3.175 \(\int \frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x^3} \, dx\)

Optimal. Leaf size=266 \[ \frac {1}{4} c^2 \log \left (1-\frac {1}{1-c x}\right ) (2 h \log (1-c x)+g)-\frac {1}{2} c^2 h \text {Li}_2\left (\frac {1}{1-c x}\right )-\frac {1}{2} c^2 h \text {Li}_3(c x)-c^2 h \text {Li}_3(1-c x)+\frac {1}{2} c^2 h \text {Li}_2(c x) \log (1-c x)+c^2 h \text {Li}_2(1-c x) \log (1-c x)+\frac {1}{2} c^2 h \log (c x) \log ^2(1-c x)-c^2 h \log (x)+\frac {1}{2} c^2 h \log (1-c x)-\frac {\text {Li}_2(c x) (h \log (1-c x)+g)}{2 x^2}+\frac {\log (1-c x) (h \log (1-c x)+g)}{4 x^2}-\frac {c (1-c x) (2 h \log (1-c x)+g)}{4 x}+\frac {c h \text {Li}_2(c x)}{2 x}-\frac {c h \log (1-c x)}{2 x} \]

[Out]

-c^2*h*ln(x)+1/2*c^2*h*ln(-c*x+1)-1/2*c*h*ln(-c*x+1)/x+1/2*c^2*h*ln(c*x)*ln(-c*x+1)^2+1/4*ln(-c*x+1)*(g+h*ln(-
c*x+1))/x^2-1/4*c*(-c*x+1)*(g+2*h*ln(-c*x+1))/x+1/4*c^2*(g+2*h*ln(-c*x+1))*ln(1-1/(-c*x+1))+1/2*c*h*polylog(2,
c*x)/x+1/2*c^2*h*ln(-c*x+1)*polylog(2,c*x)-1/2*(g+h*ln(-c*x+1))*polylog(2,c*x)/x^2-1/2*c^2*h*polylog(2,1/(-c*x
+1))+c^2*h*ln(-c*x+1)*polylog(2,-c*x+1)-1/2*c^2*h*polylog(3,c*x)-c^2*h*polylog(3,-c*x+1)

________________________________________________________________________________________

Rubi [A]  time = 0.50, antiderivative size = 278, normalized size of antiderivative = 1.05, number of steps used = 31, number of rules used = 22, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {6603, 2439, 2410, 2395, 36, 29, 31, 2391, 2390, 2301, 2411, 2347, 2344, 2316, 2315, 2314, 6591, 6589, 6596, 2396, 2433, 2374} \[ -\frac {1}{2} c^2 h \text {PolyLog}(2,c x)-\frac {1}{2} c^2 h \text {PolyLog}(3,c x)-c^2 h \text {PolyLog}(3,1-c x)+\frac {1}{2} c^2 h \log (1-c x) \text {PolyLog}(2,c x)+c^2 h \log (1-c x) \text {PolyLog}(2,1-c x)-\frac {\text {PolyLog}(2,c x) (h \log (1-c x)+g)}{2 x^2}+\frac {c h \text {PolyLog}(2,c x)}{2 x}-\frac {c^2 (h \log (1-c x)+g)^2}{8 h}+\frac {1}{4} c^2 g \log (x)-\frac {1}{8} c^2 h \log ^2(1-c x)+\frac {1}{2} c^2 h \log (c x) \log ^2(1-c x)-c^2 h \log (x)+\frac {3}{4} c^2 h \log (1-c x)+\frac {\log (1-c x) (h \log (1-c x)+g)}{4 x^2}-\frac {c (1-c x) (h \log (1-c x)+g)}{4 x}-\frac {3 c h \log (1-c x)}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[((g + h*Log[1 - c*x])*PolyLog[2, c*x])/x^3,x]

[Out]

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

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 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 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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

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

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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 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 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 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 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 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -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]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 238, normalized size = 0.89 \[ \frac {g \left (c^2 x^2 \log (x)-c^2 x^2 \log (1-c x)-2 \text {Li}_2(c x)-c x+\log (1-c x)\right )}{4 x^2}+\frac {1}{4} h \left (\frac {2 \text {Li}_2(c x) \left (\left (c^2 x^2-1\right ) \log (1-c x)+c x\right )}{x^2}-2 c^2 \text {Li}_3(c x)-4 c^2 \text {Li}_3(1-c x)+2 c^2 \text {Li}_2(1-c x) (2 \log (1-c x)+1)+2 c^2 \log (c x) \log ^2(1-c x)-c^2 \log ^2(1-c x)-2 c^2 \log (x)-2 c^2 \log (c x)+2 c^2 \log (c x) \log (1-c x)+4 c^2 \log (1-c x)+\frac {\log ^2(1-c x)}{x^2}-\frac {4 c \log (1-c x)}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*Log[1 - c*x])*PolyLog[2, c*x])/x^3,x]

[Out]

(g*(-(c*x) + c^2*x^2*Log[x] + Log[1 - c*x] - c^2*x^2*Log[1 - c*x] - 2*PolyLog[2, c*x]))/(4*x^2) + (h*(-2*c^2*L
og[x] - 2*c^2*Log[c*x] + 4*c^2*Log[1 - c*x] - (4*c*Log[1 - c*x])/x + 2*c^2*Log[c*x]*Log[1 - c*x] - c^2*Log[1 -
c*x]^2 + Log[1 - c*x]^2/x^2 + 2*c^2*Log[c*x]*Log[1 - c*x]^2 + (2*(c*x + (-1 + c^2*x^2)*Log[1 - c*x])*PolyLog[
2, c*x])/x^2 + 2*c^2*(1 + 2*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 2*c^2*PolyLog[3, c*x] - 4*c^2*PolyLog[3, 1 - c
*x]))/4

________________________________________________________________________________________

fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {h {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) + g {\rm Li}_2\left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (g +h \ln \left (-c x +1\right )\right ) \polylog \left (2, c x \right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left (c^{2} \log \relax (x) - \frac {c x + {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right ) + 2 \, {\rm Li}_2\left (c x\right )}{x^{2}}\right )} g + h \int \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c^2*log(x) - (c*x + (c^2*x^2 - 1)*log(-c*x + 1) + 2*dilog(c*x))/x^2)*g + h*integrate(dilog(c*x)*log(-c*x
+ 1)/x^3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((g + h*log(1 - c*x))*polylog(2, c*x))/x^3,x)

[Out]

int(((g + h*log(1 - c*x))*polylog(2, c*x))/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g + h \log {\left (- c x + 1 \right )}\right ) \operatorname {Li}_{2}\left (c x\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((g + h*log(-c*x + 1))*polylog(2, c*x)/x**3, x)

________________________________________________________________________________________