Integrand size = 16, antiderivative size = 245 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {7 c^2}{36 x}-\frac {3}{4} c^3 \log (x)+\frac {3}{4} c^3 \log (1-c x)-\frac {7 c \log (1-c x)}{36 x^2}-\frac {5 c^2 \log (1-c x)}{9 x}-\frac {1}{9} c^3 \log ^2(1-c x)+\frac {\log ^2(1-c x)}{9 x^3}+\frac {1}{3} c^3 \log (c x) \log ^2(1-c x)-\frac {2}{9} c^3 \operatorname {PolyLog}(2,c x)+\frac {c \operatorname {PolyLog}(2,c x)}{6 x^2}+\frac {c^2 \operatorname {PolyLog}(2,c x)}{3 x}+\frac {1}{3} c^3 \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{3 x^3}+\frac {2}{3} c^3 \log (1-c x) \operatorname {PolyLog}(2,1-c x)-\frac {1}{3} c^3 \operatorname {PolyLog}(3,c x)-\frac {2}{3} c^3 \operatorname {PolyLog}(3,1-c x) \]
7/36*c^2/x-3/4*c^3*ln(x)+3/4*c^3*ln(-c*x+1)-7/36*c*ln(-c*x+1)/x^2-5/9*c^2* ln(-c*x+1)/x-1/9*c^3*ln(-c*x+1)^2+1/9*ln(-c*x+1)^2/x^3+1/3*c^3*ln(c*x)*ln( -c*x+1)^2-2/9*c^3*polylog(2,c*x)+1/6*c*polylog(2,c*x)/x^2+1/3*c^2*polylog( 2,c*x)/x+1/3*c^3*ln(-c*x+1)*polylog(2,c*x)-1/3*ln(-c*x+1)*polylog(2,c*x)/x ^3+2/3*c^3*ln(-c*x+1)*polylog(2,-c*x+1)-1/3*c^3*polylog(3,c*x)-2/3*c^3*pol ylog(3,-c*x+1)
Time = 0.18 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {7 c^2 x^2-4 c^3 x^3-15 c^3 x^3 \log (x)-12 c^3 x^3 \log (c x)-7 c x \log (1-c x)-20 c^2 x^2 \log (1-c x)+27 c^3 x^3 \log (1-c x)+8 c^3 x^3 \log (c x) \log (1-c x)+4 \log ^2(1-c x)-4 c^3 x^3 \log ^2(1-c x)+12 c^3 x^3 \log (c x) \log ^2(1-c x)+6 \left (c x (1+2 c x)+2 \left (-1+c^3 x^3\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)+8 c^3 x^3 (1+3 \log (1-c x)) \operatorname {PolyLog}(2,1-c x)-12 c^3 x^3 \operatorname {PolyLog}(3,c x)-24 c^3 x^3 \operatorname {PolyLog}(3,1-c x)}{36 x^3} \]
(7*c^2*x^2 - 4*c^3*x^3 - 15*c^3*x^3*Log[x] - 12*c^3*x^3*Log[c*x] - 7*c*x*L og[1 - c*x] - 20*c^2*x^2*Log[1 - c*x] + 27*c^3*x^3*Log[1 - c*x] + 8*c^3*x^ 3*Log[c*x]*Log[1 - c*x] + 4*Log[1 - c*x]^2 - 4*c^3*x^3*Log[1 - c*x]^2 + 12 *c^3*x^3*Log[c*x]*Log[1 - c*x]^2 + 6*(c*x*(1 + 2*c*x) + 2*(-1 + c^3*x^3)*L og[1 - c*x])*PolyLog[2, c*x] + 8*c^3*x^3*(1 + 3*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 12*c^3*x^3*PolyLog[3, c*x] - 24*c^3*x^3*PolyLog[3, 1 - c*x])/(36 *x^3)
Time = 0.79 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.20, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 7157 |
| \(\displaystyle -\frac {1}{3} c \int \left (\frac {\operatorname {PolyLog}(2,c x) c^3}{1-c x}+\frac {\operatorname {PolyLog}(2,c x) c^2}{x}+\frac {\operatorname {PolyLog}(2,c x) c}{x^2}+\frac {\operatorname {PolyLog}(2,c x)}{x^3}\right )dx-\frac {1}{3} \int \frac {\log ^2(1-c x)}{x^4}dx-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{3} \int \frac {\log ^2(1-c x)}{x^4}dx-\frac {1}{3} c \left (c^2 \operatorname {PolyLog}(3,c x)+2 c^2 \operatorname {PolyLog}(3,1-c x)-c^2 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^2 (-\log (c x)) \log ^2(1-c x)+\frac {5}{4} c^2 \log (x)-\frac {5}{4} c^2 \log (1-c x)-\frac {\operatorname {PolyLog}(2,c x)}{2 x^2}-\frac {c \operatorname {PolyLog}(2,c x)}{x}+\frac {\log (1-c x)}{4 x^2}-\frac {c}{4 x}+\frac {c \log (1-c x)}{x}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{3 x^3}\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} c \int \frac {\log (1-c x)}{x^3 (1-c x)}dx+\frac {\log ^2(1-c x)}{3 x^3}\right )-\frac {1}{3} c \left (c^2 \operatorname {PolyLog}(3,c x)+2 c^2 \operatorname {PolyLog}(3,1-c x)-c^2 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^2 (-\log (c x)) \log ^2(1-c x)+\frac {5}{4} c^2 \log (x)-\frac {5}{4} c^2 \log (1-c x)-\frac {\operatorname {PolyLog}(2,c x)}{2 x^2}-\frac {c \operatorname {PolyLog}(2,c x)}{x}+\frac {\log (1-c x)}{4 x^2}-\frac {c}{4 x}+\frac {c \log (1-c x)}{x}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{3 x^3}\) |
| \(\Big \downarrow \) 2857 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{3} c \int \left (-\frac {\log (1-c x) c^3}{c x-1}+\frac {\log (1-c x) c^2}{x}+\frac {\log (1-c x) c}{x^2}+\frac {\log (1-c x)}{x^3}\right )dx+\frac {\log ^2(1-c x)}{3 x^3}\right )-\frac {1}{3} c \left (c^2 \operatorname {PolyLog}(3,c x)+2 c^2 \operatorname {PolyLog}(3,1-c x)-c^2 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^2 (-\log (c x)) \log ^2(1-c x)+\frac {5}{4} c^2 \log (x)-\frac {5}{4} c^2 \log (1-c x)-\frac {\operatorname {PolyLog}(2,c x)}{2 x^2}-\frac {c \operatorname {PolyLog}(2,c x)}{x}+\frac {\log (1-c x)}{4 x^2}-\frac {c}{4 x}+\frac {c \log (1-c x)}{x}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{3} c \left (c^2 \operatorname {PolyLog}(3,c x)+2 c^2 \operatorname {PolyLog}(3,1-c x)-c^2 \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)+c^2 (-\log (c x)) \log ^2(1-c x)+\frac {5}{4} c^2 \log (x)-\frac {5}{4} c^2 \log (1-c x)-\frac {\operatorname {PolyLog}(2,c x)}{2 x^2}-\frac {c \operatorname {PolyLog}(2,c x)}{x}+\frac {\log (1-c x)}{4 x^2}-\frac {c}{4 x}+\frac {c \log (1-c x)}{x}\right )+\frac {1}{3} \left (\frac {2}{3} c \left (-c^2 \operatorname {PolyLog}(2,c x)-\frac {1}{2} c^2 \log ^2(1-c x)-\frac {3}{2} c^2 \log (x)+\frac {3}{2} c^2 \log (1-c x)-\frac {\log (1-c x)}{2 x^2}+\frac {c}{2 x}-\frac {c \log (1-c x)}{x}\right )+\frac {\log ^2(1-c x)}{3 x^3}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{3 x^3}\) |
-1/3*(Log[1 - c*x]*PolyLog[2, c*x])/x^3 + (Log[1 - c*x]^2/(3*x^3) + (2*c*( c/(2*x) - (3*c^2*Log[x])/2 + (3*c^2*Log[1 - c*x])/2 - Log[1 - c*x]/(2*x^2) - (c*Log[1 - c*x])/x - (c^2*Log[1 - c*x]^2)/2 - c^2*PolyLog[2, c*x]))/3)/ 3 - (c*(-1/4*c/x + (5*c^2*Log[x])/4 - (5*c^2*Log[1 - c*x])/4 + Log[1 - c*x ]/(4*x^2) + (c*Log[1 - c*x])/x - c^2*Log[c*x]*Log[1 - c*x]^2 - PolyLog[2, c*x]/(2*x^2) - (c*PolyLog[2, c*x])/x - c^2*Log[1 - c*x]*PolyLog[2, c*x] - 2*c^2*Log[1 - c*x]*PolyLog[2, 1 - c*x] + c^2*PolyLog[3, c*x] + 2*c^2*PolyL og[3, 1 - c*x]))/3
3.2.68.3.1 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]
\[\int \frac {\ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )}{x^{4}}d x\]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{4}} \,d x } \]
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int \frac {\log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{4}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.77 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {1}{3} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} c^{3} + \frac {2}{9} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} c^{3} - \frac {3}{4} \, c^{3} \log \left (x\right ) - \frac {1}{3} \, c^{3} {\rm Li}_{3}(c x) + \frac {7 \, c^{2} x^{2} - 4 \, {\left (c^{3} x^{3} - 1\right )} \log \left (-c x + 1\right )^{2} + 6 \, {\left (2 \, c^{2} x^{2} + c x + 2 \, {\left (c^{3} x^{3} - 1\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) + {\left (27 \, c^{3} x^{3} - 20 \, c^{2} x^{2} - 7 \, c x\right )} \log \left (-c x + 1\right )}{36 \, x^{3}} \]
1/3*(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^3 + 2/9*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*c^3 - 3/4*c^3*log(x) - 1/3*c^3*polylog(3, c*x) + 1/36*(7*c^2*x^2 - 4*(c^3*x^3 - 1)*log(-c*x + 1)^2 + 6*(2*c^2*x^2 + c*x + 2*(c^3*x^3 - 1)*log(-c*x + 1))* dilog(c*x) + (27*c^3*x^3 - 20*c^2*x^2 - 7*c*x)*log(-c*x + 1))/x^3
\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )}{x^4} \,d x \]