3.185 \(\int x (a+b x) \log (1-c x) \text{PolyLog}(2,c x) \, dx\)

Optimal. Leaf size=546 \[ -\frac{x (3 a c+2 b) \text{PolyLog}(2,c x)}{6 c^2}+\frac{(3 a c+2 b) \text{PolyLog}(3,1-c x)}{3 c^3}-\frac{(3 a c+2 b) \log (1-c x) \text{PolyLog}(2,c x)}{6 c^3}-\frac{(3 a c+2 b) \log (1-c x) \text{PolyLog}(2,1-c x)}{3 c^3}-\frac{x^2 (3 a c+2 b) \text{PolyLog}(2,c x)}{12 c}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{PolyLog}(2,c x)-\frac{1}{9} b x^3 \text{PolyLog}(2,c x)+\frac{5 x (3 a c+2 b)}{24 c^2}-\frac{(3 a c+2 b) \log (c x) \log ^2(1-c x)}{6 c^3}+\frac{(3 a c+2 b) \log (1-c x)}{24 c^3}+\frac{(1-c x) (3 a c+2 b) \log (1-c x)}{6 c^3}+\frac{x^2 (3 a c+2 b)}{48 c}-\frac{x^2 (3 a c+2 b) \log (1-c x)}{24 c}+\frac{a (1-c x)^2}{8 c^2}+\frac{a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{a (1-c x) \log ^2(1-c x)}{2 c^2}-\frac{a (1-c x)^2 \log (1-c x)}{4 c^2}+\frac{a (1-c x) \log (1-c x)}{c^2}+\frac{a x}{c}+\frac{4 b x}{9 c^2}-\frac{b \log ^2(1-c x)}{9 c^3}+\frac{2 b (1-c x) \log (1-c x)}{9 c^3}+\frac{2 b \log (1-c x)}{9 c^3}+\frac{b x^2}{9 c}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{1}{9} b x^3 \log (1-c x)-\frac{b x^2 \log (1-c x)}{9 c}+\frac{b x^3}{27} \]

[Out]

(4*b*x)/(9*c^2) + (a*x)/c + (5*(2*b + 3*a*c)*x)/(24*c^2) + (b*x^2)/(9*c) + ((2*b + 3*a*c)*x^2)/(48*c) + (b*x^3
)/27 + (a*(1 - c*x)^2)/(8*c^2) + (2*b*Log[1 - c*x])/(9*c^3) + ((2*b + 3*a*c)*Log[1 - c*x])/(24*c^3) - (b*x^2*L
og[1 - c*x])/(9*c) - ((2*b + 3*a*c)*x^2*Log[1 - c*x])/(24*c) - (b*x^3*Log[1 - c*x])/9 + (2*b*(1 - c*x)*Log[1 -
c*x])/(9*c^3) + (a*(1 - c*x)*Log[1 - c*x])/c^2 + ((2*b + 3*a*c)*(1 - c*x)*Log[1 - c*x])/(6*c^3) - (a*(1 - c*x
)^2*Log[1 - c*x])/(4*c^2) - (b*Log[1 - c*x]^2)/(9*c^3) + (b*x^3*Log[1 - c*x]^2)/9 - (a*(1 - c*x)*Log[1 - c*x]^
2)/(2*c^2) + (a*(1 - c*x)^2*Log[1 - c*x]^2)/(4*c^2) - ((2*b + 3*a*c)*Log[c*x]*Log[1 - c*x]^2)/(6*c^3) - ((2*b
+ 3*a*c)*x*PolyLog[2, c*x])/(6*c^2) - ((2*b + 3*a*c)*x^2*PolyLog[2, c*x])/(12*c) - (b*x^3*PolyLog[2, c*x])/9 -
((2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, c*x])/(6*c^3) + ((3*a*x^2 + 2*b*x^3)*Log[1 - c*x]*PolyLog[2, c*x])/6 -
((2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x])/(3*c^3) + ((2*b + 3*a*c)*PolyLog[3, 1 - c*x])/(3*c^3)

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Rubi [A]  time = 0.71802, antiderivative size = 546, normalized size of antiderivative = 1., number of steps used = 40, number of rules used = 21, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.105, Rules used = {6742, 6591, 2395, 43, 6604, 2401, 2389, 2296, 2295, 2390, 2305, 2304, 2398, 2410, 2301, 6586, 6596, 2396, 2433, 2374, 6589} \[ -\frac{x (3 a c+2 b) \text{PolyLog}(2,c x)}{6 c^2}+\frac{(3 a c+2 b) \text{PolyLog}(3,1-c x)}{3 c^3}-\frac{(3 a c+2 b) \log (1-c x) \text{PolyLog}(2,c x)}{6 c^3}-\frac{(3 a c+2 b) \log (1-c x) \text{PolyLog}(2,1-c x)}{3 c^3}-\frac{x^2 (3 a c+2 b) \text{PolyLog}(2,c x)}{12 c}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{PolyLog}(2,c x)-\frac{1}{9} b x^3 \text{PolyLog}(2,c x)+\frac{5 x (3 a c+2 b)}{24 c^2}-\frac{(3 a c+2 b) \log (c x) \log ^2(1-c x)}{6 c^3}+\frac{(3 a c+2 b) \log (1-c x)}{24 c^3}+\frac{(1-c x) (3 a c+2 b) \log (1-c x)}{6 c^3}+\frac{x^2 (3 a c+2 b)}{48 c}-\frac{x^2 (3 a c+2 b) \log (1-c x)}{24 c}+\frac{a (1-c x)^2}{8 c^2}+\frac{a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{a (1-c x) \log ^2(1-c x)}{2 c^2}-\frac{a (1-c x)^2 \log (1-c x)}{4 c^2}+\frac{a (1-c x) \log (1-c x)}{c^2}+\frac{a x}{c}+\frac{4 b x}{9 c^2}-\frac{b \log ^2(1-c x)}{9 c^3}+\frac{2 b (1-c x) \log (1-c x)}{9 c^3}+\frac{2 b \log (1-c x)}{9 c^3}+\frac{b x^2}{9 c}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{1}{9} b x^3 \log (1-c x)-\frac{b x^2 \log (1-c x)}{9 c}+\frac{b x^3}{27} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*x)/(9*c^2) + (a*x)/c + (5*(2*b + 3*a*c)*x)/(24*c^2) + (b*x^2)/(9*c) + ((2*b + 3*a*c)*x^2)/(48*c) + (b*x^3
)/27 + (a*(1 - c*x)^2)/(8*c^2) + (2*b*Log[1 - c*x])/(9*c^3) + ((2*b + 3*a*c)*Log[1 - c*x])/(24*c^3) - (b*x^2*L
og[1 - c*x])/(9*c) - ((2*b + 3*a*c)*x^2*Log[1 - c*x])/(24*c) - (b*x^3*Log[1 - c*x])/9 + (2*b*(1 - c*x)*Log[1 -
c*x])/(9*c^3) + (a*(1 - c*x)*Log[1 - c*x])/c^2 + ((2*b + 3*a*c)*(1 - c*x)*Log[1 - c*x])/(6*c^3) - (a*(1 - c*x
)^2*Log[1 - c*x])/(4*c^2) - (b*Log[1 - c*x]^2)/(9*c^3) + (b*x^3*Log[1 - c*x]^2)/9 - (a*(1 - c*x)*Log[1 - c*x]^
2)/(2*c^2) + (a*(1 - c*x)^2*Log[1 - c*x]^2)/(4*c^2) - ((2*b + 3*a*c)*Log[c*x]*Log[1 - c*x]^2)/(6*c^3) - ((2*b
+ 3*a*c)*x*PolyLog[2, c*x])/(6*c^2) - ((2*b + 3*a*c)*x^2*PolyLog[2, c*x])/(12*c) - (b*x^3*PolyLog[2, c*x])/9 -
((2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, c*x])/(6*c^3) + ((3*a*x^2 + 2*b*x^3)*Log[1 - c*x]*PolyLog[2, c*x])/6 -
((2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x])/(3*c^3) + ((2*b + 3*a*c)*PolyLog[3, 1 - c*x])/(3*c^3)

Rule 6742

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

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 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 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 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 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 2295

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

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 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2410

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[Log[c*(d
+ e*x)], x^m/(f + g*x), x], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m
]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 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]

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]

Rubi steps

\begin{align*} \int x (a+b x) \log (1-c x) \text{Li}_2(c x) \, dx &=\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)+c \int \left (\frac{(-2 b-3 a c) \text{Li}_2(c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{b x^2 \text{Li}_2(c x)}{3 c}+\frac{(-2 b-3 a c) \text{Li}_2(c x)}{6 c^3 (-1+c x)}\right ) \, dx+\int \left (\frac{1}{2} a x \log ^2(1-c x)+\frac{1}{3} b x^2 \log ^2(1-c x)\right ) \, dx\\ &=\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)+\frac{1}{2} a \int x \log ^2(1-c x) \, dx+\frac{1}{3} b \int x^2 \log ^2(1-c x) \, dx-\frac{1}{3} b \int x^2 \text{Li}_2(c x) \, dx-\frac{(2 b+3 a c) \int \text{Li}_2(c x) \, dx}{6 c^2}-\frac{(2 b+3 a c) \int \frac{\text{Li}_2(c x)}{-1+c x} \, dx}{6 c^2}-\frac{(2 b+3 a c) \int x \text{Li}_2(c x) \, dx}{6 c}\\ &=\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)+\frac{1}{2} a \int \left (\frac{\log ^2(1-c x)}{c}-\frac{(1-c x) \log ^2(1-c x)}{c}\right ) \, dx-\frac{1}{9} b \int x^2 \log (1-c x) \, dx+\frac{1}{9} (2 b c) \int \frac{x^3 \log (1-c x)}{1-c x} \, dx-\frac{(2 b+3 a c) \int \frac{\log ^2(1-c x)}{x} \, dx}{6 c^3}-\frac{(2 b+3 a c) \int \log (1-c x) \, dx}{6 c^2}-\frac{(2 b+3 a c) \int x \log (1-c x) \, dx}{12 c}\\ &=-\frac{(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac{1}{27} b x^3 \log (1-c x)+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)+\frac{a \int \log ^2(1-c x) \, dx}{2 c}-\frac{a \int (1-c x) \log ^2(1-c x) \, dx}{2 c}-\frac{1}{27} (b c) \int \frac{x^3}{1-c x} \, dx+\frac{1}{9} (2 b c) \int \left (-\frac{\log (1-c x)}{c^3}-\frac{x \log (1-c x)}{c^2}-\frac{x^2 \log (1-c x)}{c}-\frac{\log (1-c x)}{c^3 (-1+c x)}\right ) \, dx-\frac{1}{24} (2 b+3 a c) \int \frac{x^2}{1-c x} \, dx+\frac{(2 b+3 a c) \operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{6 c^3}-\frac{(2 b+3 a c) \int \frac{\log (c x) \log (1-c x)}{1-c x} \, dx}{3 c^2}\\ &=\frac{(2 b+3 a c) x}{6 c^2}-\frac{(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac{1}{27} b x^3 \log (1-c x)+\frac{(2 b+3 a c) (1-c x) \log (1-c x)}{6 c^3}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)-\frac{1}{9} (2 b) \int x^2 \log (1-c x) \, dx-\frac{a \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}+\frac{a \operatorname{Subst}\left (\int x \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}-\frac{(2 b) \int \log (1-c x) \, dx}{9 c^2}-\frac{(2 b) \int \frac{\log (1-c x)}{-1+c x} \, dx}{9 c^2}-\frac{(2 b) \int x \log (1-c x) \, dx}{9 c}-\frac{1}{27} (b c) \int \left (-\frac{1}{c^3}-\frac{x}{c^2}-\frac{x^2}{c}-\frac{1}{c^3 (-1+c x)}\right ) \, dx-\frac{1}{24} (2 b+3 a c) \int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx+\frac{(2 b+3 a c) \operatorname{Subst}\left (\int \frac{\log (x) \log \left (c \left (\frac{1}{c}-\frac{x}{c}\right )\right )}{x} \, dx,x,1-c x\right )}{3 c^3}\\ &=\frac{b x}{27 c^2}+\frac{5 (2 b+3 a c) x}{24 c^2}+\frac{b x^2}{54 c}+\frac{(2 b+3 a c) x^2}{48 c}+\frac{b x^3}{81}+\frac{b \log (1-c x)}{27 c^3}+\frac{(2 b+3 a c) \log (1-c x)}{24 c^3}-\frac{b x^2 \log (1-c x)}{9 c}-\frac{(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac{1}{9} b x^3 \log (1-c x)+\frac{(2 b+3 a c) (1-c x) \log (1-c x)}{6 c^3}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{a (1-c x) \log ^2(1-c x)}{2 c^2}+\frac{a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(1-c x)}{3 c^3}-\frac{1}{9} b \int \frac{x^2}{1-c x} \, dx+\frac{(2 b) \operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{9 c^3}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1-c x\right )}{9 c^3}-\frac{a \operatorname{Subst}(\int x \log (x) \, dx,x,1-c x)}{2 c^2}+\frac{a \operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{c^2}-\frac{1}{27} (2 b c) \int \frac{x^3}{1-c x} \, dx+\frac{(2 b+3 a c) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-c x\right )}{3 c^3}\\ &=\frac{7 b x}{27 c^2}+\frac{a x}{c}+\frac{5 (2 b+3 a c) x}{24 c^2}+\frac{b x^2}{54 c}+\frac{(2 b+3 a c) x^2}{48 c}+\frac{b x^3}{81}+\frac{a (1-c x)^2}{8 c^2}+\frac{b \log (1-c x)}{27 c^3}+\frac{(2 b+3 a c) \log (1-c x)}{24 c^3}-\frac{b x^2 \log (1-c x)}{9 c}-\frac{(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac{1}{9} b x^3 \log (1-c x)+\frac{2 b (1-c x) \log (1-c x)}{9 c^3}+\frac{a (1-c x) \log (1-c x)}{c^2}+\frac{(2 b+3 a c) (1-c x) \log (1-c x)}{6 c^3}-\frac{a (1-c x)^2 \log (1-c x)}{4 c^2}-\frac{b \log ^2(1-c x)}{9 c^3}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{a (1-c x) \log ^2(1-c x)}{2 c^2}+\frac{a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(1-c x)}{3 c^3}+\frac{(2 b+3 a c) \text{Li}_3(1-c x)}{3 c^3}-\frac{1}{9} b \int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx-\frac{1}{27} (2 b c) \int \left (-\frac{1}{c^3}-\frac{x}{c^2}-\frac{x^2}{c}-\frac{1}{c^3 (-1+c x)}\right ) \, dx\\ &=\frac{4 b x}{9 c^2}+\frac{a x}{c}+\frac{5 (2 b+3 a c) x}{24 c^2}+\frac{b x^2}{9 c}+\frac{(2 b+3 a c) x^2}{48 c}+\frac{b x^3}{27}+\frac{a (1-c x)^2}{8 c^2}+\frac{2 b \log (1-c x)}{9 c^3}+\frac{(2 b+3 a c) \log (1-c x)}{24 c^3}-\frac{b x^2 \log (1-c x)}{9 c}-\frac{(2 b+3 a c) x^2 \log (1-c x)}{24 c}-\frac{1}{9} b x^3 \log (1-c x)+\frac{2 b (1-c x) \log (1-c x)}{9 c^3}+\frac{a (1-c x) \log (1-c x)}{c^2}+\frac{(2 b+3 a c) (1-c x) \log (1-c x)}{6 c^3}-\frac{a (1-c x)^2 \log (1-c x)}{4 c^2}-\frac{b \log ^2(1-c x)}{9 c^3}+\frac{1}{9} b x^3 \log ^2(1-c x)-\frac{a (1-c x) \log ^2(1-c x)}{2 c^2}+\frac{a (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{(2 b+3 a c) \log (c x) \log ^2(1-c x)}{6 c^3}-\frac{(2 b+3 a c) x \text{Li}_2(c x)}{6 c^2}-\frac{(2 b+3 a c) x^2 \text{Li}_2(c x)}{12 c}-\frac{1}{9} b x^3 \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(c x)}{6 c^3}+\frac{1}{6} \left (3 a x^2+2 b x^3\right ) \log (1-c x) \text{Li}_2(c x)-\frac{(2 b+3 a c) \log (1-c x) \text{Li}_2(1-c x)}{3 c^3}+\frac{(2 b+3 a c) \text{Li}_3(1-c x)}{3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.610532, size = 362, normalized size = 0.66 \[ \frac{12 \text{PolyLog}(2,c x) \left (6 \log (1-c x) \left (3 a c \left (c^2 x^2-1\right )+2 b \left (c^3 x^3-1\right )\right )-c x \left (9 a c (c x+2)+2 b \left (2 c^2 x^2+3 c x+6\right )\right )\right )-144 (3 a c+2 b) \log (1-c x) \text{PolyLog}(2,1-c x)+432 a c \text{PolyLog}(3,1-c x)+288 b \text{PolyLog}(3,1-c x)+81 a c^3 x^2+108 a c^3 x^2 \log ^2(1-c x)-162 a c^3 x^2 \log (1-c x)+594 a c^2 x-432 a c^2 x \log (1-c x)-108 a c \log ^2(1-c x)-216 a c \log (c x) \log ^2(1-c x)+594 a c \log (1-c x)-378 a c+16 b c^3 x^3+66 b c^2 x^2+48 b c^3 x^3 \log ^2(1-c x)-48 b c^3 x^3 \log (1-c x)-84 b c^2 x^2 \log (1-c x)+372 b c x-48 b \log ^2(1-c x)-144 b \log (c x) \log ^2(1-c x)-240 b c x \log (1-c x)+372 b \log (1-c x)}{432 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-378*a*c + 372*b*c*x + 594*a*c^2*x + 66*b*c^2*x^2 + 81*a*c^3*x^2 + 16*b*c^3*x^3 + 372*b*Log[1 - c*x] + 594*a*
c*Log[1 - c*x] - 240*b*c*x*Log[1 - c*x] - 432*a*c^2*x*Log[1 - c*x] - 84*b*c^2*x^2*Log[1 - c*x] - 162*a*c^3*x^2
*Log[1 - c*x] - 48*b*c^3*x^3*Log[1 - c*x] - 48*b*Log[1 - c*x]^2 - 108*a*c*Log[1 - c*x]^2 + 108*a*c^3*x^2*Log[1
- c*x]^2 + 48*b*c^3*x^3*Log[1 - c*x]^2 - 144*b*Log[c*x]*Log[1 - c*x]^2 - 216*a*c*Log[c*x]*Log[1 - c*x]^2 + 12
*(-(c*x*(9*a*c*(2 + c*x) + 2*b*(6 + 3*c*x + 2*c^2*x^2))) + 6*(3*a*c*(-1 + c^2*x^2) + 2*b*(-1 + c^3*x^3))*Log[1
- c*x])*PolyLog[2, c*x] - 144*(2*b + 3*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 288*b*PolyLog[3, 1 - c*x] + 43
2*a*c*PolyLog[3, 1 - c*x])/(432*c^3)

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int x \left ( bx+a \right ) \ln \left ( -cx+1 \right ){\it polylog} \left ( 2,cx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03758, size = 466, normalized size = 0.85 \begin{align*} -\frac{1}{432} \, c{\left (\frac{72 \,{\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 (3 \, a c + 2 \, b\right )}}{c^{4}} - \frac{16 \, b c^{3} x^{3} + 3 \,{\left (27 \, a c^{3} + 22 \, b c^{2}\right )} x^{2} + 6 \,{\left (99 \, a c^{2} + 62 \, b c\right )} x - 12 \,{\left (4 \, b c^{3} x^{3} + 3 \,{\left (3 \, a c^{3} + 2 \, b c^{2}\right )} x^{2} + 6 \,{\left (3 \, a c^{2} + 2 \, b c\right )} x + 6 \,{\left (3 \, a c + 2 \, b\right )} \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) - 2 \,{\left (16 \, b c^{3} x^{3} + 6 \,{\left (9 \, a c^{3} + 5 \, b c^{2}\right )} x^{2} - 297 \, a c + 6 \,{\left (27 \, a c^{2} + 16 \, b c\right )} x - 186 \, b\right )} \log \left (-c x + 1\right )}{c^{4}}\right )} + \frac{1}{216} \,{\left (\frac{27 \,{\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 )} a}{c^{2}} + \frac{4 \,{\left (18 \, c^{3} x^{3}{\rm Li}_2\left (c x\right ) - 2 \, c^{3} x^{3} - 3 \, c^{2} x^{2} - 6 \, c x + 6 \,{\left (c^{3} x^{3} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{3}}\right )} \log \left (-c x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/432*c*(72*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))*(3*a*c + 2*
b)/c^4 - (16*b*c^3*x^3 + 3*(27*a*c^3 + 22*b*c^2)*x^2 + 6*(99*a*c^2 + 62*b*c)*x - 12*(4*b*c^3*x^3 + 3*(3*a*c^3
+ 2*b*c^2)*x^2 + 6*(3*a*c^2 + 2*b*c)*x + 6*(3*a*c + 2*b)*log(-c*x + 1))*dilog(c*x) - 2*(16*b*c^3*x^3 + 6*(9*a*
c^3 + 5*b*c^2)*x^2 - 297*a*c + 6*(27*a*c^2 + 16*b*c)*x - 186*b)*log(-c*x + 1))/c^4) + 1/216*(27*(4*c^2*x^2*dil
og(c*x) - c^2*x^2 - 2*c*x + 2*(c^2*x^2 - 1)*log(-c*x + 1))*a/c^2 + 4*(18*c^3*x^3*dilog(c*x) - 2*c^3*x^3 - 3*c^
2*x^2 - 6*c*x + 6*(c^3*x^3 - 1)*log(-c*x + 1))*b/c^3)*log(-c*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)*x*dilog(c*x)*log(-c*x + 1), x)