3.161 \(\int x^3 \log (1-c x) \text {Li}_2(c x) \, dx\)

Optimal. Leaf size=300 \[ \frac {\text {Li}_3(1-c x)}{2 c^4}-\frac {\text {Li}_2(c x) \log (1-c x)}{4 c^4}-\frac {\text {Li}_2(1-c x) \log (1-c x)}{2 c^4}-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {\log ^2(1-c x)}{16 c^4}+\frac {3 (1-c x) \log (1-c x)}{8 c^4}+\frac {139 \log (1-c x)}{576 c^4}-\frac {x \text {Li}_2(c x)}{4 c^3}+\frac {355 x}{576 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}+\frac {139 x^2}{1152 c^2}-\frac {x^2 \log (1-c x)}{8 c^2}-\frac {1}{16} x^4 \text {Li}_2(c x)+\frac {1}{4} x^4 \text {Li}_2(c x) \log (1-c x)-\frac {x^3 \text {Li}_2(c x)}{12 c}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {3}{64} x^4 \log (1-c x)+\frac {67 x^3}{1728 c}-\frac {5 x^3 \log (1-c x)}{72 c}+\frac {3 x^4}{256} \]

[Out]

355/576*x/c^3+139/1152*x^2/c^2+67/1728*x^3/c+3/256*x^4+139/576*ln(-c*x+1)/c^4-1/8*x^2*ln(-c*x+1)/c^2-5/72*x^3*
ln(-c*x+1)/c-3/64*x^4*ln(-c*x+1)+3/8*(-c*x+1)*ln(-c*x+1)/c^4-1/16*ln(-c*x+1)^2/c^4+1/16*x^4*ln(-c*x+1)^2-1/4*l
n(c*x)*ln(-c*x+1)^2/c^4-1/4*x*polylog(2,c*x)/c^3-1/8*x^2*polylog(2,c*x)/c^2-1/12*x^3*polylog(2,c*x)/c-1/16*x^4
*polylog(2,c*x)-1/4*ln(-c*x+1)*polylog(2,c*x)/c^4+1/4*x^4*ln(-c*x+1)*polylog(2,c*x)-1/2*ln(-c*x+1)*polylog(2,-
c*x+1)/c^4+1/2*polylog(3,-c*x+1)/c^4

________________________________________________________________________________________

Rubi [A]  time = 0.52, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 38, number of rules used = 16, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6591, 2395, 43, 6603, 2398, 2410, 2389, 2295, 2390, 2301, 6586, 6596, 2396, 2433, 2374, 6589} \[ -\frac {x^2 \text {PolyLog}(2,c x)}{8 c^2}-\frac {x \text {PolyLog}(2,c x)}{4 c^3}+\frac {\text {PolyLog}(3,1-c x)}{2 c^4}-\frac {\log (1-c x) \text {PolyLog}(2,c x)}{4 c^4}-\frac {\log (1-c x) \text {PolyLog}(2,1-c x)}{2 c^4}-\frac {1}{16} x^4 \text {PolyLog}(2,c x)-\frac {x^3 \text {PolyLog}(2,c x)}{12 c}+\frac {1}{4} x^4 \log (1-c x) \text {PolyLog}(2,c x)+\frac {139 x^2}{1152 c^2}-\frac {x^2 \log (1-c x)}{8 c^2}+\frac {355 x}{576 c^3}-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {\log ^2(1-c x)}{16 c^4}+\frac {3 (1-c x) \log (1-c x)}{8 c^4}+\frac {139 \log (1-c x)}{576 c^4}+\frac {67 x^3}{1728 c}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {3}{64} x^4 \log (1-c x)-\frac {5 x^3 \log (1-c x)}{72 c}+\frac {3 x^4}{256} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(355*x)/(576*c^3) + (139*x^2)/(1152*c^2) + (67*x^3)/(1728*c) + (3*x^4)/256 + (139*Log[1 - c*x])/(576*c^4) - (x
^2*Log[1 - c*x])/(8*c^2) - (5*x^3*Log[1 - c*x])/(72*c) - (3*x^4*Log[1 - c*x])/64 + (3*(1 - c*x)*Log[1 - c*x])/
(8*c^4) - Log[1 - c*x]^2/(16*c^4) + (x^4*Log[1 - c*x]^2)/16 - (Log[c*x]*Log[1 - c*x]^2)/(4*c^4) - (x*PolyLog[2
, c*x])/(4*c^3) - (x^2*PolyLog[2, c*x])/(8*c^2) - (x^3*PolyLog[2, c*x])/(12*c) - (x^4*PolyLog[2, c*x])/16 - (L
og[1 - c*x]*PolyLog[2, c*x])/(4*c^4) + (x^4*Log[1 - c*x]*PolyLog[2, c*x])/4 - (Log[1 - c*x]*PolyLog[2, 1 - c*x
])/(2*c^4) + PolyLog[3, 1 - c*x]/(2*c^4)

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 2295

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

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 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 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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int x^3 \log (1-c x) \text {Li}_2(c x) \, dx &=\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)+\frac {1}{4} \int x^3 \log ^2(1-c x) \, dx+\frac {1}{4} c \int \left (-\frac {\text {Li}_2(c x)}{c^4}-\frac {x \text {Li}_2(c x)}{c^3}-\frac {x^2 \text {Li}_2(c x)}{c^2}-\frac {x^3 \text {Li}_2(c x)}{c}-\frac {\text {Li}_2(c x)}{c^4 (-1+c x)}\right ) \, dx\\ &=\frac {1}{16} x^4 \log ^2(1-c x)+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {1}{4} \int x^3 \text {Li}_2(c x) \, dx-\frac {\int \text {Li}_2(c x) \, dx}{4 c^3}-\frac {\int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{4 c^3}-\frac {\int x \text {Li}_2(c x) \, dx}{4 c^2}-\frac {\int x^2 \text {Li}_2(c x) \, dx}{4 c}+\frac {1}{8} c \int \frac {x^4 \log (1-c x)}{1-c x} \, dx\\ &=\frac {1}{16} x^4 \log ^2(1-c x)-\frac {x \text {Li}_2(c x)}{4 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}-\frac {x^3 \text {Li}_2(c x)}{12 c}-\frac {1}{16} x^4 \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {1}{16} \int x^3 \log (1-c x) \, dx-\frac {\int \frac {\log ^2(1-c x)}{x} \, dx}{4 c^4}-\frac {\int \log (1-c x) \, dx}{4 c^3}-\frac {\int x \log (1-c x) \, dx}{8 c^2}-\frac {\int x^2 \log (1-c x) \, dx}{12 c}+\frac {1}{8} c \int \left (-\frac {\log (1-c x)}{c^4}-\frac {x \log (1-c x)}{c^3}-\frac {x^2 \log (1-c x)}{c^2}-\frac {x^3 \log (1-c x)}{c}-\frac {\log (1-c x)}{c^4 (-1+c x)}\right ) \, dx\\ &=-\frac {x^2 \log (1-c x)}{16 c^2}-\frac {x^3 \log (1-c x)}{36 c}-\frac {1}{64} x^4 \log (1-c x)+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {x \text {Li}_2(c x)}{4 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}-\frac {x^3 \text {Li}_2(c x)}{12 c}-\frac {1}{16} x^4 \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {1}{36} \int \frac {x^3}{1-c x} \, dx-\frac {1}{8} \int x^3 \log (1-c x) \, dx+\frac {\operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{4 c^4}-\frac {\int \log (1-c x) \, dx}{8 c^3}-\frac {\int \frac {\log (1-c x)}{-1+c x} \, dx}{8 c^3}-\frac {\int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{2 c^3}-\frac {\int x \log (1-c x) \, dx}{8 c^2}-\frac {\int \frac {x^2}{1-c x} \, dx}{16 c}-\frac {\int x^2 \log (1-c x) \, dx}{8 c}-\frac {1}{64} c \int \frac {x^4}{1-c x} \, dx\\ &=\frac {x}{4 c^3}-\frac {x^2 \log (1-c x)}{8 c^2}-\frac {5 x^3 \log (1-c x)}{72 c}-\frac {3}{64} x^4 \log (1-c x)+\frac {(1-c x) \log (1-c x)}{4 c^4}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {x \text {Li}_2(c x)}{4 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}-\frac {x^3 \text {Li}_2(c x)}{12 c}-\frac {1}{16} x^4 \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {1}{36} \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} \int \frac {x^3}{1-c x} \, dx+\frac {\operatorname {Subst}(\int \log (x) \, dx,x,1-c x)}{8 c^4}-\frac {\operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{8 c^4}+\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 )}{2 c^4}-\frac {\int \frac {x^2}{1-c x} \, dx}{16 c}-\frac {\int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx}{16 c}-\frac {1}{64} c \int \left (-\frac {1}{c^4}-\frac {x}{c^3}-\frac {x^2}{c^2}-\frac {x^3}{c}-\frac {1}{c^4 (-1+c x)}\right ) \, dx-\frac {1}{32} c \int \frac {x^4}{1-c x} \, dx\\ &=\frac {277 x}{576 c^3}+\frac {61 x^2}{1152 c^2}+\frac {25 x^3}{1728 c}+\frac {x^4}{256}+\frac {61 \log (1-c x)}{576 c^4}-\frac {x^2 \log (1-c x)}{8 c^2}-\frac {5 x^3 \log (1-c x)}{72 c}-\frac {3}{64} x^4 \log (1-c x)+\frac {3 (1-c x) \log (1-c x)}{8 c^4}-\frac {\log ^2(1-c x)}{16 c^4}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {x \text {Li}_2(c x)}{4 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}-\frac {x^3 \text {Li}_2(c x)}{12 c}-\frac {1}{16} x^4 \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(1-c x)}{2 c^4}-\frac {1}{24} \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{2 c^4}-\frac {\int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx}{16 c}-\frac {1}{32} c \int \left (-\frac {1}{c^4}-\frac {x}{c^3}-\frac {x^2}{c^2}-\frac {x^3}{c}-\frac {1}{c^4 (-1+c x)}\right ) \, dx\\ &=\frac {355 x}{576 c^3}+\frac {139 x^2}{1152 c^2}+\frac {67 x^3}{1728 c}+\frac {3 x^4}{256}+\frac {139 \log (1-c x)}{576 c^4}-\frac {x^2 \log (1-c x)}{8 c^2}-\frac {5 x^3 \log (1-c x)}{72 c}-\frac {3}{64} x^4 \log (1-c x)+\frac {3 (1-c x) \log (1-c x)}{8 c^4}-\frac {\log ^2(1-c x)}{16 c^4}+\frac {1}{16} x^4 \log ^2(1-c x)-\frac {\log (c x) \log ^2(1-c x)}{4 c^4}-\frac {x \text {Li}_2(c x)}{4 c^3}-\frac {x^2 \text {Li}_2(c x)}{8 c^2}-\frac {x^3 \text {Li}_2(c x)}{12 c}-\frac {1}{16} x^4 \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(c x)}{4 c^4}+\frac {1}{4} x^4 \log (1-c x) \text {Li}_2(c x)-\frac {\log (1-c x) \text {Li}_2(1-c x)}{2 c^4}+\frac {\text {Li}_3(1-c x)}{2 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 223, normalized size = 0.74 \[ \frac {81 c^4 x^4+432 c^4 x^4 \log ^2(1-c x)-324 c^4 x^4 \log (1-c x)+268 c^3 x^3-480 c^3 x^3 \log (1-c x)+834 c^2 x^2-864 c^2 x^2 \log (1-c x)+144 \text {Li}_2(c x) \left (12 \left (c^4 x^4-1\right ) \log (1-c x)-c x \left (3 c^3 x^3+4 c^2 x^2+6 c x+12\right )\right )+3456 \text {Li}_3(1-c x)-3456 \text {Li}_2(1-c x) \log (1-c x)+4260 c x-1728 \log (c x) \log ^2(1-c x)-432 \log ^2(1-c x)-2592 c x \log (1-c x)+4260 \log (1-c x)}{6912 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(4260*c*x + 834*c^2*x^2 + 268*c^3*x^3 + 81*c^4*x^4 + 4260*Log[1 - c*x] - 2592*c*x*Log[1 - c*x] - 864*c^2*x^2*L
og[1 - c*x] - 480*c^3*x^3*Log[1 - c*x] - 324*c^4*x^4*Log[1 - c*x] - 432*Log[1 - c*x]^2 + 432*c^4*x^4*Log[1 - c
*x]^2 - 1728*Log[c*x]*Log[1 - c*x]^2 + 144*(-(c*x*(12 + 6*c*x + 4*c^2*x^2 + 3*c^3*x^3)) + 12*(-1 + c^4*x^4)*Lo
g[1 - c*x])*PolyLog[2, c*x] - 3456*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 3456*PolyLog[3, 1 - c*x])/(6912*c^4)

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fricas [F]  time = 2.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(-c*x+1)*polylog(2,c*x),x, algorithm="fricas")

[Out]

integral(x^3*dilog(c*x)*log(-c*x + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(-c*x+1)*polylog(2,c*x),x, algorithm="giac")

[Out]

integrate(x^3*dilog(c*x)*log(-c*x + 1), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int x^{3} \ln \left (-c x +1\right ) \polylog \left (2, c x \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*ln(-c*x+1)*polylog(2,c*x),x)

[Out]

int(x^3*ln(-c*x+1)*polylog(2,c*x),x)

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maxima [A]  time = 0.35, size = 376, normalized size = 1.25 \[ \frac {9 \, c^{4} {\left (\frac {3 \, c^{3} x^{4} + 4 \, c^{2} x^{3} + 6 \, c x^{2} + 12 \, x}{c^{4}} + \frac {12 \, \log \left (c x - 1\right )}{c^{5}}\right )} + 24 \, c^{3} {\left (\frac {2 \, c^{2} x^{3} + 3 \, c x^{2} + 6 \, x}{c^{3}} + \frac {6 \, \log \left (c x - 1\right )}{c^{4}}\right )} + 108 \, c^{2} {\left (\frac {c x^{2} + 2 \, x}{c^{2}} + \frac {2 \, \log \left (c x - 1\right )}{c^{3}}\right )} + 432 \, c {\left (\frac {x}{c} + \frac {\log \left (c x - 1\right )}{c^{2}}\right )} + \frac {2 \, {\left (27 \, c^{4} x^{4} + 92 \, c^{3} x^{3} + 300 \, c^{2} x^{2} + 1680 \, c x - 72 \, {\left (3 \, c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 12 \, c x + 12 \, \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 12 \, {\left (9 \, c^{4} x^{4} + 14 \, c^{3} x^{3} + 27 \, c^{2} x^{2} + 90 \, c x - 140\right )} \log \left (-c x + 1\right )\right )}}{c} - \frac {1728 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )}}{c}}{6912 \, c^{3}} + \frac {{\left (48 \, c^{4} x^{4} {\rm Li}_2\left (c x\right ) - 3 \, c^{4} x^{4} - 4 \, c^{3} x^{3} - 6 \, c^{2} x^{2} - 12 \, c x + 12 \, {\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{192 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(-c*x+1)*polylog(2,c*x),x, algorithm="maxima")

[Out]

1/6912*(9*c^4*((3*c^3*x^4 + 4*c^2*x^3 + 6*c*x^2 + 12*x)/c^4 + 12*log(c*x - 1)/c^5) + 24*c^3*((2*c^2*x^3 + 3*c*
x^2 + 6*x)/c^3 + 6*log(c*x - 1)/c^4) + 108*c^2*((c*x^2 + 2*x)/c^2 + 2*log(c*x - 1)/c^3) + 432*c*(x/c + log(c*x
- 1)/c^2) + 2*(27*c^4*x^4 + 92*c^3*x^3 + 300*c^2*x^2 + 1680*c*x - 72*(3*c^4*x^4 + 4*c^3*x^3 + 6*c^2*x^2 + 12*
c*x + 12*log(-c*x + 1))*dilog(c*x) - 12*(9*c^4*x^4 + 14*c^3*x^3 + 27*c^2*x^2 + 90*c*x - 140)*log(-c*x + 1))/c
- 1728*(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^3 + 1/192*(4
8*c^4*x^4*dilog(c*x) - 3*c^4*x^4 - 4*c^3*x^3 - 6*c^2*x^2 - 12*c*x + 12*(c^4*x^4 - 1)*log(-c*x + 1))*log(-c*x +
1)/c^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(1 - c*x)*polylog(2, c*x),x)

[Out]

int(x^3*log(1 - c*x)*polylog(2, c*x), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*ln(-c*x+1)*polylog(2,c*x),x)

[Out]

Timed out

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