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

Optimal. Leaf size=40 \[ x \text{PolyLog}\left (2,a x^2\right )+2 x \log \left (1-a x^2\right )+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}-4 x \]

[Out]

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

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Rubi [A]  time = 0.0174073, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6586, 2448, 321, 206} \[ x \text{PolyLog}\left (2,a x^2\right )+2 x \log \left (1-a x^2\right )+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}-4 x \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2],x]

[Out]

-4*x + (4*ArcTanh[Sqrt[a]*x])/Sqrt[a] + 2*x*Log[1 - a*x^2] + x*PolyLog[2, 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}_2\left (a x^2\right ) \, dx &=x \text{Li}_2\left (a x^2\right )+2 \int \log \left (1-a x^2\right ) \, dx\\ &=2 x \log \left (1-a x^2\right )+x \text{Li}_2\left (a x^2\right )+(4 a) \int \frac{x^2}{1-a x^2} \, dx\\ &=-4 x+2 x \log \left (1-a x^2\right )+x \text{Li}_2\left (a x^2\right )+4 \int \frac{1}{1-a x^2} \, dx\\ &=-4 x+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}}+2 x \log \left (1-a x^2\right )+x \text{Li}_2\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0278312, size = 39, normalized size = 0.98 \[ x \text{PolyLog}\left (2,a x^2\right )+2 x \left (\log \left (1-a x^2\right )-2\right )+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2],x]

[Out]

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

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Maple [A]  time = 0.046, size = 37, normalized size = 0.9 \begin{align*} -4\,x+2\,x\ln \left ( -a{x}^{2}+1 \right ) +x{\it polylog} \left ( 2,a{x}^{2} \right ) +4\,{\frac{{\it Artanh} \left ( x\sqrt{a} \right ) }{\sqrt{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(2,a*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.63136, size = 269, normalized size = 6.72 \begin{align*} \left [\frac{a x{\rm Li}_2\left (a x^{2}\right ) + 2 \, a x \log \left (-a x^{2} + 1\right ) - 4 \, a x + 2 \, \sqrt{a} \log \left (\frac{a x^{2} + 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right )}{a}, \frac{a x{\rm Li}_2\left (a x^{2}\right ) + 2 \, a x \log \left (-a x^{2} + 1\right ) - 4 \, a x - 4 \, \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(2,a*x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 23.2238, size = 60, normalized size = 1.5 \begin{align*} \begin{cases} - 2 x \operatorname{Li}_{1}\left (a x^{2}\right ) + x \operatorname{Li}_{2}\left (a x^{2}\right ) - 4 x - \frac{4 \log{\left (x - \sqrt{\frac{1}{a}} \right )}}{a \sqrt{\frac{1}{a}}} - \frac{2 \operatorname{Li}_{1}\left (a x^{2}\right )}{a \sqrt{\frac{1}{a}}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x*polylog(1, a*x**2) + x*polylog(2, a*x**2) - 4*x - 4*log(x - sqrt(1/a))/(a*sqrt(1/a)) - 2*polyl
og(1, a*x**2)/(a*sqrt(1/a)), Ne(a, 0)), (0, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^2), x)