3.8 \(\int \frac {\text {Li}_2(a x)}{x^3} \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{4} a^2 \log (x)-\frac {1}{4} a^2 \log (1-a x)-\frac {\text {Li}_2(a x)}{2 x^2}+\frac {\log (1-a x)}{4 x^2}-\frac {a}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.86 \[ \frac {a^2 x^2 \log (x)-a^2 x^2 \log (1-a x)-2 \text {Li}_2(a x)-a x+\log (1-a x)}{4 x^2} \]

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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fricas [A]  time = 0.55, size = 47, normalized size = 0.81 \[ -\frac {a^{2} x^{2} \log \left (a x - 1\right ) - a^{2} x^{2} \log \relax (x) + a x + 2 \, {\rm Li}_2\left (a x\right ) - \log \left (-a x + 1\right )}{4 \, x^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x\right )}{x^{3}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 52, normalized size = 0.90 \[ -\frac {\polylog \left (2, a x \right )}{2 x^{2}}+\frac {a^{2} \ln \left (-a x \right )}{4}-\frac {a}{4 x}-\frac {a^{2} \ln \left (-a x +1\right )}{4}+\frac {\ln \left (-a x +1\right )}{4 x^{2}} \]

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^2*ln(-a*x)-1/4*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 = 0.33, size = 40, normalized size = 0.69 \[ \frac {1}{4} \, a^{2} \log \relax (x) - \frac {a x + {\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right ) + 2 \, {\rm Li}_2\left (a x\right )}{4 \, x^{2}} \]

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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mupad [B]  time = 0.30, size = 51, normalized size = 0.88 \[ \frac {a^2\,\ln \relax (x)}{2}+\frac {\ln \left (1-a\,x\right )}{4\,x^2}-\frac {a^2\,\ln \left (a\,x^2-x\right )}{4}-\frac {a}{4\,x}-\frac {\mathrm {polylog}\left (2,a\,x\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/x^3,x)

[Out]

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

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sympy [A]  time = 1.97, size = 42, normalized size = 0.72 \[ \frac {a^{2} \log {\relax (x )}}{4} + \frac {a^{2} \operatorname {Li}_{1}\left (a x\right )}{4} - \frac {a}{4 x} - \frac {\operatorname {Li}_{1}\left (a x\right )}{4 x^{2}} - \frac {\operatorname {Li}_{2}\left (a x\right )}{2 x^{2}} \]

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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