3.2.86 \(\int (a+b x) \log (1-c x) \text {PolyLog}(2,c x) \, dx\) [186]

Optimal. Leaf size=390 \[ 2 a x+\frac {9 b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b (1-c x)^2}{8 c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {PolyLog}(2,c x)}{2 c}-\frac {1}{4} b x^2 \text {PolyLog}(2,c x)-\frac {(b+2 a c) \log (1-c x) \text {PolyLog}(2,c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {PolyLog}(2,c x)-\frac {(b+2 a c) \log (1-c x) \text {PolyLog}(2,1-c x)}{c^2}+\frac {(b+2 a c) \text {PolyLog}(3,1-c x)}{c^2} \]

[Out]

2*a*x+9/8*b*x/c+1/2*(2*a*c+b)*x/c+1/16*b*x^2+1/8*b*(-c*x+1)^2/c^2+1/8*b*ln(-c*x+1)/c^2-1/8*b*x^2*ln(-c*x+1)+b*
(-c*x+1)*ln(-c*x+1)/c^2+2*a*(-c*x+1)*ln(-c*x+1)/c+1/2*(2*a*c+b)*(-c*x+1)*ln(-c*x+1)/c^2-1/4*b*(-c*x+1)^2*ln(-c
*x+1)/c^2-1/2*b*(-c*x+1)*ln(-c*x+1)^2/c^2-a*(-c*x+1)*ln(-c*x+1)^2/c+1/4*b*(-c*x+1)^2*ln(-c*x+1)^2/c^2-1/2*(2*a
*c+b)*ln(c*x)*ln(-c*x+1)^2/c^2-1/2*(2*a*c+b)*x*polylog(2,c*x)/c-1/4*b*x^2*polylog(2,c*x)-1/2*(2*a*c+b)*ln(-c*x
+1)*polylog(2,c*x)/c^2+1/2*(b*x^2+2*a*x)*ln(-c*x+1)*polylog(2,c*x)-(2*a*c+b)*ln(-c*x+1)*polylog(2,-c*x+1)/c^2+
(2*a*c+b)*polylog(3,-c*x+1)/c^2

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 20, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.111, Rules used = {6733, 45, 2463, 2436, 2332, 2438, 2442, 6739, 2333, 2448, 2437, 2342, 2341, 6721, 6726, 6731, 2443, 2481, 2421, 6724} \begin {gather*} \frac {(2 a c+b) \text {Li}_3(1-c x)}{c^2}-\frac {(2 a c+b) \text {Li}_2(c x) \log (1-c x)}{2 c^2}-\frac {(2 a c+b) \text {Li}_2(1-c x) \log (1-c x)}{c^2}-\frac {(2 a c+b) \log (c x) \log ^2(1-c x)}{2 c^2}+\frac {(1-c x) (2 a c+b) \log (1-c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \text {Li}_2(c x) \log (1-c x)-\frac {x (2 a c+b) \text {Li}_2(c x)}{2 c}+\frac {x (2 a c+b)}{2 c}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {2 a (1-c x) \log (1-c x)}{c}+2 a x+\frac {b (1-c x)^2}{8 c^2}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {1}{8} b x^2 \log (1-c x)+\frac {9 b x}{8 c}+\frac {b x^2}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

2*a*x + (9*b*x)/(8*c) + ((b + 2*a*c)*x)/(2*c) + (b*x^2)/16 + (b*(1 - c*x)^2)/(8*c^2) + (b*Log[1 - c*x])/(8*c^2
) - (b*x^2*Log[1 - c*x])/8 + (b*(1 - c*x)*Log[1 - c*x])/c^2 + (2*a*(1 - c*x)*Log[1 - c*x])/c + ((b + 2*a*c)*(1
- c*x)*Log[1 - c*x])/(2*c^2) - (b*(1 - c*x)^2*Log[1 - c*x])/(4*c^2) - (b*(1 - c*x)*Log[1 - c*x]^2)/(2*c^2) -
(a*(1 - c*x)*Log[1 - c*x]^2)/c + (b*(1 - c*x)^2*Log[1 - c*x]^2)/(4*c^2) - ((b + 2*a*c)*Log[c*x]*Log[1 - c*x]^2
)/(2*c^2) - ((b + 2*a*c)*x*PolyLog[2, c*x])/(2*c) - (b*x^2*PolyLog[2, c*x])/4 - ((b + 2*a*c)*Log[1 - c*x]*Poly
Log[2, c*x])/(2*c^2) + ((2*a*x + b*x^2)*Log[1 - c*x]*PolyLog[2, c*x])/2 - ((b + 2*a*c)*Log[1 - c*x]*PolyLog[2,
1 - c*x])/c^2 + ((b + 2*a*c)*PolyLog[3, 1 - c*x])/c^2

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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2341

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

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

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 2437

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 2438

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 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 2448

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 0]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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 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 6733

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

Rule 6739

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(Px_)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symb
ol] :> With[{u = IntHide[Px, x]}, Simp[u*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)], x] + (Dist[b, Int
[ExpandIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], u/(a + b*x), x], x], x] - Dist[e*h*n, Int[Ex
pandIntegrand[PolyLog[2, c*(a + b*x)], u/(d + e*x), x], x], x])] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && P
olyQ[Px, x]

Rubi steps

\begin {align*} \int (a+b x) \log (1-c x) \text {Li}_2(c x) \, dx &=\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+c \int \left (\frac {(-b-2 a c) \text {Li}_2(c x)}{2 c^2}-\frac {b x \text {Li}_2(c x)}{2 c}+\frac {(-b-2 a c) \text {Li}_2(c x)}{2 c^2 (-1+c x)}\right ) \, dx+\int \left (a \log ^2(1-c x)+\frac {1}{2} b x \log ^2(1-c x)\right ) \, dx\\ &=\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+a \int \log ^2(1-c x) \, dx+\frac {1}{2} b \int x \log ^2(1-c x) \, dx-\frac {1}{2} b \int x \text {Li}_2(c x) \, dx-\frac {(b+2 a c) \int \text {Li}_2(c x) \, dx}{2 c}-\frac {(b+2 a c) \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx}{2 c}\\ &=-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {1}{4} b \int x \log (1-c x) \, dx+\frac {1}{2} b \int \left (\frac {\log ^2(1-c x)}{c}-\frac {(1-c x) \log ^2(1-c x)}{c}\right ) \, dx-\frac {a \text {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{c}-\frac {(b+2 a c) \int \frac {\log ^2(1-c x)}{x} \, dx}{2 c^2}-\frac {(b+2 a c) \int \log (1-c x) \, dx}{2 c}\\ &=-\frac {1}{8} b x^2 \log (1-c x)-\frac {a (1-c x) \log ^2(1-c x)}{c}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)+\frac {(2 a) \text {Subst}(\int \log (x) \, dx,x,1-c x)}{c}+\frac {b \int \log ^2(1-c x) \, dx}{2 c}-\frac {b \int (1-c x) \log ^2(1-c x) \, dx}{2 c}-\frac {1}{8} (b c) \int \frac {x^2}{1-c x} \, dx+\frac {(b+2 a c) \text {Subst}(\int \log (x) \, dx,x,1-c x)}{2 c^2}-\frac {(b+2 a c) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx}{c}\\ &=2 a x+\frac {(b+2 a c) x}{2 c}-\frac {1}{8} b x^2 \log (1-c x)+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {b \text {Subst}\left (\int \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}+\frac {b \text {Subst}\left (\int x \log ^2(x) \, dx,x,1-c x\right )}{2 c^2}-\frac {1}{8} (b c) \int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx+\frac {(b+2 a c) \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}\\ &=2 a x+\frac {b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(1-c x)}{c^2}-\frac {b \text {Subst}(\int x \log (x) \, dx,x,1-c x)}{2 c^2}+\frac {b \text {Subst}(\int \log (x) \, dx,x,1-c x)}{c^2}+\frac {(b+2 a c) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )}{c^2}\\ &=2 a x+\frac {9 b x}{8 c}+\frac {(b+2 a c) x}{2 c}+\frac {b x^2}{16}+\frac {b (1-c x)^2}{8 c^2}+\frac {b \log (1-c x)}{8 c^2}-\frac {1}{8} b x^2 \log (1-c x)+\frac {b (1-c x) \log (1-c x)}{c^2}+\frac {2 a (1-c x) \log (1-c x)}{c}+\frac {(b+2 a c) (1-c x) \log (1-c x)}{2 c^2}-\frac {b (1-c x)^2 \log (1-c x)}{4 c^2}-\frac {b (1-c x) \log ^2(1-c x)}{2 c^2}-\frac {a (1-c x) \log ^2(1-c x)}{c}+\frac {b (1-c x)^2 \log ^2(1-c x)}{4 c^2}-\frac {(b+2 a c) \log (c x) \log ^2(1-c x)}{2 c^2}-\frac {(b+2 a c) x \text {Li}_2(c x)}{2 c}-\frac {1}{4} b x^2 \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(c x)}{2 c^2}+\frac {1}{2} \left (2 a x+b x^2\right ) \log (1-c x) \text {Li}_2(c x)-\frac {(b+2 a c) \log (1-c x) \text {Li}_2(1-c x)}{c^2}+\frac {(b+2 a c) \text {Li}_3(1-c x)}{c^2}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 285, normalized size = 0.73 \begin {gather*} \frac {-14 b-32 a c+22 b c x+48 a c^2 x+3 b c^2 x^2+22 b \log (1-c x)+48 a c \log (1-c x)-16 b c x \log (1-c x)-48 a c^2 x \log (1-c x)-6 b c^2 x^2 \log (1-c x)-4 b \log ^2(1-c x)-16 a c \log ^2(1-c x)+16 a c^2 x \log ^2(1-c x)+4 b c^2 x^2 \log ^2(1-c x)-8 b \log (c x) \log ^2(1-c x)-16 a c \log (c x) \log ^2(1-c x)+4 (-c x (2 b+4 a c+b c x)+2 (-1+c x) (b+2 a c+b c x) \log (1-c x)) \text {PolyLog}(2,c x)-16 (b+2 a c) \log (1-c x) \text {PolyLog}(2,1-c x)+16 b \text {PolyLog}(3,1-c x)+32 a c \text {PolyLog}(3,1-c x)}{16 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

(-14*b - 32*a*c + 22*b*c*x + 48*a*c^2*x + 3*b*c^2*x^2 + 22*b*Log[1 - c*x] + 48*a*c*Log[1 - c*x] - 16*b*c*x*Log
[1 - c*x] - 48*a*c^2*x*Log[1 - c*x] - 6*b*c^2*x^2*Log[1 - c*x] - 4*b*Log[1 - c*x]^2 - 16*a*c*Log[1 - c*x]^2 +
16*a*c^2*x*Log[1 - c*x]^2 + 4*b*c^2*x^2*Log[1 - c*x]^2 - 8*b*Log[c*x]*Log[1 - c*x]^2 - 16*a*c*Log[c*x]*Log[1 -
c*x]^2 + 4*(-(c*x*(2*b + 4*a*c + b*c*x)) + 2*(-1 + c*x)*(b + 2*a*c + b*c*x)*Log[1 - c*x])*PolyLog[2, c*x] - 1
6*(b + 2*a*c)*Log[1 - c*x]*PolyLog[2, 1 - c*x] + 16*b*PolyLog[3, 1 - c*x] + 32*a*c*PolyLog[3, 1 - c*x])/(16*c^
2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*ln(-c*x+1)*polylog(2,c*x),x)

[Out]

int((b*x+a)*ln(-c*x+1)*polylog(2,c*x),x)

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Maxima [A]
time = 0.27, size = 258, normalized size = 0.66 \begin {gather*} -\frac {1}{16} \, c {\left (\frac {8 \, {\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 )} {\left (2 \, a c + b\right )}}{c^{3}} - \frac {3 \, b c^{2} x^{2} + 2 \, {\left (24 \, a c^{2} + 11 \, b c\right )} x - 4 \, {\left (b c^{2} x^{2} + 2 \, {\left (2 \, a c^{2} + b c\right )} x + 2 \, {\left (2 \, a c + b\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 2 \, {\left (2 \, b c^{2} x^{2} - 24 \, a c + 2 \, {\left (8 \, a c^{2} + 3 \, b c\right )} x - 11 \, b\right )} \log \left (-c x + 1\right )}{c^{3}}\right )} + \frac {1}{8} \, {\left (\frac {8 \, {\left (c x {\rm Li}_2\left (c x\right ) - c x + {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} a}{c} + \frac {{\left (4 \, c^{2} x^{2} {\rm Li}_2\left (c x\right ) - c^{2} x^{2} - 2 \, c x + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} b}{c^{2}}\right )} \log \left (-c x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*c*(8*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))*(2*a*c + b)/c
^3 - (3*b*c^2*x^2 + 2*(24*a*c^2 + 11*b*c)*x - 4*(b*c^2*x^2 + 2*(2*a*c^2 + b*c)*x + 2*(2*a*c + b)*log(-c*x + 1)
)*dilog(c*x) - 2*(2*b*c^2*x^2 - 24*a*c + 2*(8*a*c^2 + 3*b*c)*x - 11*b)*log(-c*x + 1))/c^3) + 1/8*(8*(c*x*dilog
(c*x) - c*x + (c*x - 1)*log(-c*x + 1))*a/c + (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))*b/c^2)*log(-c*x + 1)

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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((b*x+a)*log(-c*x+1)*polylog(2,c*x),x, algorithm="fricas")

[Out]

integral((b*x + a)*dilog(c*x)*log(-c*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)*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((b*x+a)*log(-c*x+1)*polylog(2,c*x),x, algorithm="giac")

[Out]

integrate((b*x + a)*dilog(c*x)*log(-c*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(1 - c*x)*polylog(2, c*x)*(a + b*x),x)

[Out]

int(log(1 - c*x)*polylog(2, c*x)*(a + b*x), x)

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