3.173 \(\int \frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x} \, dx\)

Optimal. Leaf size=20 \[ g \text {Li}_3(c x)-\frac {1}{2} h \text {Li}_2(c x){}^2 \]

[Out]

-1/2*h*polylog(2,c*x)^2+g*polylog(3,c*x)

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6602, 6589, 6601} \[ g \text {PolyLog}(3,c x)-\frac {1}{2} h \text {PolyLog}(2,c x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(h*PolyLog[2, c*x]^2)/2 + g*PolyLog[3, c*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 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 6602

Int[((Log[1 + (e_.)*(x_)]*(h_.) + (g_))*PolyLog[2, (c_.)*(x_)])/(x_), x_Symbol] :> Dist[g, Int[PolyLog[2, c*x]
/x, x], x] + Dist[h, Int[(Log[1 + e*x]*PolyLog[2, c*x])/x, x], x] /; FreeQ[{c, e, g, h}, x] && EqQ[c + e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \[ g \text {Li}_3(c x)-\frac {1}{2} h \text {Li}_2(c x){}^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {h {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) + g {\rm Li}_2\left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((h*dilog(c*x)*log(-c*x + 1) + g*dilog(c*x))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (h \log \left (-c x + 1\right ) + g\right )} {\rm Li}_2\left (c x\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 19, normalized size = 0.95 \[ -\frac {h \polylog \left (2, c x \right )^{2}}{2}+g \polylog \left (3, c x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*h*polylog(2,c*x)^2+g*polylog(3,c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (h \log \left (-c x + 1\right ) + g\right )} {\rm Li}_2\left (c x\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.25, size = 18, normalized size = 0.90 \[ g\,\mathrm {polylog}\left (3,c\,x\right )-\frac {h\,{\mathrm {polylog}\left (2,c\,x\right )}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((g + h*log(1 - c*x))*polylog(2, c*x))/x,x)

[Out]

g*polylog(3, c*x) - (h*polylog(2, c*x)^2)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g + h \log {\left (- c x + 1 \right )}\right ) \operatorname {Li}_{2}\left (c x\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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