3.2.97 \(\int \frac {(a+b x+c x^2) \log (1-d x) \text {PolyLog}(2,d x)}{x^4} \, dx\) [197]

Optimal. Leaf size=515 \[ \frac {7 a d^2}{36 x}-\frac {1}{2} b d^2 \log (x)-\frac {5}{12} a d^3 \log (x)-\frac {1}{6} d^2 (3 b+2 a d) \log (x)+\frac {1}{2} b d^2 \log (1-d x)+\frac {5}{12} a d^3 \log (1-d x)+\frac {1}{6} d^2 (3 b+2 a d) \log (1-d x)-\frac {7 a d \log (1-d x)}{36 x^2}-\frac {b d \log (1-d x)}{2 x}-\frac {2 a d^2 \log (1-d x)}{9 x}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}-\frac {1}{4} b d^2 \log ^2(1-d x)-\frac {1}{9} a d^3 \log ^2(1-d x)+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {PolyLog}(2,d x)-\frac {1}{2} b d^2 \text {PolyLog}(2,d x)-\frac {2}{9} a d^3 \text {PolyLog}(2,d x)+\frac {a d \text {PolyLog}(2,d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {PolyLog}(2,d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {PolyLog}(2,d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {PolyLog}(2,d x)+\frac {1}{3} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {PolyLog}(2,1-d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {PolyLog}(3,d x)-\frac {1}{3} d (6 c+d (3 b+2 a d)) \text {PolyLog}(3,1-d x) \]

[Out]

1/6*a*d*polylog(2,d*x)/x^2+1/6*d*(2*a*d+3*b)*polylog(2,d*x)/x-7/36*a*d*ln(-d*x+1)/x^2-1/2*b*d*ln(-d*x+1)/x-2/9
*a*d^2*ln(-d*x+1)/x-1/6*d*(2*a*d+3*b)*ln(-d*x+1)/x+1/6*d*(6*c+d*(2*a*d+3*b))*ln(d*x)*ln(-d*x+1)^2+1/6*d*(6*c+d
*(2*a*d+3*b))*ln(-d*x+1)*polylog(2,d*x)+1/3*d*(6*c+d*(2*a*d+3*b))*ln(-d*x+1)*polylog(2,-d*x+1)+c*(-d*x+1)*ln(-
d*x+1)^2/x+7/36*a*d^2/x-2*c*d*polylog(2,d*x)-1/2*b*d^2*polylog(2,d*x)-2/9*a*d^3*polylog(2,d*x)-1/6*d*(6*c+d*(2
*a*d+3*b))*polylog(3,d*x)-1/3*d*(6*c+d*(2*a*d+3*b))*polylog(3,-d*x+1)-1/2*b*d^2*ln(x)-5/12*a*d^3*ln(x)-1/6*d^2
*(2*a*d+3*b)*ln(x)+1/2*b*d^2*ln(-d*x+1)+5/12*a*d^3*ln(-d*x+1)+1/6*d^2*(2*a*d+3*b)*ln(-d*x+1)-1/4*b*d^2*ln(-d*x
+1)^2-1/9*a*d^3*ln(-d*x+1)^2+1/9*a*ln(-d*x+1)^2/x^3+1/4*b*ln(-d*x+1)^2/x^2-1/6*(2*a/x^3+3*b/x^2+6*c/x)*ln(-d*x
+1)*polylog(2,d*x)

________________________________________________________________________________________

Rubi [A]
time = 0.53, antiderivative size = 515, normalized size of antiderivative = 1.00, number of steps used = 43, number of rules used = 20, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {6874, 6726, 2442, 46, 36, 29, 31, 14, 6741, 2445, 2457, 2438, 2437, 2338, 2444, 6724, 6731, 2443, 2481, 2421} \begin {gather*} -\frac {1}{6} \text {Li}_2(d x) \log (1-d x) \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right )-\frac {1}{6} d \text {Li}_3(d x) (d (2 a d+3 b)+6 c)-\frac {1}{3} d \text {Li}_3(1-d x) (d (2 a d+3 b)+6 c)+\frac {1}{6} d \text {Li}_2(d x) \log (1-d x) (d (2 a d+3 b)+6 c)+\frac {1}{3} d \text {Li}_2(1-d x) \log (1-d x) (d (2 a d+3 b)+6 c)+\frac {1}{6} d \log (d x) \log ^2(1-d x) (d (2 a d+3 b)+6 c)-\frac {1}{6} d^2 \log (x) (2 a d+3 b)+\frac {1}{6} d^2 (2 a d+3 b) \log (1-d x)+\frac {d (2 a d+3 b) \text {Li}_2(d x)}{6 x}-\frac {d (2 a d+3 b) \log (1-d x)}{6 x}-\frac {2}{9} a d^3 \text {Li}_2(d x)-\frac {1}{9} a d^3 \log ^2(1-d x)-\frac {5}{12} a d^3 \log (x)+\frac {5}{12} a d^3 \log (1-d x)+\frac {7 a d^2}{36 x}-\frac {2 a d^2 \log (1-d x)}{9 x}+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {a \log ^2(1-d x)}{9 x^3}-\frac {7 a d \log (1-d x)}{36 x^2}-\frac {1}{2} b d^2 \text {Li}_2(d x)-\frac {1}{4} b d^2 \log ^2(1-d x)-\frac {1}{2} b d^2 \log (x)+\frac {1}{2} b d^2 \log (1-d x)+\frac {b \log ^2(1-d x)}{4 x^2}-\frac {b d \log (1-d x)}{2 x}-2 c d \text {Li}_2(d x)+\frac {c (1-d x) \log ^2(1-d x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a*d^2)/(36*x) - (b*d^2*Log[x])/2 - (5*a*d^3*Log[x])/12 - (d^2*(3*b + 2*a*d)*Log[x])/6 + (b*d^2*Log[1 - d*x]
)/2 + (5*a*d^3*Log[1 - d*x])/12 + (d^2*(3*b + 2*a*d)*Log[1 - d*x])/6 - (7*a*d*Log[1 - d*x])/(36*x^2) - (b*d*Lo
g[1 - d*x])/(2*x) - (2*a*d^2*Log[1 - d*x])/(9*x) - (d*(3*b + 2*a*d)*Log[1 - d*x])/(6*x) - (b*d^2*Log[1 - d*x]^
2)/4 - (a*d^3*Log[1 - d*x]^2)/9 + (a*Log[1 - d*x]^2)/(9*x^3) + (b*Log[1 - d*x]^2)/(4*x^2) + (c*(1 - d*x)*Log[1
 - d*x]^2)/x + (d*(6*c + d*(3*b + 2*a*d))*Log[d*x]*Log[1 - d*x]^2)/6 - 2*c*d*PolyLog[2, d*x] - (b*d^2*PolyLog[
2, d*x])/2 - (2*a*d^3*PolyLog[2, d*x])/9 + (a*d*PolyLog[2, d*x])/(6*x^2) + (d*(3*b + 2*a*d)*PolyLog[2, d*x])/(
6*x) + (d*(6*c + d*(3*b + 2*a*d))*Log[1 - d*x]*PolyLog[2, d*x])/6 - (((2*a)/x^3 + (3*b)/x^2 + (6*c)/x)*Log[1 -
 d*x]*PolyLog[2, d*x])/6 + (d*(6*c + d*(3*b + 2*a*d))*Log[1 - d*x]*PolyLog[2, 1 - d*x])/3 - (d*(6*c + d*(3*b +
 2*a*d))*PolyLog[3, d*x])/6 - (d*(6*c + d*(3*b + 2*a*d))*PolyLog[3, 1 - d*x])/3

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 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 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 {\left (a+b x+c x^2\right ) \log (1-d x) \text {Li}_2(d x)}{x^4} \, dx &=-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)+d \int \left (-\frac {a \text {Li}_2(d x)}{3 x^3}+\frac {(-3 b-2 a d) \text {Li}_2(d x)}{6 x^2}+\frac {\left (-6 c-3 b d-2 a d^2\right ) \text {Li}_2(d x)}{6 x}+\frac {d (-6 c-d (3 b+2 a d)) \text {Li}_2(d x)}{6 (1-d x)}\right ) \, dx+\int \left (-\frac {a \log ^2(1-d x)}{3 x^4}-\frac {b \log ^2(1-d x)}{2 x^3}-\frac {c \log ^2(1-d x)}{x^2}\right ) \, dx\\ &=-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)-\frac {1}{3} a \int \frac {\log ^2(1-d x)}{x^4} \, dx-\frac {1}{2} b \int \frac {\log ^2(1-d x)}{x^3} \, dx-c \int \frac {\log ^2(1-d x)}{x^2} \, dx-\frac {1}{3} (a d) \int \frac {\text {Li}_2(d x)}{x^3} \, dx-\frac {1}{6} (d (3 b+2 a d)) \int \frac {\text {Li}_2(d x)}{x^2} \, dx-\frac {1}{6} (d (6 c+d (3 b+2 a d))) \int \frac {\text {Li}_2(d x)}{x} \, dx-\frac {1}{6} \left (d^2 (6 c+d (3 b+2 a d))\right ) \int \frac {\text {Li}_2(d x)}{1-d x} \, dx\\ &=\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)+\frac {1}{6} (a d) \int \frac {\log (1-d x)}{x^3} \, dx+\frac {1}{9} (2 a d) \int \frac {\log (1-d x)}{x^3 (1-d x)} \, dx+\frac {1}{2} (b d) \int \frac {\log (1-d x)}{x^2 (1-d x)} \, dx+(2 c d) \int \frac {\log (1-d x)}{x} \, dx+\frac {1}{6} (d (3 b+2 a d)) \int \frac {\log (1-d x)}{x^2} \, dx+\frac {1}{6} (d (6 c+d (3 b+2 a d))) \int \frac {\log ^2(1-d x)}{x} \, dx\\ &=-\frac {a d \log (1-d x)}{12 x^2}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {Li}_2(d x)+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)+\frac {1}{9} (2 a d) \int \left (\frac {\log (1-d x)}{x^3}+\frac {d \log (1-d x)}{x^2}+\frac {d^2 \log (1-d x)}{x}-\frac {d^3 \log (1-d x)}{-1+d x}\right ) \, dx+\frac {1}{2} (b 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}{12} \left (a d^2\right ) \int \frac {1}{x^2 (1-d x)} \, dx-\frac {1}{6} \left (d^2 (3 b+2 a d)\right ) \int \frac {1}{x (1-d x)} \, dx+\frac {1}{3} \left (d^2 (6 c+d (3 b+2 a d))\right ) \int \frac {\log (d x) \log (1-d x)}{1-d x} \, dx\\ &=-\frac {a d \log (1-d x)}{12 x^2}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {Li}_2(d x)+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)+\frac {1}{9} (2 a d) \int \frac {\log (1-d x)}{x^3} \, dx+\frac {1}{2} (b d) \int \frac {\log (1-d x)}{x^2} \, dx-\frac {1}{12} \left (a d^2\right ) \int \left (\frac {1}{x^2}+\frac {d}{x}-\frac {d^2}{-1+d x}\right ) \, dx+\frac {1}{9} \left (2 a d^2\right ) \int \frac {\log (1-d x)}{x^2} \, dx+\frac {1}{2} \left (b d^2\right ) \int \frac {\log (1-d x)}{x} \, dx+\frac {1}{9} \left (2 a d^3\right ) \int \frac {\log (1-d x)}{x} \, dx-\frac {1}{2} \left (b d^3\right ) \int \frac {\log (1-d x)}{-1+d x} \, dx-\frac {1}{9} \left (2 a d^4\right ) \int \frac {\log (1-d x)}{-1+d x} \, dx-\frac {1}{6} \left (d^2 (3 b+2 a d)\right ) \int \frac {1}{x} \, dx-\frac {1}{6} \left (d^3 (3 b+2 a d)\right ) \int \frac {1}{1-d x} \, dx-\frac {1}{3} (d (6 c+d (3 b+2 a d))) \text {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 )\\ &=\frac {a d^2}{12 x}-\frac {1}{12} a d^3 \log (x)-\frac {1}{6} d^2 (3 b+2 a d) \log (x)+\frac {1}{12} a d^3 \log (1-d x)+\frac {1}{6} d^2 (3 b+2 a d) \log (1-d x)-\frac {7 a d \log (1-d x)}{36 x^2}-\frac {b d \log (1-d x)}{2 x}-\frac {2 a d^2 \log (1-d x)}{9 x}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {Li}_2(d x)-\frac {1}{2} b d^2 \text {Li}_2(d x)-\frac {2}{9} a d^3 \text {Li}_2(d x)+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)+\frac {1}{3} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(1-d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)-\frac {1}{9} \left (a d^2\right ) \int \frac {1}{x^2 (1-d x)} \, dx-\frac {1}{2} \left (b d^2\right ) \int \frac {1}{x (1-d x)} \, dx-\frac {1}{2} \left (b d^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-d x\right )-\frac {1}{9} \left (2 a d^3\right ) \int \frac {1}{x (1-d x)} \, dx-\frac {1}{9} \left (2 a d^3\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-d x\right )-\frac {1}{3} (d (6 c+d (3 b+2 a d))) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-d x\right )\\ &=\frac {a d^2}{12 x}-\frac {1}{12} a d^3 \log (x)-\frac {1}{6} d^2 (3 b+2 a d) \log (x)+\frac {1}{12} a d^3 \log (1-d x)+\frac {1}{6} d^2 (3 b+2 a d) \log (1-d x)-\frac {7 a d \log (1-d x)}{36 x^2}-\frac {b d \log (1-d x)}{2 x}-\frac {2 a d^2 \log (1-d x)}{9 x}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}-\frac {1}{4} b d^2 \log ^2(1-d x)-\frac {1}{9} a d^3 \log ^2(1-d x)+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {Li}_2(d x)-\frac {1}{2} b d^2 \text {Li}_2(d x)-\frac {2}{9} a d^3 \text {Li}_2(d x)+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)+\frac {1}{3} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(1-d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)-\frac {1}{3} d (6 c+d (3 b+2 a d)) \text {Li}_3(1-d x)-\frac {1}{9} \left (a d^2\right ) \int \left (\frac {1}{x^2}+\frac {d}{x}-\frac {d^2}{-1+d x}\right ) \, dx-\frac {1}{2} \left (b d^2\right ) \int \frac {1}{x} \, dx-\frac {1}{9} \left (2 a d^3\right ) \int \frac {1}{x} \, dx-\frac {1}{2} \left (b d^3\right ) \int \frac {1}{1-d x} \, dx-\frac {1}{9} \left (2 a d^4\right ) \int \frac {1}{1-d x} \, dx\\ &=\frac {7 a d^2}{36 x}-\frac {1}{2} b d^2 \log (x)-\frac {5}{12} a d^3 \log (x)-\frac {1}{6} d^2 (3 b+2 a d) \log (x)+\frac {1}{2} b d^2 \log (1-d x)+\frac {5}{12} a d^3 \log (1-d x)+\frac {1}{6} d^2 (3 b+2 a d) \log (1-d x)-\frac {7 a d \log (1-d x)}{36 x^2}-\frac {b d \log (1-d x)}{2 x}-\frac {2 a d^2 \log (1-d x)}{9 x}-\frac {d (3 b+2 a d) \log (1-d x)}{6 x}-\frac {1}{4} b d^2 \log ^2(1-d x)-\frac {1}{9} a d^3 \log ^2(1-d x)+\frac {a \log ^2(1-d x)}{9 x^3}+\frac {b \log ^2(1-d x)}{4 x^2}+\frac {c (1-d x) \log ^2(1-d x)}{x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (d x) \log ^2(1-d x)-2 c d \text {Li}_2(d x)-\frac {1}{2} b d^2 \text {Li}_2(d x)-\frac {2}{9} a d^3 \text {Li}_2(d x)+\frac {a d \text {Li}_2(d x)}{6 x^2}+\frac {d (3 b+2 a d) \text {Li}_2(d x)}{6 x}+\frac {1}{6} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(d x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}+\frac {6 c}{x}\right ) \log (1-d x) \text {Li}_2(d x)+\frac {1}{3} d (6 c+d (3 b+2 a d)) \log (1-d x) \text {Li}_2(1-d x)-\frac {1}{6} d (6 c+d (3 b+2 a d)) \text {Li}_3(d x)-\frac {1}{3} d (6 c+d (3 b+2 a d)) \text {Li}_3(1-d x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.21, size = 488, normalized size = 0.95 \begin {gather*} \frac {1}{36} \left (-7 a d^3+\frac {7 a d^2}{x}-36 b d^2 \log (d x)-27 a d^3 \log (d x)+36 b d^2 \log (1-d x)+27 a d^3 \log (1-d x)-\frac {7 a d \log (1-d x)}{x^2}-\frac {36 b d \log (1-d x)}{x}-\frac {20 a d^2 \log (1-d x)}{x}+72 c d \log (d x) \log (1-d x)+18 b d^2 \log (d x) \log (1-d x)+8 a d^3 \log (d x) \log (1-d x)-36 c d \log ^2(1-d x)-9 b d^2 \log ^2(1-d x)-4 a d^3 \log ^2(1-d x)+\frac {4 a \log ^2(1-d x)}{x^3}+\frac {9 b \log ^2(1-d x)}{x^2}+\frac {36 c \log ^2(1-d x)}{x}+36 c d \log (d x) \log ^2(1-d x)+18 b d^2 \log (d x) \log ^2(1-d x)+12 a d^3 \log (d x) \log ^2(1-d x)+\frac {6 \left (d x (a+3 b x+2 a d x)+(-1+d x) \left (3 x (b+2 c x+b d x)+2 a \left (1+d x+d^2 x^2\right )\right ) \log (1-d x)\right ) \text {PolyLog}(2,d x)}{x^3}+2 d \left (36 c+9 b d+4 a d^2+6 \left (6 c+3 b d+2 a d^2\right ) \log (1-d x)\right ) \text {PolyLog}(2,1-d x)-36 c d \text {PolyLog}(3,d x)-18 b d^2 \text {PolyLog}(3,d x)-12 a d^3 \text {PolyLog}(3,d x)-72 c d \text {PolyLog}(3,1-d x)-36 b d^2 \text {PolyLog}(3,1-d x)-24 a d^3 \text {PolyLog}(3,1-d x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*d^3 + (7*a*d^2)/x - 36*b*d^2*Log[d*x] - 27*a*d^3*Log[d*x] + 36*b*d^2*Log[1 - d*x] + 27*a*d^3*Log[1 - d*x
] - (7*a*d*Log[1 - d*x])/x^2 - (36*b*d*Log[1 - d*x])/x - (20*a*d^2*Log[1 - d*x])/x + 72*c*d*Log[d*x]*Log[1 - d
*x] + 18*b*d^2*Log[d*x]*Log[1 - d*x] + 8*a*d^3*Log[d*x]*Log[1 - d*x] - 36*c*d*Log[1 - d*x]^2 - 9*b*d^2*Log[1 -
 d*x]^2 - 4*a*d^3*Log[1 - d*x]^2 + (4*a*Log[1 - d*x]^2)/x^3 + (9*b*Log[1 - d*x]^2)/x^2 + (36*c*Log[1 - d*x]^2)
/x + 36*c*d*Log[d*x]*Log[1 - d*x]^2 + 18*b*d^2*Log[d*x]*Log[1 - d*x]^2 + 12*a*d^3*Log[d*x]*Log[1 - d*x]^2 + (6
*(d*x*(a + 3*b*x + 2*a*d*x) + (-1 + d*x)*(3*x*(b + 2*c*x + b*d*x) + 2*a*(1 + d*x + d^2*x^2))*Log[1 - d*x])*Pol
yLog[2, d*x])/x^3 + 2*d*(36*c + 9*b*d + 4*a*d^2 + 6*(6*c + 3*b*d + 2*a*d^2)*Log[1 - d*x])*PolyLog[2, 1 - d*x]
- 36*c*d*PolyLog[3, d*x] - 18*b*d^2*PolyLog[3, d*x] - 12*a*d^3*PolyLog[3, d*x] - 72*c*d*PolyLog[3, 1 - d*x] -
36*b*d^2*PolyLog[3, 1 - d*x] - 24*a*d^3*PolyLog[3, 1 - d*x])/36

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right ) \ln \left (-d x +1\right ) \polylog \left (2, d x \right )}{x^{4}}\, dx\]

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^4,x)

[Out]

int((c*x^2+b*x+a)*ln(-d*x+1)*polylog(2,d*x)/x^4,x)

________________________________________________________________________________________

Maxima [A]
time = 0.32, size = 319, normalized size = 0.62 \begin {gather*} \frac {1}{6} \, {\left (2 \, a d^{3} + 3 \, b d^{2} + 6 \, c d\right )} {\left (\log \left (d x\right ) \log \left (-d x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-d x + 1\right ) \log \left (-d x + 1\right ) - 2 \, {\rm Li}_{3}(-d x + 1)\right )} + \frac {1}{18} \, {\left (4 \, a d^{3} + 9 \, b d^{2} + 36 \, c d\right )} {\left (\log \left (d x\right ) \log \left (-d x + 1\right ) + {\rm Li}_2\left (-d x + 1\right )\right )} - \frac {1}{4} \, {\left (3 \, a d^{3} + 4 \, b d^{2}\right )} \log \left (x\right ) - \frac {1}{6} \, {\left (2 \, a d^{3} + 3 \, b d^{2} + 6 \, c d\right )} {\rm Li}_{3}(d x) + \frac {7 \, a d^{2} x^{2} - {\left ({\left (4 \, a d^{3} + 9 \, b d^{2} + 36 \, c d\right )} x^{3} - 36 \, c x^{2} - 9 \, b x - 4 \, a\right )} \log \left (-d x + 1\right )^{2} + 6 \, {\left (a d x + {\left (2 \, a d^{2} + 3 \, b d\right )} x^{2} + {\left ({\left (2 \, a d^{3} + 3 \, b d^{2} + 6 \, c d\right )} x^{3} - 6 \, c x^{2} - 3 \, b x - 2 \, a\right )} \log \left (-d x + 1\right )\right )} {\rm Li}_2\left (d x\right ) + {\left (9 \, {\left (3 \, a d^{3} + 4 \, b d^{2}\right )} x^{3} - 7 \, a d x - 4 \, {\left (5 \, a d^{2} + 9 \, b d\right )} x^{2}\right )} \log \left (-d x + 1\right )}{36 \, x^{3}} \end {gather*}

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^4,x, algorithm="maxima")

[Out]

1/6*(2*a*d^3 + 3*b*d^2 + 6*c*d)*(log(d*x)*log(-d*x + 1)^2 + 2*dilog(-d*x + 1)*log(-d*x + 1) - 2*polylog(3, -d*
x + 1)) + 1/18*(4*a*d^3 + 9*b*d^2 + 36*c*d)*(log(d*x)*log(-d*x + 1) + dilog(-d*x + 1)) - 1/4*(3*a*d^3 + 4*b*d^
2)*log(x) - 1/6*(2*a*d^3 + 3*b*d^2 + 6*c*d)*polylog(3, d*x) + 1/36*(7*a*d^2*x^2 - ((4*a*d^3 + 9*b*d^2 + 36*c*d
)*x^3 - 36*c*x^2 - 9*b*x - 4*a)*log(-d*x + 1)^2 + 6*(a*d*x + (2*a*d^2 + 3*b*d)*x^2 + ((2*a*d^3 + 3*b*d^2 + 6*c
*d)*x^3 - 6*c*x^2 - 3*b*x - 2*a)*log(-d*x + 1))*dilog(d*x) + (9*(3*a*d^3 + 4*b*d^2)*x^3 - 7*a*d*x - 4*(5*a*d^2
 + 9*b*d)*x^2)*log(-d*x + 1))/x^3

________________________________________________________________________________________

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

[Out]

integral((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x^4, x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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**4,x)

[Out]

Timed out

________________________________________________________________________________________

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((c*x^2+b*x+a)*log(-d*x+1)*polylog(2,d*x)/x^4,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)*dilog(d*x)*log(-d*x + 1)/x^4, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (1-d\,x\right )\,\mathrm {polylog}\left (2,d\,x\right )\,\left (c\,x^2+b\,x+a\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(1 - d*x)*polylog(2, d*x)*(a + b*x + c*x^2))/x^4,x)

[Out]

int((log(1 - d*x)*polylog(2, d*x)*(a + b*x + c*x^2))/x^4, x)

________________________________________________________________________________________