Integrand size = 16, antiderivative size = 287 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\frac {5 c^2}{144 x^2}+\frac {7 c^3}{36 x}-\frac {41}{72} c^4 \log (x)+\frac {41}{72} c^4 \log (1-c x)-\frac {5 c \log (1-c x)}{72 x^3}-\frac {c^2 \log (1-c x)}{8 x^2}-\frac {3 c^3 \log (1-c x)}{8 x}-\frac {1}{16} c^4 \log ^2(1-c x)+\frac {\log ^2(1-c x)}{16 x^4}+\frac {1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac {1}{8} c^4 \operatorname {PolyLog}(2,c x)+\frac {c \operatorname {PolyLog}(2,c x)}{12 x^3}+\frac {c^2 \operatorname {PolyLog}(2,c x)}{8 x^2}+\frac {c^3 \operatorname {PolyLog}(2,c x)}{4 x}+\frac {1}{4} c^4 \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{4 x^4}+\frac {1}{2} c^4 \log (1-c x) \operatorname {PolyLog}(2,1-c x)-\frac {1}{4} c^4 \operatorname {PolyLog}(3,c x)-\frac {1}{2} c^4 \operatorname {PolyLog}(3,1-c x) \]
5/144*c^2/x^2+7/36*c^3/x-41/72*c^4*ln(x)+41/72*c^4*ln(-c*x+1)-5/72*c*ln(-c *x+1)/x^3-1/8*c^2*ln(-c*x+1)/x^2-3/8*c^3*ln(-c*x+1)/x-1/16*c^4*ln(-c*x+1)^ 2+1/16*ln(-c*x+1)^2/x^4+1/4*c^4*ln(c*x)*ln(-c*x+1)^2-1/8*c^4*polylog(2,c*x )+1/12*c*polylog(2,c*x)/x^3+1/8*c^2*polylog(2,c*x)/x^2+1/4*c^3*polylog(2,c *x)/x+1/4*c^4*ln(-c*x+1)*polylog(2,c*x)-1/4*ln(-c*x+1)*polylog(2,c*x)/x^4+ 1/2*c^4*ln(-c*x+1)*polylog(2,-c*x+1)-1/4*c^4*polylog(3,c*x)-1/2*c^4*polylo g(3,-c*x+1)
Time = 0.19 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.97 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\frac {5 c^2 x^2+28 c^3 x^3-18 c^4 x^4-49 c^4 x^4 \log (x)-33 c^4 x^4 \log (c x)-10 c x \log (1-c x)-18 c^2 x^2 \log (1-c x)-54 c^3 x^3 \log (1-c x)+82 c^4 x^4 \log (1-c x)+18 c^4 x^4 \log (c x) \log (1-c x)+9 \log ^2(1-c x)-9 c^4 x^4 \log ^2(1-c x)+36 c^4 x^4 \log (c x) \log ^2(1-c x)+6 \left (c x \left (2+3 c x+6 c^2 x^2\right )+6 \left (-1+c^4 x^4\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)+18 c^4 x^4 (1+4 \log (1-c x)) \operatorname {PolyLog}(2,1-c x)-36 c^4 x^4 \operatorname {PolyLog}(3,c x)-72 c^4 x^4 \operatorname {PolyLog}(3,1-c x)}{144 x^4} \]
(5*c^2*x^2 + 28*c^3*x^3 - 18*c^4*x^4 - 49*c^4*x^4*Log[x] - 33*c^4*x^4*Log[ c*x] - 10*c*x*Log[1 - c*x] - 18*c^2*x^2*Log[1 - c*x] - 54*c^3*x^3*Log[1 - c*x] + 82*c^4*x^4*Log[1 - c*x] + 18*c^4*x^4*Log[c*x]*Log[1 - c*x] + 9*Log[ 1 - c*x]^2 - 9*c^4*x^4*Log[1 - c*x]^2 + 36*c^4*x^4*Log[c*x]*Log[1 - c*x]^2 + 6*(c*x*(2 + 3*c*x + 6*c^2*x^2) + 6*(-1 + c^4*x^4)*Log[1 - c*x])*PolyLog [2, c*x] + 18*c^4*x^4*(1 + 4*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 36*c^4*x^ 4*PolyLog[3, c*x] - 72*c^4*x^4*PolyLog[3, 1 - c*x])/(144*x^4)
Time = 0.93 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {7157, 2009, 2845, 2857, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x^5} \, dx\) |
| \(\Big \downarrow \) 7157 |
| \(\displaystyle -\frac {1}{4} c \int \left (\frac {\operatorname {PolyLog}(2,c x) c^4}{1-c x}+\frac {\operatorname {PolyLog}(2,c x) c^3}{x}+\frac {\operatorname {PolyLog}(2,c x) c^2}{x^2}+\frac {\operatorname {PolyLog}(2,c x) c}{x^3}+\frac {\operatorname {PolyLog}(2,c x)}{x^4}\right )dx-\frac {1}{4} \int \frac {\log ^2(1-c x)}{x^5}dx-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{4 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} \int \frac {\log ^2(1-c x)}{x^5}dx-\frac {1}{4} c \left (c^3 \operatorname {PolyLog}(3,c x)+2 c^3 \operatorname {PolyLog}(3,1-c x)-c^3 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^3 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^3 (-\log (c x)) \log ^2(1-c x)+\frac {49}{36} c^3 \log (x)-\frac {49}{36} c^3 \log (1-c x)-\frac {c^2 \operatorname {PolyLog}(2,c x)}{x}-\frac {13 c^2}{36 x}+\frac {c^2 \log (1-c x)}{x}-\frac {\operatorname {PolyLog}(2,c x)}{3 x^3}-\frac {c \operatorname {PolyLog}(2,c x)}{2 x^2}+\frac {\log (1-c x)}{9 x^3}-\frac {c}{18 x^2}+\frac {c \log (1-c x)}{4 x^2}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{4 x^4}\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} c \int \frac {\log (1-c x)}{x^4 (1-c x)}dx+\frac {\log ^2(1-c x)}{4 x^4}\right )-\frac {1}{4} c \left (c^3 \operatorname {PolyLog}(3,c x)+2 c^3 \operatorname {PolyLog}(3,1-c x)-c^3 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^3 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^3 (-\log (c x)) \log ^2(1-c x)+\frac {49}{36} c^3 \log (x)-\frac {49}{36} c^3 \log (1-c x)-\frac {c^2 \operatorname {PolyLog}(2,c x)}{x}-\frac {13 c^2}{36 x}+\frac {c^2 \log (1-c x)}{x}-\frac {\operatorname {PolyLog}(2,c x)}{3 x^3}-\frac {c \operatorname {PolyLog}(2,c x)}{2 x^2}+\frac {\log (1-c x)}{9 x^3}-\frac {c}{18 x^2}+\frac {c \log (1-c x)}{4 x^2}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{4 x^4}\) |
| \(\Big \downarrow \) 2857 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} c \int \left (-\frac {\log (1-c x) c^4}{c x-1}+\frac {\log (1-c x) c^3}{x}+\frac {\log (1-c x) c^2}{x^2}+\frac {\log (1-c x) c}{x^3}+\frac {\log (1-c x)}{x^4}\right )dx+\frac {\log ^2(1-c x)}{4 x^4}\right )-\frac {1}{4} c \left (c^3 \operatorname {PolyLog}(3,c x)+2 c^3 \operatorname {PolyLog}(3,1-c x)-c^3 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^3 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^3 (-\log (c x)) \log ^2(1-c x)+\frac {49}{36} c^3 \log (x)-\frac {49}{36} c^3 \log (1-c x)-\frac {c^2 \operatorname {PolyLog}(2,c x)}{x}-\frac {13 c^2}{36 x}+\frac {c^2 \log (1-c x)}{x}-\frac {\operatorname {PolyLog}(2,c x)}{3 x^3}-\frac {c \operatorname {PolyLog}(2,c x)}{2 x^2}+\frac {\log (1-c x)}{9 x^3}-\frac {c}{18 x^2}+\frac {c \log (1-c x)}{4 x^2}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{4 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} c \left (c^3 \operatorname {PolyLog}(3,c x)+2 c^3 \operatorname {PolyLog}(3,1-c x)-c^3 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^3 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^3 (-\log (c x)) \log ^2(1-c x)+\frac {49}{36} c^3 \log (x)-\frac {49}{36} c^3 \log (1-c x)-\frac {c^2 \operatorname {PolyLog}(2,c x)}{x}-\frac {13 c^2}{36 x}+\frac {c^2 \log (1-c x)}{x}-\frac {\operatorname {PolyLog}(2,c x)}{3 x^3}-\frac {c \operatorname {PolyLog}(2,c x)}{2 x^2}+\frac {\log (1-c x)}{9 x^3}-\frac {c}{18 x^2}+\frac {c \log (1-c x)}{4 x^2}\right )+\frac {1}{4} \left (\frac {1}{2} c \left (-c^3 \operatorname {PolyLog}(2,c x)-\frac {1}{2} c^3 \log ^2(1-c x)-\frac {11}{6} c^3 \log (x)+\frac {11}{6} c^3 \log (1-c x)+\frac {5 c^2}{6 x}-\frac {c^2 \log (1-c x)}{x}-\frac {\log (1-c x)}{3 x^3}+\frac {c}{6 x^2}-\frac {c \log (1-c x)}{2 x^2}\right )+\frac {\log ^2(1-c x)}{4 x^4}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{4 x^4}\) |
-1/4*(Log[1 - c*x]*PolyLog[2, c*x])/x^4 + (Log[1 - c*x]^2/(4*x^4) + (c*(c/ (6*x^2) + (5*c^2)/(6*x) - (11*c^3*Log[x])/6 + (11*c^3*Log[1 - c*x])/6 - Lo g[1 - c*x]/(3*x^3) - (c*Log[1 - c*x])/(2*x^2) - (c^2*Log[1 - c*x])/x - (c^ 3*Log[1 - c*x]^2)/2 - c^3*PolyLog[2, c*x]))/2)/4 - (c*(-1/18*c/x^2 - (13*c ^2)/(36*x) + (49*c^3*Log[x])/36 - (49*c^3*Log[1 - c*x])/36 + Log[1 - c*x]/ (9*x^3) + (c*Log[1 - c*x])/(4*x^2) + (c^2*Log[1 - c*x])/x - c^3*Log[c*x]*L og[1 - c*x]^2 - PolyLog[2, c*x]/(3*x^3) - (c*PolyLog[2, c*x])/(2*x^2) - (c ^2*PolyLog[2, c*x])/x - c^3*Log[1 - c*x]*PolyLog[2, c*x] - 2*c^3*Log[1 - c *x]*PolyLog[2, 1 - c*x] + c^3*PolyLog[3, c*x] + 2*c^3*PolyLog[3, 1 - c*x]) )/4
3.2.69.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^ n])^p/(g*(q + 1))), x] - Simp[b*e*n*(p/(g*(q + 1))) Int[(f + g*x)^(q + 1) *((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && In tegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))
Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symb ol] :> Int[ExpandIntegrand[Log[c*(d + e*x)], x^m/(f + g*x), x], x] /; FreeQ [{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m]
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]
\[\int \frac {\ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )}{x^{5}}d x\]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{5}} \,d x } \]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\int \frac {\log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{5}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.75 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\frac {1}{4} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} c^{4} + \frac {1}{8} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} c^{4} - \frac {41}{72} \, c^{4} \log \left (x\right ) - \frac {1}{4} \, c^{4} {\rm Li}_{3}(c x) + \frac {28 \, c^{3} x^{3} + 5 \, c^{2} x^{2} - 9 \, {\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )^{2} + 6 \, {\left (6 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + 2 \, c x + 6 \, {\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) + 2 \, {\left (41 \, c^{4} x^{4} - 27 \, c^{3} x^{3} - 9 \, c^{2} x^{2} - 5 \, c x\right )} \log \left (-c x + 1\right )}{144 \, x^{4}} \]
1/4*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylo g(3, -c*x + 1))*c^4 + 1/8*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*c^4 - 41/72*c^4*log(x) - 1/4*c^4*polylog(3, c*x) + 1/144*(28*c^3*x^3 + 5*c^2*x^ 2 - 9*(c^4*x^4 - 1)*log(-c*x + 1)^2 + 6*(6*c^3*x^3 + 3*c^2*x^2 + 2*c*x + 6 *(c^4*x^4 - 1)*log(-c*x + 1))*dilog(c*x) + 2*(41*c^4*x^4 - 27*c^3*x^3 - 9* c^2*x^2 - 5*c*x)*log(-c*x + 1))/x^4
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^5} \, dx=\int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )}{x^5} \,d x \]