3.196 \(\int \frac{(a+b x+c x^2) \log (1-d x) \text{PolyLog}(2,d x)}{x^3} \, dx\)

Optimal. Leaf size=343 \[ \frac{b^2 \text{PolyLog}(3,1-d x)}{a}-\frac{b^2 \log (1-d x) \text{PolyLog}(2,1-d x)}{a}-\frac{(a+b x)^2 \log (1-d x) \text{PolyLog}(2,d x)}{2 a x^2}-\frac{1}{2} d (a d+2 b) \text{PolyLog}(3,d x)-\frac{(a d+b)^2 \text{PolyLog}(3,1-d x)}{a}+\frac{(a d+b)^2 \log (1-d x) \text{PolyLog}(2,d x)}{2 a}+\frac{(a d+b)^2 \log (1-d x) \text{PolyLog}(2,1-d x)}{a}-\frac{1}{2} a d^2 \text{PolyLog}(2,d x)+\frac{a d \text{PolyLog}(2,d x)}{2 x}-2 b d \text{PolyLog}(2,d x)-\frac{1}{2} c \text{PolyLog}(2,d x)^2-\frac{b^2 \log (d x) \log ^2(1-d x)}{2 a}+\frac{(a d+b)^2 \log (d x) \log ^2(1-d x)}{2 a}-\frac{1}{4} a d^2 \log ^2(1-d x)-a d^2 \log (x)+a d^2 \log (1-d x)+\frac{a \log ^2(1-d x)}{4 x^2}-\frac{a d \log (1-d x)}{x}+\frac{b (1-d x) \log ^2(1-d x)}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.740443, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 22, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.846, Rules used = {6742, 6591, 2395, 44, 36, 29, 31, 6589, 6605, 6601, 37, 6606, 2398, 2410, 2391, 2390, 2301, 2397, 2396, 2433, 2374, 6596} \[ \frac{b^2 \text{PolyLog}(3,1-d x)}{a}-\frac{b^2 \log (1-d x) \text{PolyLog}(2,1-d x)}{a}-\frac{(a+b x)^2 \log (1-d x) \text{PolyLog}(2,d x)}{2 a x^2}-\frac{1}{2} d (a d+2 b) \text{PolyLog}(3,d x)-\frac{(a d+b)^2 \text{PolyLog}(3,1-d x)}{a}+\frac{(a d+b)^2 \log (1-d x) \text{PolyLog}(2,d x)}{2 a}+\frac{(a d+b)^2 \log (1-d x) \text{PolyLog}(2,1-d x)}{a}-\frac{1}{2} a d^2 \text{PolyLog}(2,d x)+\frac{a d \text{PolyLog}(2,d x)}{2 x}-2 b d \text{PolyLog}(2,d x)-\frac{1}{2} c \text{PolyLog}(2,d x)^2-\frac{b^2 \log (d x) \log ^2(1-d x)}{2 a}+\frac{(a d+b)^2 \log (d x) \log ^2(1-d x)}{2 a}-\frac{1}{4} a d^2 \log ^2(1-d x)-a d^2 \log (x)+a d^2 \log (1-d x)+\frac{a \log ^2(1-d x)}{4 x^2}-\frac{a d \log (1-d x)}{x}+\frac{b (1-d x) \log ^2(1-d x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

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 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 44

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] && L
tQ[m + n + 2, 0])

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 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 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 6605

Int[((g_.) + Log[1 + (e_.)*(x_)]*(h_.))*(Px_)*(x_)^(m_)*PolyLog[2, (c_.)*(x_)], x_Symbol] :> Dist[Coeff[Px, x,
-m - 1], Int[((g + h*Log[1 + e*x])*PolyLog[2, c*x])/x, x], x] + Int[x^m*(Px - Coeff[Px, x, -m - 1]*x^(-m - 1)
)*(g + h*Log[1 + e*x])*PolyLog[2, c*x], x] /; FreeQ[{c, e, g, h}, x] && PolyQ[Px, x] && ILtQ[m, 0] && EqQ[c +
e, 0] && NeQ[Coeff[Px, x, -m - 1], 0]

Rule 6601

Int[(Log[1 + (e_.)*(x_)]*PolyLog[2, (c_.)*(x_)])/(x_), x_Symbol] :> -Simp[PolyLog[2, c*x]^2/2, x] /; FreeQ[{c,
e}, x] && EqQ[c + e, 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 6606

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 2398

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 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 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 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 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 2397

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 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 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 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 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]

Rubi steps

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

Mathematica [F]  time = 1.98638, size = 0, normalized size = 0. \[ \int \frac{\left (a+b x+c x^2\right ) \log (1-d x) \text{PolyLog}(2,d x)}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) \ln \left ( -dx+1 \right ){\it polylog} \left ( 2,dx \right ) }{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}{\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\rm Li}_2\left (d x\right ) \log \left (-d x + 1\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x^3, x)