3.31 \(\int \frac{\text{PolyLog}(2,a x^2)}{x^6} \, dx\)

Optimal. Leaf size=66 \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{5 x^5}-\frac{4 a^2}{25 x}+\frac{4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt{a} x\right )-\frac{4 a}{75 x^3}+\frac{2 \log \left (1-a x^2\right )}{25 x^5} \]

[Out]

(-4*a)/(75*x^3) - (4*a^2)/(25*x) + (4*a^(5/2)*ArcTanh[Sqrt[a]*x])/25 + (2*Log[1 - a*x^2])/(25*x^5) - PolyLog[2
, a*x^2]/(5*x^5)

________________________________________________________________________________________

Rubi [A]  time = 0.0376341, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac{\text{PolyLog}\left (2,a x^2\right )}{5 x^5}-\frac{4 a^2}{25 x}+\frac{4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt{a} x\right )-\frac{4 a}{75 x^3}+\frac{2 \log \left (1-a x^2\right )}{25 x^5} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^6,x]

[Out]

(-4*a)/(75*x^3) - (4*a^2)/(25*x) + (4*a^(5/2)*ArcTanh[Sqrt[a]*x])/25 + (2*Log[1 - a*x^2])/(25*x^5) - PolyLog[2
, a*x^2]/(5*x^5)

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^6} \, dx &=-\frac{\text{Li}_2\left (a x^2\right )}{5 x^5}-\frac{2}{5} \int \frac{\log \left (1-a x^2\right )}{x^6} \, dx\\ &=\frac{2 \log \left (1-a x^2\right )}{25 x^5}-\frac{\text{Li}_2\left (a x^2\right )}{5 x^5}+\frac{1}{25} (4 a) \int \frac{1}{x^4 \left (1-a x^2\right )} \, dx\\ &=-\frac{4 a}{75 x^3}+\frac{2 \log \left (1-a x^2\right )}{25 x^5}-\frac{\text{Li}_2\left (a x^2\right )}{5 x^5}+\frac{1}{25} \left (4 a^2\right ) \int \frac{1}{x^2 \left (1-a x^2\right )} \, dx\\ &=-\frac{4 a}{75 x^3}-\frac{4 a^2}{25 x}+\frac{2 \log \left (1-a x^2\right )}{25 x^5}-\frac{\text{Li}_2\left (a x^2\right )}{5 x^5}+\frac{1}{25} \left (4 a^3\right ) \int \frac{1}{1-a x^2} \, dx\\ &=-\frac{4 a}{75 x^3}-\frac{4 a^2}{25 x}+\frac{4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt{a} x\right )+\frac{2 \log \left (1-a x^2\right )}{25 x^5}-\frac{\text{Li}_2\left (a x^2\right )}{5 x^5}\\ \end{align*}

Mathematica [C]  time = 0.0151209, size = 47, normalized size = 0.71 \[ -\frac{4 a x^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},a x^2\right )+15 \text{PolyLog}\left (2,a x^2\right )-6 \log \left (1-a x^2\right )}{75 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^6,x]

[Out]

-(4*a*x^2*Hypergeometric2F1[-3/2, 1, -1/2, a*x^2] - 6*Log[1 - a*x^2] + 15*PolyLog[2, a*x^2])/(75*x^5)

________________________________________________________________________________________

Maple [A]  time = 0.052, size = 53, normalized size = 0.8 \begin{align*} -{\frac{4\,a}{75\,{x}^{3}}}-{\frac{4\,{a}^{2}}{25\,x}}+{\frac{4}{25}{a}^{{\frac{5}{2}}}{\it Artanh} \left ( x\sqrt{a} \right ) }+{\frac{2\,\ln \left ( -a{x}^{2}+1 \right ) }{25\,{x}^{5}}}-{\frac{{\it polylog} \left ( 2,a{x}^{2} \right ) }{5\,{x}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/x^6,x)

[Out]

-4/75*a/x^3-4/25*a^2/x+4/25*a^(5/2)*arctanh(x*a^(1/2))+2/25*ln(-a*x^2+1)/x^5-1/5*polylog(2,a*x^2)/x^5

________________________________________________________________________________________

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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.66061, size = 333, normalized size = 5.05 \begin{align*} \left [\frac{6 \, a^{\frac{5}{2}} x^{5} \log \left (\frac{a x^{2} + 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right ) - 12 \, a^{2} x^{4} - 4 \, a x^{2} - 15 \,{\rm Li}_2\left (a x^{2}\right ) + 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}, -\frac{12 \, \sqrt{-a} a^{2} x^{5} \arctan \left (\sqrt{-a} x\right ) + 12 \, a^{2} x^{4} + 4 \, a x^{2} + 15 \,{\rm Li}_2\left (a x^{2}\right ) - 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^6,x, algorithm="fricas")

[Out]

[1/75*(6*a^(5/2)*x^5*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 12*a^2*x^4 - 4*a*x^2 - 15*dilog(a*x^2) + 6*l
og(-a*x^2 + 1))/x^5, -1/75*(12*sqrt(-a)*a^2*x^5*arctan(sqrt(-a)*x) + 12*a^2*x^4 + 4*a*x^2 + 15*dilog(a*x^2) -
6*log(-a*x^2 + 1))/x^5]

________________________________________________________________________________________

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**6,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (a x^{2}\right )}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^6,x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/x^6, x)