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

Optimal. Leaf size=153 \[ -\frac{1}{2} a \text{PolyLog}(2,c x)^2-b x \text{PolyLog}(2,c x)+\frac{2 b \text{PolyLog}(3,1-c x)}{c}+b x \log (1-c x) \text{PolyLog}(2,c x)-\frac{b \log (1-c x) \text{PolyLog}(2,c x)}{c}-\frac{2 b \log (1-c x) \text{PolyLog}(2,1-c x)}{c}-\frac{b (1-c x) \log ^2(1-c x)}{c}-\frac{b \log (c x) \log ^2(1-c x)}{c}+\frac{3 b (1-c x) \log (1-c x)}{c}+3 b x \]

[Out]

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

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Rubi [A]  time = 0.338379, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 15, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {6742, 6586, 2389, 2295, 6589, 6605, 6601, 12, 6600, 2296, 6688, 6596, 2396, 2433, 2374} \[ -\frac{1}{2} a \text{PolyLog}(2,c x)^2-b x \text{PolyLog}(2,c x)+\frac{2 b \text{PolyLog}(3,1-c x)}{c}+b x \log (1-c x) \text{PolyLog}(2,c x)-\frac{b \log (1-c x) \text{PolyLog}(2,c x)}{c}-\frac{2 b \log (1-c x) \text{PolyLog}(2,1-c x)}{c}-\frac{b (1-c x) \log ^2(1-c x)}{c}-\frac{b \log (c x) \log ^2(1-c x)}{c}+\frac{3 b (1-c x) \log (1-c x)}{c}+3 b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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]

Rubi steps

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

Mathematica [A]  time = 0.234598, size = 137, normalized size = 0.9 \[ -\frac{1}{2} a \text{PolyLog}(2,c x)^2+\frac{b \left (2 \text{PolyLog}(3,1-c x)-2 \log (1-c x) \text{PolyLog}(2,1-c x)+3 c x+c x \log ^2(1-c x)-\log (c x) \log ^2(1-c x)-\log ^2(1-c x)-3 c x \log (1-c x)+3 \log (1-c x)-2\right )}{c}+\frac{b ((c x-1) \log (1-c x)-c x) \text{PolyLog}(2,c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.005, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) \ln \left ( -cx+1 \right ){\it polylog} \left ( 2,cx \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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