Integrand size = 19, antiderivative size = 546 \[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {4 b x}{9 c^2}+\frac {a x}{c}+\frac {5 (2 b+3 a c) x}{24 c^2}+\frac {b x^2}{9 c}+\frac {(2 b+3 a c) x^2}{48 c}+\frac {b x^3}{27}+\frac {a (1-c x)^2}{8 c^2}+\frac {2 b \log (1-c x)}{9 c^3}+\frac {(2 b+3 a c) \log (1-c x)}{24 c^3}-\frac {b x^2 \log (1-c x)}{9 c}-\frac {(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac {1}{9} b x^3 \log (1-c x)+\frac {2 b (1-c x) \log (1-c x)}{9 c^3}+\frac {a (1-c x) \log (1-c x)}{c^2}+\frac {(2 b+3 a c) (1-c x) \log (1-c x)}{6 c^3}-\frac {a (1-c x)^2 \log (1-c x)}{4 c^2}-\frac {b \log ^2(1-c x)}{9 c^3}+\frac {1}{9} b x^3 \log ^2(1-c x)-\frac {a (1-c x) \log ^2(1-c x)}{2 c^2}+\frac {a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac {(2 b+3 a c) x \operatorname {PolyLog}(2,c x)}{6 c^2}-\frac {(2 b+3 a c) x^2 \operatorname {PolyLog}(2,c x)}{12 c}-\frac {1}{9} b x^3 \operatorname {PolyLog}(2,c x)-\frac {(2 b+3 a c) \log (1-c x) \operatorname {PolyLog}(2,c x)}{6 c^3}+\frac {1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {(2 b+3 a c) \log (1-c x) \operatorname {PolyLog}(2,1-c x)}{3 c^3}+\frac {(2 b+3 a c) \operatorname {PolyLog}(3,1-c x)}{3 c^3} \]
-1/6*(3*a*c+2*b)*x*polylog(2,c*x)/c^2-1/12*(3*a*c+2*b)*x^2*polylog(2,c*x)/ c-1/9*b*x^2*ln(-c*x+1)/c-1/24*(3*a*c+2*b)*x^2*ln(-c*x+1)/c+2/9*b*(-c*x+1)* ln(-c*x+1)/c^3+1/6*(3*a*c+2*b)*(-c*x+1)*ln(-c*x+1)/c^3-1/4*a*(-c*x+1)^2*ln (-c*x+1)/c^2-1/2*a*(-c*x+1)*ln(-c*x+1)^2/c^2+1/4*a*(-c*x+1)^2*ln(-c*x+1)^2 /c^2-1/6*(3*a*c+2*b)*ln(c*x)*ln(-c*x+1)^2/c^3-1/6*(3*a*c+2*b)*ln(-c*x+1)*p olylog(2,c*x)/c^3-1/3*(3*a*c+2*b)*ln(-c*x+1)*polylog(2,-c*x+1)/c^3+1/27*b* x^3-1/9*b*ln(-c*x+1)^2/c^3+1/9*b*x^3*ln(-c*x+1)^2+1/6*(2*b*x^3+3*a*x^2)*ln (-c*x+1)*polylog(2,c*x)+a*x/c+a*(-c*x+1)*ln(-c*x+1)/c^2+4/9*b*x/c^2+5/24*( 3*a*c+2*b)*x/c^2+1/9*b*x^2/c+1/48*(3*a*c+2*b)*x^2/c+1/8*a*(-c*x+1)^2/c^2-1 /9*b*x^3*polylog(2,c*x)+1/3*(3*a*c+2*b)*polylog(3,-c*x+1)/c^3+2/9*b*ln(-c* x+1)/c^3+1/24*(3*a*c+2*b)*ln(-c*x+1)/c^3-1/9*b*x^3*ln(-c*x+1)
Time = 0.47 (sec) , antiderivative size = 362, normalized size of antiderivative = 0.66 \[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {-378 a c+372 b c x+594 a c^2 x+66 b c^2 x^2+81 a c^3 x^2+16 b c^3 x^3+372 b \log (1-c x)+594 a c \log (1-c x)-240 b c x \log (1-c x)-432 a c^2 x \log (1-c x)-84 b c^2 x^2 \log (1-c x)-162 a c^3 x^2 \log (1-c x)-48 b c^3 x^3 \log (1-c x)-48 b \log ^2(1-c x)-108 a c \log ^2(1-c x)+108 a c^3 x^2 \log ^2(1-c x)+48 b c^3 x^3 \log ^2(1-c x)-144 b \log (c x) \log ^2(1-c x)-216 a c \log (c x) \log ^2(1-c x)+12 \left (-c x \left (9 a c (2+c x)+2 b \left (6+3 c x+2 c^2 x^2\right )\right )+6 \left (3 a c \left (-1+c^2 x^2\right )+2 b \left (-1+c^3 x^3\right )\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)-144 (2 b+3 a c) \log (1-c x) \operatorname {PolyLog}(2,1-c x)+288 b \operatorname {PolyLog}(3,1-c x)+432 a c \operatorname {PolyLog}(3,1-c x)}{432 c^3} \]
(-378*a*c + 372*b*c*x + 594*a*c^2*x + 66*b*c^2*x^2 + 81*a*c^3*x^2 + 16*b*c ^3*x^3 + 372*b*Log[1 - c*x] + 594*a*c*Log[1 - c*x] - 240*b*c*x*Log[1 - c*x ] - 432*a*c^2*x*Log[1 - c*x] - 84*b*c^2*x^2*Log[1 - c*x] - 162*a*c^3*x^2*L og[1 - c*x] - 48*b*c^3*x^3*Log[1 - c*x] - 48*b*Log[1 - c*x]^2 - 108*a*c*Lo g[1 - c*x]^2 + 108*a*c^3*x^2*Log[1 - c*x]^2 + 48*b*c^3*x^3*Log[1 - c*x]^2 - 144*b*Log[c*x]*Log[1 - c*x]^2 - 216*a*c*Log[c*x]*Log[1 - c*x]^2 + 12*(-( c*x*(9*a*c*(2 + c*x) + 2*b*(6 + 3*c*x + 2*c^2*x^2))) + 6*(3*a*c*(-1 + c^2* x^2) + 2*b*(-1 + c^3*x^3))*Log[1 - c*x])*PolyLog[2, c*x] - 144*(2*b + 3*a* c)*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 288*b*PolyLog[3, 1 - c*x] + 432*a*c* PolyLog[3, 1 - c*x])/(432*c^3)
Time = 0.98 (sec) , antiderivative size = 616, normalized size of antiderivative = 1.13, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {7158, 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.
| \(\displaystyle \int x (a+b x) \operatorname {PolyLog}(2,c x) \log (1-c x) \, dx\) |
| \(\Big \downarrow \) 7158 |
| \(\displaystyle c \int \left (-\frac {b \operatorname {PolyLog}(2,c x) x^2}{3 c}-\frac {(2 b+3 a c) \operatorname {PolyLog}(2,c x) x}{6 c^2}-\frac {(2 b+3 a c) \operatorname {PolyLog}(2,c x)}{6 c^3}+\frac {(2 b+3 a c) \operatorname {PolyLog}(2,c x)}{6 c^3 (1-c x)}\right )dx+\int \left (\frac {1}{3} b x^2 \log ^2(1-c x)+\frac {1}{2} a x \log ^2(1-c x)\right )dx+\frac {1}{6} \left (3 a x^2+2 b x^3\right ) \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle c \left (\frac {(3 a c+2 b) \operatorname {PolyLog}(3,1-c x)}{3 c^4}-\frac {(3 a c+2 b) \operatorname {PolyLog}(2,c x) \log (1-c x)}{6 c^4}-\frac {(3 a c+2 b) \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{3 c^4}-\frac {(3 a c+2 b) \log (c x) \log ^2(1-c x)}{6 c^4}+\frac {(3 a c+2 b) \log (1-c x)}{24 c^4}+\frac {(1-c x) (3 a c+2 b) \log (1-c x)}{6 c^4}-\frac {x (3 a c+2 b) \operatorname {PolyLog}(2,c x)}{6 c^3}+\frac {5 x (3 a c+2 b)}{24 c^3}-\frac {x^2 (3 a c+2 b) \operatorname {PolyLog}(2,c x)}{12 c^2}+\frac {x^2 (3 a c+2 b)}{48 c^2}-\frac {x^2 (3 a c+2 b) \log (1-c x)}{24 c^2}+\frac {b \log (1-c x)}{27 c^4}+\frac {b x}{27 c^3}+\frac {b x^2}{54 c^2}-\frac {b x^3 \operatorname {PolyLog}(2,c x)}{9 c}+\frac {b x^3}{81 c}-\frac {b x^3 \log (1-c x)}{27 c}\right )+\frac {1}{6} \left (3 a x^2+2 b x^3\right ) \operatorname {PolyLog}(2,c x) \log (1-c x)+\frac {a (1-c x)^2}{8 c^2}+\frac {a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x)^2 \log (1-c x)}{4 c^2}+\frac {a (1-c x) \log (1-c x)}{c^2}+\frac {a x}{c}-\frac {b \log ^2(1-c x)}{9 c^3}+\frac {2 b (1-c x) \log (1-c x)}{9 c^3}+\frac {5 b \log (1-c x)}{27 c^3}+\frac {11 b x}{27 c^2}+\frac {1}{9} b x^3 \log ^2(1-c x)-\frac {2}{27} b x^3 \log (1-c x)+\frac {5 b x^2}{54 c}-\frac {b x^2 \log (1-c x)}{9 c}+\frac {2 b x^3}{81}\) |
(11*b*x)/(27*c^2) + (a*x)/c + (5*b*x^2)/(54*c) + (2*b*x^3)/81 + (a*(1 - c* x)^2)/(8*c^2) + (5*b*Log[1 - c*x])/(27*c^3) - (b*x^2*Log[1 - c*x])/(9*c) - (2*b*x^3*Log[1 - c*x])/27 + (2*b*(1 - c*x)*Log[1 - c*x])/(9*c^3) + (a*(1 - c*x)*Log[1 - c*x])/c^2 - (a*(1 - c*x)^2*Log[1 - c*x])/(4*c^2) - (b*Log[1 - c*x]^2)/(9*c^3) + (b*x^3*Log[1 - c*x]^2)/9 - (a*(1 - c*x)*Log[1 - c*x]^ 2)/(2*c^2) + (a*(1 - c*x)^2*Log[1 - c*x]^2)/(4*c^2) + ((3*a*x^2 + 2*b*x^3) *Log[1 - c*x]*PolyLog[2, c*x])/6 + c*((b*x)/(27*c^3) + (5*(2*b + 3*a*c)*x) /(24*c^3) + (b*x^2)/(54*c^2) + ((2*b + 3*a*c)*x^2)/(48*c^2) + (b*x^3)/(81* c) + (b*Log[1 - c*x])/(27*c^4) + ((2*b + 3*a*c)*Log[1 - c*x])/(24*c^4) - ( (2*b + 3*a*c)*x^2*Log[1 - c*x])/(24*c^2) - (b*x^3*Log[1 - c*x])/(27*c) + ( (2*b + 3*a*c)*(1 - c*x)*Log[1 - c*x])/(6*c^4) - ((2*b + 3*a*c)*Log[c*x]*Lo g[1 - c*x]^2)/(6*c^4) - ((2*b + 3*a*c)*x*PolyLog[2, c*x])/(6*c^3) - ((2*b + 3*a*c)*x^2*PolyLog[2, c*x])/(12*c^2) - (b*x^3*PolyLog[2, c*x])/(9*c) - ( (2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, c*x])/(6*c^4) - ((2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x])/(3*c^4) + ((2*b + 3*a*c)*PolyLog[3, 1 - c*x]) /(3*c^4))
3.2.85.3.1 Defintions of rubi rules used
Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(Px_)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{u = IntHide[Px, x]}, Simp[u *(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)], x] + (Simp[b Int[Exp andIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], u/(a + b*x), x], x], x] - Simp[e*h*n Int[ExpandIntegrand[PolyLog[2, c*(a + b*x)], u/(d + e*x), x], x], x])] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && PolyQ[Px, x]
\[\int x \left (b x +a \right ) \ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )d x\]
\[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { {\left (b x + a\right )} x {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
\[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int x \left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )\, dx \]
Time = 0.21 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.63 \[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=-\frac {1}{432} \, c {\left (\frac {72 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} {\left (3 \, a c + 2 \, b\right )}}{c^{4}} - \frac {16 \, b c^{3} x^{3} + 3 \, {\left (27 \, a c^{3} + 22 \, b c^{2}\right )} x^{2} + 6 \, {\left (99 \, a c^{2} + 62 \, b c\right )} x - 12 \, {\left (4 \, b c^{3} x^{3} + 3 \, {\left (3 \, a c^{3} + 2 \, b c^{2}\right )} x^{2} + 6 \, {\left (3 \, a c^{2} + 2 \, b c\right )} x + 6 \, {\left (3 \, a c + 2 \, b\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 2 \, {\left (16 \, b c^{3} x^{3} + 6 \, {\left (9 \, a c^{3} + 5 \, b c^{2}\right )} x^{2} - 297 \, a c + 6 \, {\left (27 \, a c^{2} + 16 \, b c\right )} x - 186 \, b\right )} \log \left (-c x + 1\right )}{c^{4}}\right )} + \frac {1}{216} \, {\left (\frac {27 \, {\left (4 \, c^{2} x^{2} {\rm Li}_2\left (c x\right ) - c^{2} x^{2} - 2 \, c x + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} a}{c^{2}} + \frac {4 \, {\left (18 \, c^{3} x^{3} {\rm Li}_2\left (c x\right ) - 2 \, c^{3} x^{3} - 3 \, c^{2} x^{2} - 6 \, c x + 6 \, {\left (c^{3} x^{3} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{3}}\right )} \log \left (-c x + 1\right ) \]
-1/432*c*(72*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))*(3*a*c + 2*b)/c^4 - (16*b*c^3*x^3 + 3*(27*a*c^3 + 22*b*c^2)*x^2 + 6*(99*a*c^2 + 62*b*c)*x - 12*(4*b*c^3*x^3 + 3*(3*a*c^3 + 2*b*c^2)*x^2 + 6*(3*a*c^2 + 2*b*c)*x + 6*(3*a*c + 2*b)*log(-c*x + 1))*dilo g(c*x) - 2*(16*b*c^3*x^3 + 6*(9*a*c^3 + 5*b*c^2)*x^2 - 297*a*c + 6*(27*a*c ^2 + 16*b*c)*x - 186*b)*log(-c*x + 1))/c^4) + 1/216*(27*(4*c^2*x^2*dilog(c *x) - c^2*x^2 - 2*c*x + 2*(c^2*x^2 - 1)*log(-c*x + 1))*a/c^2 + 4*(18*c^3*x ^3*dilog(c*x) - 2*c^3*x^3 - 3*c^2*x^2 - 6*c*x + 6*(c^3*x^3 - 1)*log(-c*x + 1))*b/c^3)*log(-c*x + 1)
\[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { {\left (b x + a\right )} x {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
Timed out. \[ \int x (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int x\,\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right ) \,d x \]