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

Optimal. Leaf size=258 \[ -\frac{x \text{PolyLog}(2,c x)}{3 c^2}+\frac{2 \text{PolyLog}(3,1-c x)}{3 c^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{3 c^3}-\frac{2 \log (1-c x) \text{PolyLog}(2,1-c x)}{3 c^3}-\frac{1}{9} x^3 \text{PolyLog}(2,c x)-\frac{x^2 \text{PolyLog}(2,c x)}{6 c}+\frac{1}{3} x^3 \log (1-c x) \text{PolyLog}(2,c x)+\frac{31 x}{36 c^2}-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{\log ^2(1-c x)}{9 c^3}+\frac{5 (1-c x) \log (1-c x)}{9 c^3}+\frac{11 \log (1-c x)}{36 c^3}+\frac{11 x^2}{72 c}+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{1}{9} x^3 \log (1-c x)-\frac{7 x^2 \log (1-c x)}{36 c}+\frac{x^3}{27} \]

[Out]

(31*x)/(36*c^2) + (11*x^2)/(72*c) + x^3/27 + (11*Log[1 - c*x])/(36*c^3) - (7*x^2*Log[1 - c*x])/(36*c) - (x^3*L
og[1 - c*x])/9 + (5*(1 - c*x)*Log[1 - c*x])/(9*c^3) - Log[1 - c*x]^2/(9*c^3) + (x^3*Log[1 - c*x]^2)/9 - (Log[c
*x]*Log[1 - c*x]^2)/(3*c^3) - (x*PolyLog[2, c*x])/(3*c^2) - (x^2*PolyLog[2, c*x])/(6*c) - (x^3*PolyLog[2, c*x]
)/9 - (Log[1 - c*x]*PolyLog[2, c*x])/(3*c^3) + (x^3*Log[1 - c*x]*PolyLog[2, c*x])/3 - (2*Log[1 - c*x]*PolyLog[
2, 1 - c*x])/(3*c^3) + (2*PolyLog[3, 1 - c*x])/(3*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.404814, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 16, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6591, 2395, 43, 6603, 2398, 2410, 2389, 2295, 2390, 2301, 6586, 6596, 2396, 2433, 2374, 6589} \[ -\frac{x \text{PolyLog}(2,c x)}{3 c^2}+\frac{2 \text{PolyLog}(3,1-c x)}{3 c^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{3 c^3}-\frac{2 \log (1-c x) \text{PolyLog}(2,1-c x)}{3 c^3}-\frac{1}{9} x^3 \text{PolyLog}(2,c x)-\frac{x^2 \text{PolyLog}(2,c x)}{6 c}+\frac{1}{3} x^3 \log (1-c x) \text{PolyLog}(2,c x)+\frac{31 x}{36 c^2}-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{\log ^2(1-c x)}{9 c^3}+\frac{5 (1-c x) \log (1-c x)}{9 c^3}+\frac{11 \log (1-c x)}{36 c^3}+\frac{11 x^2}{72 c}+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{1}{9} x^3 \log (1-c x)-\frac{7 x^2 \log (1-c x)}{36 c}+\frac{x^3}{27} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(31*x)/(36*c^2) + (11*x^2)/(72*c) + x^3/27 + (11*Log[1 - c*x])/(36*c^3) - (7*x^2*Log[1 - c*x])/(36*c) - (x^3*L
og[1 - c*x])/9 + (5*(1 - c*x)*Log[1 - c*x])/(9*c^3) - Log[1 - c*x]^2/(9*c^3) + (x^3*Log[1 - c*x]^2)/9 - (Log[c
*x]*Log[1 - c*x]^2)/(3*c^3) - (x*PolyLog[2, c*x])/(3*c^2) - (x^2*PolyLog[2, c*x])/(6*c) - (x^3*PolyLog[2, c*x]
)/9 - (Log[1 - c*x]*PolyLog[2, c*x])/(3*c^3) + (x^3*Log[1 - c*x]*PolyLog[2, c*x])/3 - (2*Log[1 - c*x]*PolyLog[
2, 1 - c*x])/(3*c^3) + (2*PolyLog[3, 1 - c*x])/(3*c^3)

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 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 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 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 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^2 \log (1-c x) \text{Li}_2(c x) \, dx &=\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)+\frac{1}{3} \int x^2 \log ^2(1-c x) \, dx+\frac{1}{3} c \int \left (-\frac{\text{Li}_2(c x)}{c^3}-\frac{x \text{Li}_2(c x)}{c^2}-\frac{x^2 \text{Li}_2(c x)}{c}-\frac{\text{Li}_2(c x)}{c^3 (-1+c x)}\right ) \, dx\\ &=\frac{1}{9} x^3 \log ^2(1-c x)+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{1}{3} \int x^2 \text{Li}_2(c x) \, dx-\frac{\int \text{Li}_2(c x) \, dx}{3 c^2}-\frac{\int \frac{\text{Li}_2(c x)}{-1+c x} \, dx}{3 c^2}-\frac{\int x \text{Li}_2(c x) \, dx}{3 c}+\frac{1}{9} (2 c) \int \frac{x^3 \log (1-c x)}{1-c x} \, dx\\ &=\frac{1}{9} x^3 \log ^2(1-c x)-\frac{x \text{Li}_2(c x)}{3 c^2}-\frac{x^2 \text{Li}_2(c x)}{6 c}-\frac{1}{9} x^3 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{3 c^3}+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{1}{9} \int x^2 \log (1-c x) \, dx-\frac{\int \frac{\log ^2(1-c x)}{x} \, dx}{3 c^3}-\frac{\int \log (1-c x) \, dx}{3 c^2}-\frac{\int x \log (1-c x) \, dx}{6 c}+\frac{1}{9} (2 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{x^2 \log (1-c x)}{12 c}-\frac{1}{27} x^3 \log (1-c x)+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{x \text{Li}_2(c x)}{3 c^2}-\frac{x^2 \text{Li}_2(c x)}{6 c}-\frac{1}{9} x^3 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{3 c^3}+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{1}{12} \int \frac{x^2}{1-c x} \, dx-\frac{2}{9} \int x^2 \log (1-c x) \, dx+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{3 c^3}-\frac{2 \int \log (1-c x) \, dx}{9 c^2}-\frac{2 \int \frac{\log (1-c x)}{-1+c x} \, dx}{9 c^2}-\frac{2 \int \frac{\log (c x) \log (1-c x)}{1-c x} \, dx}{3 c^2}-\frac{2 \int x \log (1-c x) \, dx}{9 c}-\frac{1}{27} c \int \frac{x^3}{1-c x} \, dx\\ &=\frac{x}{3 c^2}-\frac{7 x^2 \log (1-c x)}{36 c}-\frac{1}{9} x^3 \log (1-c x)+\frac{(1-c x) \log (1-c x)}{3 c^3}+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{x \text{Li}_2(c x)}{3 c^2}-\frac{x^2 \text{Li}_2(c x)}{6 c}-\frac{1}{9} x^3 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{3 c^3}+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{1}{12} \int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx-\frac{1}{9} \int \frac{x^2}{1-c x} \, dx+\frac{2 \operatorname{Subst}(\int \log (x) \, dx,x,1-c x)}{9 c^3}-\frac{2 \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1-c x\right )}{9 c^3}+\frac{2 \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{1}{27} 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}{27} (2 c) \int \frac{x^3}{1-c x} \, dx\\ &=\frac{73 x}{108 c^2}+\frac{13 x^2}{216 c}+\frac{x^3}{81}+\frac{13 \log (1-c x)}{108 c^3}-\frac{7 x^2 \log (1-c x)}{36 c}-\frac{1}{9} x^3 \log (1-c x)+\frac{5 (1-c x) \log (1-c x)}{9 c^3}-\frac{\log ^2(1-c x)}{9 c^3}+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{x \text{Li}_2(c x)}{3 c^2}-\frac{x^2 \text{Li}_2(c x)}{6 c}-\frac{1}{9} x^3 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{3 c^3}+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{2 \log (1-c x) \text{Li}_2(1-c x)}{3 c^3}-\frac{1}{9} \int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx+\frac{2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-c x\right )}{3 c^3}-\frac{1}{27} (2 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{31 x}{36 c^2}+\frac{11 x^2}{72 c}+\frac{x^3}{27}+\frac{11 \log (1-c x)}{36 c^3}-\frac{7 x^2 \log (1-c x)}{36 c}-\frac{1}{9} x^3 \log (1-c x)+\frac{5 (1-c x) \log (1-c x)}{9 c^3}-\frac{\log ^2(1-c x)}{9 c^3}+\frac{1}{9} x^3 \log ^2(1-c x)-\frac{\log (c x) \log ^2(1-c x)}{3 c^3}-\frac{x \text{Li}_2(c x)}{3 c^2}-\frac{x^2 \text{Li}_2(c x)}{6 c}-\frac{1}{9} x^3 \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{3 c^3}+\frac{1}{3} x^3 \log (1-c x) \text{Li}_2(c x)-\frac{2 \log (1-c x) \text{Li}_2(1-c x)}{3 c^3}+\frac{2 \text{Li}_3(1-c x)}{3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.357232, size = 192, normalized size = 0.74 \[ \frac{12 \left (6 \left (c^3 x^3-1\right ) \log (1-c x)-c x \left (2 c^2 x^2+3 c x+6\right )\right ) \text{PolyLog}(2,c x)+144 \text{PolyLog}(3,1-c x)-144 \log (1-c x) \text{PolyLog}(2,1-c x)+8 c^3 x^3+33 c^2 x^2+24 c^3 x^3 \log ^2(1-c x)-24 c^3 x^3 \log (1-c x)-42 c^2 x^2 \log (1-c x)+186 c x-72 \log (c x) \log ^2(1-c x)-24 \log ^2(1-c x)-120 c x \log (1-c x)+186 \log (1-c x)}{216 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(186*c*x + 33*c^2*x^2 + 8*c^3*x^3 + 186*Log[1 - c*x] - 120*c*x*Log[1 - c*x] - 42*c^2*x^2*Log[1 - c*x] - 24*c^3
*x^3*Log[1 - c*x] - 24*Log[1 - c*x]^2 + 24*c^3*x^3*Log[1 - c*x]^2 - 72*Log[c*x]*Log[1 - c*x]^2 + 12*(-(c*x*(6
+ 3*c*x + 2*c^2*x^2)) + 6*(-1 + c^3*x^3)*Log[1 - c*x])*PolyLog[2, c*x] - 144*Log[1 - c*x]*PolyLog[2, 1 - c*x]
+ 144*PolyLog[3, 1 - c*x])/(216*c^3)

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\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^2*ln(-c*x+1)*polylog(2,c*x),x)

[Out]

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

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Maxima [A]  time = 1.11877, size = 400, normalized size = 1.55 \begin{align*} \frac{4 \, c^{3}{\left (\frac{2 \, c^{2} x^{3} + 3 \, c x^{2} + 6 \, x}{c^{3}} + \frac{6 \, \log \left (c x - 1\right )}{c^{4}}\right )} + 18 \, c^{2}{\left (\frac{c x^{2} + 2 \, x}{c^{2}} + \frac{2 \, \log \left (c x - 1\right )}{c^{3}}\right )} + 72 \, c{\left (\frac{x}{c} + \frac{\log \left (c x - 1\right )}{c^{2}}\right )} + \frac{16 \, c^{3} x^{3} + 69 \, c^{2} x^{2} + 426 \, c x - 36 \,{\left (2 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + 6 \, c x + 6 \, \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) - 6 \,{\left (8 \, c^{3} x^{3} + 15 \, c^{2} x^{2} + 48 \, c x - 71\right )} \log \left (-c x + 1\right )}{c} - \frac{216 \,{\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}}{648 \, c^{2}} + \frac{{\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 )} \log \left (-c x + 1\right )}{54 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/648*(4*c^3*((2*c^2*x^3 + 3*c*x^2 + 6*x)/c^3 + 6*log(c*x - 1)/c^4) + 18*c^2*((c*x^2 + 2*x)/c^2 + 2*log(c*x -
1)/c^3) + 72*c*(x/c + log(c*x - 1)/c^2) + (16*c^3*x^3 + 69*c^2*x^2 + 426*c*x - 36*(2*c^3*x^3 + 3*c^2*x^2 + 6*c
*x + 6*log(-c*x + 1))*dilog(c*x) - 6*(8*c^3*x^3 + 15*c^2*x^2 + 48*c*x - 71)*log(-c*x + 1))/c - 216*(log(c*x)*l
og(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))/c)/c^2 + 1/54*(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))*log(-c*x + 1)/c^3

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

[Out]

integral(x^2*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**2*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 x^{2}{\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^2*log(-c*x+1)*polylog(2,c*x),x, algorithm="giac")

[Out]

integrate(x^2*dilog(c*x)*log(-c*x + 1), x)