Integrand size = 16, antiderivative size = 300 \[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {355 x}{576 c^3}+\frac {139 x^2}{1152 c^2}+\frac {67 x^3}{1728 c}+\frac {3 x^4}{256}+\frac {139 \log (1-c x)}{576 c^4}-\frac {x^2 \log (1-c x)}{8 c^2}-\frac {5 x^3 \log (1-c x)}{72 c}-\frac {3}{64} x^4 \log (1-c x)+\frac {3 (1-c x) \log (1-c x)}{8 c^4}-\frac {\log ^2(1-c x)}{16 c^4}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {x \operatorname {PolyLog}(2,c x)}{4 c^3}-\frac {x^2 \operatorname {PolyLog}(2,c x)}{8 c^2}-\frac {x^3 \operatorname {PolyLog}(2,c x)}{12 c}-\frac {1}{16} x^4 \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,1-c x)}{2 c^4}+\frac {\operatorname {PolyLog}(3,1-c x)}{2 c^4} \]
355/576*x/c^3+139/1152*x^2/c^2+67/1728*x^3/c+3/256*x^4+139/576*ln(-c*x+1)/ c^4-1/8*x^2*ln(-c*x+1)/c^2-5/72*x^3*ln(-c*x+1)/c-3/64*x^4*ln(-c*x+1)+3/8*( -c*x+1)*ln(-c*x+1)/c^4-1/16*ln(-c*x+1)^2/c^4+1/16*x^4*ln(-c*x+1)^2-1/4*ln( c*x)*ln(-c*x+1)^2/c^4-1/4*x*polylog(2,c*x)/c^3-1/8*x^2*polylog(2,c*x)/c^2- 1/12*x^3*polylog(2,c*x)/c-1/16*x^4*polylog(2,c*x)-1/4*ln(-c*x+1)*polylog(2 ,c*x)/c^4+1/4*x^4*ln(-c*x+1)*polylog(2,c*x)-1/2*ln(-c*x+1)*polylog(2,-c*x+ 1)/c^4+1/2*polylog(3,-c*x+1)/c^4
Time = 0.45 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.74 \[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {4260 c x+834 c^2 x^2+268 c^3 x^3+81 c^4 x^4+4260 \log (1-c x)-2592 c x \log (1-c x)-864 c^2 x^2 \log (1-c x)-480 c^3 x^3 \log (1-c x)-324 c^4 x^4 \log (1-c x)-432 \log ^2(1-c x)+432 c^4 x^4 \log ^2(1-c x)-1728 \log (c x) \log ^2(1-c x)+144 \left (-c x \left (12+6 c x+4 c^2 x^2+3 c^3 x^3\right )+12 \left (-1+c^4 x^4\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)-3456 \log (1-c x) \operatorname {PolyLog}(2,1-c x)+3456 \operatorname {PolyLog}(3,1-c x)}{6912 c^4} \]
(4260*c*x + 834*c^2*x^2 + 268*c^3*x^3 + 81*c^4*x^4 + 4260*Log[1 - c*x] - 2 592*c*x*Log[1 - c*x] - 864*c^2*x^2*Log[1 - c*x] - 480*c^3*x^3*Log[1 - c*x] - 324*c^4*x^4*Log[1 - c*x] - 432*Log[1 - c*x]^2 + 432*c^4*x^4*Log[1 - c*x ]^2 - 1728*Log[c*x]*Log[1 - c*x]^2 + 144*(-(c*x*(12 + 6*c*x + 4*c^2*x^2 + 3*c^3*x^3)) + 12*(-1 + c^4*x^4)*Log[1 - c*x])*PolyLog[2, c*x] - 3456*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 3456*PolyLog[3, 1 - c*x])/(6912*c^4)
Time = 1.00 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.44, 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.
| \(\displaystyle \int x^3 \operatorname {PolyLog}(2,c x) \log (1-c x) \, dx\) |
| \(\Big \downarrow \) 7157 |
| \(\displaystyle \frac {1}{4} c \int \left (-\frac {\operatorname {PolyLog}(2,c x) x^3}{c}-\frac {\operatorname {PolyLog}(2,c x) x^2}{c^2}-\frac {\operatorname {PolyLog}(2,c x) x}{c^3}+\frac {\operatorname {PolyLog}(2,c x)}{c^4 (1-c x)}-\frac {\operatorname {PolyLog}(2,c x)}{c^4}\right )dx+\frac {1}{4} \int x^3 \log ^2(1-c x)dx+\frac {1}{4} x^4 \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \int x^3 \log ^2(1-c x)dx+\frac {1}{4} c \left (\frac {2 \operatorname {PolyLog}(3,1-c x)}{c^5}-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{c^5}-\frac {2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{c^5}-\frac {\log (c x) \log ^2(1-c x)}{c^5}+\frac {(1-c x) \log (1-c x)}{c^5}+\frac {61 \log (1-c x)}{144 c^5}-\frac {x \operatorname {PolyLog}(2,c x)}{c^4}+\frac {205 x}{144 c^4}-\frac {x^2 \operatorname {PolyLog}(2,c x)}{2 c^3}+\frac {61 x^2}{288 c^3}-\frac {x^2 \log (1-c x)}{4 c^3}-\frac {x^3 \operatorname {PolyLog}(2,c x)}{3 c^2}+\frac {25 x^3}{432 c^2}-\frac {x^3 \log (1-c x)}{9 c^2}-\frac {x^4 \operatorname {PolyLog}(2,c x)}{4 c}+\frac {x^4}{64 c}-\frac {x^4 \log (1-c x)}{16 c}\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} c \int \frac {x^4 \log (1-c x)}{1-c x}dx+\frac {1}{4} x^4 \log ^2(1-c x)\right )+\frac {1}{4} c \left (\frac {2 \operatorname {PolyLog}(3,1-c x)}{c^5}-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{c^5}-\frac {2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{c^5}-\frac {\log (c x) \log ^2(1-c x)}{c^5}+\frac {(1-c x) \log (1-c x)}{c^5}+\frac {61 \log (1-c x)}{144 c^5}-\frac {x \operatorname {PolyLog}(2,c x)}{c^4}+\frac {205 x}{144 c^4}-\frac {x^2 \operatorname {PolyLog}(2,c x)}{2 c^3}+\frac {61 x^2}{288 c^3}-\frac {x^2 \log (1-c x)}{4 c^3}-\frac {x^3 \operatorname {PolyLog}(2,c x)}{3 c^2}+\frac {25 x^3}{432 c^2}-\frac {x^3 \log (1-c x)}{9 c^2}-\frac {x^4 \operatorname {PolyLog}(2,c x)}{4 c}+\frac {x^4}{64 c}-\frac {x^4 \log (1-c x)}{16 c}\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2857 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} c \int \left (-\frac {\log (1-c x) x^3}{c}-\frac {\log (1-c x) x^2}{c^2}-\frac {\log (1-c x) x}{c^3}-\frac {\log (1-c x)}{c^4 (c x-1)}-\frac {\log (1-c x)}{c^4}\right )dx+\frac {1}{4} x^4 \log ^2(1-c x)\right )+\frac {1}{4} c \left (\frac {2 \operatorname {PolyLog}(3,1-c x)}{c^5}-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{c^5}-\frac {2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{c^5}-\frac {\log (c x) \log ^2(1-c x)}{c^5}+\frac {(1-c x) \log (1-c x)}{c^5}+\frac {61 \log (1-c x)}{144 c^5}-\frac {x \operatorname {PolyLog}(2,c x)}{c^4}+\frac {205 x}{144 c^4}-\frac {x^2 \operatorname {PolyLog}(2,c x)}{2 c^3}+\frac {61 x^2}{288 c^3}-\frac {x^2 \log (1-c x)}{4 c^3}-\frac {x^3 \operatorname {PolyLog}(2,c x)}{3 c^2}+\frac {25 x^3}{432 c^2}-\frac {x^3 \log (1-c x)}{9 c^2}-\frac {x^4 \operatorname {PolyLog}(2,c x)}{4 c}+\frac {x^4}{64 c}-\frac {x^4 \log (1-c x)}{16 c}\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} c \left (\frac {2 \operatorname {PolyLog}(3,1-c x)}{c^5}-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{c^5}-\frac {2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{c^5}-\frac {\log (c x) \log ^2(1-c x)}{c^5}+\frac {(1-c x) \log (1-c x)}{c^5}+\frac {61 \log (1-c x)}{144 c^5}-\frac {x \operatorname {PolyLog}(2,c x)}{c^4}+\frac {205 x}{144 c^4}-\frac {x^2 \operatorname {PolyLog}(2,c x)}{2 c^3}+\frac {61 x^2}{288 c^3}-\frac {x^2 \log (1-c x)}{4 c^3}-\frac {x^3 \operatorname {PolyLog}(2,c x)}{3 c^2}+\frac {25 x^3}{432 c^2}-\frac {x^3 \log (1-c x)}{9 c^2}-\frac {x^4 \operatorname {PolyLog}(2,c x)}{4 c}+\frac {x^4}{64 c}-\frac {x^4 \log (1-c x)}{16 c}\right )+\frac {1}{4} \left (\frac {1}{2} c \left (-\frac {\log ^2(1-c x)}{2 c^5}+\frac {(1-c x) \log (1-c x)}{c^5}+\frac {13 \log (1-c x)}{12 c^5}+\frac {25 x}{12 c^4}+\frac {13 x^2}{24 c^3}-\frac {x^2 \log (1-c x)}{2 c^3}+\frac {7 x^3}{36 c^2}-\frac {x^3 \log (1-c x)}{3 c^2}+\frac {x^4}{16 c}-\frac {x^4 \log (1-c x)}{4 c}\right )+\frac {1}{4} x^4 \log ^2(1-c x)\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
((x^4*Log[1 - c*x]^2)/4 + (c*((25*x)/(12*c^4) + (13*x^2)/(24*c^3) + (7*x^3 )/(36*c^2) + x^4/(16*c) + (13*Log[1 - c*x])/(12*c^5) - (x^2*Log[1 - c*x])/ (2*c^3) - (x^3*Log[1 - c*x])/(3*c^2) - (x^4*Log[1 - c*x])/(4*c) + ((1 - c* x)*Log[1 - c*x])/c^5 - Log[1 - c*x]^2/(2*c^5)))/2)/4 + (x^4*Log[1 - c*x]*P olyLog[2, c*x])/4 + (c*((205*x)/(144*c^4) + (61*x^2)/(288*c^3) + (25*x^3)/ (432*c^2) + x^4/(64*c) + (61*Log[1 - c*x])/(144*c^5) - (x^2*Log[1 - c*x])/ (4*c^3) - (x^3*Log[1 - c*x])/(9*c^2) - (x^4*Log[1 - c*x])/(16*c) + ((1 - c *x)*Log[1 - c*x])/c^5 - (Log[c*x]*Log[1 - c*x]^2)/c^5 - (x*PolyLog[2, c*x] )/c^4 - (x^2*PolyLog[2, c*x])/(2*c^3) - (x^3*PolyLog[2, c*x])/(3*c^2) - (x ^4*PolyLog[2, c*x])/(4*c) - (Log[1 - c*x]*PolyLog[2, c*x])/c^5 - (2*Log[1 - c*x]*PolyLog[2, 1 - c*x])/c^5 + (2*PolyLog[3, 1 - c*x])/c^5))/4
3.2.61.3.1 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]
\[\int x^{3} \ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )d x\]
\[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { x^{3} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
\[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int x^{3} \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )\, dx \]
Time = 0.22 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.25 \[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {9 \, c^{4} {\left (\frac {3 \, c^{3} x^{4} + 4 \, c^{2} x^{3} + 6 \, c x^{2} + 12 \, x}{c^{4}} + \frac {12 \, \log \left (c x - 1\right )}{c^{5}}\right )} + 24 \, c^{3} {\left (\frac {2 \, c^{2} x^{3} + 3 \, c x^{2} + 6 \, x}{c^{3}} + \frac {6 \, \log \left (c x - 1\right )}{c^{4}}\right )} + 108 \, c^{2} {\left (\frac {c x^{2} + 2 \, x}{c^{2}} + \frac {2 \, \log \left (c x - 1\right )}{c^{3}}\right )} + 432 \, c {\left (\frac {x}{c} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} + \frac {2 \, {\left (27 \, c^{4} x^{4} + 92 \, c^{3} x^{3} + 300 \, c^{2} x^{2} + 1680 \, c x - 72 \, {\left (3 \, c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 12 \, c x + 12 \, \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 12 \, {\left (9 \, c^{4} x^{4} + 14 \, c^{3} x^{3} + 27 \, c^{2} x^{2} + 90 \, c x - 140\right )} \log \left (-c x + 1\right )\right )}}{c} - \frac {1728 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )}}{c}}{6912 \, c^{3}} + \frac {{\left (48 \, c^{4} x^{4} {\rm Li}_2\left (c x\right ) - 3 \, c^{4} x^{4} - 4 \, c^{3} x^{3} - 6 \, c^{2} x^{2} - 12 \, c x + 12 \, {\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{192 \, c^{4}} \]
1/6912*(9*c^4*((3*c^3*x^4 + 4*c^2*x^3 + 6*c*x^2 + 12*x)/c^4 + 12*log(c*x - 1)/c^5) + 24*c^3*((2*c^2*x^3 + 3*c*x^2 + 6*x)/c^3 + 6*log(c*x - 1)/c^4) + 108*c^2*((c*x^2 + 2*x)/c^2 + 2*log(c*x - 1)/c^3) + 432*c*(x/c + log(c*x - 1)/c^2) + 2*(27*c^4*x^4 + 92*c^3*x^3 + 300*c^2*x^2 + 1680*c*x - 72*(3*c^4 *x^4 + 4*c^3*x^3 + 6*c^2*x^2 + 12*c*x + 12*log(-c*x + 1))*dilog(c*x) - 12* (9*c^4*x^4 + 14*c^3*x^3 + 27*c^2*x^2 + 90*c*x - 140)*log(-c*x + 1))/c - 17 28*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog (3, -c*x + 1))/c)/c^3 + 1/192*(48*c^4*x^4*dilog(c*x) - 3*c^4*x^4 - 4*c^3*x ^3 - 6*c^2*x^2 - 12*c*x + 12*(c^4*x^4 - 1)*log(-c*x + 1))*log(-c*x + 1)/c^ 4
\[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { x^{3} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
Timed out. \[ \int x^3 \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int x^3\,\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right ) \,d x \]