3.2.94 \(\int \frac {(a+b x+c x^2) \log (1-d x) \text {PolyLog}(2,d x)}{x} \, dx\) [194]

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

[Out]

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)*ln(-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*polylog(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

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 402, normalized size of antiderivative = 1.00, 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 = {6874, 6721, 2436, 2332, 6724, 6726, 2442, 45, 6740, 6736, 1598, 6733, 2463, 2438, 6739, 2333, 2448, 2437, 2342, 2341, 6731, 2443, 2481, 2421} \begin {gather*} -\frac {1}{2} a \text {Li}_2(d x){}^2+\frac {(2 b d+c) \text {Li}_3(1-d x)}{d^2}-\frac {(2 b d+c) \text {Li}_2(d x) \log (1-d x)}{2 d^2}-\frac {(2 b d+c) \text {Li}_2(1-d x) \log (1-d x)}{d^2}-\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 {1}{2} \left (2 b x+c x^2\right ) \text {Li}_2(d x) \log (1-d x)-\frac {x (2 b d+c) \text {Li}_2(d x)}{2 d}+\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}{4} c x^2 \text {Li}_2(d x)-\frac {1}{8} c x^2 \log (1-d x)+\frac {9 c x}{8 d}+\frac {c x^2}{16} \end {gather*}

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 45

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 1598

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 2332

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

Rule 2333

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 2341

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 2342

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 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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]

Rule 2436

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 2437

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 2438

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 2442

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 2443

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 2448

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 2463

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 2481

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 6721

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 6724

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 6726

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 6731

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 6733

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 6736

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 6739

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 6740

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 6874

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

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 \text {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) \text {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) \text {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 \text {Subst}\left (\int \log ^2(x) \, dx,x,1-d x\right )}{2 d^2}+\frac {c \text {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) \text {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 \text {Subst}(\int x \log (x) \, dx,x,1-d x)}{2 d^2}+\frac {c \text {Subst}(\int \log (x) \, dx,x,1-d x)}{d^2}+\frac {(c+2 b d) \text {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 = 0.24, size = 298, normalized size = 0.74 \begin {gather*} \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)) \text {PolyLog}(2,d x)-8 a d^2 \text {PolyLog}(2,d x)^2-16 (c+2 b d) \log (1-d x) \text {PolyLog}(2,1-d x)+16 c \text {PolyLog}(3,1-d x)+32 b d \text {PolyLog}(3,1-d x)}{16 d^2} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right ) \ln \left (-d x +1\right ) \polylog \left (2, d x \right )}{x}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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]

Integral((a + b*x + c*x**2)*log(-d*x + 1)*polylog(2, d*x)/x, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________