Integrand size = 18, antiderivative size = 390 \[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=2 a x+\frac {9 b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b (1-c x)^2}{8 c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \operatorname {PolyLog}(2,c x)}{2 c}-\frac {1}{4} b x^2 \operatorname {PolyLog}(2,c x)-\frac {(b+2 a c) \log (1-c x) \operatorname {PolyLog}(2,c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {(b+2 a c) \log (1-c x) \operatorname {PolyLog}(2,1-c x)}{c^2}+\frac {(b+2 a c) \operatorname {PolyLog}(3,1-c x)}{c^2} \]
2*a*x+9/8*b*x/c+1/2*(2*a*c+b)*x/c+1/16*b*x^2+1/8*b*(-c*x+1)^2/c^2+1/8*b*ln (-c*x+1)/c^2-1/8*b*x^2*ln(-c*x+1)+b*(-c*x+1)*ln(-c*x+1)/c^2+2*a*(-c*x+1)*l n(-c*x+1)/c+1/2*(2*a*c+b)*(-c*x+1)*ln(-c*x+1)/c^2-1/4*b*(-c*x+1)^2*ln(-c*x +1)/c^2-1/2*b*(-c*x+1)*ln(-c*x+1)^2/c^2-a*(-c*x+1)*ln(-c*x+1)^2/c+1/4*b*(- c*x+1)^2*ln(-c*x+1)^2/c^2-1/2*(2*a*c+b)*ln(c*x)*ln(-c*x+1)^2/c^2-1/2*(2*a* c+b)*x*polylog(2,c*x)/c-1/4*b*x^2*polylog(2,c*x)-1/2*(2*a*c+b)*ln(-c*x+1)* polylog(2,c*x)/c^2+1/2*(b*x^2+2*a*x)*ln(-c*x+1)*polylog(2,c*x)-(2*a*c+b)*l n(-c*x+1)*polylog(2,-c*x+1)/c^2+(2*a*c+b)*polylog(3,-c*x+1)/c^2
Time = 0.38 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.73 \[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\frac {-14 b-32 a c+22 b c x+48 a c^2 x+3 b c^2 x^2+22 b \log (1-c x)+48 a c \log (1-c x)-16 b c x \log (1-c x)-48 a c^2 x \log (1-c x)-6 b c^2 x^2 \log (1-c x)-4 b \log ^2(1-c x)-16 a c \log ^2(1-c x)+16 a c^2 x \log ^2(1-c x)+4 b c^2 x^2 \log ^2(1-c x)-8 b \log (c x) \log ^2(1-c x)-16 a c \log (c x) \log ^2(1-c x)+4 (-c x (2 b+4 a c+b c x)+2 (-1+c x) (b+2 a c+b c x) \log (1-c x)) \operatorname {PolyLog}(2,c x)-16 (b+2 a c) \log (1-c x) \operatorname {PolyLog}(2,1-c x)+16 b \operatorname {PolyLog}(3,1-c x)+32 a c \operatorname {PolyLog}(3,1-c x)}{16 c^2} \]
(-14*b - 32*a*c + 22*b*c*x + 48*a*c^2*x + 3*b*c^2*x^2 + 22*b*Log[1 - c*x] + 48*a*c*Log[1 - c*x] - 16*b*c*x*Log[1 - c*x] - 48*a*c^2*x*Log[1 - c*x] - 6*b*c^2*x^2*Log[1 - c*x] - 4*b*Log[1 - c*x]^2 - 16*a*c*Log[1 - c*x]^2 + 16 *a*c^2*x*Log[1 - c*x]^2 + 4*b*c^2*x^2*Log[1 - c*x]^2 - 8*b*Log[c*x]*Log[1 - c*x]^2 - 16*a*c*Log[c*x]*Log[1 - c*x]^2 + 4*(-(c*x*(2*b + 4*a*c + b*c*x) ) + 2*(-1 + c*x)*(b + 2*a*c + b*c*x)*Log[1 - c*x])*PolyLog[2, c*x] - 16*(b + 2*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 16*b*PolyLog[3, 1 - c*x] + 32 *a*c*PolyLog[3, 1 - c*x])/(16*c^2)
Time = 0.82 (sec) , antiderivative size = 485, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
| \(\displaystyle \int (a+b x) \operatorname {PolyLog}(2,c x) \log (1-c x) \, dx\) |
| \(\Big \downarrow \) 7158 |
| \(\displaystyle \int \left (a \log ^2(1-c x)+\frac {1}{2} b x \log ^2(1-c x)+\frac {a^2 \log ^2(1-c x)}{2 b x}\right )dx+c \int \left (\frac {\operatorname {PolyLog}(2,c x) (b+a c)^2}{2 b c^2 (1-c x)}-\frac {(b+2 a c) \operatorname {PolyLog}(2,c x)}{2 c^2}-\frac {b x \operatorname {PolyLog}(2,c x)}{2 c}\right )dx+\frac {(a+b x)^2 \operatorname {PolyLog}(2,c x) \log (1-c x)}{2 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \operatorname {PolyLog}(3,1-c x)}{b}+\frac {a^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{b}+\frac {a^2 \log (c x) \log ^2(1-c x)}{2 b}+c \left (\frac {(a c+b)^2 \operatorname {PolyLog}(3,1-c x)}{b c^3}-\frac {(a c+b)^2 \operatorname {PolyLog}(2,c x) \log (1-c x)}{2 b c^3}-\frac {(a c+b)^2 \operatorname {PolyLog}(2,1-c x) \log (1-c x)}{b c^3}-\frac {(a c+b)^2 \log (c x) \log ^2(1-c x)}{2 b c^3}+\frac {(1-c x) (2 a c+b) \log (1-c x)}{2 c^3}-\frac {x (2 a c+b) \operatorname {PolyLog}(2,c x)}{2 c^2}+\frac {x (2 a c+b)}{2 c^2}+\frac {b \log (1-c x)}{8 c^3}+\frac {b x}{8 c^2}-\frac {b x^2 \operatorname {PolyLog}(2,c x)}{4 c}+\frac {b x^2}{16 c}-\frac {b x^2 \log (1-c x)}{8 c}\right )+\frac {(a+b x)^2 \operatorname {PolyLog}(2,c x) \log (1-c x)}{2 b}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {2 a (1-c x) \log (1-c x)}{c}+2 a x+\frac {b (1-c x)^2}{8 c^2}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {b x}{c}\) |
2*a*x + (b*x)/c + (b*(1 - c*x)^2)/(8*c^2) + (b*(1 - c*x)*Log[1 - c*x])/c^2 + (2*a*(1 - c*x)*Log[1 - c*x])/c - (b*(1 - c*x)^2*Log[1 - c*x])/(4*c^2) - (b*(1 - c*x)*Log[1 - c*x]^2)/(2*c^2) - (a*(1 - c*x)*Log[1 - c*x]^2)/c + ( b*(1 - c*x)^2*Log[1 - c*x]^2)/(4*c^2) + (a^2*Log[c*x]*Log[1 - c*x]^2)/(2*b ) + ((a + b*x)^2*Log[1 - c*x]*PolyLog[2, c*x])/(2*b) + (a^2*Log[1 - c*x]*P olyLog[2, 1 - c*x])/b - (a^2*PolyLog[3, 1 - c*x])/b + c*((b*x)/(8*c^2) + ( (b + 2*a*c)*x)/(2*c^2) + (b*x^2)/(16*c) + (b*Log[1 - c*x])/(8*c^3) - (b*x^ 2*Log[1 - c*x])/(8*c) + ((b + 2*a*c)*(1 - c*x)*Log[1 - c*x])/(2*c^3) - ((b + a*c)^2*Log[c*x]*Log[1 - c*x]^2)/(2*b*c^3) - ((b + 2*a*c)*x*PolyLog[2, c *x])/(2*c^2) - (b*x^2*PolyLog[2, c*x])/(4*c) - ((b + a*c)^2*Log[1 - c*x]*P olyLog[2, c*x])/(2*b*c^3) - ((b + a*c)^2*Log[1 - c*x]*PolyLog[2, 1 - c*x]) /(b*c^3) + ((b + a*c)^2*PolyLog[3, 1 - c*x])/(b*c^3))
3.2.86.3.1 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]
\[\int \left (b x +a \right ) \ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )d x\]
\[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { {\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
\[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int \left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )\, dx \]
Time = 0.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.66 \[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=-\frac {1}{16} \, c {\left (\frac {8 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} {\left (2 \, a c + b\right )}}{c^{3}} - \frac {3 \, b c^{2} x^{2} + 2 \, {\left (24 \, a c^{2} + 11 \, b c\right )} x - 4 \, {\left (b c^{2} x^{2} + 2 \, {\left (2 \, a c^{2} + b c\right )} x + 2 \, {\left (2 \, a c + b\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 2 \, {\left (2 \, b c^{2} x^{2} - 24 \, a c + 2 \, {\left (8 \, a c^{2} + 3 \, b c\right )} x - 11 \, b\right )} \log \left (-c x + 1\right )}{c^{3}}\right )} + \frac {1}{8} \, {\left (\frac {8 \, {\left (c x {\rm Li}_2\left (c x\right ) - c x + {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} a}{c} + \frac {{\left (4 \, c^{2} x^{2} {\rm Li}_2\left (c x\right ) - c^{2} x^{2} - 2 \, c x + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{2}}\right )} \log \left (-c x + 1\right ) \]
-1/16*c*(8*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2 *polylog(3, -c*x + 1))*(2*a*c + b)/c^3 - (3*b*c^2*x^2 + 2*(24*a*c^2 + 11*b *c)*x - 4*(b*c^2*x^2 + 2*(2*a*c^2 + b*c)*x + 2*(2*a*c + b)*log(-c*x + 1))* dilog(c*x) - 2*(2*b*c^2*x^2 - 24*a*c + 2*(8*a*c^2 + 3*b*c)*x - 11*b)*log(- c*x + 1))/c^3) + 1/8*(8*(c*x*dilog(c*x) - c*x + (c*x - 1)*log(-c*x + 1))*a /c + (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))*b/c^2)*log(-c*x + 1)
\[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int { {\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) \,d x } \]
Timed out. \[ \int (a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x) \, dx=\int \ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right ) \,d x \]