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

Optimal. Leaf size=49 \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{2 x^2}-\frac{1}{2} a \log \left (1-a x^2\right )+\frac{\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

[Out]

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

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Rubi [A]  time = 0.040811, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {6591, 2454, 2395, 36, 29, 31} \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{2 x^2}-\frac{1}{2} a \log \left (1-a x^2\right )+\frac{\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^3,x]

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0123956, size = 49, normalized size = 1. \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{2 x^2}-\frac{1}{2} a \log \left (1-a x^2\right )+\frac{\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^3,x]

[Out]

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

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Maple [A]  time = 0.05, size = 43, normalized size = 0.9 \begin{align*} -{\frac{{\it polylog} \left ( 2,a{x}^{2} \right ) }{2\,{x}^{2}}}+{\frac{\ln \left ( -a{x}^{2}+1 \right ) }{2\,{x}^{2}}}+a\ln \left ( x \right ) -{\frac{a\ln \left ( a{x}^{2}-1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960105, size = 46, normalized size = 0.94 \begin{align*} a \log \left (x\right ) - \frac{{\left (a x^{2} - 1\right )} \log \left (-a x^{2} + 1\right ) +{\rm Li}_2\left (a x^{2}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.68007, size = 112, normalized size = 2.29 \begin{align*} -\frac{a x^{2} \log \left (a x^{2} - 1\right ) - 2 \, a x^{2} \log \left (x\right ) +{\rm Li}_2\left (a x^{2}\right ) - \log \left (-a x^{2} + 1\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.87995, size = 37, normalized size = 0.76 \begin{align*} a \log{\left (x \right )} + \frac{a \operatorname{Li}_{1}\left (a x^{2}\right )}{2} - \frac{\operatorname{Li}_{1}\left (a x^{2}\right )}{2 x^{2}} - \frac{\operatorname{Li}_{2}\left (a x^{2}\right )}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (a x^{2}\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^2)/x^3, x)