3.37 \(\int \frac{\text{PolyLog}(3,a x^2)}{x^5} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{8 x^4}-\frac{\text{PolyLog}\left (3,a x^2\right )}{4 x^4}-\frac{1}{16} a^2 \log \left (1-a x^2\right )+\frac{1}{8} a^2 \log (x)-\frac{a}{16 x^2}+\frac{\log \left (1-a x^2\right )}{16 x^4} \]

[Out]

-a/(16*x^2) + (a^2*Log[x])/8 - (a^2*Log[1 - a*x^2])/16 + Log[1 - a*x^2]/(16*x^4) - PolyLog[2, a*x^2]/(8*x^4) -
PolyLog[3, a*x^2]/(4*x^4)

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Rubi [A]  time = 0.0637554, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6591, 2454, 2395, 44} \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{8 x^4}-\frac{\text{PolyLog}\left (3,a x^2\right )}{4 x^4}-\frac{1}{16} a^2 \log \left (1-a x^2\right )+\frac{1}{8} a^2 \log (x)-\frac{a}{16 x^2}+\frac{\log \left (1-a x^2\right )}{16 x^4} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/x^5,x]

[Out]

-a/(16*x^2) + (a^2*Log[x])/8 - (a^2*Log[1 - a*x^2])/16 + Log[1 - a*x^2]/(16*x^4) - PolyLog[2, a*x^2]/(8*x^4) -
PolyLog[3, a*x^2]/(4*x^4)

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]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 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 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])

Rubi steps

\begin{align*} \int \frac{\text{Li}_3\left (a x^2\right )}{x^5} \, dx &=-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}+\frac{1}{2} \int \frac{\text{Li}_2\left (a x^2\right )}{x^5} \, dx\\ &=-\frac{\text{Li}_2\left (a x^2\right )}{8 x^4}-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}-\frac{1}{4} \int \frac{\log \left (1-a x^2\right )}{x^5} \, dx\\ &=-\frac{\text{Li}_2\left (a x^2\right )}{8 x^4}-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{\log (1-a x)}{x^3} \, dx,x,x^2\right )\\ &=\frac{\log \left (1-a x^2\right )}{16 x^4}-\frac{\text{Li}_2\left (a x^2\right )}{8 x^4}-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}+\frac{1}{16} a \operatorname{Subst}\left (\int \frac{1}{x^2 (1-a x)} \, dx,x,x^2\right )\\ &=\frac{\log \left (1-a x^2\right )}{16 x^4}-\frac{\text{Li}_2\left (a x^2\right )}{8 x^4}-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}+\frac{1}{16} a \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{a}{x}-\frac{a^2}{-1+a x}\right ) \, dx,x,x^2\right )\\ &=-\frac{a}{16 x^2}+\frac{1}{8} a^2 \log (x)-\frac{1}{16} a^2 \log \left (1-a x^2\right )+\frac{\log \left (1-a x^2\right )}{16 x^4}-\frac{\text{Li}_2\left (a x^2\right )}{8 x^4}-\frac{\text{Li}_3\left (a x^2\right )}{4 x^4}\\ \end{align*}

Mathematica [C]  time = 0.0118626, size = 30, normalized size = 0.38 \[ \frac{G_{5,5}^{2,4}\left (-a x^2|\begin{array}{c} 1,1,1,1,3 \\ 1,2,0,0,0 \\\end{array}\right )}{2 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^2]/x^5,x]

[Out]

MeijerG[{{1, 1, 1, 1}, {3}}, {{1, 2}, {0, 0, 0}}, -(a*x^2)]/(2*x^4)

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Maple [A]  time = 0.065, size = 98, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{2} \left ({\frac{1}{a{x}^{2}}}+{\frac{3}{16}}-{\frac{\ln \left ( x \right ) }{4}}-{\frac{\ln \left ( -a \right ) }{8}}-{\frac{81\,a{x}^{2}+378}{432\,a{x}^{2}}}-{\frac{ \left ( -27\,{a}^{2}{x}^{4}+27 \right ) \ln \left ( -a{x}^{2}+1 \right ) }{216\,{a}^{2}{x}^{4}}}+{\frac{{\it polylog} \left ( 2,a{x}^{2} \right ) }{4\,{a}^{2}{x}^{4}}}+{\frac{{\it polylog} \left ( 3,a{x}^{2} \right ) }{2\,{a}^{2}{x}^{4}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/x^5,x)

[Out]

-1/2*a^2*(1/a/x^2+3/16-1/4*ln(x)-1/8*ln(-a)-1/432/a/x^2*(81*a*x^2+378)-1/216/a^2/x^4*(-27*a^2*x^4+27)*ln(-a*x^
2+1)+1/4/a^2/x^4*polylog(2,a*x^2)+1/2/a^2/x^4*polylog(3,a*x^2))

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Maxima [A]  time = 1.0028, size = 74, normalized size = 0.95 \begin{align*} \frac{1}{8} \, a^{2} \log \left (x\right ) - \frac{a x^{2} +{\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right ) + 2 \,{\rm Li}_2\left (a x^{2}\right ) + 4 \,{\rm Li}_{3}(a x^{2})}{16 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.71783, size = 217, normalized size = 2.78 \begin{align*} -\frac{a^{2} x^{4} \log \left (a x^{2} - 1\right ) - 2 \, a^{2} x^{4} \log \left (x\right ) + a x^{2} + 2 \,{\rm \%iint}\left (a, x, -\frac{\log \left (-a x^{2} + 1\right )}{a}, -\frac{2 \, \log \left (-a x^{2} + 1\right )}{x}\right ) - \log \left (-a x^{2} + 1\right ) + 4 \,{\rm polylog}\left (3, a x^{2}\right )}{16 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^5,x, algorithm="fricas")

[Out]

-1/16*(a^2*x^4*log(a*x^2 - 1) - 2*a^2*x^4*log(x) + a*x^2 + 2*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1
)/x) - log(-a*x^2 + 1) + 4*polylog(3, a*x^2))/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{Li}_{3}\left (a x^{2}\right )}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/x**5,x)

[Out]

Integral(polylog(3, a*x**2)/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_{3}(a x^{2})}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^5,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/x^5, x)