Integrand size = 16, antiderivative size = 111 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=\frac {(1-c x) \log ^2(1-c x)}{x}+c \log (c x) \log ^2(1-c x)-2 c \operatorname {PolyLog}(2,c x)+c \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x}+2 c \log (1-c x) \operatorname {PolyLog}(2,1-c x)-c \operatorname {PolyLog}(3,c x)-2 c \operatorname {PolyLog}(3,1-c x) \]
(-c*x+1)*ln(-c*x+1)^2/x+c*ln(c*x)*ln(-c*x+1)^2-2*c*polylog(2,c*x)+c*ln(-c* x+1)*polylog(2,c*x)-ln(-c*x+1)*polylog(2,c*x)/x+2*c*ln(-c*x+1)*polylog(2,- c*x+1)-c*polylog(3,c*x)-2*c*polylog(3,-c*x+1)
Time = 0.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=2 c \log (c x) \log (1-c x)-c \log ^2(1-c x)+\frac {\log ^2(1-c x)}{x}+c \log (c x) \log ^2(1-c x)+\frac {(-1+c x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x}+2 c (1+\log (1-c x)) \operatorname {PolyLog}(2,1-c x)-c \operatorname {PolyLog}(3,c x)-2 c \operatorname {PolyLog}(3,1-c x) \]
2*c*Log[c*x]*Log[1 - c*x] - c*Log[1 - c*x]^2 + Log[1 - c*x]^2/x + c*Log[c* x]*Log[1 - c*x]^2 + ((-1 + c*x)*Log[1 - c*x]*PolyLog[2, c*x])/x + 2*c*(1 + Log[1 - c*x])*PolyLog[2, 1 - c*x] - c*PolyLog[3, c*x] - 2*c*PolyLog[3, 1 - c*x]
Time = 0.47 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {7157, 2009, 2844, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x^2} \, dx\) |
\(\Big \downarrow \) 7157 |
\(\displaystyle -c \int \left (\frac {\operatorname {PolyLog}(2,c x)}{x}+\frac {c \operatorname {PolyLog}(2,c x)}{1-c x}\right )dx-\int \frac {\log ^2(1-c x)}{x^2}dx-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log ^2(1-c x)}{x^2}dx-c \left (\operatorname {PolyLog}(3,c x)+2 \operatorname {PolyLog}(3,1-c x)-\operatorname {PolyLog}(2,c x) \log (1-c x)-2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)-\log (c x) \log ^2(1-c x)\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x}\) |
\(\Big \downarrow \) 2844 |
\(\displaystyle 2 c \int \frac {\log (1-c x)}{x}dx-c \left (\operatorname {PolyLog}(3,c x)+2 \operatorname {PolyLog}(3,1-c x)-\operatorname {PolyLog}(2,c x) \log (1-c x)-2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)-\log (c x) \log ^2(1-c x)\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x}+\frac {(1-c x) \log ^2(1-c x)}{x}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 c \operatorname {PolyLog}(2,c x)-c \left (\operatorname {PolyLog}(3,c x)+2 \operatorname {PolyLog}(3,1-c x)-\operatorname {PolyLog}(2,c x) \log (1-c x)-2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)-\log (c x) \log ^2(1-c x)\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x}+\frac {(1-c x) \log ^2(1-c x)}{x}\) |
((1 - c*x)*Log[1 - c*x]^2)/x - 2*c*PolyLog[2, c*x] - (Log[1 - c*x]*PolyLog [2, c*x])/x - c*(-(Log[c*x]*Log[1 - c*x]^2) - Log[1 - c*x]*PolyLog[2, c*x] - 2*Log[1 - c*x]*PolyLog[2, 1 - c*x] + PolyLog[3, c*x] + 2*PolyLog[3, 1 - c*x])
3.2.66.3.1 Defintions of rubi rules used
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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. )*(x_))^2, x_Symbol] :> Simp[(d + e*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f - d*g)*(f + g*x))), x] - Simp[b*e*n*(p/(e*f - d*g)) Int[(a + b*Log[c*(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] & & NeQ[e*f - d*g, 0] && GtQ[p, 0]
Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(x_)^(m_.)*PolyLo g[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> Simp[x^(m + 1)*(g + h*Log[f* (d + e*x)^n])*(PolyLog[2, c*(a + b*x)]/(m + 1)), x] + (Simp[b/(m + 1) Int [ExpandIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], x^(m + 1) /(a + b*x), x], x], x] - Simp[e*h*(n/(m + 1)) Int[ExpandIntegrand[PolyLog [2, c*(a + b*x)], x^(m + 1)/(d + e*x), x], x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && IntegerQ[m] && NeQ[m, -1]
\[\int \frac {\ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )}{x^{2}}d x\]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{2}} \,d x } \]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=\int \frac {\log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{2}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx={\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} c + 2 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} c - c {\rm Li}_{3}(c x) + \frac {{\left (c x - 1\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) - {\left (c x - 1\right )} \log \left (-c x + 1\right )^{2}}{x} \]
(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))*c + 2*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*c - c*polylog (3, c*x) + ((c*x - 1)*dilog(c*x)*log(-c*x + 1) - (c*x - 1)*log(-c*x + 1)^2 )/x
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^2} \, dx=\int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )}{x^2} \,d x \]