3.145 \(\int \frac {\text {Li}_2(x)}{-1+x} \, dx\)

Optimal. Leaf size=46 \[ -2 \text {Li}_3(1-x)+2 \text {Li}_2(1-x) \log (1-x)+\text {Li}_2(x) \log (1-x)+\log (x) \log ^2(1-x) \]

[Out]

ln(1-x)^2*ln(x)+2*ln(1-x)*polylog(2,1-x)+ln(1-x)*polylog(2,x)-2*polylog(3,1-x)

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Rubi [A]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6596, 2396, 2433, 2374, 6589} \[ -2 \text {PolyLog}(3,1-x)+2 \log (1-x) \text {PolyLog}(2,1-x)+\log (1-x) \text {PolyLog}(2,x)+\log (x) \log ^2(1-x) \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, x]/(-1 + x),x]

[Out]

Log[1 - x]^2*Log[x] + 2*Log[1 - x]*PolyLog[2, 1 - x] + Log[1 - x]*PolyLog[2, x] - 2*PolyLog[3, 1 - x]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 46, normalized size = 1.00 \[ -2 \text {Li}_3(1-x)+2 \text {Li}_2(1-x) \log (1-x)+\text {Li}_2(x) \log (1-x)+\log (x) \log ^2(1-x) \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, x]/(-1 + x),x]

[Out]

Log[1 - x]^2*Log[x] + 2*Log[1 - x]*PolyLog[2, 1 - x] + Log[1 - x]*PolyLog[2, x] - 2*PolyLog[3, 1 - x]

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm Li}_2\relax (x)}{x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,x)/(-1+x),x, algorithm="fricas")

[Out]

integral(dilog(x)/(x - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\relax (x)}{x - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,x)/(-1+x),x, algorithm="giac")

[Out]

integrate(dilog(x)/(x - 1), x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\polylog \left (2, x\right )}{-1+x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,x)/(-1+x),x)

[Out]

int(polylog(2,x)/(-1+x),x)

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maxima [A]  time = 0.31, size = 44, normalized size = 0.96 \[ \log \relax (x) \log \left (-x + 1\right )^{2} + {\rm Li}_2\relax (x) \log \left (-x + 1\right ) + 2 \, {\rm Li}_2\left (-x + 1\right ) \log \left (-x + 1\right ) - 2 \, {\rm Li}_{3}(-x + 1) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,x)/(-1+x),x, algorithm="maxima")

[Out]

log(x)*log(-x + 1)^2 + dilog(x)*log(-x + 1) + 2*dilog(-x + 1)*log(-x + 1) - 2*polylog(3, -x + 1)

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mupad [B]  time = 0.04, size = 46, normalized size = 1.00 \[ {\ln \left (1-x\right )}^2\,\ln \relax (x)-2\,\mathrm {polylog}\left (3,1-x\right )+2\,\ln \left (1-x\right )\,\mathrm {polylog}\left (2,1-x\right )+\ln \left (1-x\right )\,\mathrm {polylog}\left (2,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, x)/(x - 1),x)

[Out]

log(1 - x)^2*log(x) - 2*polylog(3, 1 - x) + 2*log(1 - x)*polylog(2, 1 - x) + log(1 - x)*polylog(2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{2}\relax (x)}{x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,x)/(-1+x),x)

[Out]

Integral(polylog(2, x)/(x - 1), x)

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