3.2.47 \(\int \frac {\text {PolyLog}(2,x)}{(-1+x) x} \, dx\) [147]

Optimal. Leaf size=51 \[ \log ^2(1-x) \log (x)+2 \log (1-x) \text {PolyLog}(2,1-x)+\log (1-x) \text {PolyLog}(2,x)-2 \text {PolyLog}(3,1-x)-\text {PolyLog}(3,x) \]

[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)-polylog(3,x)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6874, 6731, 2443, 2481, 2421, 6724} \begin {gather*} -2 \text {Li}_3(1-x)-\text {Li}_3(x)+2 \text {Li}_2(1-x) \log (1-x)+\text {Li}_2(x) \log (1-x)+\log (x) \log ^2(1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, x]/((-1 + x)*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] - PolyLo
g[3, x]

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 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 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 6874

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 51, normalized size = 1.00 \begin {gather*} \log ^2(1-x) \log (x)+2 \log (1-x) \text {PolyLog}(2,1-x)+\log (1-x) \text {PolyLog}(2,x)-2 \text {PolyLog}(3,1-x)-\text {PolyLog}(3,x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, x]/((-1 + x)*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] - PolyLo
g[3, x]

________________________________________________________________________________________

Maple [A]
time = 0.73, size = 82, normalized size = 1.61

method	result	size
default	\(\ln \left (-1+x \right ) \polylog \left (2, x\right )+\ln \left (-1+x \right )^{2} \ln \left (x \right )+2 \ln \left (-1+x \right ) \polylog \left (2, 1-x \right )-2 \polylog \left (3, 1-x \right )+\ln \left (x \right ) \ln \left (-1+x \right ) \left (\ln \left (1-x \right )-\ln \left (-1+x \right )\right )+\dilog \left (x \right ) \left (\ln \left (1-x \right )-\ln \left (-1+x \right )\right )-\polylog \left (3, x\right )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,x)/(-1+x)/x,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 49, normalized size = 0.96 \begin {gather*} \log \left (x\right ) \log \left (-x + 1\right )^{2} + {\rm Li}_2\left (x\right ) \log \left (-x + 1\right ) + 2 \, {\rm Li}_2\left (-x + 1\right ) \log \left (-x + 1\right ) - {\rm Li}_{3}(x) - 2 \, {\rm Li}_{3}(-x + 1) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

integral(dilog(x)/(x^2 - x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(dilog(x)/((x - 1)*x), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {polylog}\left (2,x\right )}{x\,\left (x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, x)/(x*(x - 1)),x)

[Out]

int(polylog(2, x)/(x*(x - 1)), x)

________________________________________________________________________________________