3.131 \(\int x^2 \text{PolyLog}(3,c (a+b x)) \, dx\)

Optimal. Leaf size=347 \[ -\frac{\left (a^3-b^3 x^3\right ) \text{PolyLog}(3,c (a+b x))}{3 b^3}-\frac{11 a^3 \text{PolyLog}(2,c (a+b x))}{18 b^3}+\frac{2 a^3 \text{PolyLog}(3,c (a+b x))}{3 b^3}-\frac{a^2 x \text{PolyLog}(2,c (a+b x))}{3 b^2}+\frac{a x^2 \text{PolyLog}(2,c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{PolyLog}(2,c (a+b x))+\frac{11 a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{18 b^3 c}+\frac{11 a^2 x}{18 b^2}+\frac{x (1-a c)^2}{27 b^2 c^2}-\frac{5 a (1-a c)^2 \log (-a c-b c x+1)}{36 b^3 c^2}+\frac{(1-a c)^3 \log (-a c-b c x+1)}{27 b^3 c^3}-\frac{5 a x (1-a c)}{36 b^2 c}+\frac{x^2 (1-a c)}{54 b c}+\frac{5 a x^2 \log (-a c-b c x+1)}{36 b}-\frac{1}{27} x^3 \log (-a c-b c x+1)-\frac{5 a x^2}{72 b}+\frac{x^3}{81} \]

[Out]

(11*a^2*x)/(18*b^2) - (5*a*(1 - a*c)*x)/(36*b^2*c) + ((1 - a*c)^2*x)/(27*b^2*c^2) - (5*a*x^2)/(72*b) + ((1 - a
*c)*x^2)/(54*b*c) + x^3/81 - (5*a*(1 - a*c)^2*Log[1 - a*c - b*c*x])/(36*b^3*c^2) + ((1 - a*c)^3*Log[1 - a*c -
b*c*x])/(27*b^3*c^3) + (5*a*x^2*Log[1 - a*c - b*c*x])/(36*b) - (x^3*Log[1 - a*c - b*c*x])/27 + (11*a^2*(1 - a*
c - b*c*x)*Log[1 - a*c - b*c*x])/(18*b^3*c) - (11*a^3*PolyLog[2, c*(a + b*x)])/(18*b^3) - (a^2*x*PolyLog[2, c*
(a + b*x)])/(3*b^2) + (a*x^2*PolyLog[2, c*(a + b*x)])/(6*b) - (x^3*PolyLog[2, c*(a + b*x)])/9 + (2*a^3*PolyLog
[3, c*(a + b*x)])/(3*b^3) - ((a^3 - b^3*x^3)*PolyLog[3, c*(a + b*x)])/(3*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.637739, antiderivative size = 347, normalized size of antiderivative = 1., number of steps used = 33, number of rules used = 13, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6599, 6595, 2444, 2389, 2295, 2421, 2393, 2391, 6598, 43, 2416, 2395, 6589} \[ -\frac{\left (a^3-b^3 x^3\right ) \text{PolyLog}(3,c (a+b x))}{3 b^3}-\frac{11 a^3 \text{PolyLog}(2,c (a+b x))}{18 b^3}+\frac{2 a^3 \text{PolyLog}(3,c (a+b x))}{3 b^3}-\frac{a^2 x \text{PolyLog}(2,c (a+b x))}{3 b^2}+\frac{a x^2 \text{PolyLog}(2,c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{PolyLog}(2,c (a+b x))+\frac{11 a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{18 b^3 c}+\frac{11 a^2 x}{18 b^2}+\frac{x (1-a c)^2}{27 b^2 c^2}-\frac{5 a (1-a c)^2 \log (-a c-b c x+1)}{36 b^3 c^2}+\frac{(1-a c)^3 \log (-a c-b c x+1)}{27 b^3 c^3}-\frac{5 a x (1-a c)}{36 b^2 c}+\frac{x^2 (1-a c)}{54 b c}+\frac{5 a x^2 \log (-a c-b c x+1)}{36 b}-\frac{1}{27} x^3 \log (-a c-b c x+1)-\frac{5 a x^2}{72 b}+\frac{x^3}{81} \]

Antiderivative was successfully verified.

[In]

Int[x^2*PolyLog[3, c*(a + b*x)],x]

[Out]

(11*a^2*x)/(18*b^2) - (5*a*(1 - a*c)*x)/(36*b^2*c) + ((1 - a*c)^2*x)/(27*b^2*c^2) - (5*a*x^2)/(72*b) + ((1 - a
*c)*x^2)/(54*b*c) + x^3/81 - (5*a*(1 - a*c)^2*Log[1 - a*c - b*c*x])/(36*b^3*c^2) + ((1 - a*c)^3*Log[1 - a*c -
b*c*x])/(27*b^3*c^3) + (5*a*x^2*Log[1 - a*c - b*c*x])/(36*b) - (x^3*Log[1 - a*c - b*c*x])/27 + (11*a^2*(1 - a*
c - b*c*x)*Log[1 - a*c - b*c*x])/(18*b^3*c) - (11*a^3*PolyLog[2, c*(a + b*x)])/(18*b^3) - (a^2*x*PolyLog[2, c*
(a + b*x)])/(3*b^2) + (a*x^2*PolyLog[2, c*(a + b*x)])/(6*b) - (x^3*PolyLog[2, c*(a + b*x)])/9 + (2*a^3*PolyLog
[3, c*(a + b*x)])/(3*b^3) - ((a^3 - b^3*x^3)*PolyLog[3, c*(a + b*x)])/(3*b^3)

Rule 6599

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

Rule 6595

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

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 2389

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rule 2393

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

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 6598

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 43

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 2416

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 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 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]

Rubi steps

\begin{align*} \int x^2 \text{Li}_3(c (a+b x)) \, dx &=-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}+\frac{\int \left (-a^2 \text{Li}_2(c (a+b x))+a b x \text{Li}_2(c (a+b x))-b^2 x^2 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_2(c (a+b x))}{a+b x}\right ) \, dx}{3 b^2}\\ &=-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}-\frac{1}{3} \int x^2 \text{Li}_2(c (a+b x)) \, dx-\frac{a^2 \int \text{Li}_2(c (a+b x)) \, dx}{3 b^2}+\frac{\left (2 a^3\right ) \int \frac{\text{Li}_2(c (a+b x))}{a+b x} \, dx}{3 b^2}+\frac{a \int x \text{Li}_2(c (a+b x)) \, dx}{3 b}\\ &=-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}+\frac{1}{6} a \int \frac{x^2 \log (1-a c-b c x)}{a+b x} \, dx-\frac{a^2 \int \log (1-c (a+b x)) \, dx}{3 b^2}+\frac{a^3 \int \frac{\log (1-c (a+b x))}{a+b x} \, dx}{3 b^2}-\frac{1}{9} b \int \frac{x^3 \log (1-a c-b c x)}{a+b x} \, dx\\ &=-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}+\frac{1}{6} a \int \left (-\frac{a \log (1-a c-b c x)}{b^2}+\frac{x \log (1-a c-b c x)}{b}+\frac{a^2 \log (1-a c-b c x)}{b^2 (a+b x)}\right ) \, dx-\frac{a^2 \int \log (1-a c-b c x) \, dx}{3 b^2}+\frac{a^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{3 b^2}-\frac{1}{9} b \int \left (\frac{a^2 \log (1-a c-b c x)}{b^3}-\frac{a x \log (1-a c-b c x)}{b^2}+\frac{x^2 \log (1-a c-b c x)}{b}-\frac{a^3 \log (1-a c-b c x)}{b^3 (a+b x)}\right ) \, dx\\ &=-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}-\frac{1}{9} \int x^2 \log (1-a c-b c x) \, dx+\frac{a^3 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{3 b^3}-\frac{a^2 \int \log (1-a c-b c x) \, dx}{9 b^2}-\frac{a^2 \int \log (1-a c-b c x) \, dx}{6 b^2}+\frac{a^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{9 b^2}+\frac{a^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{6 b^2}+\frac{a \int x \log (1-a c-b c x) \, dx}{9 b}+\frac{a \int x \log (1-a c-b c x) \, dx}{6 b}+\frac{a^2 \operatorname{Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{3 b^3 c}\\ &=\frac{a^2 x}{3 b^2}+\frac{5 a x^2 \log (1-a c-b c x)}{36 b}-\frac{1}{27} x^3 \log (1-a c-b c x)+\frac{a^2 (1-a c-b c x) \log (1-a c-b c x)}{3 b^3 c}-\frac{a^3 \text{Li}_2(c (a+b x))}{3 b^3}-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}+\frac{a^3 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{9 b^3}+\frac{a^3 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{6 b^3}+\frac{a^2 \operatorname{Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{9 b^3 c}+\frac{a^2 \operatorname{Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{6 b^3 c}+\frac{1}{18} (a c) \int \frac{x^2}{1-a c-b c x} \, dx+\frac{1}{12} (a c) \int \frac{x^2}{1-a c-b c x} \, dx-\frac{1}{27} (b c) \int \frac{x^3}{1-a c-b c x} \, dx\\ &=\frac{11 a^2 x}{18 b^2}+\frac{5 a x^2 \log (1-a c-b c x)}{36 b}-\frac{1}{27} x^3 \log (1-a c-b c x)+\frac{11 a^2 (1-a c-b c x) \log (1-a c-b c x)}{18 b^3 c}-\frac{11 a^3 \text{Li}_2(c (a+b x))}{18 b^3}-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}+\frac{1}{18} (a c) \int \left (\frac{-1+a c}{b^2 c^2}-\frac{x}{b c}-\frac{(-1+a c)^2}{b^2 c^2 (-1+a c+b c x)}\right ) \, dx+\frac{1}{12} (a c) \int \left (\frac{-1+a c}{b^2 c^2}-\frac{x}{b c}-\frac{(-1+a c)^2}{b^2 c^2 (-1+a c+b c x)}\right ) \, dx-\frac{1}{27} (b c) \int \left (-\frac{(-1+a c)^2}{b^3 c^3}+\frac{(-1+a c) x}{b^2 c^2}-\frac{x^2}{b c}+\frac{(-1+a c)^3}{b^3 c^3 (-1+a c+b c x)}\right ) \, dx\\ &=\frac{11 a^2 x}{18 b^2}-\frac{5 a (1-a c) x}{36 b^2 c}+\frac{(1-a c)^2 x}{27 b^2 c^2}-\frac{5 a x^2}{72 b}+\frac{(1-a c) x^2}{54 b c}+\frac{x^3}{81}-\frac{5 a (1-a c)^2 \log (1-a c-b c x)}{36 b^3 c^2}+\frac{(1-a c)^3 \log (1-a c-b c x)}{27 b^3 c^3}+\frac{5 a x^2 \log (1-a c-b c x)}{36 b}-\frac{1}{27} x^3 \log (1-a c-b c x)+\frac{11 a^2 (1-a c-b c x) \log (1-a c-b c x)}{18 b^3 c}-\frac{11 a^3 \text{Li}_2(c (a+b x))}{18 b^3}-\frac{a^2 x \text{Li}_2(c (a+b x))}{3 b^2}+\frac{a x^2 \text{Li}_2(c (a+b x))}{6 b}-\frac{1}{9} x^3 \text{Li}_2(c (a+b x))+\frac{2 a^3 \text{Li}_3(c (a+b x))}{3 b^3}-\frac{\left (a^3-b^3 x^3\right ) \text{Li}_3(c (a+b x))}{3 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0681564, size = 296, normalized size = 0.85 \[ \frac{-36 c^3 \left (6 a^2 b x+11 a^3-3 a b^2 x^2+2 b^3 x^3\right ) \text{PolyLog}(2,c (a+b x))+216 c^3 \left (a^3+b^3 x^3\right ) \text{PolyLog}(3,c (a+b x))+510 a^2 b c^3 x-510 a^3 c^3 \log (-a c-b c x+1)-396 a^2 b c^3 x \log (-a c-b c x+1)+648 a^2 c^2 \log (-a c-b c x+1)+575 a^3 c^3-150 a^2 c^2-57 a b^2 c^3 x^2-24 b^3 c^3 x^3 \log (-a c-b c x+1)+90 a b^2 c^3 x^2 \log (-a c-b c x+1)-138 a b c^2 x-162 a c \log (-a c-b c x+1)+24 \log (-a c-b c x+1)+24 a c+8 b^3 c^3 x^3+12 b^2 c^2 x^2+24 b c x}{648 b^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*c - 150*a^2*c^2 + 575*a^3*c^3 + 24*b*c*x - 138*a*b*c^2*x + 510*a^2*b*c^3*x + 12*b^2*c^2*x^2 - 57*a*b^2*c
^3*x^2 + 8*b^3*c^3*x^3 + 24*Log[1 - a*c - b*c*x] - 162*a*c*Log[1 - a*c - b*c*x] + 648*a^2*c^2*Log[1 - a*c - b*
c*x] - 510*a^3*c^3*Log[1 - a*c - b*c*x] - 396*a^2*b*c^3*x*Log[1 - a*c - b*c*x] + 90*a*b^2*c^3*x^2*Log[1 - a*c
- b*c*x] - 24*b^3*c^3*x^3*Log[1 - a*c - b*c*x] - 36*c^3*(11*a^3 + 6*a^2*b*x - 3*a*b^2*x^2 + 2*b^3*x^3)*PolyLog
[2, c*(a + b*x)] + 216*c^3*(a^3 + b^3*x^3)*PolyLog[3, c*(a + b*x)])/(648*b^3*c^3)

________________________________________________________________________________________

Maple [F]  time = 0.005, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}{\it polylog} \left ( 3,c \left ( bx+a \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*polylog(3,c*(b*x+a)),x)

[Out]

int(x^2*polylog(3,c*(b*x+a)),x)

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Maxima [A]  time = 1.03005, size = 356, normalized size = 1.03 \begin{align*} \frac{11 \,{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) +{\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{3}}{18 \, b^{3}} + \frac{a^{3}{\rm Li}_{3}(b c x + a c)}{3 \, b^{3}} + \frac{216 \, b^{3} c^{3} x^{3}{\rm Li}_{3}(b c x + a c) + 8 \, b^{3} c^{3} x^{3} - 3 \,{\left (19 \, a b^{2} c^{3} - 4 \, b^{2} c^{2}\right )} x^{2} + 6 \,{\left (85 \, a^{2} b c^{3} - 23 \, a b c^{2} + 4 \, b c\right )} x - 36 \,{\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x\right )}{\rm Li}_2\left (b c x + a c\right ) - 6 \,{\left (4 \, b^{3} c^{3} x^{3} - 15 \, a b^{2} c^{3} x^{2} + 66 \, a^{2} b c^{3} x + 85 \, a^{3} c^{3} - 108 \, a^{2} c^{2} + 27 \, a c - 4\right )} \log \left (-b c x - a c + 1\right )}{648 \, b^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,c*(b*x+a)),x, algorithm="maxima")

[Out]

11/18*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^3/b^3 + 1/3*a^3*polylog(3, b*c*x +
a*c)/b^3 + 1/648*(216*b^3*c^3*x^3*polylog(3, b*c*x + a*c) + 8*b^3*c^3*x^3 - 3*(19*a*b^2*c^3 - 4*b^2*c^2)*x^2 +
6*(85*a^2*b*c^3 - 23*a*b*c^2 + 4*b*c)*x - 36*(2*b^3*c^3*x^3 - 3*a*b^2*c^3*x^2 + 6*a^2*b*c^3*x)*dilog(b*c*x +
a*c) - 6*(4*b^3*c^3*x^3 - 15*a*b^2*c^3*x^2 + 66*a^2*b*c^3*x + 85*a^3*c^3 - 108*a^2*c^2 + 27*a*c - 4)*log(-b*c*
x - a*c + 1))/(b^3*c^3)

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Fricas [C]  time = 2.34524, size = 675, normalized size = 1.95 \begin{align*} \frac{8 \, b^{3} c^{3} x^{3} - 3 \,{\left (19 \, a b^{2} c^{3} - 4 \, b^{2} c^{2}\right )} x^{2} + 6 \,{\left (85 \, a^{2} b c^{3} - 23 \, a b c^{2} + 4 \, b c\right )} x - 36 \,{\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3}\right )}{\rm \%iint}\left (a, b, c, x, -\frac{\log \left (-b c x - a c + 1\right )}{b x + a}, -\frac{x \log \left (-b c x - a c + 1\right )}{b x + a}, -\frac{\log \left (-b c x - a c + 1\right )}{c}, -\frac{b \log \left (-b c x - a c + 1\right )}{b x + a}\right ) - 6 \,{\left (4 \, b^{3} c^{3} x^{3} - 15 \, a b^{2} c^{3} x^{2} + 66 \, a^{2} b c^{3} x + 85 \, a^{3} c^{3} - 108 \, a^{2} c^{2} + 27 \, a c - 4\right )} \log \left (-b c x - a c + 1\right ) + 216 \,{\left (b^{3} c^{3} x^{3} + a^{3} c^{3}\right )}{\rm polylog}\left (3, b c x + a c\right )}{648 \, b^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,c*(b*x+a)),x, algorithm="fricas")

[Out]

1/648*(8*b^3*c^3*x^3 - 3*(19*a*b^2*c^3 - 4*b^2*c^2)*x^2 + 6*(85*a^2*b*c^3 - 23*a*b*c^2 + 4*b*c)*x - 36*(2*b^3*
c^3*x^3 - 3*a*b^2*c^3*x^2 + 6*a^2*b*c^3*x + 11*a^3*c^3)*\%iint(a, b, c, x, -log(-b*c*x - a*c + 1)/(b*x + a), -x
*log(-b*c*x - a*c + 1)/(b*x + a), -log(-b*c*x - a*c + 1)/c, -b*log(-b*c*x - a*c + 1)/(b*x + a)) - 6*(4*b^3*c^3
*x^3 - 15*a*b^2*c^3*x^2 + 66*a^2*b*c^3*x + 85*a^3*c^3 - 108*a^2*c^2 + 27*a*c - 4)*log(-b*c*x - a*c + 1) + 216*
(b^3*c^3*x^3 + a^3*c^3)*polylog(3, b*c*x + a*c))/(b^3*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{Li}_{3}\left (a c + b c x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*polylog(3,c*(b*x+a)),x)

[Out]

Integral(x**2*polylog(3, a*c + b*c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,c*(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*polylog(3, (b*x + a)*c), x)