3.164 \(\int \log (1-c x) \text{PolyLog}(2,c x) \, dx\)

Optimal. Leaf size=132 \[ -x \text{PolyLog}(2,c x)+\frac{2 \text{PolyLog}(3,1-c x)}{c}+x \log (1-c x) \text{PolyLog}(2,c x)-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{c}-\frac{2 \log (1-c x) \text{PolyLog}(2,1-c x)}{c}-\frac{(1-c x) \log ^2(1-c x)}{c}-\frac{\log (c x) \log ^2(1-c x)}{c}+\frac{3 (1-c x) \log (1-c x)}{c}+3 x \]

[Out]

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

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Rubi [A]  time = 0.211916, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.923, Rules used = {6586, 2389, 2295, 6600, 2296, 6688, 6742, 6596, 2396, 2433, 2374, 6589} \[ -x \text{PolyLog}(2,c x)+\frac{2 \text{PolyLog}(3,1-c x)}{c}+x \log (1-c x) \text{PolyLog}(2,c x)-\frac{\log (1-c x) \text{PolyLog}(2,c x)}{c}-\frac{2 \log (1-c x) \text{PolyLog}(2,1-c x)}{c}-\frac{(1-c x) \log ^2(1-c x)}{c}-\frac{\log (c x) \log ^2(1-c x)}{c}+\frac{3 (1-c x) \log (1-c x)}{c}+3 x \]

Antiderivative was successfully verified.

[In]

Int[Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

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

Rule 6586

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 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 6600

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :>
Simp[x*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)], x] + (Dist[b, Int[(g + h*Log[f*(d + e*x)^n])*Log[1
- a*c - b*c*x]*ExpandIntegrand[x/(a + b*x), x], x], x] - Dist[e*h*n, Int[PolyLog[2, c*(a + b*x)]*ExpandIntegr
and[x/(d + e*x), x], x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, n}, x]

Rule 2296

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 6688

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

Rule 6742

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

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]

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

Mathematica [A]  time = 0.0230911, size = 119, normalized size = 0.9 \[ \frac{2 \text{PolyLog}(3,1-c x)-2 \log (1-c x) \text{PolyLog}(2,1-c x)+((c x-1) \log (1-c x)-c x) \text{PolyLog}(2,c x)+3 c x+c x \log ^2(1-c x)-\log (c x) \log ^2(1-c x)-\log ^2(1-c x)-3 c x \log (1-c x)+3 \log (1-c x)-2}{c} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 - c*x]*PolyLog[2, c*x],x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10237, size = 190, normalized size = 1.44 \begin{align*} c{\left (\frac{x}{c} + \frac{\log \left (c x - 1\right )}{c^{2}}\right )} + \frac{{\left (c x{\rm Li}_2\left (c x\right ) - c x +{\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{c} - \frac{\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)}{c} + \frac{2 \, c x -{\left (c x + \log \left (-c x + 1\right )\right )}{\rm Li}_2\left (c x\right ) - 2 \,{\left (c x - 1\right )} \log \left (-c x + 1\right )}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(x/c + log(c*x - 1)/c^2) + (c*x*dilog(c*x) - c*x + (c*x - 1)*log(-c*x + 1))*log(-c*x + 1)/c - (log(c*x)*log(
-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylog(3, -c*x + 1))/c + (2*c*x - (c*x + log(-c*x + 1))*dil
og(c*x) - 2*(c*x - 1)*log(-c*x + 1))/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(dilog(c*x)*log(-c*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(-c*x + 1)*polylog(2, c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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