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

Optimal. Leaf size=50 \[ -2 x \text{PolyLog}\left (2,a x^2\right )+x \text{PolyLog}\left (3,a x^2\right )-4 x \log \left (1-a x^2\right )-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}+8 x \]

[Out]

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

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Rubi [A]  time = 0.0233193, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6586, 2448, 321, 206} \[ -2 x \text{PolyLog}\left (2,a x^2\right )+x \text{PolyLog}\left (3,a x^2\right )-4 x \log \left (1-a x^2\right )-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}+8 x \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2],x]

[Out]

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

Rule 6586

Int[PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[x*PolyLog[n, a*(b*x^p)^q], x] - Dist[p*q, I
nt[PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, b, p, q}, x] && GtQ[n, 0]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 \text{Li}_3\left (a x^2\right ) \, dx &=x \text{Li}_3\left (a x^2\right )-2 \int \text{Li}_2\left (a x^2\right ) \, dx\\ &=-2 x \text{Li}_2\left (a x^2\right )+x \text{Li}_3\left (a x^2\right )-4 \int \log \left (1-a x^2\right ) \, dx\\ &=-4 x \log \left (1-a x^2\right )-2 x \text{Li}_2\left (a x^2\right )+x \text{Li}_3\left (a x^2\right )-(8 a) \int \frac{x^2}{1-a x^2} \, dx\\ &=8 x-4 x \log \left (1-a x^2\right )-2 x \text{Li}_2\left (a x^2\right )+x \text{Li}_3\left (a x^2\right )-8 \int \frac{1}{1-a x^2} \, dx\\ &=8 x-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}-4 x \log \left (1-a x^2\right )-2 x \text{Li}_2\left (a x^2\right )+x \text{Li}_3\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.087684, size = 50, normalized size = 1. \[ -2 x \text{PolyLog}\left (2,a x^2\right )+x \text{PolyLog}\left (3,a x^2\right )-4 x \log \left (1-a x^2\right )-\frac{8 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}+8 x \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^2],x]

[Out]

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

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Maple [B]  time = 0.173, size = 119, normalized size = 2.4 \begin{align*} -{\frac{1}{2} \left ( 16\,{\frac{x \left ( -a \right ) ^{3/2}}{a}}+8\,{\frac{x \left ( -a \right ) ^{3/2} \left ( \ln \left ( 1-\sqrt{a{x}^{2}} \right ) -\ln \left ( 1+\sqrt{a{x}^{2}} \right ) \right ) }{a\sqrt{a{x}^{2}}}}-8\,{\frac{x \left ( -a \right ) ^{3/2}\ln \left ( -a{x}^{2}+1 \right ) }{a}}-4\,{\frac{x \left ( -a \right ) ^{3/2}{\it polylog} \left ( 2,a{x}^{2} \right ) }{a}}+2\,{\frac{x \left ( -a \right ) ^{3/2}{\it polylog} \left ( 3,a{x}^{2} \right ) }{a}} \right ){\frac{1}{\sqrt{-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(-a)^(1/2)*(16*x*(-a)^(3/2)/a+8*x*(-a)^(3/2)/a/(a*x^2)^(1/2)*(ln(1-(a*x^2)^(1/2))-ln(1+(a*x^2)^(1/2)))-8*
x*(-a)^(3/2)/a*ln(-a*x^2+1)-4*x*(-a)^(3/2)*polylog(2,a*x^2)/a+2*x*(-a)^(3/2)/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(polylog(3,a*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.69418, size = 452, normalized size = 9.04 \begin{align*} \left [-\frac{2 \, a x{\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 ) + 4 \, a x \log \left (-a x^{2} + 1\right ) - a x{\rm polylog}\left (3, a x^{2}\right ) - 8 \, a x - 4 \, \sqrt{a} \log \left (\frac{a x^{2} - 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right )}{a}, -\frac{2 \, a x{\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 ) + 4 \, a x \log \left (-a x^{2} + 1\right ) - a x{\rm polylog}\left (3, a x^{2}\right ) - 8 \, a x - 8 \, \sqrt{-a} \arctan \left (\sqrt{-a} x\right )}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*a*x*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) + 4*a*x*log(-a*x^2 + 1) - a*x*polylog(3, a*x^2)
- 8*a*x - 4*sqrt(a)*log((a*x^2 - 2*sqrt(a)*x + 1)/(a*x^2 - 1)))/a, -(2*a*x*\%iint(a, x, -log(-a*x^2 + 1)/a, -2
*log(-a*x^2 + 1)/x) + 4*a*x*log(-a*x^2 + 1) - a*x*polylog(3, a*x^2) - 8*a*x - 8*sqrt(-a)*arctan(sqrt(-a)*x))/a
]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, a*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(polylog(3, a*x^2), x)