3.2.89 \(\int \frac {(a+b x) \log (1-c x) \text {PolyLog}(2,c x)}{x^3} \, dx\) [189]

Optimal. Leaf size=331 \[ -a c^2 \log (x)+a c^2 \log (1-c x)-\frac {a c \log (1-c x)}{x}-\frac {1}{4} a c^2 \log ^2(1-c x)+\frac {a \log ^2(1-c x)}{4 x^2}+\frac {b (1-c x) \log ^2(1-c x)}{x}-\frac {b^2 \log (c x) \log ^2(1-c x)}{2 a}+\frac {(b+a c)^2 \log (c x) \log ^2(1-c x)}{2 a}-2 b c \text {PolyLog}(2,c x)-\frac {1}{2} a c^2 \text {PolyLog}(2,c x)+\frac {a c \text {PolyLog}(2,c x)}{2 x}+\frac {(b+a c)^2 \log (1-c x) \text {PolyLog}(2,c x)}{2 a}-\frac {(a+b x)^2 \log (1-c x) \text {PolyLog}(2,c x)}{2 a x^2}-\frac {b^2 \log (1-c x) \text {PolyLog}(2,1-c x)}{a}+\frac {(b+a c)^2 \log (1-c x) \text {PolyLog}(2,1-c x)}{a}-\frac {1}{2} c (2 b+a c) \text {PolyLog}(3,c x)+\frac {b^2 \text {PolyLog}(3,1-c x)}{a}-\frac {(b+a c)^2 \text {PolyLog}(3,1-c x)}{a} \]

[Out]

-a*c^2*ln(x)+a*c^2*ln(-c*x+1)-a*c*ln(-c*x+1)/x-1/4*a*c^2*ln(-c*x+1)^2+1/4*a*ln(-c*x+1)^2/x^2+b*(-c*x+1)*ln(-c*
x+1)^2/x-1/2*b^2*ln(c*x)*ln(-c*x+1)^2/a+1/2*(a*c+b)^2*ln(c*x)*ln(-c*x+1)^2/a-2*b*c*polylog(2,c*x)-1/2*a*c^2*po
lylog(2,c*x)+1/2*a*c*polylog(2,c*x)/x+1/2*(a*c+b)^2*ln(-c*x+1)*polylog(2,c*x)/a-1/2*(b*x+a)^2*ln(-c*x+1)*polyl
og(2,c*x)/a/x^2-b^2*ln(-c*x+1)*polylog(2,-c*x+1)/a+(a*c+b)^2*ln(-c*x+1)*polylog(2,-c*x+1)/a-1/2*c*(a*c+2*b)*po
lylog(3,c*x)+b^2*polylog(3,-c*x+1)/a-(a*c+b)^2*polylog(3,-c*x+1)/a

________________________________________________________________________________________

Rubi [A]
time = 0.35, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 20, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.952, Rules used = {6874, 6726, 2442, 46, 36, 29, 31, 37, 6741, 2445, 2457, 2438, 2437, 2338, 2444, 2443, 2481, 2421, 6724, 6731} \begin {gather*} \frac {b^2 \text {Li}_3(1-c x)}{a}-\frac {b^2 \text {Li}_2(1-c x) \log (1-c x)}{a}-\frac {b^2 \log (c x) \log ^2(1-c x)}{2 a}-\frac {(a+b x)^2 \text {Li}_2(c x) \log (1-c x)}{2 a x^2}-\frac {1}{2} c (a c+2 b) \text {Li}_3(c x)-\frac {(a c+b)^2 \text {Li}_3(1-c x)}{a}+\frac {(a c+b)^2 \text {Li}_2(c x) \log (1-c x)}{2 a}+\frac {(a c+b)^2 \text {Li}_2(1-c x) \log (1-c x)}{a}+\frac {(a c+b)^2 \log (c x) \log ^2(1-c x)}{2 a}-\frac {1}{2} a c^2 \text {Li}_2(c x)-\frac {1}{4} a c^2 \log ^2(1-c x)-a c^2 \log (x)+a c^2 \log (1-c x)+\frac {a c \text {Li}_2(c x)}{2 x}+\frac {a \log ^2(1-c x)}{4 x^2}-\frac {a c \log (1-c x)}{x}-2 b c \text {Li}_2(c x)+\frac {b (1-c x) \log ^2(1-c x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 46

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] && Lt
Q[m + n + 2, 0])

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 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 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 2444

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[(d + e
*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f - d*g)*(f + g*x))), x] - Dist[b*e*n*(p/(e*f - d*g)), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rule 2445

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 2457

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 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 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 6741

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(Px_)*(x_)^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_
))], x_Symbol] :> With[{u = IntHide[x^m*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] - Dis
t[e*h*n, Int[ExpandIntegrand[PolyLog[2, c*(a + b*x)], u/(d + e*x), x], x], x])] /; FreeQ[{a, b, c, d, e, f, g,
h, n}, x] && PolyQ[Px, x] && IntegerQ[m]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.82, size = 285, normalized size = 0.86 \begin {gather*} \frac {1}{4} \left (-4 a c^2 \log (c x)+4 a c^2 \log (1-c x)-\frac {4 a c \log (1-c x)}{x}+8 b c \log (c x) \log (1-c x)+2 a c^2 \log (c x) \log (1-c x)-4 b c \log ^2(1-c x)-a c^2 \log ^2(1-c x)+\frac {a \log ^2(1-c x)}{x^2}+\frac {4 b \log ^2(1-c x)}{x}+4 b c \log (c x) \log ^2(1-c x)+2 a c^2 \log (c x) \log ^2(1-c x)+\frac {2 (a c x+(-1+c x) (a+2 b x+a c x) \log (1-c x)) \text {PolyLog}(2,c x)}{x^2}+2 c (4 b+a c+2 (2 b+a c) \log (1-c x)) \text {PolyLog}(2,1-c x)-4 b c \text {PolyLog}(3,c x)-2 a c^2 \text {PolyLog}(3,c x)-8 b c \text {PolyLog}(3,1-c x)-4 a c^2 \text {PolyLog}(3,1-c x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right ) \ln \left (-c x +1\right ) \polylog \left (2, c x \right )}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 213, normalized size = 0.64 \begin {gather*} -a c^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (a c^{2} + 2 \, b c\right )} {\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 )} + \frac {1}{2} \, {\left (a c^{2} + 4 \, b c\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} - \frac {1}{2} \, {\left (a c^{2} + 2 \, b c\right )} {\rm Li}_{3}(c x) - \frac {{\left ({\left (a c^{2} + 4 \, b c\right )} x^{2} - 4 \, b x - a\right )} \log \left (-c x + 1\right )^{2} - 2 \, {\left (a c x + {\left ({\left (a c^{2} + 2 \, b c\right )} x^{2} - 2 \, b x - a\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 4 \, {\left (a c^{2} x^{2} - a c x\right )} \log \left (-c x + 1\right )}{4 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

-a*c^2*log(x) + 1/2*(a*c^2 + 2*b*c)*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3,
-c*x + 1)) + 1/2*(a*c^2 + 4*b*c)*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1)) - 1/2*(a*c^2 + 2*b*c)*polylog(3,
c*x) - 1/4*(((a*c^2 + 4*b*c)*x^2 - 4*b*x - a)*log(-c*x + 1)^2 - 2*(a*c*x + ((a*c^2 + 2*b*c)*x^2 - 2*b*x - a)*l
og(-c*x + 1))*dilog(c*x) - 4*(a*c^2*x^2 - a*c*x)*log(-c*x + 1))/x^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((b*x+a)*log(-c*x+1)*polylog(2,c*x)/x^3,x, algorithm="fricas")

[Out]

integral((b*x + a)*dilog(c*x)*log(-c*x + 1)/x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{3}}\, 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**3,x)

[Out]

Integral((a + b*x)*log(-c*x + 1)*polylog(2, c*x)/x**3, 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^3,x, algorithm="giac")

[Out]

integrate((b*x + a)*dilog(c*x)*log(-c*x + 1)/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right )}{x^3} \,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^3,x)

[Out]

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

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