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

Optimal. Leaf size=77 \[ -\frac{2}{9} x^3 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{3} x^3 \text{PolyLog}\left (3,a x^2\right )-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{27 a^{3/2}}-\frac{4}{27} x^3 \log \left (1-a x^2\right )+\frac{8 x}{27 a}+\frac{8 x^3}{81} \]

[Out]

(8*x)/(27*a) + (8*x^3)/81 - (8*ArcTanh[Sqrt[a]*x])/(27*a^(3/2)) - (4*x^3*Log[1 - a*x^2])/27 - (2*x^3*PolyLog[2
, a*x^2])/9 + (x^3*PolyLog[3, a*x^2])/3

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Rubi [A]  time = 0.0489431, antiderivative size = 77, 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, 2455, 302, 206} \[ -\frac{2}{9} x^3 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{3} x^3 \text{PolyLog}\left (3,a x^2\right )-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{27 a^{3/2}}-\frac{4}{27} x^3 \log \left (1-a x^2\right )+\frac{8 x}{27 a}+\frac{8 x^3}{81} \]

Antiderivative was successfully verified.

[In]

Int[x^2*PolyLog[3, a*x^2],x]

[Out]

(8*x)/(27*a) + (8*x^3)/81 - (8*ArcTanh[Sqrt[a]*x])/(27*a^(3/2)) - (4*x^3*Log[1 - a*x^2])/27 - (2*x^3*PolyLog[2
, a*x^2])/9 + (x^3*PolyLog[3, a*x^2])/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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 x^2 \text{Li}_3\left (a x^2\right ) \, dx &=\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )-\frac{2}{3} \int x^2 \text{Li}_2\left (a x^2\right ) \, dx\\ &=-\frac{2}{9} x^3 \text{Li}_2\left (a x^2\right )+\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )-\frac{4}{9} \int x^2 \log \left (1-a x^2\right ) \, dx\\ &=-\frac{4}{27} x^3 \log \left (1-a x^2\right )-\frac{2}{9} x^3 \text{Li}_2\left (a x^2\right )+\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )-\frac{1}{27} (8 a) \int \frac{x^4}{1-a x^2} \, dx\\ &=-\frac{4}{27} x^3 \log \left (1-a x^2\right )-\frac{2}{9} x^3 \text{Li}_2\left (a x^2\right )+\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )-\frac{1}{27} (8 a) \int \left (-\frac{1}{a^2}-\frac{x^2}{a}+\frac{1}{a^2 \left (1-a x^2\right )}\right ) \, dx\\ &=\frac{8 x}{27 a}+\frac{8 x^3}{81}-\frac{4}{27} x^3 \log \left (1-a x^2\right )-\frac{2}{9} x^3 \text{Li}_2\left (a x^2\right )+\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )-\frac{8 \int \frac{1}{1-a x^2} \, dx}{27 a}\\ &=\frac{8 x}{27 a}+\frac{8 x^3}{81}-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{27 a^{3/2}}-\frac{4}{27} x^3 \log \left (1-a x^2\right )-\frac{2}{9} x^3 \text{Li}_2\left (a x^2\right )+\frac{1}{3} x^3 \text{Li}_3\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.138001, size = 69, normalized size = 0.9 \[ \frac{1}{81} \left (-18 x^3 \text{PolyLog}\left (2,a x^2\right )+27 x^3 \text{PolyLog}\left (3,a x^2\right )-\frac{24 \tanh ^{-1}\left (\sqrt{a} x\right )}{a^{3/2}}-12 x^3 \log \left (1-a x^2\right )+\frac{24 x}{a}+8 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*PolyLog[3, a*x^2],x]

[Out]

((24*x)/a + 8*x^3 - (24*ArcTanh[Sqrt[a]*x])/a^(3/2) - 12*x^3*Log[1 - a*x^2] - 18*x^3*PolyLog[2, a*x^2] + 27*x^
3*PolyLog[3, a*x^2])/81

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Maple [B]  time = 0.175, size = 136, normalized size = 1.8 \begin{align*}{\frac{1}{2\,a} \left ({\frac{2\,x \left ( 40\,a{x}^{2}+120 \right ) }{405\,{a}^{2}} \left ( -a \right ) ^{{\frac{5}{2}}}}+{\frac{8\,x}{27\,{a}^{2}} \left ( -a \right ) ^{{\frac{5}{2}}} \left ( \ln \left ( 1-\sqrt{a{x}^{2}} \right ) -\ln \left ( 1+\sqrt{a{x}^{2}} \right ) \right ){\frac{1}{\sqrt{a{x}^{2}}}}}-{\frac{8\,{x}^{3}\ln \left ( -a{x}^{2}+1 \right ) }{27\,a} \left ( -a \right ) ^{{\frac{5}{2}}}}-{\frac{4\,{x}^{3}{\it polylog} \left ( 2,a{x}^{2} \right ) }{9\,a} \left ( -a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{x}^{3}{\it polylog} \left ( 3,a{x}^{2} \right ) }{3\,a} \left ( -a \right ) ^{{\frac{5}{2}}}} \right ){\frac{1}{\sqrt{-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a/(-a)^(1/2)*(2/405*x*(-a)^(5/2)*(40*a*x^2+120)/a^2+8/27*x*(-a)^(5/2)/a^2/(a*x^2)^(1/2)*(ln(1-(a*x^2)^(1/2
))-ln(1+(a*x^2)^(1/2)))-8/27*x^3*(-a)^(5/2)/a*ln(-a*x^2+1)-4/9*x^3*(-a)^(5/2)*polylog(2,a*x^2)/a+2/3*x^3*(-a)^
(5/2)/a*polylog(3,a*x^2))

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.74317, size = 555, normalized size = 7.21 \begin{align*} \left [-\frac{18 \, a^{2} x^{3}{\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 ) + 12 \, a^{2} x^{3} \log \left (-a x^{2} + 1\right ) - 27 \, a^{2} x^{3}{\rm polylog}\left (3, a x^{2}\right ) - 8 \, a^{2} x^{3} - 24 \, a x - 12 \, \sqrt{a} \log \left (\frac{a x^{2} - 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right )}{81 \, a^{2}}, -\frac{18 \, a^{2} x^{3}{\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 ) + 12 \, a^{2} x^{3} \log \left (-a x^{2} + 1\right ) - 27 \, a^{2} x^{3}{\rm polylog}\left (3, a x^{2}\right ) - 8 \, a^{2} x^{3} - 24 \, a x - 24 \, \sqrt{-a} \arctan \left (\sqrt{-a} x\right )}{81 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/81*(18*a^2*x^3*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) + 12*a^2*x^3*log(-a*x^2 + 1) - 27*a^2
*x^3*polylog(3, a*x^2) - 8*a^2*x^3 - 24*a*x - 12*sqrt(a)*log((a*x^2 - 2*sqrt(a)*x + 1)/(a*x^2 - 1)))/a^2, -1/8
1*(18*a^2*x^3*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) + 12*a^2*x^3*log(-a*x^2 + 1) - 27*a^2*x^3*
polylog(3, a*x^2) - 8*a^2*x^3 - 24*a*x - 24*sqrt(-a)*arctan(sqrt(-a)*x))/a^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*polylog(3, a*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*polylog(3, a*x^2), x)