3.1.16 \(\int \frac {\text {PolyLog}(3,a x)}{x^2} \, dx\) [16]

Optimal. Leaf size=46 \[ a \log (x)-a \log (1-a x)+\frac {\log (1-a x)}{x}-\frac {\text {PolyLog}(2,a x)}{x}-\frac {\text {PolyLog}(3,a x)}{x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6726, 2442, 36, 29, 31} \begin {gather*} -\frac {\text {Li}_2(a x)}{x}-\frac {\text {Li}_3(a x)}{x}+a \log (x)-a \log (1-a x)+\frac {\log (1-a x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x]/x^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2442

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 6726

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 44, normalized size = 0.96 \begin {gather*} -\frac {-a x \log (-a x)-\log (1-a x)+a x \log (1-a x)+\text {PolyLog}(2,a x)+\text {PolyLog}(3,a x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x]/x^2,x]

[Out]

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

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Maple [A]
time = 0.12, size = 57, normalized size = 1.24

method	result	size
meijerg	\(a \left (\frac {\left (-8 a x +8\right ) \ln \left (-a x +1\right )}{8 a x}-\frac {\polylog \left (2, a x \right )}{a x}-\frac {\polylog \left (3, a x \right )}{a x}+\ln \left (x \right )+\ln \left (-a \right )\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(1/8/a/x*(-8*a*x+8)*ln(-a*x+1)-1/a/x*polylog(2,a*x)-1/a/x*polylog(3,a*x)+ln(x)+ln(-a))

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Maxima [A]
time = 0.26, size = 33, normalized size = 0.72 \begin {gather*} a \log \left (x\right ) - \frac {{\left (a x - 1\right )} \log \left (-a x + 1\right ) + {\rm Li}_2\left (a x\right ) + {\rm Li}_{3}(a x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - ((a*x - 1)*log(-a*x + 1) + dilog(a*x) + polylog(3, a*x))/x

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Fricas [A]
time = 0.40, size = 39, normalized size = 0.85 \begin {gather*} -\frac {a x \log \left (a x - 1\right ) - a x \log \left (x\right ) + {\rm Li}_2\left (a x\right ) - \log \left (-a x + 1\right ) + {\rm polylog}\left (3, a x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x*log(a*x - 1) - a*x*log(x) + dilog(a*x) - log(-a*x + 1) + polylog(3, a*x))/x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Li}_{3}\left (a x\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, a*x)/x**2, x)

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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(3,a*x)/x^2,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x)/x^2, x)

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Mupad [B]
time = 0.89, size = 36, normalized size = 0.78 \begin {gather*} 2\,a\,\mathrm {atanh}\left (2\,a\,x-1\right )-\frac {\mathrm {polylog}\left (2,a\,x\right )-\ln \left (1-a\,x\right )+\mathrm {polylog}\left (3,a\,x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x)/x^2,x)

[Out]

2*a*atanh(2*a*x - 1) - (polylog(2, a*x) - log(1 - a*x) + polylog(3, a*x))/x

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