3.2.71 \(\int x (g+h \log (1-c x)) \text {PolyLog}(2,c x) \, dx\) [171]

Optimal. Leaf size=330 \[ \frac {13 h x}{8 c}+\frac {h x^2}{16}+\frac {h (1-c x)^2}{8 c^2}+\frac {h \log (1-c x)}{8 c^2}-\frac {1}{8} h x^2 \log (1-c x)+\frac {h (1-c x) \log (1-c x)}{2 c^2}+\frac {h \log ^2(1-c x)}{4 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {(1-c x) (g+2 h \log (1-c x))}{2 c^2}-\frac {(1-c x)^2 (g+2 h \log (1-c x))}{8 c^2}-\frac {\log (1-c x) (g+2 h \log (1-c x))}{4 c^2}-\frac {h x \text {PolyLog}(2,c x)}{2 c}-\frac {1}{4} h x^2 \text {PolyLog}(2,c x)-\frac {h \log (1-c x) \text {PolyLog}(2,c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {PolyLog}(2,c x)-\frac {h \log (1-c x) \text {PolyLog}(2,1-c x)}{c^2}+\frac {h \text {PolyLog}(3,1-c x)}{c^2} \]

[Out]

13/8*h*x/c+1/16*h*x^2+1/8*h*(-c*x+1)^2/c^2+1/8*h*ln(-c*x+1)/c^2-1/8*h*x^2*ln(-c*x+1)+1/2*h*(-c*x+1)*ln(-c*x+1)
/c^2+1/4*h*ln(-c*x+1)^2/c^2-1/2*h*ln(c*x)*ln(-c*x+1)^2/c^2+1/4*x^2*ln(-c*x+1)*(g+h*ln(-c*x+1))+1/2*(-c*x+1)*(g
+2*h*ln(-c*x+1))/c^2-1/8*(-c*x+1)^2*(g+2*h*ln(-c*x+1))/c^2-1/4*ln(-c*x+1)*(g+2*h*ln(-c*x+1))/c^2-1/2*h*x*polyl
og(2,c*x)/c-1/4*h*x^2*polylog(2,c*x)-1/2*h*ln(-c*x+1)*polylog(2,c*x)/c^2+1/2*x^2*(g+h*ln(-c*x+1))*polylog(2,c*
x)-h*ln(-c*x+1)*polylog(2,-c*x+1)/c^2+h*polylog(3,-c*x+1)/c^2

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 18, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6738, 2483, 2458, 45, 2372, 12, 14, 2338, 6721, 2436, 2332, 6726, 2442, 6731, 2443, 2481, 2421, 6724} \begin {gather*} -\frac {(1-c x)^2 (2 h \log (1-c x)+g)}{8 c^2}+\frac {(1-c x) (2 h \log (1-c x)+g)}{2 c^2}-\frac {\log (1-c x) (2 h \log (1-c x)+g)}{4 c^2}+\frac {h \text {Li}_3(1-c x)}{c^2}-\frac {h \text {Li}_2(c x) \log (1-c x)}{2 c^2}-\frac {h \text {Li}_2(1-c x) \log (1-c x)}{c^2}+\frac {h (1-c x)^2}{8 c^2}+\frac {h \log ^2(1-c x)}{4 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {h \log (1-c x)}{8 c^2}+\frac {h (1-c x) \log (1-c x)}{2 c^2}+\frac {1}{2} x^2 \text {Li}_2(c x) (h \log (1-c x)+g)+\frac {1}{4} x^2 \log (1-c x) (h \log (1-c x)+g)-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{8} h x^2 \log (1-c x)+\frac {13 h x}{8 c}+\frac {h x^2}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

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 2332

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

Rule 2338

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 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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 2436

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 2442

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 2443

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 2458

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 2481

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 2483

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

Rule 6721

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 6724

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 6726

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 6731

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 6738

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 (g+h \log (1-c x)) \text {Li}_2(c x) \, dx &=\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)+\frac {1}{2} \int x \log (1-c x) (g+h \log (1-c x)) \, dx+\frac {1}{2} (c h) \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}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)+\frac {1}{4} c \int \frac {x^2 (g+h \log (1-c x))}{1-c x} \, dx-\frac {1}{2} h \int x \text {Li}_2(c x) \, dx-\frac {h \int \text {Li}_2(c x) \, dx}{2 c}-\frac {h \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{2 c}+\frac {1}{4} (c h) \int \frac {x^2 \log (1-c x)}{1-c x} \, dx\\ &=\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {1}{4} \text {Subst}\left (\int \frac {\left (\frac {1}{c}-\frac {x}{c}\right )^2 (g+h \log (x))}{x} \, dx,x,1-c x\right )-\frac {1}{4} h \int x \log (1-c x) \, dx-\frac {h \int \frac {\log ^2(1-c x)}{x} \, dx}{2 c^2}-\frac {h \int \log (1-c x) \, dx}{2 c}+\frac {1}{4} (c h) \int \left (-\frac {\log (1-c x)}{c^2}-\frac {x \log (1-c x)}{c}-\frac {\log (1-c x)}{c^2 (-1+c x)}\right ) \, dx\\ &=-\frac {1}{8} h x^2 \log (1-c x)-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{8} \left (\frac {4 (1-c x)}{c^2}-\frac {(1-c x)^2}{c^2}-\frac {2 \log (1-c x)}{c^2}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {1}{4} h \int x \log (1-c x) \, dx+\frac {1}{4} h \text {Subst}\left (\int \frac {(-4+x) x+2 \log (x)}{2 c^2 x} \, dx,x,1-c x\right )+\frac {h \text {Subst}(\int \log (x) \, dx,x,1-c x)}{2 c^2}-\frac {h \int \log (1-c x) \, dx}{4 c}-\frac {h \int \frac {\log (1-c x)}{-1+c x} \, dx}{4 c}-\frac {h \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{c}-\frac {1}{8} (c h) \int \frac {x^2}{1-c x} \, dx\\ &=\frac {h x}{2 c}-\frac {1}{4} h x^2 \log (1-c x)+\frac {h (1-c x) \log (1-c x)}{2 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{8} \left (\frac {4 (1-c x)}{c^2}-\frac {(1-c x)^2}{c^2}-\frac {2 \log (1-c x)}{c^2}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)+\frac {h \text {Subst}\left (\int \frac {(-4+x) x+2 \log (x)}{x} \, dx,x,1-c x\right )}{8 c^2}+\frac {h \text {Subst}(\int \log (x) \, dx,x,1-c x)}{4 c^2}-\frac {h \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{4 c^2}+\frac {h \text {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 h) \int \frac {x^2}{1-c x} \, dx-\frac {1}{8} (c h) \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx\\ &=\frac {7 h x}{8 c}+\frac {h x^2}{16}+\frac {h \log (1-c x)}{8 c^2}-\frac {1}{4} h x^2 \log (1-c x)+\frac {3 h (1-c x) \log (1-c x)}{4 c^2}-\frac {h \log ^2(1-c x)}{8 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{8} \left (\frac {4 (1-c x)}{c^2}-\frac {(1-c x)^2}{c^2}-\frac {2 \log (1-c x)}{c^2}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(1-c x)}{c^2}+\frac {h \text {Subst}\left (\int \left (-4+x+\frac {2 \log (x)}{x}\right ) \, dx,x,1-c x\right )}{8 c^2}+\frac {h \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{c^2}-\frac {1}{8} (c h) \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx\\ &=\frac {3 h x}{2 c}+\frac {h x^2}{8}+\frac {h (1-c x)^2}{16 c^2}+\frac {h \log (1-c x)}{4 c^2}-\frac {1}{4} h x^2 \log (1-c x)+\frac {3 h (1-c x) \log (1-c x)}{4 c^2}-\frac {h \log ^2(1-c x)}{8 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{8} \left (\frac {4 (1-c x)}{c^2}-\frac {(1-c x)^2}{c^2}-\frac {2 \log (1-c x)}{c^2}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(1-c x)}{c^2}+\frac {h \text {Li}_3(1-c x)}{c^2}+\frac {h \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{4 c^2}\\ &=\frac {3 h x}{2 c}+\frac {h x^2}{8}+\frac {h (1-c x)^2}{16 c^2}+\frac {h \log (1-c x)}{4 c^2}-\frac {1}{4} h x^2 \log (1-c x)+\frac {3 h (1-c x) \log (1-c x)}{4 c^2}-\frac {h \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {1}{4} x^2 \log (1-c x) (g+h \log (1-c x))+\frac {1}{8} \left (\frac {4 (1-c x)}{c^2}-\frac {(1-c x)^2}{c^2}-\frac {2 \log (1-c x)}{c^2}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{2 c}-\frac {1}{4} h x^2 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} x^2 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(1-c x)}{c^2}+\frac {h \text {Li}_3(1-c x)}{c^2}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 211, normalized size = 0.64 \begin {gather*} \frac {g \left (-c x (2+c x)+2 \left (-1+c^2 x^2\right ) \log (1-c x)+4 c^2 x^2 \text {PolyLog}(2,c x)\right )}{8 c^2}+\frac {h \left (-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 ^2(1-c x)+4 c^2 x^2 \log ^2(1-c x)-8 \log (c x) \log ^2(1-c x)+\left (-4 c x (2+c x)+8 \left (-1+c^2 x^2\right ) \log (1-c x)\right ) \text {PolyLog}(2,c x)-16 \log (1-c x) \text {PolyLog}(2,1-c x)+16 \text {PolyLog}(3,1-c x)\right )}{16 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int x \left (g +h \ln \left (-c x +1\right )\right ) \polylog \left (2, c x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (g + h \log {\left (- c x + 1 \right )}\right ) \operatorname {Li}_{2}\left (c x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x\,\left (g+h\,\ln \left (1-c\,x\right )\right )\,\mathrm {polylog}\left (2,c\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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