3.194 \(\int \frac{(a+b x+c x^2) \log (1-d x) \text{PolyLog}(2,d x)}{x} \, dx\)

Optimal. Leaf size=402 \[ -\frac{1}{2} a \text{PolyLog}(2,d x)^2+\frac{(2 b d+c) \text{PolyLog}(3,1-d x)}{d^2}-\frac{(2 b d+c) \log (1-d x) \text{PolyLog}(2,d x)}{2 d^2}-\frac{(2 b d+c) \log (1-d x) \text{PolyLog}(2,1-d x)}{d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{PolyLog}(2,d x)-\frac{x (2 b d+c) \text{PolyLog}(2,d x)}{2 d}-\frac{1}{4} c x^2 \text{PolyLog}(2,d x)-\frac{(2 b d+c) \log (d x) \log ^2(1-d x)}{2 d^2}+\frac{(1-d x) (2 b d+c) \log (1-d x)}{2 d^2}+\frac{x (2 b d+c)}{2 d}-\frac{b (1-d x) \log ^2(1-d x)}{d}+\frac{2 b (1-d x) \log (1-d x)}{d}+2 b x+\frac{c (1-d x)^2}{8 d^2}+\frac{c (1-d x)^2 \log ^2(1-d x)}{4 d^2}-\frac{c (1-d x) \log ^2(1-d x)}{2 d^2}-\frac{c (1-d x)^2 \log (1-d x)}{4 d^2}+\frac{c (1-d x) \log (1-d x)}{d^2}+\frac{c \log (1-d x)}{8 d^2}-\frac{1}{8} c x^2 \log (1-d x)+\frac{9 c x}{8 d}+\frac{c x^2}{16} \]

[Out]

2*b*x + (9*c*x)/(8*d) + ((c + 2*b*d)*x)/(2*d) + (c*x^2)/16 + (c*(1 - d*x)^2)/(8*d^2) + (c*Log[1 - d*x])/(8*d^2
) - (c*x^2*Log[1 - d*x])/8 + (c*(1 - d*x)*Log[1 - d*x])/d^2 + (2*b*(1 - d*x)*Log[1 - d*x])/d + ((c + 2*b*d)*(1
- d*x)*Log[1 - d*x])/(2*d^2) - (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) - ((c + 2*b*d)*Log[d*x]*Log[1 - d*x]^2
)/(2*d^2) - ((c + 2*b*d)*x*PolyLog[2, d*x])/(2*d) - (c*x^2*PolyLog[2, d*x])/4 - ((c + 2*b*d)*Log[1 - d*x]*Poly
Log[2, d*x])/(2*d^2) + ((2*b*x + c*x^2)*Log[1 - d*x]*PolyLog[2, d*x])/2 - (a*PolyLog[2, d*x]^2)/2 - ((c + 2*b*
d)*Log[1 - d*x]*PolyLog[2, 1 - d*x])/d^2 + ((c + 2*b*d)*PolyLog[3, 1 - d*x])/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.618594, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 29, number of rules used = 24, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.923, Rules used = {6742, 6586, 2389, 2295, 6589, 6591, 2395, 43, 6605, 6601, 1584, 6598, 2416, 2391, 6604, 2296, 2401, 2390, 2305, 2304, 6596, 2396, 2433, 2374} \[ -\frac{1}{2} a \text{PolyLog}(2,d x)^2+\frac{(2 b d+c) \text{PolyLog}(3,1-d x)}{d^2}-\frac{(2 b d+c) \log (1-d x) \text{PolyLog}(2,d x)}{2 d^2}-\frac{(2 b d+c) \log (1-d x) \text{PolyLog}(2,1-d x)}{d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{PolyLog}(2,d x)-\frac{x (2 b d+c) \text{PolyLog}(2,d x)}{2 d}-\frac{1}{4} c x^2 \text{PolyLog}(2,d x)-\frac{(2 b d+c) \log (d x) \log ^2(1-d x)}{2 d^2}+\frac{(1-d x) (2 b d+c) \log (1-d x)}{2 d^2}+\frac{x (2 b d+c)}{2 d}-\frac{b (1-d x) \log ^2(1-d x)}{d}+\frac{2 b (1-d x) \log (1-d x)}{d}+2 b x+\frac{c (1-d x)^2}{8 d^2}+\frac{c (1-d x)^2 \log ^2(1-d x)}{4 d^2}-\frac{c (1-d x) \log ^2(1-d x)}{2 d^2}-\frac{c (1-d x)^2 \log (1-d x)}{4 d^2}+\frac{c (1-d x) \log (1-d x)}{d^2}+\frac{c \log (1-d x)}{8 d^2}-\frac{1}{8} c x^2 \log (1-d x)+\frac{9 c x}{8 d}+\frac{c x^2}{16} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x + c*x^2)*Log[1 - d*x]*PolyLog[2, d*x])/x,x]

[Out]

2*b*x + (9*c*x)/(8*d) + ((c + 2*b*d)*x)/(2*d) + (c*x^2)/16 + (c*(1 - d*x)^2)/(8*d^2) + (c*Log[1 - d*x])/(8*d^2
) - (c*x^2*Log[1 - d*x])/8 + (c*(1 - d*x)*Log[1 - d*x])/d^2 + (2*b*(1 - d*x)*Log[1 - d*x])/d + ((c + 2*b*d)*(1
- d*x)*Log[1 - d*x])/(2*d^2) - (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) - ((c + 2*b*d)*Log[d*x]*Log[1 - d*x]^2
)/(2*d^2) - ((c + 2*b*d)*x*PolyLog[2, d*x])/(2*d) - (c*x^2*PolyLog[2, d*x])/4 - ((c + 2*b*d)*Log[1 - d*x]*Poly
Log[2, d*x])/(2*d^2) + ((2*b*x + c*x^2)*Log[1 - d*x]*PolyLog[2, d*x])/2 - (a*PolyLog[2, d*x]^2)/2 - ((c + 2*b*
d)*Log[1 - d*x]*PolyLog[2, 1 - d*x])/d^2 + ((c + 2*b*d)*PolyLog[3, 1 - d*x])/d^2

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 6586

Int[PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[x*PolyLog[n, a*(b*x^p)^q], x] - Dist[p*q, I
nt[PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, b, p, q}, x] && GtQ[n, 0]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6591

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[((d*x)^(m + 1)*PolyLog[n
, a*(b*x^p)^q])/(d*(m + 1)), x] - Dist[(p*q)/(m + 1), Int[(d*x)^m*PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ
[{a, b, d, m, p, q}, x] && NeQ[m, -1] && GtQ[n, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6605

Int[((g_.) + Log[1 + (e_.)*(x_)]*(h_.))*(Px_)*(x_)^(m_)*PolyLog[2, (c_.)*(x_)], x_Symbol] :> Dist[Coeff[Px, x,
-m - 1], Int[((g + h*Log[1 + e*x])*PolyLog[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[Px, x] && ILtQ[m, 0] && EqQ[c +
e, 0] && NeQ[Coeff[Px, x, -m - 1], 0]

Rule 6601

Int[(Log[1 + (e_.)*(x_)]*PolyLog[2, (c_.)*(x_)])/(x_), x_Symbol] :> -Simp[PolyLog[2, c*x]^2/2, x] /; FreeQ[{c,
e}, x] && EqQ[c + e, 0]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6598

Int[((d_.) + (e_.)*(x_))^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> Simp[((d + e*x)^(m + 1)*Po
lyLog[2, c*(a + b*x)])/(e*(m + 1)), x] + Dist[b/(e*(m + 1)), Int[((d + e*x)^(m + 1)*Log[1 - a*c - b*c*x])/(a +
b*x), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6604

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(Px_)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symb
ol] :> With[{u = IntHide[Px, x]}, Simp[u*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)], x] + (Dist[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] - Dist[e*h*n, Int[Ex
pandIntegrand[PolyLog[2, c*(a + b*x)], u/(d + e*x), x], x], x])] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && P
olyQ[Px, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 0]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 6596

Int[PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 - a*c - b*c*x]*PolyL
og[2, c*(a + b*x)])/e, x] + Dist[b/e, Int[Log[1 - a*c - b*c*x]^2/(a + b*x), x], x] /; FreeQ[{a, b, c, d, e}, x
] && EqQ[c*(b*d - a*e) + e, 0]

Rule 2396

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*
(f + g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n])^p)/g, x] - Dist[(b*e*n*p)/g, Int[(Log[(e*(f + g*x))/(e*f -
d*g)]*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*
f - d*g, 0] && IGtQ[p, 1]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)}{x} \, dx &=a \int \frac{\log (1-d x) \text{Li}_2(d x)}{x} \, dx+\int \frac{\left (b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)}{x} \, dx\\ &=-\frac{1}{2} a \text{Li}_2(d x){}^2+\int (b+c x) \log (1-d x) \text{Li}_2(d x) \, dx\\ &=\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2+d \int \left (\frac{(-c-2 b d) \text{Li}_2(d x)}{2 d^2}-\frac{c x \text{Li}_2(d x)}{2 d}+\frac{(-c-2 b d) \text{Li}_2(d x)}{2 d^2 (-1+d x)}\right ) \, dx+\int \left (b \log ^2(1-d x)+\frac{1}{2} c x \log ^2(1-d x)\right ) \, dx\\ &=\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2+b \int \log ^2(1-d x) \, dx+\frac{1}{2} c \int x \log ^2(1-d x) \, dx-\frac{1}{2} c \int x \text{Li}_2(d x) \, dx-\frac{(c+2 b d) \int \text{Li}_2(d x) \, dx}{2 d}-\frac{(c+2 b d) \int \frac{\text{Li}_2(d x)}{-1+d x} \, dx}{2 d}\\ &=-\frac{(c+2 b d) x \text{Li}_2(d x)}{2 d}-\frac{1}{4} c x^2 \text{Li}_2(d x)-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(d x)}{2 d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2-\frac{1}{4} c \int x \log (1-d x) \, dx+\frac{1}{2} c \int \left (\frac{\log ^2(1-d x)}{d}-\frac{(1-d x) \log ^2(1-d x)}{d}\right ) \, dx-\frac{b \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1-d x\right )}{d}-\frac{(c+2 b d) \int \frac{\log ^2(1-d x)}{x} \, dx}{2 d^2}-\frac{(c+2 b d) \int \log (1-d x) \, dx}{2 d}\\ &=-\frac{1}{8} c x^2 \log (1-d x)-\frac{b (1-d x) \log ^2(1-d x)}{d}-\frac{(c+2 b d) \log (d x) \log ^2(1-d x)}{2 d^2}-\frac{(c+2 b d) x \text{Li}_2(d x)}{2 d}-\frac{1}{4} c x^2 \text{Li}_2(d x)-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(d x)}{2 d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2+\frac{(2 b) \operatorname{Subst}(\int \log (x) \, dx,x,1-d x)}{d}+\frac{c \int \log ^2(1-d x) \, dx}{2 d}-\frac{c \int (1-d x) \log ^2(1-d x) \, dx}{2 d}-\frac{1}{8} (c d) \int \frac{x^2}{1-d x} \, dx+\frac{(c+2 b d) \operatorname{Subst}(\int \log (x) \, dx,x,1-d x)}{2 d^2}-\frac{(c+2 b d) \int \frac{\log (d x) \log (1-d x)}{1-d x} \, dx}{d}\\ &=2 b x+\frac{(c+2 b d) x}{2 d}-\frac{1}{8} c x^2 \log (1-d x)+\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{b (1-d x) \log ^2(1-d x)}{d}-\frac{(c+2 b d) \log (d x) \log ^2(1-d x)}{2 d^2}-\frac{(c+2 b d) x \text{Li}_2(d x)}{2 d}-\frac{1}{4} c x^2 \text{Li}_2(d x)-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(d x)}{2 d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2-\frac{c \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1-d x\right )}{2 d^2}+\frac{c \operatorname{Subst}\left (\int x \log ^2(x) \, dx,x,1-d x\right )}{2 d^2}-\frac{1}{8} (c d) \int \left (-\frac{1}{d^2}-\frac{x}{d}-\frac{1}{d^2 (-1+d x)}\right ) \, dx+\frac{(c+2 b d) \operatorname{Subst}\left (\int \frac{\log (x) \log \left (d \left (\frac{1}{d}-\frac{x}{d}\right )\right )}{x} \, dx,x,1-d x\right )}{d^2}\\ &=2 b x+\frac{c x}{8 d}+\frac{(c+2 b d) x}{2 d}+\frac{c x^2}{16}+\frac{c \log (1-d x)}{8 d^2}-\frac{1}{8} c x^2 \log (1-d x)+\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) \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 \text{Li}_2(d x)}{2 d}-\frac{1}{4} c x^2 \text{Li}_2(d x)-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(d x)}{2 d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(1-d x)}{d^2}-\frac{c \operatorname{Subst}(\int x \log (x) \, dx,x,1-d x)}{2 d^2}+\frac{c \operatorname{Subst}(\int \log (x) \, dx,x,1-d x)}{d^2}+\frac{(c+2 b d) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-d x\right )}{d^2}\\ &=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 \text{Li}_2(d x)}{2 d}-\frac{1}{4} c x^2 \text{Li}_2(d x)-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(d x)}{2 d^2}+\frac{1}{2} \left (2 b x+c x^2\right ) \log (1-d x) \text{Li}_2(d x)-\frac{1}{2} a \text{Li}_2(d x){}^2-\frac{(c+2 b d) \log (1-d x) \text{Li}_2(1-d x)}{d^2}+\frac{(c+2 b d) \text{Li}_3(1-d x)}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.355676, size = 298, normalized size = 0.74 \[ \frac{-8 a d^2 \text{PolyLog}(2,d x)^2+4 \text{PolyLog}(2,d x) (2 (d x-1) \log (1-d x) (2 b d+c d x+c)-d x (4 b d+c d x+2 c))-16 (2 b d+c) \log (1-d x) \text{PolyLog}(2,1-d x)+32 b d \text{PolyLog}(3,1-d x)+16 c \text{PolyLog}(3,1-d x)+48 b d^2 x+16 b d^2 x \log ^2(1-d x)-48 b d^2 x \log (1-d x)-16 b d \log ^2(1-d x)-16 b d \log (d x) \log ^2(1-d x)+48 b d \log (1-d x)-32 b d+3 c d^2 x^2+4 c d^2 x^2 \log ^2(1-d x)-6 c d^2 x^2 \log (1-d x)+22 c d x-4 c \log ^2(1-d x)-8 c \log (d x) \log ^2(1-d x)-16 c d x \log (1-d x)+22 c \log (1-d x)-14 c}{16 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x + c*x^2)*Log[1 - d*x]*PolyLog[2, d*x])/x,x]

[Out]

(-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] + 16*c*PolyLog[3, 1 - d*x] + 32*b*d
*PolyLog[3, 1 - d*x])/(16*d^2)

________________________________________________________________________________________

Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) \ln \left ( -dx+1 \right ){\it polylog} \left ( 2,dx \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)*ln(-d*x+1)*polylog(2,d*x)/x,x)

[Out]

int((c*x^2+b*x+a)*ln(-d*x+1)*polylog(2,d*x)/x,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)*log(-d*x+1)*polylog(2,d*x)/x,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}{\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)*log(-d*x+1)*polylog(2,d*x)/x,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)*ln(-d*x+1)*polylog(2,d*x)/x,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)*log(-d*x+1)*polylog(2,d*x)/x,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x, x)