3.17 \(\int \frac{\text{PolyLog}(3,a x)}{x^3} \, dx\)

Optimal. Leaf size=70 \[ -\frac{\text{PolyLog}(2,a x)}{4 x^2}-\frac{\text{PolyLog}(3,a x)}{2 x^2}+\frac{1}{8} a^2 \log (x)-\frac{1}{8} a^2 \log (1-a x)+\frac{\log (1-a x)}{8 x^2}-\frac{a}{8 x} \]

[Out]

-a/(8*x) + (a^2*Log[x])/8 - (a^2*Log[1 - a*x])/8 + Log[1 - a*x]/(8*x^2) - PolyLog[2, a*x]/(4*x^2) - PolyLog[3,
a*x]/(2*x^2)

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Rubi [A]  time = 0.0431429, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6591, 2395, 44} \[ -\frac{\text{PolyLog}(2,a x)}{4 x^2}-\frac{\text{PolyLog}(3,a x)}{2 x^2}+\frac{1}{8} a^2 \log (x)-\frac{1}{8} a^2 \log (1-a x)+\frac{\log (1-a x)}{8 x^2}-\frac{a}{8 x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x]/x^3,x]

[Out]

-a/(8*x) + (a^2*Log[x])/8 - (a^2*Log[1 - a*x])/8 + Log[1 - a*x]/(8*x^2) - PolyLog[2, a*x]/(4*x^2) - PolyLog[3,
a*x]/(2*x^2)

Rule 6591

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[((d*x)^(m + 1)*PolyLog[n
, a*(b*x^p)^q])/(d*(m + 1)), x] - Dist[(p*q)/(m + 1), Int[(d*x)^m*PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ
[{a, b, d, m, p, q}, x] && NeQ[m, -1] && GtQ[n, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\text{Li}_3(a x)}{x^3} \, dx &=-\frac{\text{Li}_3(a x)}{2 x^2}+\frac{1}{2} \int \frac{\text{Li}_2(a x)}{x^3} \, dx\\ &=-\frac{\text{Li}_2(a x)}{4 x^2}-\frac{\text{Li}_3(a x)}{2 x^2}-\frac{1}{4} \int \frac{\log (1-a x)}{x^3} \, dx\\ &=\frac{\log (1-a x)}{8 x^2}-\frac{\text{Li}_2(a x)}{4 x^2}-\frac{\text{Li}_3(a x)}{2 x^2}+\frac{1}{8} a \int \frac{1}{x^2 (1-a x)} \, dx\\ &=\frac{\log (1-a x)}{8 x^2}-\frac{\text{Li}_2(a x)}{4 x^2}-\frac{\text{Li}_3(a x)}{2 x^2}+\frac{1}{8} a \int \left (\frac{1}{x^2}+\frac{a}{x}-\frac{a^2}{-1+a x}\right ) \, dx\\ &=-\frac{a}{8 x}+\frac{1}{8} a^2 \log (x)-\frac{1}{8} a^2 \log (1-a x)+\frac{\log (1-a x)}{8 x^2}-\frac{\text{Li}_2(a x)}{4 x^2}-\frac{\text{Li}_3(a x)}{2 x^2}\\ \end{align*}

Mathematica [C]  time = 0.0090975, size = 25, normalized size = 0.36 \[ \frac{G_{5,5}^{2,4}\left (-a x\left |\begin{array}{c} 1,1,1,1,3 \\ 1,2,0,0,0 \\\end{array}\right .\right )}{x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x]/x^3,x]

[Out]

MeijerG[{{1, 1, 1, 1}, {3}}, {{1, 2}, {0, 0, 0}}, -(a*x)]/x^2

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Maple [A]  time = 0.165, size = 90, normalized size = 1.3 \begin{align*} -{a}^{2} \left ({\frac{1}{ax}}+{\frac{3}{16}}-{\frac{\ln \left ( x \right ) }{8}}-{\frac{\ln \left ( -a \right ) }{8}}-{\frac{81\,ax+378}{432\,ax}}-{\frac{ \left ( -27\,{a}^{2}{x}^{2}+27 \right ) \ln \left ( -ax+1 \right ) }{216\,{a}^{2}{x}^{2}}}+{\frac{{\it polylog} \left ( 2,ax \right ) }{4\,{a}^{2}{x}^{2}}}+{\frac{{\it polylog} \left ( 3,ax \right ) }{2\,{a}^{2}{x}^{2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x)/x^3,x)

[Out]

-a^2*(1/a/x+3/16-1/8*ln(x)-1/8*ln(-a)-1/432/a/x*(81*a*x+378)-1/216/a^2/x^2*(-27*a^2*x^2+27)*ln(-a*x+1)+1/4/a^2
/x^2*polylog(2,a*x)+1/2/a^2/x^2*polylog(3,a*x))

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Maxima [A]  time = 0.992882, size = 63, normalized size = 0.9 \begin{align*} \frac{1}{8} \, a^{2} \log \left (x\right ) - \frac{a x +{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right ) + 2 \,{\rm Li}_2\left (a x\right ) + 4 \,{\rm Li}_{3}(a x)}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x)/x^3,x, algorithm="maxima")

[Out]

1/8*a^2*log(x) - 1/8*(a*x + (a^2*x^2 - 1)*log(-a*x + 1) + 2*dilog(a*x) + 4*polylog(3, a*x))/x^2

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Fricas [C]  time = 2.71799, size = 194, normalized size = 2.77 \begin{align*} -\frac{a^{2} x^{2} \log \left (a x - 1\right ) - a^{2} x^{2} \log \left (x\right ) + a x + 2 \,{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - \log \left (-a x + 1\right ) + 4 \,{\rm polylog}\left (3, a x\right )}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x)/x^3,x, algorithm="fricas")

[Out]

-1/8*(a^2*x^2*log(a*x - 1) - a^2*x^2*log(x) + a*x + 2*\%iint(a, x, -log(-a*x + 1)/a, -log(-a*x + 1)/x) - log(-a
*x + 1) + 4*polylog(3, a*x))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{Li}_{3}\left (a x\right )}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x)/x**3,x)

[Out]

Integral(polylog(3, a*x)/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_{3}(a x)}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x)/x^3,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x)/x^3, x)