3.30 \(\int \frac{\text{PolyLog}(2,a x^2)}{x^4} \, dx\)

Optimal. Leaf size=56 \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{3 x^3}+\frac{4}{9} a^{3/2} \tanh ^{-1}\left (\sqrt{a} x\right )+\frac{2 \log \left (1-a x^2\right )}{9 x^3}-\frac{4 a}{9 x} \]

[Out]

(-4*a)/(9*x) + (4*a^(3/2)*ArcTanh[Sqrt[a]*x])/9 + (2*Log[1 - a*x^2])/(9*x^3) - PolyLog[2, a*x^2]/(3*x^3)

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Rubi [A]  time = 0.0321837, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{3 x^3}+\frac{4}{9} a^{3/2} \tanh ^{-1}\left (\sqrt{a} x\right )+\frac{2 \log \left (1-a x^2\right )}{9 x^3}-\frac{4 a}{9 x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^4,x]

[Out]

(-4*a)/(9*x) + (4*a^(3/2)*ArcTanh[Sqrt[a]*x])/9 + (2*Log[1 - a*x^2])/(9*x^3) - PolyLog[2, a*x^2]/(3*x^3)

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 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0136452, size = 47, normalized size = 0.84 \[ -\frac{4 a x^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},a x^2\right )+3 \text{PolyLog}\left (2,a x^2\right )-2 \log \left (1-a x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^4,x]

[Out]

-(4*a*x^2*Hypergeometric2F1[-1/2, 1, 1/2, a*x^2] - 2*Log[1 - a*x^2] + 3*PolyLog[2, a*x^2])/(9*x^3)

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Maple [A]  time = 0.053, size = 45, normalized size = 0.8 \begin{align*} -{\frac{4\,a}{9\,x}}+{\frac{4}{9}{a}^{{\frac{3}{2}}}{\it Artanh} \left ( x\sqrt{a} \right ) }+{\frac{2\,\ln \left ( -a{x}^{2}+1 \right ) }{9\,{x}^{3}}}-{\frac{{\it polylog} \left ( 2,a{x}^{2} \right ) }{3\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/9*a/x+4/9*a^(3/2)*arctanh(x*a^(1/2))+2/9*ln(-a*x^2+1)/x^3-1/3*polylog(2,a*x^2)/x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.74604, size = 289, normalized size = 5.16 \begin{align*} \left [\frac{2 \, a^{\frac{3}{2}} x^{3} \log \left (\frac{a x^{2} + 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right ) - 4 \, a x^{2} - 3 \,{\rm Li}_2\left (a x^{2}\right ) + 2 \, \log \left (-a x^{2} + 1\right )}{9 \, x^{3}}, -\frac{4 \, \sqrt{-a} a x^{3} \arctan \left (\sqrt{-a} x\right ) + 4 \, a x^{2} + 3 \,{\rm Li}_2\left (a x^{2}\right ) - 2 \, \log \left (-a x^{2} + 1\right )}{9 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/9*(2*a^(3/2)*x^3*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 4*a*x^2 - 3*dilog(a*x^2) + 2*log(-a*x^2 + 1))
/x^3, -1/9*(4*sqrt(-a)*a*x^3*arctan(sqrt(-a)*x) + 4*a*x^2 + 3*dilog(a*x^2) - 2*log(-a*x^2 + 1))/x^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^2)/x^4, x)