Integrand size = 25, antiderivative size = 179 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=-\frac {b e n \log (d+e x)}{f (e f-d g)}+\frac {a+b \log \left (c (d+e x)^n\right )}{f (f+g x)}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {b e n \log (f+g x)}{f (e f-d g)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^2}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^2}+\frac {b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \]
-b*e*n*ln(e*x+d)/f/(-d*g+e*f)+(a+b*ln(c*(e*x+d)^n))/f/(g*x+f)+ln(-e*x/d)*( a+b*ln(c*(e*x+d)^n))/f^2+b*e*n*ln(g*x+f)/f/(-d*g+e*f)-(a+b*ln(c*(e*x+d)^n) )*ln(e*(g*x+f)/(-d*g+e*f))/f^2-b*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/f^2+b* n*polylog(2,1+e*x/d)/f^2
Time = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\frac {\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}-\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )+b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \]
((f*(a + b*Log[c*(d + e*x)^n]))/(f + g*x) + Log[-((e*x)/d)]*(a + b*Log[c*( d + e*x)^n]) - (b*e*f*n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g) - (a + b*Log[c*(d + e*x)^n])*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^2
Time = 0.41 (sec) , antiderivative size = 179, 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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 (f+g x)}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f (f+g x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f (f+g x)}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^2}+\frac {b n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}-\frac {b e n \log (d+e x)}{f (e f-d g)}+\frac {b e n \log (f+g x)}{f (e f-d g)}\) |
-((b*e*n*Log[d + e*x])/(f*(e*f - d*g))) + (a + b*Log[c*(d + e*x)^n])/(f*(f + g*x)) + (Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^2 + (b*e*n*Log[f + g*x])/(f*(e*f - d*g)) - ((a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/( e*f - d*g)])/f^2 - (b*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/f^2 + (b *n*PolyLog[2, 1 + (e*x)/d])/f^2
3.3.53.3.1 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 = 0.74 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.98
method | result | size |
risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f \left (g x +f \right )}-\frac {b n \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {b e n \ln \left (e x +d \right )}{f \left (d g -e f \right )}-\frac {b e n \ln \left (g x +f \right )}{f \left (d g -e f \right )}+\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}+\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (x \right )}{f^{2}}-\frac {\ln \left (g x +f \right )}{f^{2}}+\frac {1}{f \left (g x +f \right )}\right )\) | \(354\) |
b*ln((e*x+d)^n)/f^2*ln(x)-b*ln((e*x+d)^n)/f^2*ln(g*x+f)+b*ln((e*x+d)^n)/f/ (g*x+f)-b*n/f^2*dilog((e*x+d)/d)-b*n/f^2*ln(x)*ln((e*x+d)/d)+b*e*n/f/(d*g- e*f)*ln(e*x+d)-b*e*n/f/(d*g-e*f)*ln(g*x+f)+b*n/f^2*dilog(((g*x+f)*e+d*g-e* f)/(d*g-e*f))+b*n/f^2*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^2*ln(x)-1/f^2*ln(g*x+f)+1/ f/(g*x+f))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x} \,d x } \]
a*(1/(f*g*x + f^2) - log(g*x + f)/f^2 + log(x)/f^2) + b*integrate((log((e* x + d)^n) + log(c))/(g^2*x^3 + 2*f*g*x^2 + f^2*x), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x\,{\left (f+g\,x\right )}^2} \,d x \]