Integrand size = 20, antiderivative size = 107 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=b e n \log (x)-\frac {1}{2} b e n \log ^2(x)+e \log (x) \left (a+b \log \left (c x^n\right )\right )-b e n \log (1+e x)-\frac {b n \log (1+e x)}{x}-e \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x}-b e n \operatorname {PolyLog}(2,-e x) \]
b*e*n*ln(x)-1/2*b*e*n*ln(x)^2+e*ln(x)*(a+b*ln(c*x^n))-b*e*n*ln(e*x+1)-b*n* ln(e*x+1)/x-e*(a+b*ln(c*x^n))*ln(e*x+1)-(a+b*ln(c*x^n))*ln(e*x+1)/x-b*e*n* polylog(2,-e*x)
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=-\frac {1}{2} b e n \log ^2(x)+e \log (x) \left (a+b n+b \log \left (c x^n\right )\right )-\frac {(1+e x) \left (a+b n+b \log \left (c x^n\right )\right ) \log (1+e x)}{x}-b e n \operatorname {PolyLog}(2,-e x) \]
-1/2*(b*e*n*Log[x]^2) + e*Log[x]*(a + b*n + b*Log[c*x^n]) - ((1 + e*x)*(a + b*n + b*Log[c*x^n])*Log[1 + e*x])/x - b*e*n*PolyLog[2, -(e*x)]
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int \frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (\frac {e \log (x)}{x}-\frac {e \log (e x+1)}{x}-\frac {\log (e x+1)}{x^2}\right )dx+e \log (x) \left (a+b \log \left (c x^n\right )\right )-e \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e \log (x) \left (a+b \log \left (c x^n\right )\right )-e \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x}-b n \left (e \operatorname {PolyLog}(2,-e x)+\frac {1}{2} e \log ^2(x)-e \log (x)+e \log (e x+1)+\frac {\log (e x+1)}{x}\right )\) |
e*Log[x]*(a + b*Log[c*x^n]) - e*(a + b*Log[c*x^n])*Log[1 + e*x] - ((a + b* Log[c*x^n])*Log[1 + e*x])/x - b*n*(-(e*Log[x]) + (e*Log[x]^2)/2 + e*Log[1 + e*x] + Log[1 + e*x]/x + e*PolyLog[2, -(e*x)])
3.1.7.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.07 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.88
method | result | size |
risch | \(\left (-\frac {b \ln \left (e x +1\right )}{x}-b e \ln \left (e x +1\right )+b e \ln \left (x \right )\right ) \ln \left (x^{n}\right )-\frac {b e n \ln \left (x \right )^{2}}{2}-n b e \operatorname {dilog}\left (e x +1\right )+n b e \ln \left (e x \right )-b e n \ln \left (e x +1\right )-\frac {b n \ln \left (e x +1\right )}{x}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) e \left (\ln \left (e x \right )-\frac {\ln \left (e x +1\right ) \left (e x +1\right )}{x e}\right )\) | \(201\) |
(-b/x*ln(e*x+1)-b*e*ln(e*x+1)+b*e*ln(x))*ln(x^n)-1/2*b*e*n*ln(x)^2-n*b*e*d ilog(e*x+1)+n*b*e*ln(e*x)-b*e*n*ln(e*x+1)-b*n*ln(e*x+1)/x+(-1/2*I*b*Pi*csg n(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)*e *(ln(e*x)-ln(e*x+1)/x/e*(e*x+1))
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{2}} \,d x } \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \log {\left (e x + 1 \right )}}{x^{2}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=-{\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b e n - {\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (e x + 1\right ) + {\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (x\right ) - \frac {b e n x \log \left (x\right )^{2} - 2 \, {\left (b e n x \log \left (x\right ) - b {\left (n + \log \left (c\right )\right )} - a\right )} \log \left (e x + 1\right ) - 2 \, {\left (b e x \log \left (x\right ) - {\left (b e x + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{2 \, x} \]
-(log(e*x + 1)*log(x) + dilog(-e*x))*b*e*n - ((e*n + e*log(c))*b + a*e)*lo g(e*x + 1) + ((e*n + e*log(c))*b + a*e)*log(x) - 1/2*(b*e*n*x*log(x)^2 - 2 *(b*e*n*x*log(x) - b*(n + log(c)) - a)*log(e*x + 1) - 2*(b*e*x*log(x) - (b *e*x + b)*log(e*x + 1))*log(x^n))/x
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \]