Integrand size = 26, antiderivative size = 402 \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=2 b x+\frac {9 c x}{8 d}+\frac {(c+2 b d) x}{2 d}+\frac {c x^2}{16}+\frac {c (1-d x)^2}{8 d^2}+\frac {c \log (1-d x)}{8 d^2}-\frac {1}{8} c x^2 \log (1-d x)+\frac {c (1-d x) \log (1-d x)}{d^2}+\frac {2 b (1-d x) \log (1-d x)}{d}+\frac {(c+2 b d) (1-d x) \log (1-d x)}{2 d^2}-\frac {c (1-d x)^2 \log (1-d x)}{4 d^2}-\frac {c (1-d x) \log ^2(1-d x)}{2 d^2}-\frac {b (1-d x) \log ^2(1-d x)}{d}+\frac {c (1-d x)^2 \log ^2(1-d x)}{4 d^2}-\frac {(c+2 b d) \log (d x) \log ^2(1-d x)}{2 d^2}-\frac {(c+2 b d) x \operatorname {PolyLog}(2,d x)}{2 d}-\frac {1}{4} c x^2 \operatorname {PolyLog}(2,d x)-\frac {(c+2 b d) \log (1-d x) \operatorname {PolyLog}(2,d x)}{2 d^2}+\frac {1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)-\frac {1}{2} a \operatorname {PolyLog}(2,d x)^2-\frac {(c+2 b d) \log (1-d x) \operatorname {PolyLog}(2,1-d x)}{d^2}+\frac {(c+2 b d) \operatorname {PolyLog}(3,1-d x)}{d^2} \]
2*b*x+9/8*c*x/d+1/2*(2*b*d+c)*x/d+1/16*c*x^2+1/8*c*(-d*x+1)^2/d^2+1/8*c*ln (-d*x+1)/d^2-1/8*c*x^2*ln(-d*x+1)+c*(-d*x+1)*ln(-d*x+1)/d^2+2*b*(-d*x+1)*l n(-d*x+1)/d+1/2*(2*b*d+c)*(-d*x+1)*ln(-d*x+1)/d^2-1/4*c*(-d*x+1)^2*ln(-d*x +1)/d^2-1/2*c*(-d*x+1)*ln(-d*x+1)^2/d^2-b*(-d*x+1)*ln(-d*x+1)^2/d+1/4*c*(- d*x+1)^2*ln(-d*x+1)^2/d^2-1/2*(2*b*d+c)*ln(d*x)*ln(-d*x+1)^2/d^2-1/2*(2*b* d+c)*x*polylog(2,d*x)/d-1/4*c*x^2*polylog(2,d*x)-1/2*(2*b*d+c)*ln(-d*x+1)* polylog(2,d*x)/d^2+1/2*(c*x^2+2*b*x)*ln(-d*x+1)*polylog(2,d*x)-1/2*a*polyl og(2,d*x)^2-(2*b*d+c)*ln(-d*x+1)*polylog(2,-d*x+1)/d^2+(2*b*d+c)*polylog(3 ,-d*x+1)/d^2
Time = 0.26 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\frac {-14 c-32 b d+22 c d x+48 b d^2 x+3 c d^2 x^2+22 c \log (1-d x)+48 b d \log (1-d x)-16 c d x \log (1-d x)-48 b d^2 x \log (1-d x)-6 c d^2 x^2 \log (1-d x)-4 c \log ^2(1-d x)-16 b d \log ^2(1-d x)+16 b d^2 x \log ^2(1-d x)+4 c d^2 x^2 \log ^2(1-d x)-8 c \log (d x) \log ^2(1-d x)-16 b d \log (d x) \log ^2(1-d x)+4 (-d x (2 c+4 b d+c d x)+2 (-1+d x) (c+2 b d+c d x) \log (1-d x)) \operatorname {PolyLog}(2,d x)-8 a d^2 \operatorname {PolyLog}(2,d x)^2-16 (c+2 b d) \log (1-d x) \operatorname {PolyLog}(2,1-d x)+16 c \operatorname {PolyLog}(3,1-d x)+32 b d \operatorname {PolyLog}(3,1-d x)}{16 d^2} \]
(-14*c - 32*b*d + 22*c*d*x + 48*b*d^2*x + 3*c*d^2*x^2 + 22*c*Log[1 - d*x] + 48*b*d*Log[1 - d*x] - 16*c*d*x*Log[1 - d*x] - 48*b*d^2*x*Log[1 - d*x] - 6*c*d^2*x^2*Log[1 - d*x] - 4*c*Log[1 - d*x]^2 - 16*b*d*Log[1 - d*x]^2 + 16 *b*d^2*x*Log[1 - d*x]^2 + 4*c*d^2*x^2*Log[1 - d*x]^2 - 8*c*Log[d*x]*Log[1 - d*x]^2 - 16*b*d*Log[d*x]*Log[1 - d*x]^2 + 4*(-(d*x*(2*c + 4*b*d + c*d*x) ) + 2*(-1 + d*x)*(c + 2*b*d + c*d*x)*Log[1 - d*x])*PolyLog[2, d*x] - 8*a*d ^2*PolyLog[2, d*x]^2 - 16*(c + 2*b*d)*Log[1 - d*x]*PolyLog[2, 1 - d*x] + 1 6*c*PolyLog[3, 1 - d*x] + 32*b*d*PolyLog[3, 1 - d*x])/(16*d^2)
Time = 1.15 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {7159, 9, 7155, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}(2,d x) \log (1-d x) \left (a+b x+c x^2\right )}{x} \, dx\) |
| \(\Big \downarrow \) 7159 |
| \(\displaystyle a \int \frac {\log (1-d x) \operatorname {PolyLog}(2,d x)}{x}dx+\int \frac {\left (c x^2+b x\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x}dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle a \int \frac {\log (1-d x) \operatorname {PolyLog}(2,d x)}{x}dx+\int (b+c x) \log (1-d x) \operatorname {PolyLog}(2,d x)dx\) |
| \(\Big \downarrow \) 7155 |
| \(\displaystyle \int (b+c x) \log (1-d x) \operatorname {PolyLog}(2,d x)dx-\frac {1}{2} a \operatorname {PolyLog}(2,d x)^2\) |
| \(\Big \downarrow \) 7158 |
| \(\displaystyle \int \left (b \log ^2(1-d x)+\frac {1}{2} c x \log ^2(1-d x)+\frac {b^2 \log ^2(1-d x)}{2 c x}\right )dx+d \int \left (\frac {\operatorname {PolyLog}(2,d x) (c+b d)^2}{2 c d^2 (1-d x)}-\frac {(c+2 b d) \operatorname {PolyLog}(2,d x)}{2 d^2}-\frac {c x \operatorname {PolyLog}(2,d x)}{2 d}\right )dx-\frac {1}{2} a \operatorname {PolyLog}(2,d x)^2+\frac {(b+c x)^2 \operatorname {PolyLog}(2,d x) \log (1-d x)}{2 c}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{2} a \operatorname {PolyLog}(2,d x)^2-\frac {b^2 \operatorname {PolyLog}(3,1-d x)}{c}+\frac {b^2 \operatorname {PolyLog}(2,1-d x) \log (1-d x)}{c}+\frac {b^2 \log (d x) \log ^2(1-d x)}{2 c}+d \left (\frac {(b d+c)^2 \operatorname {PolyLog}(3,1-d x)}{c d^3}-\frac {(b d+c)^2 \operatorname {PolyLog}(2,d x) \log (1-d x)}{2 c d^3}-\frac {(b d+c)^2 \operatorname {PolyLog}(2,1-d x) \log (1-d x)}{c d^3}-\frac {(b d+c)^2 \log (d x) \log ^2(1-d x)}{2 c d^3}+\frac {(1-d x) (2 b d+c) \log (1-d x)}{2 d^3}-\frac {x (2 b d+c) \operatorname {PolyLog}(2,d x)}{2 d^2}+\frac {x (2 b d+c)}{2 d^2}+\frac {c \log (1-d x)}{8 d^3}+\frac {c x}{8 d^2}-\frac {c x^2 \operatorname {PolyLog}(2,d x)}{4 d}+\frac {c x^2}{16 d}-\frac {c x^2 \log (1-d x)}{8 d}\right )+\frac {(b+c x)^2 \operatorname {PolyLog}(2,d x) \log (1-d x)}{2 c}-\frac {b (1-d x) \log ^2(1-d x)}{d}+\frac {2 b (1-d x) \log (1-d x)}{d}+2 b x+\frac {c (1-d x)^2}{8 d^2}+\frac {c (1-d x)^2 \log ^2(1-d x)}{4 d^2}-\frac {c (1-d x) \log ^2(1-d x)}{2 d^2}-\frac {c (1-d x)^2 \log (1-d x)}{4 d^2}+\frac {c (1-d x) \log (1-d x)}{d^2}+\frac {c x}{d}\) |
2*b*x + (c*x)/d + (c*(1 - d*x)^2)/(8*d^2) + (c*(1 - d*x)*Log[1 - d*x])/d^2 + (2*b*(1 - d*x)*Log[1 - d*x])/d - (c*(1 - d*x)^2*Log[1 - d*x])/(4*d^2) - (c*(1 - d*x)*Log[1 - d*x]^2)/(2*d^2) - (b*(1 - d*x)*Log[1 - d*x]^2)/d + ( c*(1 - d*x)^2*Log[1 - d*x]^2)/(4*d^2) + (b^2*Log[d*x]*Log[1 - d*x]^2)/(2*c ) + ((b + c*x)^2*Log[1 - d*x]*PolyLog[2, d*x])/(2*c) - (a*PolyLog[2, d*x]^ 2)/2 + (b^2*Log[1 - d*x]*PolyLog[2, 1 - d*x])/c - (b^2*PolyLog[3, 1 - d*x] )/c + d*((c*x)/(8*d^2) + ((c + 2*b*d)*x)/(2*d^2) + (c*x^2)/(16*d) + (c*Log [1 - d*x])/(8*d^3) - (c*x^2*Log[1 - d*x])/(8*d) + ((c + 2*b*d)*(1 - d*x)*L og[1 - d*x])/(2*d^3) - ((c + b*d)^2*Log[d*x]*Log[1 - d*x]^2)/(2*c*d^3) - ( (c + 2*b*d)*x*PolyLog[2, d*x])/(2*d^2) - (c*x^2*PolyLog[2, d*x])/(4*d) - ( (c + b*d)^2*Log[1 - d*x]*PolyLog[2, d*x])/(2*c*d^3) - ((c + b*d)^2*Log[1 - d*x]*PolyLog[2, 1 - d*x])/(c*d^3) + ((c + b*d)^2*PolyLog[3, 1 - d*x])/(c* d^3))
3.2.94.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(Log[1 + (e_.)*(x_)]*PolyLog[2, (c_.)*(x_)])/(x_), x_Symbol] :> Simp[-P olyLog[2, c*x]^2/2, x] /; FreeQ[{c, e}, x] && EqQ[c + e, 0]
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[((g_.) + Log[1 + (e_.)*(x_)]*(h_.))*(Px_)*(x_)^(m_)*PolyLog[2, (c_.)*(x _)], x_Symbol] :> Simp[Coeff[Px, x, -m - 1] Int[(g + h*Log[1 + e*x])*(Pol yLog[2, c*x]/x), x], x] + Int[x^m*(Px - Coeff[Px, x, -m - 1]*x^(-m - 1))*(g + h*Log[1 + e*x])*PolyLog[2, c*x], x] /; FreeQ[{c, e, g, h}, x] && PolyQ[P x, x] && ILtQ[m, 0] && EqQ[c + e, 0] && NeQ[Coeff[Px, x, -m - 1], 0]
\[\int \frac {\left (c \,x^{2}+b x +a \right ) \ln \left (-d x +1\right ) \operatorname {polylog}\left (2, d x \right )}{x}d x\]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\int \frac {\left (a + b x + c x^{2}\right ) \log {\left (- d x + 1 \right )} \operatorname {Li}_{2}\left (d x\right )}{x}\, dx \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x} \, dx=\int \frac {\ln \left (1-d\,x\right )\,\mathrm {polylog}\left (2,d\,x\right )\,\left (c\,x^2+b\,x+a\right )}{x} \,d x \]