3.9 \(\int \frac {\text {Li}_2(a x)}{x^4} \, dx\)

Optimal. Leaf size=68 \[ \frac {1}{9} a^3 \log (x)-\frac {1}{9} a^3 \log (1-a x)-\frac {a^2}{9 x}-\frac {\text {Li}_2(a x)}{3 x^3}+\frac {\log (1-a x)}{9 x^3}-\frac {a}{18 x^2} \]

[Out]

-1/18*a/x^2-1/9*a^2/x+1/9*a^3*ln(x)-1/9*a^3*ln(-a*x+1)+1/9*ln(-a*x+1)/x^3-1/3*polylog(2,a*x)/x^3

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Rubi [A]  time = 0.04, antiderivative size = 68, 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} \[ -\frac {\text {PolyLog}(2,a x)}{3 x^3}-\frac {a^2}{9 x}+\frac {1}{9} a^3 \log (x)-\frac {1}{9} a^3 \log (1-a x)-\frac {a}{18 x^2}+\frac {\log (1-a x)}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/x^4,x]

[Out]

-a/(18*x^2) - a^2/(9*x) + (a^3*Log[x])/9 - (a^3*Log[1 - a*x])/9 + Log[1 - a*x]/(9*x^3) - PolyLog[2, a*x]/(3*x^
3)

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^4} \, dx &=-\frac {\text {Li}_2(a x)}{3 x^3}-\frac {1}{3} \int \frac {\log (1-a x)}{x^4} \, dx\\ &=\frac {\log (1-a x)}{9 x^3}-\frac {\text {Li}_2(a x)}{3 x^3}+\frac {1}{9} a \int \frac {1}{x^3 (1-a x)} \, dx\\ &=\frac {\log (1-a x)}{9 x^3}-\frac {\text {Li}_2(a x)}{3 x^3}+\frac {1}{9} a \int \left (\frac {1}{x^3}+\frac {a}{x^2}+\frac {a^2}{x}-\frac {a^3}{-1+a x}\right ) \, dx\\ &=-\frac {a}{18 x^2}-\frac {a^2}{9 x}+\frac {1}{9} a^3 \log (x)-\frac {1}{9} a^3 \log (1-a x)+\frac {\log (1-a x)}{9 x^3}-\frac {\text {Li}_2(a x)}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 52, normalized size = 0.76 \[ -\frac {-2 a^3 x^3 \log (x)+2 \left (a^3 x^3-1\right ) \log (1-a x)+6 \text {Li}_2(a x)+a x (2 a x+1)}{18 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/x^4,x]

[Out]

-1/18*(a*x*(1 + 2*a*x) - 2*a^3*x^3*Log[x] + 2*(-1 + a^3*x^3)*Log[1 - a*x] + 6*PolyLog[2, a*x])/x^3

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fricas [A]  time = 0.53, size = 56, normalized size = 0.82 \[ -\frac {2 \, a^{3} x^{3} \log \left (a x - 1\right ) - 2 \, a^{3} x^{3} \log \relax (x) + 2 \, a^{2} x^{2} + a x + 6 \, {\rm Li}_2\left (a x\right ) - 2 \, \log \left (-a x + 1\right )}{18 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^4,x, algorithm="fricas")

[Out]

-1/18*(2*a^3*x^3*log(a*x - 1) - 2*a^3*x^3*log(x) + 2*a^2*x^2 + a*x + 6*dilog(a*x) - 2*log(-a*x + 1))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^4,x, algorithm="giac")

[Out]

integrate(dilog(a*x)/x^4, x)

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maple [A]  time = 0.02, size = 60, normalized size = 0.88 \[ -\frac {\polylog \left (2, a x \right )}{3 x^{3}}-\frac {a}{18 x^{2}}+\frac {a^{3} \ln \left (-a x \right )}{9}-\frac {a^{2}}{9 x}-\frac {a^{3} \ln \left (-a x +1\right )}{9}+\frac {\ln \left (-a x +1\right )}{9 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x)/x^4,x)

[Out]

-1/3*polylog(2,a*x)/x^3-1/18*a/x^2+1/9*a^3*ln(-a*x)-1/9*a^2/x-1/9*a^3*ln(-a*x+1)+1/9*ln(-a*x+1)/x^3

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maxima [A]  time = 0.31, size = 49, normalized size = 0.72 \[ \frac {1}{9} \, a^{3} \log \relax (x) - \frac {2 \, a^{2} x^{2} + a x + 2 \, {\left (a^{3} x^{3} - 1\right )} \log \left (-a x + 1\right ) + 6 \, {\rm Li}_2\left (a x\right )}{18 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^4,x, algorithm="maxima")

[Out]

1/9*a^3*log(x) - 1/18*(2*a^2*x^2 + a*x + 2*(a^3*x^3 - 1)*log(-a*x + 1) + 6*dilog(a*x))/x^3

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mupad [B]  time = 0.32, size = 57, normalized size = 0.84 \[ \frac {2\,a^3\,\ln \relax (x)}{9}-\frac {\frac {a\,x}{18}-\frac {\ln \left (1-a\,x\right )}{9}+\frac {\mathrm {polylog}\left (2,a\,x\right )}{3}+\frac {a^2\,x^2}{9}}{x^3}-\frac {a^3\,\ln \left (a\,x^2-x\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/x^4,x)

[Out]

(2*a^3*log(x))/9 - ((a*x)/18 - log(1 - a*x)/9 + polylog(2, a*x)/3 + (a^2*x^2)/9)/x^3 - (a^3*log(a*x^2 - x))/9

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sympy [A]  time = 3.30, size = 51, normalized size = 0.75 \[ \frac {a^{3} \log {\relax (x )}}{9} + \frac {a^{3} \operatorname {Li}_{1}\left (a x\right )}{9} - \frac {a^{2}}{9 x} - \frac {a}{18 x^{2}} - \frac {\operatorname {Li}_{1}\left (a x\right )}{9 x^{3}} - \frac {\operatorname {Li}_{2}\left (a x\right )}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x**4,x)

[Out]

a**3*log(x)/9 + a**3*polylog(1, a*x)/9 - a**2/(9*x) - a/(18*x**2) - polylog(1, a*x)/(9*x**3) - polylog(2, a*x)
/(3*x**3)

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