3.41 \(\int \text {Li}_3(a x^2) \, dx\)

Optimal. Leaf size=50 \[ -2 x \text {Li}_2\left (a x^2\right )+x \text {Li}_3\left (a x^2\right )-4 x \log \left (1-a x^2\right )-\frac {8 \tanh ^{-1}\left (\sqrt {a} x\right )}{\sqrt {a}}+8 x \]

[Out]

8*x-4*x*ln(-a*x^2+1)-2*x*polylog(2,a*x^2)+x*polylog(3,a*x^2)-8*arctanh(x*a^(1/2))/a^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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}_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*}

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Mathematica [A]  time = 0.08, size = 50, normalized size = 1.00 \[ -2 x \text {Li}_2\left (a x^2\right )+x \text {Li}_3\left (a x^2\right )-4 x \log \left (1-a x^2\right )-\frac {8 \tanh ^{-1}\left (\sqrt {a} x\right )}{\sqrt {a}}+8 x \]

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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fricas [C]  time = 0.73, size = 133, normalized size = 2.66 \[ \left [-\frac {2 \, a x {\rm Li}_2\left (a x^{2}\right ) + 4 \, a x \log \left (-a x^{2} + 1\right ) - a x {\rm polylog}\left (3, a x^{2}\right ) - 8 \, a x - 4 \, \sqrt {a} \log \left (\frac {a x^{2} - 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right )}{a}, -\frac {2 \, a x {\rm Li}_2\left (a x^{2}\right ) + 4 \, a x \log \left (-a x^{2} + 1\right ) - a x {\rm polylog}\left (3, a x^{2}\right ) - 8 \, a x - 8 \, \sqrt {-a} \arctan \left (\sqrt {-a} x\right )}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*a*x*dilog(a*x^2) + 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*dilog(a*x^2) + 4*a*x*log(-a*x^2 + 1) - a*x*polylog(3, a*x^2) - 8*a*x - 8*sq
rt(-a)*arctan(sqrt(-a)*x))/a]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm Li}_{3}(a x^{2})\,{d x} \]

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)

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maple [B]  time = 0.17, size = 119, normalized size = 2.38 \[ -\frac {\frac {16 x \left (-a \right )^{\frac {3}{2}}}{a}+\frac {8 x \left (-a \right )^{\frac {3}{2}} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{a \sqrt {a \,x^{2}}}-\frac {8 x \left (-a \right )^{\frac {3}{2}} \ln \left (-a \,x^{2}+1\right )}{a}-\frac {4 x \left (-a \right )^{\frac {3}{2}} \polylog \left (2, a \,x^{2}\right )}{a}+\frac {2 x \left (-a \right )^{\frac {3}{2}} \polylog \left (3, a \,x^{2}\right )}{a}}{2 \sqrt {-a}} \]

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 [A]  time = 0.41, size = 59, normalized size = 1.18 \[ -2 \, x {\rm Li}_2\left (a x^{2}\right ) - 4 \, x \log \left (-a x^{2} + 1\right ) + x {\rm Li}_{3}(a x^{2}) + 8 \, x + \frac {4 \, \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right )}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*dilog(a*x^2) - 4*x*log(-a*x^2 + 1) + x*polylog(3, a*x^2) + 8*x + 4*log((a*x - sqrt(a))/(a*x + sqrt(a)))/s
qrt(a)

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mupad [B]  time = 0.36, size = 49, normalized size = 0.98 \[ 8\,x-4\,x\,\ln \left (1-a\,x^2\right )-2\,x\,\mathrm {polylog}\left (2,a\,x^2\right )+x\,\mathrm {polylog}\left (3,a\,x^2\right )+\frac {\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2),x)

[Out]

8*x + (atan(a^(1/2)*x*1i)*8i)/a^(1/2) - 4*x*log(1 - a*x^2) - 2*x*polylog(2, a*x^2) + x*polylog(3, a*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {Li}_{3}\left (a x^{2}\right )\, dx \]

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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