3.1.7 \(\int \frac {\text {PolyLog}(2,a x)}{x^2} \, dx\) [7]

Optimal. Leaf size=36 \[ a \log (x)-a \log (1-a x)+\frac {\log (1-a x)}{x}-\frac {\text {PolyLog}(2,a x)}{x} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, 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}+a \log (x)-a \log (1-a x)+\frac {\log (1-a x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/x^2,x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 36, normalized size = 1.00 \begin {gather*} a \log (x)-a \log (1-a x)+\frac {\log (1-a x)}{x}-\frac {\text {PolyLog}(2,a x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/x^2,x]

[Out]

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

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Maple [A]
time = 0.39, size = 42, normalized size = 1.17

method	result	size
derivativedivides	\(a \left (-\frac {\polylog \left (2, a x \right )}{a x}+\ln \left (-a x \right )+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{a x}\right )\)	\(42\)
default	\(a \left (-\frac {\polylog \left (2, a x \right )}{a x}+\ln \left (-a x \right )+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{a x}\right )\)	\(42\)
meijerg	\(a \left (\frac {\left (-4 a x +4\right ) \ln \left (-a x +1\right )}{4 a x}-\frac {\polylog \left (2, a x \right )}{a x}+\ln \left (x \right )+\ln \left (-a \right )\right )\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 28, normalized size = 0.78 \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 )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - ((a*x - 1)*log(-a*x + 1) + dilog(a*x))/x

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Fricas [A]
time = 0.35, size = 34, normalized size = 0.94 \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 )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,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))/x

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Sympy [A]
time = 0.37, size = 24, normalized size = 0.67 \begin {gather*} a \log {\left (x \right )} + a \operatorname {Li}_{1}\left (a x\right ) - \frac {\operatorname {Li}_{1}\left (a x\right )}{x} - \frac {\operatorname {Li}_{2}\left (a x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) + a*polylog(1, a*x) - polylog(1, a*x)/x - polylog(2, a*x)/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(2,a*x)/x^2,x, algorithm="giac")

[Out]

integrate(dilog(a*x)/x^2, x)

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Mupad [B]
time = 0.20, size = 34, normalized size = 0.94 \begin {gather*} \frac {\ln \left (1-a\,x\right )-\mathrm {polylog}\left (2,a\,x\right )}{x}+a\,\ln \left (x\right )-a\,\ln \left (1-a\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/x^2,x)

[Out]

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

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