3.2.87 \(\int \frac {(a+b x) \log (1-c x) \text {PolyLog}(2,c x)}{x} \, dx\) [187]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.22, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {6874, 6721, 2436, 2332, 6724, 6740, 6736, 12, 6735, 2333, 6820, 6731, 2443, 2481, 2421} \begin {gather*} -\frac {1}{2} a \text {Li}_2(c x){}^2-b x \text {Li}_2(c x)+\frac {2 b \text {Li}_3(1-c x)}{c}+b x \text {Li}_2(c x) \log (1-c x)-\frac {b \text {Li}_2(c x) \log (1-c x)}{c}-\frac {2 b \text {Li}_2(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}+\frac {3 b (1-c x) \log (1-c x)}{c}+3 b x \end {gather*}

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 12

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

Rule 2332

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

Rule 2333

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 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 2436

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 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 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 6721

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 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 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 6735

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 6736

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 6740

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 6820

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

Rule 6874

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

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 \text {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) \text {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 \text {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) \text {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) \text {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 = 0.15, size = 137, normalized size = 0.90 \begin {gather*} \frac {b (-c x+(-1+c x) \log (1-c x)) \text {PolyLog}(2,c x)}{c}-\frac {1}{2} a \text {PolyLog}(2,c x)^2+\frac {b \left (-2+3 c x+3 \log (1-c x)-3 c x \log (1-c x)-\log ^2(1-c x)+c x \log ^2(1-c x)-\log (c x) \log ^2(1-c x)-2 \log (1-c x) \text {PolyLog}(2,1-c x)+2 \text {PolyLog}(3,1-c x)\right )}{c} \end {gather*}

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

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right ) \ln \left (-c x +1\right ) \polylog \left (2, c x \right )}{x}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x}\, dx \end {gather*}

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)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________