3.8 \(\int \frac{\text{PolyLog}(2,a x)}{x^3} \, dx\)

Optimal. Leaf size=58 \[ -\frac{\text{PolyLog}(2,a x)}{2 x^2}+\frac{1}{4} a^2 \log (x)-\frac{1}{4} a^2 \log (1-a x)+\frac{\log (1-a x)}{4 x^2}-\frac{a}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.0332573, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6591, 2395, 44} \[ -\frac{\text{PolyLog}(2,a x)}{2 x^2}+\frac{1}{4} a^2 \log (x)-\frac{1}{4} a^2 \log (1-a x)+\frac{\log (1-a x)}{4 x^2}-\frac{a}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/x^3,x]

[Out]

-a/(4*x) + (a^2*Log[x])/4 - (a^2*Log[1 - a*x])/4 + Log[1 - a*x]/(4*x^2) - PolyLog[2, 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}_2(a x)}{x^3} \, dx &=-\frac{\text{Li}_2(a x)}{2 x^2}-\frac{1}{2} \int \frac{\log (1-a x)}{x^3} \, dx\\ &=\frac{\log (1-a x)}{4 x^2}-\frac{\text{Li}_2(a x)}{2 x^2}+\frac{1}{4} a \int \frac{1}{x^2 (1-a x)} \, dx\\ &=\frac{\log (1-a x)}{4 x^2}-\frac{\text{Li}_2(a x)}{2 x^2}+\frac{1}{4} a \int \left (\frac{1}{x^2}+\frac{a}{x}-\frac{a^2}{-1+a x}\right ) \, dx\\ &=-\frac{a}{4 x}+\frac{1}{4} a^2 \log (x)-\frac{1}{4} a^2 \log (1-a x)+\frac{\log (1-a x)}{4 x^2}-\frac{\text{Li}_2(a x)}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0221772, size = 50, normalized size = 0.86 \[ \frac{-2 \text{PolyLog}(2,a x)+a^2 x^2 \log (x)-a^2 x^2 \log (1-a x)-a x+\log (1-a x)}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/x^3,x]

[Out]

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

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Maple [A]  time = 0.118, size = 52, normalized size = 0.9 \begin{align*} -{\frac{{\it polylog} \left ( 2,ax \right ) }{2\,{x}^{2}}}-{\frac{a}{4\,x}}+{\frac{{a}^{2}\ln \left ( -ax \right ) }{4}}-{\frac{{a}^{2}\ln \left ( -ax+1 \right ) }{4}}+{\frac{\ln \left ( -ax+1 \right ) }{4\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x)/x^3,x)

[Out]

-1/2*polylog(2,a*x)/x^2-1/4*a/x+1/4*a^2*ln(-a*x)-1/4*a^2*ln(-a*x+1)+1/4*ln(-a*x+1)/x^2

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Maxima [A]  time = 0.981041, size = 54, normalized size = 0.93 \begin{align*} \frac{1}{4} \, 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 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^3,x, algorithm="maxima")

[Out]

1/4*a^2*log(x) - 1/4*(a*x + (a^2*x^2 - 1)*log(-a*x + 1) + 2*dilog(a*x))/x^2

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Fricas [A]  time = 2.62905, size = 117, normalized size = 2.02 \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 Li}_2\left (a x\right ) - \log \left (-a x + 1\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^3,x, algorithm="fricas")

[Out]

-1/4*(a^2*x^2*log(a*x - 1) - a^2*x^2*log(x) + a*x + 2*dilog(a*x) - log(-a*x + 1))/x^2

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Sympy [A]  time = 2.96484, size = 42, normalized size = 0.72 \begin{align*} \frac{a^{2} \log{\left (x \right )}}{4} + \frac{a^{2} \operatorname{Li}_{1}\left (a x\right )}{4} - \frac{a}{4 x} - \frac{\operatorname{Li}_{1}\left (a x\right )}{4 x^{2}} - \frac{\operatorname{Li}_{2}\left (a x\right )}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x**3,x)

[Out]

a**2*log(x)/4 + a**2*polylog(1, a*x)/4 - a/(4*x) - polylog(1, a*x)/(4*x**2) - polylog(2, a*x)/(2*x**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^3,x, algorithm="giac")

[Out]

integrate(dilog(a*x)/x^3, x)