Integrand size = 25, antiderivative size = 107 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f}+\frac {b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f} \]
ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f-(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d* g+e*f))/f-b*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/f+b*n*polylog(2,1+e*x/d)/f
Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (\log \left (-\frac {e x}{d}\right )-\log \left (\frac {e (f+g x)}{e f-d g}\right )\right )-b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )+b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f} \]
((a + b*Log[c*(d + e*x)^n])*(Log[-((e*x)/d)] - Log[(e*(f + g*x))/(e*f - d* g)]) - b*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)] + b*n*PolyLog[2, 1 + ( e*x)/d])/f
Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f x}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f (f+g x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f}+\frac {b n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f}\) |
(Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f - ((a + b*Log[c*(d + e*x)^n ])*Log[(e*(f + g*x))/(e*f - d*g)])/f - (b*n*PolyLog[2, -((g*(d + e*x))/(e* f - d*g))])/f + (b*n*PolyLog[2, 1 + (e*x)/d])/f
3.3.46.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.64 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.58
method | result | size |
risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{f}-\frac {b n \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f}+\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f}+\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (x \right )}{f}-\frac {\ln \left (g x +f \right )}{f}\right )\) | \(276\) |
b*ln((e*x+d)^n)/f*ln(x)-b*ln((e*x+d)^n)/f*ln(g*x+f)-b*n/f*dilog((e*x+d)/d) -b*n/f*ln(x)*ln((e*x+d)/d)+b*n/f*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+b*n/ f*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+(-1/2*I*b*Pi*csgn(I*c)*csgn( I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^ 2+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*c*( e*x+d)^n)^3+b*ln(c)+a)*(1/f*ln(x)-1/f*ln(g*x+f))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )} x} \,d x } \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x \left (f + g x\right )}\, dx \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )} x} \,d x } \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x\,\left (f+g\,x\right )} \,d x \]