3.169 \(\int \frac{\log (1-c x) \text{PolyLog}(2,c x)}{x^5} \, dx\)

Optimal. Leaf size=287 \[ \frac{c^2 \text{PolyLog}(2,c x)}{8 x^2}-\frac{1}{8} c^4 \text{PolyLog}(2,c x)-\frac{1}{4} c^4 \text{PolyLog}(3,c x)-\frac{1}{2} c^4 \text{PolyLog}(3,1-c x)+\frac{c^3 \text{PolyLog}(2,c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{PolyLog}(2,c x)+\frac{1}{2} c^4 \log (1-c x) \text{PolyLog}(2,1-c x)+\frac{c \text{PolyLog}(2,c x)}{12 x^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{4 x^4}+\frac{5 c^2}{144 x^2}-\frac{c^2 \log (1-c x)}{8 x^2}+\frac{7 c^3}{36 x}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{16} c^4 \log ^2(1-c x)-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}-\frac{5 c \log (1-c x)}{72 x^3} \]

[Out]

(5*c^2)/(144*x^2) + (7*c^3)/(36*x) - (41*c^4*Log[x])/72 + (41*c^4*Log[1 - c*x])/72 - (5*c*Log[1 - c*x])/(72*x^
3) - (c^2*Log[1 - c*x])/(8*x^2) - (3*c^3*Log[1 - c*x])/(8*x) - (c^4*Log[1 - c*x]^2)/16 + Log[1 - c*x]^2/(16*x^
4) + (c^4*Log[c*x]*Log[1 - c*x]^2)/4 - (c^4*PolyLog[2, c*x])/8 + (c*PolyLog[2, c*x])/(12*x^3) + (c^2*PolyLog[2
, c*x])/(8*x^2) + (c^3*PolyLog[2, c*x])/(4*x) + (c^4*Log[1 - c*x]*PolyLog[2, c*x])/4 - (Log[1 - c*x]*PolyLog[2
, c*x])/(4*x^4) + (c^4*Log[1 - c*x]*PolyLog[2, 1 - c*x])/2 - (c^4*PolyLog[3, c*x])/4 - (c^4*PolyLog[3, 1 - c*x
])/2

________________________________________________________________________________________

Rubi [A]  time = 0.449191, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 37, number of rules used = 17, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.062, Rules used = {6591, 2395, 44, 6603, 2398, 2410, 36, 29, 31, 2391, 2390, 2301, 6589, 6596, 2396, 2433, 2374} \[ \frac{c^2 \text{PolyLog}(2,c x)}{8 x^2}-\frac{1}{8} c^4 \text{PolyLog}(2,c x)-\frac{1}{4} c^4 \text{PolyLog}(3,c x)-\frac{1}{2} c^4 \text{PolyLog}(3,1-c x)+\frac{c^3 \text{PolyLog}(2,c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{PolyLog}(2,c x)+\frac{1}{2} c^4 \log (1-c x) \text{PolyLog}(2,1-c x)+\frac{c \text{PolyLog}(2,c x)}{12 x^3}-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{4 x^4}+\frac{5 c^2}{144 x^2}-\frac{c^2 \log (1-c x)}{8 x^2}+\frac{7 c^3}{36 x}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{16} c^4 \log ^2(1-c x)-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}-\frac{5 c \log (1-c x)}{72 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Log[1 - c*x]*PolyLog[2, c*x])/x^5,x]

[Out]

(5*c^2)/(144*x^2) + (7*c^3)/(36*x) - (41*c^4*Log[x])/72 + (41*c^4*Log[1 - c*x])/72 - (5*c*Log[1 - c*x])/(72*x^
3) - (c^2*Log[1 - c*x])/(8*x^2) - (3*c^3*Log[1 - c*x])/(8*x) - (c^4*Log[1 - c*x]^2)/16 + Log[1 - c*x]^2/(16*x^
4) + (c^4*Log[c*x]*Log[1 - c*x]^2)/4 - (c^4*PolyLog[2, c*x])/8 + (c*PolyLog[2, c*x])/(12*x^3) + (c^2*PolyLog[2
, c*x])/(8*x^2) + (c^3*PolyLog[2, c*x])/(4*x) + (c^4*Log[1 - c*x]*PolyLog[2, c*x])/4 - (Log[1 - c*x]*PolyLog[2
, c*x])/(4*x^4) + (c^4*Log[1 - c*x]*PolyLog[2, 1 - c*x])/2 - (c^4*PolyLog[3, c*x])/4 - (c^4*PolyLog[3, 1 - c*x
])/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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2391

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\log (1-c x) \text{Li}_2(c x)}{x^5} \, dx &=-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} \int \frac{\log ^2(1-c x)}{x^5} \, dx-\frac{1}{4} c \int \left (\frac{\text{Li}_2(c x)}{x^4}+\frac{c \text{Li}_2(c x)}{x^3}+\frac{c^2 \text{Li}_2(c x)}{x^2}+\frac{c^3 \text{Li}_2(c x)}{x}-\frac{c^4 \text{Li}_2(c x)}{-1+c x}\right ) \, dx\\ &=\frac{\log ^2(1-c x)}{16 x^4}-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{8} c \int \frac{\log (1-c x)}{x^4 (1-c x)} \, dx-\frac{1}{4} c \int \frac{\text{Li}_2(c x)}{x^4} \, dx-\frac{1}{4} c^2 \int \frac{\text{Li}_2(c x)}{x^3} \, dx-\frac{1}{4} c^3 \int \frac{\text{Li}_2(c x)}{x^2} \, dx-\frac{1}{4} c^4 \int \frac{\text{Li}_2(c x)}{x} \, dx+\frac{1}{4} c^5 \int \frac{\text{Li}_2(c x)}{-1+c x} \, dx\\ &=\frac{\log ^2(1-c x)}{16 x^4}+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)+\frac{1}{12} c \int \frac{\log (1-c x)}{x^4} \, dx+\frac{1}{8} c \int \left (\frac{\log (1-c x)}{x^4}+\frac{c \log (1-c x)}{x^3}+\frac{c^2 \log (1-c x)}{x^2}+\frac{c^3 \log (1-c x)}{x}-\frac{c^4 \log (1-c x)}{-1+c x}\right ) \, dx+\frac{1}{8} c^2 \int \frac{\log (1-c x)}{x^3} \, dx+\frac{1}{4} c^3 \int \frac{\log (1-c x)}{x^2} \, dx+\frac{1}{4} c^4 \int \frac{\log ^2(1-c x)}{x} \, dx\\ &=-\frac{c \log (1-c x)}{36 x^3}-\frac{c^2 \log (1-c x)}{16 x^2}-\frac{c^3 \log (1-c x)}{4 x}+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)+\frac{1}{8} c \int \frac{\log (1-c x)}{x^4} \, dx-\frac{1}{36} c^2 \int \frac{1}{x^3 (1-c x)} \, dx+\frac{1}{8} c^2 \int \frac{\log (1-c x)}{x^3} \, dx-\frac{1}{16} c^3 \int \frac{1}{x^2 (1-c x)} \, dx+\frac{1}{8} c^3 \int \frac{\log (1-c x)}{x^2} \, dx+\frac{1}{8} c^4 \int \frac{\log (1-c x)}{x} \, dx-\frac{1}{4} c^4 \int \frac{1}{x (1-c x)} \, dx-\frac{1}{8} c^5 \int \frac{\log (1-c x)}{-1+c x} \, dx+\frac{1}{2} c^5 \int \frac{\log (c x) \log (1-c x)}{1-c x} \, dx\\ &=-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{36} c^2 \int \left (\frac{1}{x^3}+\frac{c}{x^2}+\frac{c^2}{x}-\frac{c^3}{-1+c x}\right ) \, dx-\frac{1}{24} c^2 \int \frac{1}{x^3 (1-c x)} \, dx-\frac{1}{16} c^3 \int \frac{1}{x^2 (1-c x)} \, dx-\frac{1}{16} c^3 \int \left (\frac{1}{x^2}+\frac{c}{x}-\frac{c^2}{-1+c x}\right ) \, dx-\frac{1}{8} c^4 \int \frac{1}{x (1-c x)} \, dx-\frac{1}{8} c^4 \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1-c x\right )-\frac{1}{4} c^4 \int \frac{1}{x} \, dx-\frac{1}{2} c^4 \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 )-\frac{1}{4} c^5 \int \frac{1}{1-c x} \, dx\\ &=\frac{c^2}{72 x^2}+\frac{13 c^3}{144 x}-\frac{49}{144} c^4 \log (x)+\frac{49}{144} c^4 \log (1-c x)-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}-\frac{1}{16} c^4 \log ^2(1-c x)+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{2} c^4 \log (1-c x) \text{Li}_2(1-c x)-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{24} c^2 \int \left (\frac{1}{x^3}+\frac{c}{x^2}+\frac{c^2}{x}-\frac{c^3}{-1+c x}\right ) \, dx-\frac{1}{16} c^3 \int \left (\frac{1}{x^2}+\frac{c}{x}-\frac{c^2}{-1+c x}\right ) \, dx-\frac{1}{8} c^4 \int \frac{1}{x} \, dx-\frac{1}{2} c^4 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-c x\right )-\frac{1}{8} c^5 \int \frac{1}{1-c x} \, dx\\ &=\frac{5 c^2}{144 x^2}+\frac{7 c^3}{36 x}-\frac{41}{72} c^4 \log (x)+\frac{41}{72} c^4 \log (1-c x)-\frac{5 c \log (1-c x)}{72 x^3}-\frac{c^2 \log (1-c x)}{8 x^2}-\frac{3 c^3 \log (1-c x)}{8 x}-\frac{1}{16} c^4 \log ^2(1-c x)+\frac{\log ^2(1-c x)}{16 x^4}+\frac{1}{4} c^4 \log (c x) \log ^2(1-c x)-\frac{1}{8} c^4 \text{Li}_2(c x)+\frac{c \text{Li}_2(c x)}{12 x^3}+\frac{c^2 \text{Li}_2(c x)}{8 x^2}+\frac{c^3 \text{Li}_2(c x)}{4 x}+\frac{1}{4} c^4 \log (1-c x) \text{Li}_2(c x)-\frac{\log (1-c x) \text{Li}_2(c x)}{4 x^4}+\frac{1}{2} c^4 \log (1-c x) \text{Li}_2(1-c x)-\frac{1}{4} c^4 \text{Li}_3(c x)-\frac{1}{2} c^4 \text{Li}_3(1-c x)\\ \end{align*}

Mathematica [A]  time = 0.235598, size = 277, normalized size = 0.97 \[ \frac{-36 c^4 x^4 \text{PolyLog}(3,c x)-72 c^4 x^4 \text{PolyLog}(3,1-c x)+18 c^4 x^4 (4 \log (1-c x)+1) \text{PolyLog}(2,1-c x)+6 \left (c x \left (6 c^2 x^2+3 c x+2\right )+6 \left (c^4 x^4-1\right ) \log (1-c x)\right ) \text{PolyLog}(2,c x)-18 c^4 x^4+28 c^3 x^3+5 c^2 x^2-9 c^4 x^4 \log ^2(1-c x)+36 c^4 x^4 \log (c x) \log ^2(1-c x)-49 c^4 x^4 \log (x)-33 c^4 x^4 \log (c x)+82 c^4 x^4 \log (1-c x)+18 c^4 x^4 \log (c x) \log (1-c x)-54 c^3 x^3 \log (1-c x)-18 c^2 x^2 \log (1-c x)+9 \log ^2(1-c x)-10 c x \log (1-c x)}{144 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Log[1 - c*x]*PolyLog[2, c*x])/x^5,x]

[Out]

(5*c^2*x^2 + 28*c^3*x^3 - 18*c^4*x^4 - 49*c^4*x^4*Log[x] - 33*c^4*x^4*Log[c*x] - 10*c*x*Log[1 - c*x] - 18*c^2*
x^2*Log[1 - c*x] - 54*c^3*x^3*Log[1 - c*x] + 82*c^4*x^4*Log[1 - c*x] + 18*c^4*x^4*Log[c*x]*Log[1 - c*x] + 9*Lo
g[1 - c*x]^2 - 9*c^4*x^4*Log[1 - c*x]^2 + 36*c^4*x^4*Log[c*x]*Log[1 - c*x]^2 + 6*(c*x*(2 + 3*c*x + 6*c^2*x^2)
+ 6*(-1 + c^4*x^4)*Log[1 - c*x])*PolyLog[2, c*x] + 18*c^4*x^4*(1 + 4*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 36*c^
4*x^4*PolyLog[3, c*x] - 72*c^4*x^4*PolyLog[3, 1 - c*x])/(144*x^4)

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( -cx+1 \right ){\it polylog} \left ( 2,cx \right ) }{{x}^{5}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-c*x+1)*polylog(2,c*x)/x^5,x)

[Out]

int(ln(-c*x+1)*polylog(2,c*x)/x^5,x)

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Maxima [A]  time = 1.73984, size = 289, normalized size = 1.01 \begin{align*} \frac{1}{4} \,{\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^{4} + \frac{1}{8} \,{\left (\log \left (c x\right ) \log \left (-c x + 1\right ) +{\rm Li}_2\left (-c x + 1\right )\right )} c^{4} - \frac{41}{72} \, c^{4} \log \left (x\right ) - \frac{1}{4} \, c^{4}{\rm Li}_{3}(c x) + \frac{28 \, c^{3} x^{3} + 5 \, c^{2} x^{2} - 9 \,{\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )^{2} + 6 \,{\left (6 \, c^{3} x^{3} + 3 \, c^{2} x^{2} + 2 \, c x + 6 \,{\left (c^{4} x^{4} - 1\right )} \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) + 2 \,{\left (41 \, c^{4} x^{4} - 27 \, c^{3} x^{3} - 9 \, c^{2} x^{2} - 5 \, c x\right )} \log \left (-c x + 1\right )}{144 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-c*x+1)*polylog(2,c*x)/x^5,x, algorithm="maxima")

[Out]

1/4*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))*c^4 + 1/8*(log(c*x)*
log(-c*x + 1) + dilog(-c*x + 1))*c^4 - 41/72*c^4*log(x) - 1/4*c^4*polylog(3, c*x) + 1/144*(28*c^3*x^3 + 5*c^2*
x^2 - 9*(c^4*x^4 - 1)*log(-c*x + 1)^2 + 6*(6*c^3*x^3 + 3*c^2*x^2 + 2*c*x + 6*(c^4*x^4 - 1)*log(-c*x + 1))*dilo
g(c*x) + 2*(41*c^4*x^4 - 27*c^3*x^3 - 9*c^2*x^2 - 5*c*x)*log(-c*x + 1))/x^4

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-c*x+1)*polylog(2,c*x)/x^5,x, algorithm="fricas")

[Out]

integral(dilog(c*x)*log(-c*x + 1)/x^5, 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(ln(-c*x+1)*polylog(2,c*x)/x**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-c*x+1)*polylog(2,c*x)/x^5,x, algorithm="giac")

[Out]

integrate(dilog(c*x)*log(-c*x + 1)/x^5, x)