3.10 \(\int \frac {\text {Li}_2(a x)}{x^5} \, dx\)

Optimal. Leaf size=78 \[ \frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)-\frac {a^3}{16 x}-\frac {a^2}{32 x^2}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {\log (1-a x)}{16 x^4}-\frac {a}{48 x^3} \]

[Out]

-1/48*a/x^3-1/32*a^2/x^2-1/16*a^3/x+1/16*a^4*ln(x)-1/16*a^4*ln(-a*x+1)+1/16*ln(-a*x+1)/x^4-1/4*polylog(2,a*x)/
x^4

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Rubi [A]  time = 0.04, antiderivative size = 78, 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} \[ -\frac {\text {PolyLog}(2,a x)}{4 x^4}-\frac {a^2}{32 x^2}-\frac {a^3}{16 x}+\frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)-\frac {a}{48 x^3}+\frac {\log (1-a x)}{16 x^4} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/x^5,x]

[Out]

-a/(48*x^3) - a^2/(32*x^2) - a^3/(16*x) + (a^4*Log[x])/16 - (a^4*Log[1 - a*x])/16 + Log[1 - a*x]/(16*x^4) - Po
lyLog[2, a*x]/(4*x^4)

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^5} \, dx &=-\frac {\text {Li}_2(a x)}{4 x^4}-\frac {1}{4} \int \frac {\log (1-a x)}{x^5} \, dx\\ &=\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {1}{16} a \int \frac {1}{x^4 (1-a x)} \, dx\\ &=\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {1}{16} a \int \left (\frac {1}{x^4}+\frac {a}{x^3}+\frac {a^2}{x^2}+\frac {a^3}{x}-\frac {a^4}{-1+a x}\right ) \, dx\\ &=-\frac {a}{48 x^3}-\frac {a^2}{32 x^2}-\frac {a^3}{16 x}+\frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)+\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.77 \[ -\frac {-6 a^4 x^4 \log (x)+6 \left (a^4 x^4-1\right ) \log (1-a x)+a x \left (6 a^2 x^2+3 a x+2\right )+24 \text {Li}_2(a x)}{96 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/x^5,x]

[Out]

-1/96*(a*x*(2 + 3*a*x + 6*a^2*x^2) - 6*a^4*x^4*Log[x] + 6*(-1 + a^4*x^4)*Log[1 - a*x] + 24*PolyLog[2, a*x])/x^
4

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fricas [A]  time = 0.75, size = 65, normalized size = 0.83 \[ -\frac {6 \, a^{4} x^{4} \log \left (a x - 1\right ) - 6 \, a^{4} x^{4} \log \relax (x) + 6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 24 \, {\rm Li}_2\left (a x\right ) - 6 \, \log \left (-a x + 1\right )}{96 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^5,x, algorithm="fricas")

[Out]

-1/96*(6*a^4*x^4*log(a*x - 1) - 6*a^4*x^4*log(x) + 6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 24*dilog(a*x) - 6*log(-a*x
+ 1))/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^5,x, algorithm="giac")

[Out]

integrate(dilog(a*x)/x^5, x)

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maple [A]  time = 0.01, size = 68, normalized size = 0.87 \[ -\frac {\polylog \left (2, a x \right )}{4 x^{4}}-\frac {a^{2}}{32 x^{2}}+\frac {a^{4} \ln \left (-a x \right )}{16}-\frac {a}{48 x^{3}}-\frac {a^{3}}{16 x}-\frac {a^{4} \ln \left (-a x +1\right )}{16}+\frac {\ln \left (-a x +1\right )}{16 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x)/x^5,x)

[Out]

-1/4*polylog(2,a*x)/x^4-1/32*a^2/x^2+1/16*a^4*ln(-a*x)-1/48*a/x^3-1/16*a^3/x-1/16*a^4*ln(-a*x+1)+1/16*ln(-a*x+
1)/x^4

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maxima [A]  time = 0.31, size = 58, normalized size = 0.74 \[ \frac {1}{16} \, a^{4} \log \relax (x) - \frac {6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 6 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right ) + 24 \, {\rm Li}_2\left (a x\right )}{96 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x^5,x, algorithm="maxima")

[Out]

1/16*a^4*log(x) - 1/96*(6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 6*(a^4*x^4 - 1)*log(-a*x + 1) + 24*dilog(a*x))/x^4

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mupad [B]  time = 0.44, size = 60, normalized size = 0.77 \[ \frac {\ln \left (1-a\,x\right )}{16\,x^4}-\frac {\mathrm {polylog}\left (2,a\,x\right )}{4\,x^4}-\frac {a^3\,x^2+\frac {a^2\,x}{2}+\frac {a}{3}}{16\,x^3}-\frac {a^4\,\mathrm {atan}\left (a\,x\,2{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/x^5,x)

[Out]

log(1 - a*x)/(16*x^4) - (a^4*atan(a*x*2i - 1i)*1i)/8 - polylog(2, a*x)/(4*x^4) - (a/3 + (a^2*x)/2 + a^3*x^2)/(
16*x^3)

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sympy [A]  time = 5.53, size = 60, normalized size = 0.77 \[ \frac {a^{4} \log {\relax (x )}}{16} + \frac {a^{4} \operatorname {Li}_{1}\left (a x\right )}{16} - \frac {a^{3}}{16 x} - \frac {a^{2}}{32 x^{2}} - \frac {a}{48 x^{3}} - \frac {\operatorname {Li}_{1}\left (a x\right )}{16 x^{4}} - \frac {\operatorname {Li}_{2}\left (a x\right )}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x)/x**5,x)

[Out]

a**4*log(x)/16 + a**4*polylog(1, a*x)/16 - a**3/(16*x) - a**2/(32*x**2) - a/(48*x**3) - polylog(1, a*x)/(16*x*
*4) - polylog(2, a*x)/(4*x**4)

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