Integrand size = 26, antiderivative size = 218 \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=3 c x+\frac {3 c (1-d x) \log (1-d x)}{d}-\frac {c (1-d x) \log ^2(1-d x)}{d}+\frac {a (1-d x) \log ^2(1-d x)}{x}+\left (a-\frac {c}{d^2}\right ) d \log (d x) \log ^2(1-d x)-2 a d \operatorname {PolyLog}(2,d x)-c x \operatorname {PolyLog}(2,d x)+\left (a-\frac {c}{d^2}\right ) d \log (1-d x) \operatorname {PolyLog}(2,d x)-\left (\frac {a}{x}-c x\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)-\frac {1}{2} b \operatorname {PolyLog}(2,d x)^2+2 \left (a-\frac {c}{d^2}\right ) d \log (1-d x) \operatorname {PolyLog}(2,1-d x)-a d \operatorname {PolyLog}(3,d x)-2 \left (a-\frac {c}{d^2}\right ) d \operatorname {PolyLog}(3,1-d x) \]
3*c*x+3*c*(-d*x+1)*ln(-d*x+1)/d-c*(-d*x+1)*ln(-d*x+1)^2/d+a*(-d*x+1)*ln(-d *x+1)^2/x+(a-c/d^2)*d*ln(d*x)*ln(-d*x+1)^2-2*a*d*polylog(2,d*x)-c*x*polylo g(2,d*x)+(a-c/d^2)*d*ln(-d*x+1)*polylog(2,d*x)-(a/x-c*x)*ln(-d*x+1)*polylo g(2,d*x)-1/2*b*polylog(2,d*x)^2+2*(a-c/d^2)*d*ln(-d*x+1)*polylog(2,-d*x+1) -a*d*polylog(3,d*x)-2*(a-c/d^2)*d*polylog(3,-d*x+1)
Time = 0.65 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\frac {2 \left (-c d x^2+(a d+c x) (-1+d x) \log (1-d x)\right ) \operatorname {PolyLog}(2,d x)-b d x \operatorname {PolyLog}(2,d x)^2+2 \left (-2 c x+3 c d x^2+3 c x \log (1-d x)-3 c d x^2 \log (1-d x)+2 a d^2 x \log (d x) \log (1-d x)+a d \log ^2(1-d x)-c x \log ^2(1-d x)-a d^2 x \log ^2(1-d x)+c d x^2 \log ^2(1-d x)-c x \log (d x) \log ^2(1-d x)+a d^2 x \log (d x) \log ^2(1-d x)+2 x \left (a d^2+\left (-c+a d^2\right ) \log (1-d x)\right ) \operatorname {PolyLog}(2,1-d x)-a d^2 x \operatorname {PolyLog}(3,d x)+2 c x \operatorname {PolyLog}(3,1-d x)-2 a d^2 x \operatorname {PolyLog}(3,1-d x)\right )}{2 d x} \]
(2*(-(c*d*x^2) + (a*d + c*x)*(-1 + d*x)*Log[1 - d*x])*PolyLog[2, d*x] - b* d*x*PolyLog[2, d*x]^2 + 2*(-2*c*x + 3*c*d*x^2 + 3*c*x*Log[1 - d*x] - 3*c*d *x^2*Log[1 - d*x] + 2*a*d^2*x*Log[d*x]*Log[1 - d*x] + a*d*Log[1 - d*x]^2 - c*x*Log[1 - d*x]^2 - a*d^2*x*Log[1 - d*x]^2 + c*d*x^2*Log[1 - d*x]^2 - c* x*Log[d*x]*Log[1 - d*x]^2 + a*d^2*x*Log[d*x]*Log[1 - d*x]^2 + 2*x*(a*d^2 + (-c + a*d^2)*Log[1 - d*x])*PolyLog[2, 1 - d*x] - a*d^2*x*PolyLog[3, d*x] + 2*c*x*PolyLog[3, 1 - d*x] - 2*a*d^2*x*PolyLog[3, 1 - d*x]))/(2*d*x)
Time = 0.97 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {7159, 7155, 7160, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}(2,d x) \log (1-d x) \left (a+b x+c x^2\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7159 |
\(\displaystyle \int \frac {\left (c x^2+a\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2}dx+b \int \frac {\log (1-d x) \operatorname {PolyLog}(2,d x)}{x}dx\) |
\(\Big \downarrow \) 7155 |
\(\displaystyle \int \frac {\left (c x^2+a\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2}dx-\frac {1}{2} b \operatorname {PolyLog}(2,d x)^2\) |
\(\Big \downarrow \) 7160 |
\(\displaystyle d \int \left (-\frac {c \operatorname {PolyLog}(2,d x)}{d}-\frac {a \operatorname {PolyLog}(2,d x)}{x}+\frac {\left (c-a d^2\right ) \operatorname {PolyLog}(2,d x)}{d (1-d x)}\right )dx+\int \left (c \log ^2(1-d x)-\frac {a \log ^2(1-d x)}{x^2}\right )dx-\left (\frac {a}{x}-c x\right ) \operatorname {PolyLog}(2,d x) \log (1-d x)-\frac {1}{2} b \operatorname {PolyLog}(2,d x)^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle d \left (\frac {2 \left (c-a d^2\right ) \operatorname {PolyLog}(3,1-d x)}{d^2}-\frac {\left (c-a d^2\right ) \operatorname {PolyLog}(2,d x) \log (1-d x)}{d^2}-\frac {2 \left (c-a d^2\right ) \operatorname {PolyLog}(2,1-d x) \log (1-d x)}{d^2}-\frac {\left (c-a d^2\right ) \log (d x) \log ^2(1-d x)}{d^2}-a \operatorname {PolyLog}(3,d x)+\frac {c (1-d x) \log (1-d x)}{d^2}-\frac {c x \operatorname {PolyLog}(2,d x)}{d}+\frac {c x}{d}\right )-\left (\frac {a}{x}-c x\right ) \operatorname {PolyLog}(2,d x) \log (1-d x)-2 a d \operatorname {PolyLog}(2,d x)+\frac {a (1-d x) \log ^2(1-d x)}{x}-\frac {1}{2} b \operatorname {PolyLog}(2,d x)^2-\frac {c (1-d x) \log ^2(1-d x)}{d}+\frac {2 c (1-d x) \log (1-d x)}{d}+2 c x\) |
2*c*x + (2*c*(1 - d*x)*Log[1 - d*x])/d - (c*(1 - d*x)*Log[1 - d*x]^2)/d + (a*(1 - d*x)*Log[1 - d*x]^2)/x - 2*a*d*PolyLog[2, d*x] - (a/x - c*x)*Log[1 - d*x]*PolyLog[2, d*x] - (b*PolyLog[2, d*x]^2)/2 + d*((c*x)/d + (c*(1 - d *x)*Log[1 - d*x])/d^2 - ((c - a*d^2)*Log[d*x]*Log[1 - d*x]^2)/d^2 - (c*x*P olyLog[2, d*x])/d - ((c - a*d^2)*Log[1 - d*x]*PolyLog[2, d*x])/d^2 - (2*(c - a*d^2)*Log[1 - d*x]*PolyLog[2, 1 - d*x])/d^2 - a*PolyLog[3, d*x] + (2*( c - a*d^2)*PolyLog[3, 1 - d*x])/d^2)
3.2.95.3.1 Defintions of rubi rules used
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[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[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(Px_)*(x_)^(m_.)* PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{u = IntHide[x^m* Px, x]}, Simp[u*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)], x] + (S imp[b Int[ExpandIntegrand[(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] && IntegerQ[m]
\[\int \frac {\left (c \,x^{2}+b x +a \right ) \ln \left (-d x +1\right ) \operatorname {polylog}\left (2, d x \right )}{x^{2}}d x\]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{2}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\int \frac {\left (a + b x + c x^{2}\right ) \log {\left (- d x + 1 \right )} \operatorname {Li}_{2}\left (d x\right )}{x^{2}}\, dx \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{2}} \,d x } \]
\[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )} {\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right ) \log (1-d x) \operatorname {PolyLog}(2,d x)}{x^2} \, dx=\int \frac {\ln \left (1-d\,x\right )\,\mathrm {polylog}\left (2,d\,x\right )\,\left (c\,x^2+b\,x+a\right )}{x^2} \,d x \]