-\frac {b e k n x^{-m} (g x)^m \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{f g m^2}+\frac {e k x^{-m} (g x)^m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{f g m}-\frac {b n (g x)^m \log \left (d \left (e+f x^m\right )^k\right )}{g m^2}+\frac {(g x)^m \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{g m}-\frac {b e k n x^{-m} (g x)^m \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )}{f g m^2} \]
2*b*k*n*(g*x)^m/g/m^2-k*(g*x)^m*(a+b*ln(c*x^n))/g/m-b*e*k*n*(g*x)^m*ln(e+f *x^m)/f/g/m^2/(x^m)-b*e*k*n*(g*x)^m*ln(-f*x^m/e)*ln(e+f*x^m)/f/g/m^2/(x^m) +e*k*(g*x)^m*(a+b*ln(c*x^n))*ln(e+f*x^m)/f/g/m/(x^m)-b*n*(g*x)^m*ln(d*(e+f *x^m)^k)/g/m^2+(g*x)^m*(a+b*ln(c*x^n))*ln(d*(e+f*x^m)^k)/g/m-b*e*k*n*(g*x) ^m*polylog(2,1+f*x^m/e)/f/g/m^2/(x^m)
Time = 0.29 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.05 \[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=-\frac {x^{-m} (g x)^m \left (a f k m x^m-2 b f k n x^m+b e k m^2 n \log ^2(x)+b f k m x^m \log \left (c x^n\right )-a e k m \log \left (e-e x^m\right )+b e k n \log \left (e-e x^m\right )-b e k m \log \left (c x^n\right ) \log \left (e-e x^m\right )+b e k n \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )-e k m \log (x) \left (a m-b n+b m \log \left (c x^n\right )-b n \log \left (e-e x^m\right )+b n \log \left (e+f x^m\right )\right )-a f m x^m \log \left (d \left (e+f x^m\right )^k\right )+b f n x^m \log \left (d \left (e+f x^m\right )^k\right )-b f m x^m \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )+b e k n \operatorname {PolyLog}\left (2,1+\frac {f x^m}{e}\right )\right )}{f g m^2} \]
-(((g*x)^m*(a*f*k*m*x^m - 2*b*f*k*n*x^m + b*e*k*m^2*n*Log[x]^2 + b*f*k*m*x ^m*Log[c*x^n] - a*e*k*m*Log[e - e*x^m] + b*e*k*n*Log[e - e*x^m] - b*e*k*m* Log[c*x^n]*Log[e - e*x^m] + b*e*k*n*Log[-((f*x^m)/e)]*Log[e + f*x^m] - e*k *m*Log[x]*(a*m - b*n + b*m*Log[c*x^n] - b*n*Log[e - e*x^m] + b*n*Log[e + f *x^m]) - a*f*m*x^m*Log[d*(e + f*x^m)^k] + b*f*n*x^m*Log[d*(e + f*x^m)^k] - b*f*m*x^m*Log[c*x^n]*Log[d*(e + f*x^m)^k] + b*e*k*n*PolyLog[2, 1 + (f*x^m )/e]))/(f*g*m^2*x^m))
Time = 0.48 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2823, 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.
| \(\displaystyle \int (g x)^{m-1} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {e k (g x)^m \log \left (f x^m+e\right ) x^{-m-1}}{f g m}-\frac {k (g x)^m}{g m x}+\frac {(g x)^m \log \left (d \left (f x^m+e\right )^k\right )}{g m x}\right )dx+\frac {(g x)^m \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{g m}+\frac {e k x^{-m} (g x)^m \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{f g m}-\frac {k (g x)^m \left (a+b \log \left (c x^n\right )\right )}{g m}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(g x)^m \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{g m}+\frac {e k x^{-m} (g x)^m \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{f g m}-\frac {k (g x)^m \left (a+b \log \left (c x^n\right )\right )}{g m}-b n \left (\frac {(g x)^m \log \left (d \left (e+f x^m\right )^k\right )}{g m^2}+\frac {e k x^{-m} (g x)^m \operatorname {PolyLog}\left (2,\frac {f x^m}{e}+1\right )}{f g m^2}+\frac {e k x^{-m} (g x)^m \log \left (e+f x^m\right )}{f g m^2}+\frac {e k x^{-m} (g x)^m \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{f g m^2}-\frac {2 k (g x)^m}{g m^2}\right )\) |
-((k*(g*x)^m*(a + b*Log[c*x^n]))/(g*m)) + (e*k*(g*x)^m*(a + b*Log[c*x^n])* Log[e + f*x^m])/(f*g*m*x^m) + ((g*x)^m*(a + b*Log[c*x^n])*Log[d*(e + f*x^m )^k])/(g*m) - b*n*((-2*k*(g*x)^m)/(g*m^2) + (e*k*(g*x)^m*Log[e + f*x^m])/( f*g*m^2*x^m) + (e*k*(g*x)^m*Log[-((f*x^m)/e)]*Log[e + f*x^m])/(f*g*m^2*x^m ) + ((g*x)^m*Log[d*(e + f*x^m)^k])/(g*m^2) + (e*k*(g*x)^m*PolyLog[2, 1 + ( f*x^m)/e])/(f*g*m^2*x^m))
3.2.52.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
\[\int \left (g x \right )^{m -1} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \,x^{m}\right )^{k}\right )d x\]
Time = 0.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.77 \[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\frac {b e g^{m - 1} k m n \log \left (x\right ) \log \left (\frac {f x^{m} + e}{e}\right ) + b e g^{m - 1} k n {\rm Li}_2\left (-\frac {f x^{m} + e}{e} + 1\right ) - {\left (b f k m \log \left (c\right ) + a f k m - 2 \, b f k n - {\left (b f m \log \left (c\right ) + a f m - b f n\right )} \log \left (d\right ) + {\left (b f k m n - b f m n \log \left (d\right )\right )} \log \left (x\right )\right )} g^{m - 1} x^{m} + {\left ({\left (b f k m n \log \left (x\right ) + b f k m \log \left (c\right ) + a f k m - b f k n\right )} g^{m - 1} x^{m} + {\left (b e k m \log \left (c\right ) + a e k m - b e k n\right )} g^{m - 1}\right )} \log \left (f x^{m} + e\right )}{f m^{2}} \]
(b*e*g^(m - 1)*k*m*n*log(x)*log((f*x^m + e)/e) + b*e*g^(m - 1)*k*n*dilog(- (f*x^m + e)/e + 1) - (b*f*k*m*log(c) + a*f*k*m - 2*b*f*k*n - (b*f*m*log(c) + a*f*m - b*f*n)*log(d) + (b*f*k*m*n - b*f*m*n*log(d))*log(x))*g^(m - 1)* x^m + ((b*f*k*m*n*log(x) + b*f*k*m*log(c) + a*f*k*m - b*f*k*n)*g^(m - 1)*x ^m + (b*e*k*m*log(c) + a*e*k*m - b*e*k*n)*g^(m - 1))*log(f*x^m + e))/(f*m^ 2)
Timed out. \[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\text {Timed out} \]
\[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \]
(b*g^m*m*x^m*log(x^n) + (a*g^m*m + (g^m*m*log(c) - g^m*n)*b)*x^m)*log((f*x ^m + e)^k)/(g*m^2) + integrate(-(((f*g^m*k*m - f*g^m*m*log(d))*a - (f*g^m* k*n - (f*g^m*k*m - f*g^m*m*log(d))*log(c))*b)*x^(2*m) - (b*e*g^m*m*log(c)* log(d) + a*e*g^m*m*log(d))*x^m - (b*e*g^m*m*x^m*log(d) - (f*g^m*k*m - f*g^ m*m*log(d))*b*x^(2*m))*log(x^n))/(f*g*m*x*x^m + e*g*m*x), x)
\[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int (g x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx=\int \ln \left (d\,{\left (e+f\,x^m\right )}^k\right )\,{\left (g\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]