3.163 \(\int x \log (1-c x) \text{PolyLog}(2,c x) \, dx\)

Optimal. Leaf size=262 \[ \frac{\text{PolyLog}(3,1-c x)}{c^2}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{2 c^2}-\frac{\log (1-c x) \text{PolyLog}(2,1-c x)}{c^2}-\frac{1}{4} x^2 \text{PolyLog}(2,c x)+\frac{1}{2} x^2 \log (1-c x) \text{PolyLog}(2,c x)-\frac{x \text{PolyLog}(2,c x)}{2 c}+\frac{(1-c x)^2}{8 c^2}+\frac{(1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{(1-c x) \log ^2(1-c x)}{2 c^2}-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{(1-c x)^2 \log (1-c x)}{4 c^2}+\frac{3 (1-c x) \log (1-c x)}{2 c^2}+\frac{\log (1-c x)}{8 c^2}-\frac{1}{8} x^2 \log (1-c x)+\frac{13 x}{8 c}+\frac{x^2}{16} \]

[Out]

(13*x)/(8*c) + x^2/16 + (1 - c*x)^2/(8*c^2) + Log[1 - c*x]/(8*c^2) - (x^2*Log[1 - c*x])/8 + (3*(1 - c*x)*Log[1
- c*x])/(2*c^2) - ((1 - c*x)^2*Log[1 - c*x])/(4*c^2) - ((1 - c*x)*Log[1 - c*x]^2)/(2*c^2) + ((1 - c*x)^2*Log[
1 - c*x]^2)/(4*c^2) - (Log[c*x]*Log[1 - c*x]^2)/(2*c^2) - (x*PolyLog[2, c*x])/(2*c) - (x^2*PolyLog[2, c*x])/4
- (Log[1 - c*x]*PolyLog[2, c*x])/(2*c^2) + (x^2*Log[1 - c*x]*PolyLog[2, c*x])/2 - (Log[1 - c*x]*PolyLog[2, 1 -
c*x])/c^2 + PolyLog[3, 1 - c*x]/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.254646, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 17, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.214, Rules used = {6591, 2395, 43, 6603, 2401, 2389, 2296, 2295, 2390, 2305, 2304, 6586, 6596, 2396, 2433, 2374, 6589} \[ \frac{\text{PolyLog}(3,1-c x)}{c^2}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{2 c^2}-\frac{\log (1-c x) \text{PolyLog}(2,1-c x)}{c^2}-\frac{1}{4} x^2 \text{PolyLog}(2,c x)+\frac{1}{2} x^2 \log (1-c x) \text{PolyLog}(2,c x)-\frac{x \text{PolyLog}(2,c x)}{2 c}+\frac{(1-c x)^2}{8 c^2}+\frac{(1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{(1-c x) \log ^2(1-c x)}{2 c^2}-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{(1-c x)^2 \log (1-c x)}{4 c^2}+\frac{3 (1-c x) \log (1-c x)}{2 c^2}+\frac{\log (1-c x)}{8 c^2}-\frac{1}{8} x^2 \log (1-c x)+\frac{13 x}{8 c}+\frac{x^2}{16} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(13*x)/(8*c) + x^2/16 + (1 - c*x)^2/(8*c^2) + Log[1 - c*x]/(8*c^2) - (x^2*Log[1 - c*x])/8 + (3*(1 - c*x)*Log[1
- c*x])/(2*c^2) - ((1 - c*x)^2*Log[1 - c*x])/(4*c^2) - ((1 - c*x)*Log[1 - c*x]^2)/(2*c^2) + ((1 - c*x)^2*Log[
1 - c*x]^2)/(4*c^2) - (Log[c*x]*Log[1 - c*x]^2)/(2*c^2) - (x*PolyLog[2, c*x])/(2*c) - (x^2*PolyLog[2, c*x])/4
- (Log[1 - c*x]*PolyLog[2, c*x])/(2*c^2) + (x^2*Log[1 - c*x]*PolyLog[2, c*x])/2 - (Log[1 - c*x]*PolyLog[2, 1 -
c*x])/c^2 + PolyLog[3, 1 - c*x]/c^2

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 6603

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(x_)^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x
_Symbol] :> Simp[(x^(m + 1)*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)])/(m + 1), x] + (Dist[b/(m + 1),
Int[ExpandIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], x^(m + 1)/(a + b*x), x], x], x] - Dist[(
e*h*n)/(m + 1), Int[ExpandIntegrand[PolyLog[2, c*(a + b*x)], x^(m + 1)/(d + e*x), x], x], x]) /; FreeQ[{a, b,
c, d, e, f, g, h, n}, x] && IntegerQ[m] && NeQ[m, -1]

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 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 \log (1-c x) \text{Li}_2(c x) \, dx &=\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)+\frac{1}{2} \int x \log ^2(1-c x) \, dx+\frac{1}{2} c \int \left (-\frac{\text{Li}_2(c x)}{c^2}-\frac{x \text{Li}_2(c x)}{c}-\frac{\text{Li}_2(c x)}{c^2 (-1+c x)}\right ) \, dx\\ &=\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)+\frac{1}{2} \int \left (\frac{\log ^2(1-c x)}{c}-\frac{(1-c x) \log ^2(1-c x)}{c}\right ) \, dx-\frac{1}{2} \int x \text{Li}_2(c x) \, dx-\frac{\int \text{Li}_2(c x) \, dx}{2 c}-\frac{\int \frac{\text{Li}_2(c x)}{-1+c x} \, dx}{2 c}\\ &=-\frac{x \text{Li}_2(c x)}{2 c}-\frac{1}{4} x^2 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{2 c^2}+\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)-\frac{1}{4} \int x \log (1-c x) \, dx-\frac{\int \frac{\log ^2(1-c x)}{x} \, dx}{2 c^2}-\frac{\int \log (1-c x) \, dx}{2 c}+\frac{\int \log ^2(1-c x) \, dx}{2 c}-\frac{\int (1-c x) \log ^2(1-c x) \, dx}{2 c}\\ &=-\frac{1}{8} x^2 \log (1-c x)-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{x \text{Li}_2(c x)}{2 c}-\frac{1}{4} x^2 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{2 c^2}+\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{2 c^2}-\frac{\operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}+\frac{\operatorname{Subst}\left (\int x \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}-\frac{\int \frac{\log (c x) \log (1-c x)}{1-c x} \, dx}{c}-\frac{1}{8} c \int \frac{x^2}{1-c x} \, dx\\ &=\frac{x}{2 c}-\frac{1}{8} x^2 \log (1-c x)+\frac{(1-c x) \log (1-c x)}{2 c^2}-\frac{(1-c x) \log ^2(1-c x)}{2 c^2}+\frac{(1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{x \text{Li}_2(c x)}{2 c}-\frac{1}{4} x^2 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{2 c^2}+\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)-\frac{\operatorname{Subst}(\int x \log (x) \, dx,x,1-c x)}{2 c^2}+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{c^2}+\frac{\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 )}{c^2}-\frac{1}{8} c \int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx\\ &=\frac{13 x}{8 c}+\frac{x^2}{16}+\frac{(1-c x)^2}{8 c^2}+\frac{\log (1-c x)}{8 c^2}-\frac{1}{8} x^2 \log (1-c x)+\frac{3 (1-c x) \log (1-c x)}{2 c^2}-\frac{(1-c x)^2 \log (1-c x)}{4 c^2}-\frac{(1-c x) \log ^2(1-c x)}{2 c^2}+\frac{(1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{x \text{Li}_2(c x)}{2 c}-\frac{1}{4} x^2 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{2 c^2}+\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(1-c x)}{c^2}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-c x\right )}{c^2}\\ &=\frac{13 x}{8 c}+\frac{x^2}{16}+\frac{(1-c x)^2}{8 c^2}+\frac{\log (1-c x)}{8 c^2}-\frac{1}{8} x^2 \log (1-c x)+\frac{3 (1-c x) \log (1-c x)}{2 c^2}-\frac{(1-c x)^2 \log (1-c x)}{4 c^2}-\frac{(1-c x) \log ^2(1-c x)}{2 c^2}+\frac{(1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac{\log (c x) \log ^2(1-c x)}{2 c^2}-\frac{x \text{Li}_2(c x)}{2 c}-\frac{1}{4} x^2 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{2 c^2}+\frac{1}{2} x^2 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(1-c x)}{c^2}+\frac{\text{Li}_3(1-c x)}{c^2}\\ \end{align*}

Mathematica [A]  time = 0.30023, size = 160, normalized size = 0.61 \[ \frac{\left (8 \left (c^2 x^2-1\right ) \log (1-c x)-4 c x (c x+2)\right ) \text{PolyLog}(2,c x)+16 \text{PolyLog}(3,1-c x)-16 \log (1-c x) \text{PolyLog}(2,1-c x)+3 c^2 x^2+4 c^2 x^2 \log ^2(1-c x)-6 c^2 x^2 \log (1-c x)+22 c x-8 \log (c x) \log ^2(1-c x)-4 \log ^2(1-c x)-16 c x \log (1-c x)+22 \log (1-c x)-14}{16 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(-14 + 22*c*x + 3*c^2*x^2 + 22*Log[1 - c*x] - 16*c*x*Log[1 - c*x] - 6*c^2*x^2*Log[1 - c*x] - 4*Log[1 - c*x]^2
+ 4*c^2*x^2*Log[1 - c*x]^2 - 8*Log[c*x]*Log[1 - c*x]^2 + (-4*c*x*(2 + c*x) + 8*(-1 + c^2*x^2)*Log[1 - c*x])*Po
lyLog[2, c*x] - 16*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 16*PolyLog[3, 1 - c*x])/(16*c^2)

________________________________________________________________________________________

Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int x\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*ln(-c*x+1)*polylog(2,c*x),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.15223, size = 300, normalized size = 1.15 \begin{align*} \frac{c^{2}{\left (\frac{c x^{2} + 2 \, x}{c^{2}} + \frac{2 \, \log \left (c x - 1\right )}{c^{3}}\right )} + 4 \, c{\left (\frac{x}{c} + \frac{\log \left (c x - 1\right )}{c^{2}}\right )} + \frac{2 \,{\left (c^{2} x^{2} + 8 \, c x - 2 \,{\left (c^{2} x^{2} + 2 \, c x + 2 \, \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) - 2 \,{\left (c^{2} x^{2} + 3 \, c x - 4\right )} \log \left (-c x + 1\right )\right )}}{c} - \frac{8 \,{\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 )}}{c}}{16 \, c} + \frac{{\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 )} \log \left (-c x + 1\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(c^2*((c*x^2 + 2*x)/c^2 + 2*log(c*x - 1)/c^3) + 4*c*(x/c + log(c*x - 1)/c^2) + 2*(c^2*x^2 + 8*c*x - 2*(c^
2*x^2 + 2*c*x + 2*log(-c*x + 1))*dilog(c*x) - 2*(c^2*x^2 + 3*c*x - 4)*log(-c*x + 1))/c - 8*(log(c*x)*log(-c*x
+ 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))/c)/c + 1/8*(4*c^2*x^2*dilog(c*x) - c^2*x^2
- 2*c*x + 2*(c^2*x^2 - 1)*log(-c*x + 1))*log(-c*x + 1)/c^2

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x{\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*log(-c*x+1)*polylog(2,c*x),x, algorithm="fricas")

[Out]

integral(x*dilog(c*x)*log(-c*x + 1), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \log{\left (- c x + 1 \right )} \operatorname{Li}_{2}\left (c x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*log(-c*x + 1)*polylog(2, c*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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*log(-c*x+1)*polylog(2,c*x),x, algorithm="giac")

[Out]

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