Integrand size = 19, antiderivative size = 65 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {\left (a+b n+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \]
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {\frac {d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-b n (\log (x)-\log (d+e x))+\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \]
((d*(a + b*Log[c*x^n]))/(d + e*x) - b*n*(Log[x] - Log[d + e*x]) + (a + b*L og[c*x^n])*Log[1 + (e*x)/d] + b*n*PolyLog[2, -((e*x)/d)])/e^2
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2784, 2754, 2838}
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.
\(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx\) |
\(\Big \downarrow \) 2784 |
\(\displaystyle \frac {\int \frac {a+b n+b \log \left (c x^n\right )}{d+e x}dx}{e}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}\) |
\(\Big \downarrow \) 2754 |
\(\displaystyle \frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )+b n\right )}{e}-\frac {b n \int \frac {\log \left (\frac {e x}{d}+1\right )}{x}dx}{e}}{e}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )+b n\right )}{e}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e}}{e}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}\) |
-((x*(a + b*Log[c*x^n]))/(e*(d + e*x))) + (((a + b*n + b*Log[c*x^n])*Log[1 + (e*x)/d])/e + (b*n*PolyLog[2, -((e*x)/d)])/e)/e
3.1.41.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symb ol] :> Simp[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^p/e), x] - Simp[b*n*(p/e) Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] )/(e*(q + 1))), x] - Simp[f/(e*(q + 1)) Int[(f*x)^(m - 1)*(d + e*x)^(q + 1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && GtQ[m, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.78 (sec) , antiderivative size = 205, normalized size of antiderivative = 3.15
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{2}}+\frac {b \ln \left (x^{n}\right ) d}{e^{2} \left (e x +d \right )}-\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{2}}-\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{2}}+\frac {b n \ln \left (e x +d \right )}{e^{2}}-\frac {b n \ln \left (e x \right )}{e^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (e x +d \right )}{e^{2}}+\frac {d}{e^{2} \left (e x +d \right )}\right )\) | \(205\) |
b*ln(x^n)/e^2*ln(e*x+d)+b*ln(x^n)/e^2*d/(e*x+d)-b*n/e^2*ln(e*x+d)*ln(-e*x/ d)-b*n/e^2*dilog(-e*x/d)+b*n/e^2*ln(e*x+d)-b*n/e^2*ln(e*x)+(-1/2*I*b*Pi*cs gn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2 *I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)* (1/e^2*ln(e*x+d)+1/e^2*d/(e*x+d))
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{{\left (e x + d\right )}^{2}} \,d x } \]
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int \frac {x \left (a + b \log {\left (c x^{n} \right )}\right )}{\left (d + e x\right )^{2}}\, dx \]
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{{\left (e x + d\right )}^{2}} \,d x } \]
a*(d/(e^3*x + d*e^2) + log(e*x + d)/e^2) + b*integrate((x*log(c) + x*log(x ^n))/(e^2*x^2 + 2*d*e*x + d^2), x)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{{\left (e x + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \]