3.2.70 \(\int x^2 (g+h \log (1-c x)) \text {PolyLog}(2,c x) \, dx\) [170]

Optimal. Leaf size=423 \[ \frac {121 h x}{108 c^2}+\frac {13 h x^2}{216 c}+\frac {h x^3}{81}+\frac {h (1-c x)^2}{6 c^3}-\frac {2 h (1-c x)^3}{81 c^3}+\frac {13 h \log (1-c x)}{108 c^3}-\frac {h x^2 \log (1-c x)}{12 c}-\frac {1}{27} h x^3 \log (1-c x)+\frac {h (1-c x) \log (1-c x)}{3 c^3}+\frac {h \log ^2(1-c x)}{9 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {(1-c x) (g+2 h \log (1-c x))}{3 c^3}-\frac {(1-c x)^2 (g+2 h \log (1-c x))}{6 c^3}+\frac {(1-c x)^3 (g+2 h \log (1-c x))}{27 c^3}-\frac {\log (1-c x) (g+2 h \log (1-c x))}{9 c^3}-\frac {h x \text {PolyLog}(2,c x)}{3 c^2}-\frac {h x^2 \text {PolyLog}(2,c x)}{6 c}-\frac {1}{9} h x^3 \text {PolyLog}(2,c x)-\frac {h \log (1-c x) \text {PolyLog}(2,c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {PolyLog}(2,c x)-\frac {2 h \log (1-c x) \text {PolyLog}(2,1-c x)}{3 c^3}+\frac {2 h \text {PolyLog}(3,1-c x)}{3 c^3} \]

[Out]

121/108*h*x/c^2+13/216*h*x^2/c+1/81*h*x^3+1/6*h*(-c*x+1)^2/c^3-2/81*h*(-c*x+1)^3/c^3+13/108*h*ln(-c*x+1)/c^3-1
/12*h*x^2*ln(-c*x+1)/c-1/27*h*x^3*ln(-c*x+1)+1/3*h*(-c*x+1)*ln(-c*x+1)/c^3+1/9*h*ln(-c*x+1)^2/c^3-1/3*h*ln(c*x
)*ln(-c*x+1)^2/c^3+1/9*x^3*ln(-c*x+1)*(g+h*ln(-c*x+1))+1/3*(-c*x+1)*(g+2*h*ln(-c*x+1))/c^3-1/6*(-c*x+1)^2*(g+2
*h*ln(-c*x+1))/c^3+1/27*(-c*x+1)^3*(g+2*h*ln(-c*x+1))/c^3-1/9*ln(-c*x+1)*(g+2*h*ln(-c*x+1))/c^3-1/3*h*x*polylo
g(2,c*x)/c^2-1/6*h*x^2*polylog(2,c*x)/c-1/9*h*x^3*polylog(2,c*x)-1/3*h*ln(-c*x+1)*polylog(2,c*x)/c^3+1/3*x^3*(
g+h*ln(-c*x+1))*polylog(2,c*x)-2/3*h*ln(-c*x+1)*polylog(2,-c*x+1)/c^3+2/3*h*polylog(3,-c*x+1)/c^3

________________________________________________________________________________________

Rubi [A]
time = 0.31, antiderivative size = 423, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 18, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, 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)^3 (2 h \log (1-c x)+g)}{27 c^3}-\frac {(1-c x)^2 (2 h \log (1-c x)+g)}{6 c^3}+\frac {(1-c x) (2 h \log (1-c x)+g)}{3 c^3}-\frac {\log (1-c x) (2 h \log (1-c x)+g)}{9 c^3}+\frac {2 h \text {Li}_3(1-c x)}{3 c^3}-\frac {h \text {Li}_2(c x) \log (1-c x)}{3 c^3}-\frac {2 h \text {Li}_2(1-c x) \log (1-c x)}{3 c^3}-\frac {2 h (1-c x)^3}{81 c^3}+\frac {h (1-c x)^2}{6 c^3}+\frac {h \log ^2(1-c x)}{9 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {13 h \log (1-c x)}{108 c^3}+\frac {h (1-c x) \log (1-c x)}{3 c^3}-\frac {h x \text {Li}_2(c x)}{3 c^2}+\frac {121 h x}{108 c^2}+\frac {1}{3} x^3 \text {Li}_2(c x) (h \log (1-c x)+g)+\frac {1}{9} x^3 \log (1-c x) (h \log (1-c x)+g)-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{27} h x^3 \log (1-c x)+\frac {13 h x^2}{216 c}-\frac {h x^2 \log (1-c x)}{12 c}+\frac {h x^3}{81} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(121*h*x)/(108*c^2) + (13*h*x^2)/(216*c) + (h*x^3)/81 + (h*(1 - c*x)^2)/(6*c^3) - (2*h*(1 - c*x)^3)/(81*c^3) +
(13*h*Log[1 - c*x])/(108*c^3) - (h*x^2*Log[1 - c*x])/(12*c) - (h*x^3*Log[1 - c*x])/27 + (h*(1 - c*x)*Log[1 -
c*x])/(3*c^3) + (h*Log[1 - c*x]^2)/(9*c^3) - (h*Log[c*x]*Log[1 - c*x]^2)/(3*c^3) + (x^3*Log[1 - c*x]*(g + h*Lo
g[1 - c*x]))/9 + ((1 - c*x)*(g + 2*h*Log[1 - c*x]))/(3*c^3) - ((1 - c*x)^2*(g + 2*h*Log[1 - c*x]))/(6*c^3) + (
(1 - c*x)^3*(g + 2*h*Log[1 - c*x]))/(27*c^3) - (Log[1 - c*x]*(g + 2*h*Log[1 - c*x]))/(9*c^3) - (h*x*PolyLog[2,
c*x])/(3*c^2) - (h*x^2*PolyLog[2, c*x])/(6*c) - (h*x^3*PolyLog[2, c*x])/9 - (h*Log[1 - c*x]*PolyLog[2, c*x])/
(3*c^3) + (x^3*(g + h*Log[1 - c*x])*PolyLog[2, c*x])/3 - (2*h*Log[1 - c*x]*PolyLog[2, 1 - c*x])/(3*c^3) + (2*h
*PolyLog[3, 1 - c*x])/(3*c^3)

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^2 (g+h \log (1-c x)) \text {Li}_2(c x) \, dx &=\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)+\frac {1}{3} \int x^2 \log (1-c x) (g+h \log (1-c x)) \, dx+\frac {1}{3} (c h) \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 (1-c x) (g+h \log (1-c x))+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)+\frac {1}{9} c \int \frac {x^3 (g+h \log (1-c x))}{1-c x} \, dx-\frac {1}{3} h \int x^2 \text {Li}_2(c x) \, dx-\frac {h \int \text {Li}_2(c x) \, dx}{3 c^2}-\frac {h \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{3 c^2}-\frac {h \int x \text {Li}_2(c x) \, dx}{3 c}+\frac {1}{9} (c h) \int \frac {x^3 \log (1-c x)}{1-c x} \, dx\\ &=\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {1}{9} \text {Subst}\left (\int \frac {\left (\frac {1}{c}-\frac {x}{c}\right )^3 (g+h \log (x))}{x} \, dx,x,1-c x\right )-\frac {1}{9} h \int x^2 \log (1-c x) \, dx-\frac {h \int \frac {\log ^2(1-c x)}{x} \, dx}{3 c^3}-\frac {h \int \log (1-c x) \, dx}{3 c^2}-\frac {h \int x \log (1-c x) \, dx}{6 c}+\frac {1}{9} (c h) \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 {h x^2 \log (1-c x)}{12 c}-\frac {1}{27} h x^3 \log (1-c x)-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {1}{54} \left (\frac {18 (1-c x)}{c^3}-\frac {9 (1-c x)^2}{c^3}+\frac {2 (1-c x)^3}{c^3}-\frac {6 \log (1-c x)}{c^3}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {1}{12} h \int \frac {x^2}{1-c x} \, dx-\frac {1}{9} h \int x^2 \log (1-c x) \, dx+\frac {1}{9} h \text {Subst}\left (\int \frac {x \left (-18+9 x-2 x^2\right )+6 \log (x)}{6 c^3 x} \, dx,x,1-c x\right )+\frac {h \text {Subst}(\int \log (x) \, dx,x,1-c x)}{3 c^3}-\frac {h \int \log (1-c x) \, dx}{9 c^2}-\frac {h \int \frac {\log (1-c x)}{-1+c x} \, dx}{9 c^2}-\frac {(2 h) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{3 c^2}-\frac {h \int x \log (1-c x) \, dx}{9 c}-\frac {1}{27} (c h) \int \frac {x^3}{1-c x} \, dx\\ &=\frac {h x}{3 c^2}-\frac {5 h x^2 \log (1-c x)}{36 c}-\frac {2}{27} h x^3 \log (1-c x)+\frac {h (1-c x) \log (1-c x)}{3 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {1}{54} \left (\frac {18 (1-c x)}{c^3}-\frac {9 (1-c x)^2}{c^3}+\frac {2 (1-c x)^3}{c^3}-\frac {6 \log (1-c x)}{c^3}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {1}{18} h \int \frac {x^2}{1-c x} \, dx-\frac {1}{12} h \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx+\frac {h \text {Subst}\left (\int \frac {x \left (-18+9 x-2 x^2\right )+6 \log (x)}{x} \, dx,x,1-c x\right )}{54 c^3}+\frac {h \text {Subst}(\int \log (x) \, dx,x,1-c x)}{9 c^3}-\frac {h \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{9 c^3}+\frac {(2 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 )}{3 c^3}-\frac {1}{27} (c h) \int \frac {x^3}{1-c x} \, dx-\frac {1}{27} (c h) \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx\\ &=\frac {61 h x}{108 c^2}+\frac {13 h x^2}{216 c}+\frac {h x^3}{81}+\frac {13 h \log (1-c x)}{108 c^3}-\frac {5 h x^2 \log (1-c x)}{36 c}-\frac {2}{27} h x^3 \log (1-c x)+\frac {4 h (1-c x) \log (1-c x)}{9 c^3}-\frac {h \log ^2(1-c x)}{18 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {1}{54} \left (\frac {18 (1-c x)}{c^3}-\frac {9 (1-c x)^2}{c^3}+\frac {2 (1-c x)^3}{c^3}-\frac {6 \log (1-c x)}{c^3}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {2 h \log (1-c x) \text {Li}_2(1-c x)}{3 c^3}-\frac {1}{18} h \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx+\frac {h \text {Subst}\left (\int \left (-18+9 x-2 x^2+\frac {6 \log (x)}{x}\right ) \, dx,x,1-c x\right )}{54 c^3}+\frac {(2 h) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{3 c^3}-\frac {1}{27} (c h) \int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx\\ &=\frac {107 h x}{108 c^2}+\frac {23 h x^2}{216 c}+\frac {2 h x^3}{81}+\frac {h (1-c x)^2}{12 c^3}-\frac {h (1-c x)^3}{81 c^3}+\frac {23 h \log (1-c x)}{108 c^3}-\frac {5 h x^2 \log (1-c x)}{36 c}-\frac {2}{27} h x^3 \log (1-c x)+\frac {4 h (1-c x) \log (1-c x)}{9 c^3}-\frac {h \log ^2(1-c x)}{18 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {1}{54} \left (\frac {18 (1-c x)}{c^3}-\frac {9 (1-c x)^2}{c^3}+\frac {2 (1-c x)^3}{c^3}-\frac {6 \log (1-c x)}{c^3}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {2 h \log (1-c x) \text {Li}_2(1-c x)}{3 c^3}+\frac {2 h \text {Li}_3(1-c x)}{3 c^3}+\frac {h \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )}{9 c^3}\\ &=\frac {107 h x}{108 c^2}+\frac {23 h x^2}{216 c}+\frac {2 h x^3}{81}+\frac {h (1-c x)^2}{12 c^3}-\frac {h (1-c x)^3}{81 c^3}+\frac {23 h \log (1-c x)}{108 c^3}-\frac {5 h x^2 \log (1-c x)}{36 c}-\frac {2}{27} h x^3 \log (1-c x)+\frac {4 h (1-c x) \log (1-c x)}{9 c^3}-\frac {h \log (c x) \log ^2(1-c x)}{3 c^3}+\frac {1}{9} x^3 \log (1-c x) (g+h \log (1-c x))+\frac {1}{54} \left (\frac {18 (1-c x)}{c^3}-\frac {9 (1-c x)^2}{c^3}+\frac {2 (1-c x)^3}{c^3}-\frac {6 \log (1-c x)}{c^3}\right ) (g+h \log (1-c x))-\frac {h x \text {Li}_2(c x)}{3 c^2}-\frac {h x^2 \text {Li}_2(c x)}{6 c}-\frac {1}{9} h x^3 \text {Li}_2(c x)-\frac {h \log (1-c x) \text {Li}_2(c x)}{3 c^3}+\frac {1}{3} x^3 (g+h \log (1-c x)) \text {Li}_2(c x)-\frac {2 h \log (1-c x) \text {Li}_2(1-c x)}{3 c^3}+\frac {2 h \text {Li}_3(1-c x)}{3 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 252, normalized size = 0.60 \begin {gather*} \frac {g \left (-c x \left (6+3 c x+2 c^2 x^2\right )+6 \left (-1+c^3 x^3\right ) \log (1-c x)+18 c^3 x^3 \text {PolyLog}(2,c x)\right )}{54 c^3}+\frac {h \left (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 ^2(1-c x)+24 c^3 x^3 \log ^2(1-c x)-72 \log (c x) \log ^2(1-c x)+12 \left (-c x \left (6+3 c x+2 c^2 x^2\right )+6 \left (-1+c^3 x^3\right ) \log (1-c x)\right ) \text {PolyLog}(2,c x)-144 \log (1-c x) \text {PolyLog}(2,1-c x)+144 \text {PolyLog}(3,1-c x)\right )}{216 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(g*(-(c*x*(6 + 3*c*x + 2*c^2*x^2)) + 6*(-1 + c^3*x^3)*Log[1 - c*x] + 18*c^3*x^3*PolyLog[2, c*x]))/(54*c^3) + (
h*(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 = 0.17, size = 0, normalized size = 0.00 \[\int x^{2} \left (g +h \ln \left (-c x +1\right )\right ) \polylog \left (2, c x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-1/18*h*((2*c^3*x^3 + 3*c^2*x^2 + 6*c*x - 6*(c^3*x^3 - 1)*log(-c*x + 1))*dilog(c*x)/c^3 - integrate((6*(c^3*x^
3 - 1)*log(-c*x + 1)^2 - (2*c^3*x^3 + 3*c^2*x^2 + 6*c*x)*log(-c*x + 1))/x, x)/c^3) + 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))*g/c^3

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

[Out]

integral(h*x^2*dilog(c*x)*log(-c*x + 1) + g*x^2*dilog(c*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \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**2*(g+h*ln(-c*x+1))*polylog(2,c*x),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\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^2*(g + h*log(1 - c*x))*polylog(2, c*x),x)

[Out]

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

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