Integrand size = 20, antiderivative size = 107 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=b e n \log (x)-\frac {1}{2} b e n \log ^2(x)+e \log (x) \left (a+b \log \left (c x^n\right )\right )-b e n \log (1+e x)-\frac {b n \log (1+e x)}{x}-e \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x}-b e n \operatorname {PolyLog}(2,-e x) \] Output:
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 = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=-\frac {1}{2} b e n \log ^2(x)+e \log (x) \left (a+b n+b \log \left (c x^n\right )\right )-\frac {(1+e x) \left (a+b n+b \log \left (c x^n\right )\right ) \log (1+e x)}{x}-b e n \operatorname {PolyLog}(2,-e x) \] Input:
Integrate[((a + b*Log[c*x^n])*Log[1 + e*x])/x^2,x]
Output:
-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 = 0.30 (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 |
| \(\displaystyle e \log (x) \left (a+b \log \left (c x^n\right )\right )-e \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x}-b n \left (e \operatorname {PolyLog}(2,-e x)+\frac {1}{2} e \log ^2(x)-e \log (x)+e \log (e x+1)+\frac {\log (e x+1)}{x}\right )\) |
Input:
Int[((a + b*Log[c*x^n])*Log[1 + e*x])/x^2,x]
Output:
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)])
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 = 2.17 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.88
| method | result | size |
| risch | \(\left (-\frac {b \ln \left (e x +1\right )}{x}+b e \ln \left (x \right )-b e \ln \left (e x +1\right )\right ) \ln \left (x^{n}\right )-\frac {b e n \ln \left (x \right )^{2}}{2}-n b e \operatorname {dilog}\left (e x +1\right )+n b e \ln \left (e x \right )-b e n \ln \left (e x +1\right )-\frac {b n \ln \left (e x +1\right )}{x}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) e \left (\ln \left (e x \right )-\frac {\ln \left (e x +1\right ) \left (e x +1\right )}{x e}\right )\) | \(201\) |
Input:
int((a+b*ln(c*x^n))*ln(e*x+1)/x^2,x,method=_RETURNVERBOSE)
Output:
(-b/x*ln(e*x+1)+b*e*ln(x)-b*e*ln(e*x+1))*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*Pi*b*csgn (I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2 *I*Pi*b*csgn(I*c*x^n)^3+1/2*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)*e* (ln(e*x)-ln(e*x+1)/x/e*(e*x+1))
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^2,x, algorithm="fricas")
Output:
integral((b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1))/x^2, x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \log {\left (e x + 1 \right )}}{x^{2}}\, dx \] Input:
integrate((a+b*ln(c*x**n))*ln(e*x+1)/x**2,x)
Output:
Integral((a + b*log(c*x**n))*log(e*x + 1)/x**2, x)
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=-{\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b e n - {\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (e x + 1\right ) + {\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (x\right ) - \frac {b e n x \log \left (x\right )^{2} - 2 \, {\left (b e n x \log \left (x\right ) - b {\left (n + \log \left (c\right )\right )} - a\right )} \log \left (e x + 1\right ) - 2 \, {\left (b e x \log \left (x\right ) - {\left (b e x + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{2 \, x} \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^2,x, algorithm="maxima")
Output:
-(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
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^2,x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*log(e*x + 1)/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \] Input:
int((log(e*x + 1)*(a + b*log(c*x^n)))/x^2,x)
Output:
int((log(e*x + 1)*(a + b*log(c*x^n)))/x^2, x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^2} \, dx=\frac {-\left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{3}+x^{2}}d x \right ) b x -\mathrm {log}\left (e x +1\right ) \mathrm {log}\left (x^{n} c \right ) b -\mathrm {log}\left (e x +1\right ) a e x -\mathrm {log}\left (e x +1\right ) a -\mathrm {log}\left (e x +1\right ) b e n x -\mathrm {log}\left (e x +1\right ) b n -\mathrm {log}\left (x^{n} c \right ) b +\mathrm {log}\left (x \right ) a e x +\mathrm {log}\left (x \right ) b e n x -b n}{x} \] Input:
int((a+b*log(c*x^n))*log(e*x+1)/x^2,x)
Output:
( - int(log(x**n*c)/(e*x**3 + x**2),x)*b*x - log(e*x + 1)*log(x**n*c)*b - log(e*x + 1)*a*e*x - log(e*x + 1)*a - log(e*x + 1)*b*e*n*x - log(e*x + 1)* b*n - log(x**n*c)*b + log(x)*a*e*x + log(x)*b*e*n*x - b*n)/x