3.28 \(\int \text {Li}_2(a x^2) \, dx\)

Optimal. Leaf size=40 \[ x \text {Li}_2\left (a x^2\right )+2 x \log \left (1-a x^2\right )+\frac {4 \tanh ^{-1}\left (\sqrt {a} x\right )}{\sqrt {a}}-4 x \]

[Out]

-4*x+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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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 39, normalized size = 0.98 \[ x \text {Li}_2\left (a x^2\right )+2 x \left (\log \left (1-a x^2\right )-2\right )+\frac {4 \tanh ^{-1}\left (\sqrt {a} x\right )}{\sqrt {a}} \]

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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fricas [A]  time = 0.56, size = 107, normalized size = 2.68 \[ \left [\frac {a x {\rm Li}_2\left (a x^{2}\right ) + 2 \, a x \log \left (-a x^{2} + 1\right ) - 4 \, a x + 2 \, \sqrt {a} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right )}{a}, \frac {a x {\rm Li}_2\left (a x^{2}\right ) + 2 \, a x \log \left (-a x^{2} + 1\right ) - 4 \, a x - 4 \, \sqrt {-a} \arctan \left (\sqrt {-a} x\right )}{a}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm Li}_2\left (a x^{2}\right )\,{d x} \]

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)

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maple [A]  time = 0.01, size = 37, normalized size = 0.92 \[ -4 x +2 x \ln \left (-a \,x^{2}+1\right )+x \polylog \left (2, a \,x^{2}\right )+\frac {4 \arctanh \left (x \sqrt {a}\right )}{\sqrt {a}} \]

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 [A]  time = 0.41, size = 49, normalized size = 1.22 \[ x {\rm Li}_2\left (a x^{2}\right ) + 2 \, x \log \left (-a x^{2} + 1\right ) - 4 \, x - \frac {2 \, \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*dilog(a*x^2) + 2*x*log(-a*x^2 + 1) - 4*x - 2*log((a*x - sqrt(a))/(a*x + sqrt(a)))/sqrt(a)

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mupad [B]  time = 0.24, size = 39, normalized size = 0.98 \[ 2\,x\,\ln \left (1-a\,x^2\right )-4\,x+x\,\mathrm {polylog}\left (2,a\,x^2\right )-\frac {\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2),x)

[Out]

2*x*log(1 - a*x^2) - (atan(a^(1/2)*x*1i)*4i)/a^(1/2) - 4*x + x*polylog(2, a*x^2)

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sympy [A]  time = 9.92, size = 60, normalized size = 1.50 \[ \begin {cases} - 2 x \operatorname {Li}_{1}\left (a x^{2}\right ) + x \operatorname {Li}_{2}\left (a x^{2}\right ) - 4 x - \frac {4 \log {\left (x - \sqrt {\frac {1}{a}} \right )}}{a \sqrt {\frac {1}{a}}} - \frac {2 \operatorname {Li}_{1}\left (a x^{2}\right )}{a \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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