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

Optimal. Leaf size=54 \[ -\frac {2 \text {Li}_2\left (a x^2\right )}{x}-\frac {\text {Li}_3\left (a x^2\right )}{x}+\frac {4 \log \left (1-a x^2\right )}{x}+8 \sqrt {a} \tanh ^{-1}\left (\sqrt {a} x\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6591, 2455, 206} \[ -\frac {2 \text {PolyLog}\left (2,a x^2\right )}{x}-\frac {\text {PolyLog}\left (3,a x^2\right )}{x}+\frac {4 \log \left (1-a x^2\right )}{x}+8 \sqrt {a} \tanh ^{-1}\left (\sqrt {a} x\right ) \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/x^2,x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 50, normalized size = 0.93 \[ \frac {-2 \text {Li}_2\left (a x^2\right )-\text {Li}_3\left (a x^2\right )+4 \log \left (1-a x^2\right )+8 \sqrt {a} x \tanh ^{-1}\left (\sqrt {a} x\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^2]/x^2,x]

[Out]

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

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fricas [C]  time = 0.55, size = 112, normalized size = 2.07 \[ \left [\frac {4 \, \sqrt {a} x \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 2 \, {\rm Li}_2\left (a x^{2}\right ) + 4 \, \log \left (-a x^{2} + 1\right ) - {\rm polylog}\left (3, a x^{2}\right )}{x}, -\frac {8 \, \sqrt {-a} x \arctan \left (\sqrt {-a} x\right ) + 2 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right ) + {\rm polylog}\left (3, a x^{2}\right )}{x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(polylog(3, a*x^2)/x^2, x)

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maple [B]  time = 0.16, size = 112, normalized size = 2.07 \[ \frac {a \left (-\frac {8 x \sqrt {-a}\, \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{\sqrt {a \,x^{2}}}+\frac {8 \sqrt {-a}\, \ln \left (-a \,x^{2}+1\right )}{x a}-\frac {4 \sqrt {-a}\, \polylog \left (2, a \,x^{2}\right )}{x a}-\frac {2 \sqrt {-a}\, \polylog \left (3, a \,x^{2}\right )}{x a}\right )}{2 \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 58, normalized size = 1.07 \[ -4 \, \sqrt {a} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {2 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right ) + {\rm Li}_{3}(a x^{2})}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.55, size = 53, normalized size = 0.98 \[ \frac {4\,\ln \left (1-a\,x^2\right )}{x}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{x}-\frac {2\,\mathrm {polylog}\left (2,a\,x^2\right )}{x}-\sqrt {a}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/x^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, a*x**2)/x**2, x)

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