3.43 \(\int \frac {\text {Li}_3(a x^2)}{x^4} \, dx\)

Optimal. Leaf size=70 \[ \frac {8}{27} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {2 \text {Li}_2\left (a x^2\right )}{9 x^3}-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}+\frac {4 \log \left (1-a x^2\right )}{27 x^3}-\frac {8 a}{27 x} \]

[Out]

-8/27*a/x+8/27*a^(3/2)*arctanh(x*a^(1/2))+4/27*ln(-a*x^2+1)/x^3-2/9*polylog(2,a*x^2)/x^3-1/3*polylog(3,a*x^2)/
x^3

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac {2 \text {PolyLog}\left (2,a x^2\right )}{9 x^3}-\frac {\text {PolyLog}\left (3,a x^2\right )}{3 x^3}+\frac {8}{27} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {4 \log \left (1-a x^2\right )}{27 x^3}-\frac {8 a}{27 x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/x^4,x]

[Out]

(-8*a)/(27*x) + (8*a^(3/2)*ArcTanh[Sqrt[a]*x])/27 + (4*Log[1 - a*x^2])/(27*x^3) - (2*PolyLog[2, a*x^2])/(9*x^3
) - PolyLog[3, a*x^2]/(3*x^3)

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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^4} \, dx &=-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}+\frac {2}{3} \int \frac {\text {Li}_2\left (a x^2\right )}{x^4} \, dx\\ &=-\frac {2 \text {Li}_2\left (a x^2\right )}{9 x^3}-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}-\frac {4}{9} \int \frac {\log \left (1-a x^2\right )}{x^4} \, dx\\ &=\frac {4 \log \left (1-a x^2\right )}{27 x^3}-\frac {2 \text {Li}_2\left (a x^2\right )}{9 x^3}-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}+\frac {1}{27} (8 a) \int \frac {1}{x^2 \left (1-a x^2\right )} \, dx\\ &=-\frac {8 a}{27 x}+\frac {4 \log \left (1-a x^2\right )}{27 x^3}-\frac {2 \text {Li}_2\left (a x^2\right )}{9 x^3}-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}+\frac {1}{27} \left (8 a^2\right ) \int \frac {1}{1-a x^2} \, dx\\ &=-\frac {8 a}{27 x}+\frac {8}{27} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {4 \log \left (1-a x^2\right )}{27 x^3}-\frac {2 \text {Li}_2\left (a x^2\right )}{9 x^3}-\frac {\text {Li}_3\left (a x^2\right )}{3 x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 61, normalized size = 0.87 \[ -\frac {-8 a^{3/2} x^3 \tanh ^{-1}\left (\sqrt {a} x\right )+6 \text {Li}_2\left (a x^2\right )+9 \text {Li}_3\left (a x^2\right )+8 a x^2-4 \log \left (1-a x^2\right )}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^2]/x^4,x]

[Out]

-1/27*(8*a*x^2 - 8*a^(3/2)*x^3*ArcTanh[Sqrt[a]*x] - 4*Log[1 - a*x^2] + 6*PolyLog[2, a*x^2] + 9*PolyLog[3, a*x^
2])/x^3

________________________________________________________________________________________

fricas [C]  time = 0.59, size = 132, normalized size = 1.89 \[ \left [\frac {4 \, a^{\frac {3}{2}} x^{3} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 8 \, a x^{2} - 6 \, {\rm Li}_2\left (a x^{2}\right ) + 4 \, \log \left (-a x^{2} + 1\right ) - 9 \, {\rm polylog}\left (3, a x^{2}\right )}{27 \, x^{3}}, -\frac {8 \, \sqrt {-a} a x^{3} \arctan \left (\sqrt {-a} x\right ) + 8 \, a x^{2} + 6 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right ) + 9 \, {\rm polylog}\left (3, a x^{2}\right )}{27 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^4,x, algorithm="fricas")

[Out]

[1/27*(4*a^(3/2)*x^3*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 8*a*x^2 - 6*dilog(a*x^2) + 4*log(-a*x^2 + 1)
- 9*polylog(3, a*x^2))/x^3, -1/27*(8*sqrt(-a)*a*x^3*arctan(sqrt(-a)*x) + 8*a*x^2 + 6*dilog(a*x^2) - 4*log(-a*
x^2 + 1) + 9*polylog(3, a*x^2))/x^3]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}(a x^{2})}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^4,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/x^4, x)

________________________________________________________________________________________

maple [B]  time = 0.16, size = 125, normalized size = 1.79 \[ -\frac {a^{2} \left (-\frac {16}{27 x \sqrt {-a}}-\frac {8 x a \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{27 \sqrt {-a}\, \sqrt {a \,x^{2}}}+\frac {8 \ln \left (-a \,x^{2}+1\right )}{27 x^{3} \sqrt {-a}\, a}-\frac {4 \polylog \left (2, a \,x^{2}\right )}{9 x^{3} \sqrt {-a}\, a}-\frac {2 \polylog \left (3, a \,x^{2}\right )}{3 x^{3} \sqrt {-a}\, a}\right )}{2 \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/x^4,x)

[Out]

-1/2*a^2/(-a)^(1/2)*(-16/27/x/(-a)^(1/2)-8/27*x/(-a)^(1/2)*a/(a*x^2)^(1/2)*(ln(1-(a*x^2)^(1/2))-ln(1+(a*x^2)^(
1/2)))+8/27/x^3/(-a)^(1/2)/a*ln(-a*x^2+1)-4/9/x^3/(-a)^(1/2)*polylog(2,a*x^2)/a-2/3/x^3/(-a)^(1/2)/a*polylog(3
,a*x^2))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 66, normalized size = 0.94 \[ -\frac {4}{27} \, a^{\frac {3}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {8 \, a x^{2} + 6 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right ) + 9 \, {\rm Li}_{3}(a x^{2})}{27 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^4,x, algorithm="maxima")

[Out]

-4/27*a^(3/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/27*(8*a*x^2 + 6*dilog(a*x^2) - 4*log(-a*x^2 + 1) + 9*po
lylog(3, a*x^2))/x^3

________________________________________________________________________________________

mupad [B]  time = 0.77, size = 59, normalized size = 0.84 \[ \frac {4\,\ln \left (1-a\,x^2\right )}{27\,x^3}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{3\,x^3}-\frac {8\,a}{27\,x}-\frac {2\,\mathrm {polylog}\left (2,a\,x^2\right )}{9\,x^3}-\frac {a^{3/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{27} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/x^4,x)

[Out]

(4*log(1 - a*x^2))/(27*x^3) - (2*polylog(2, a*x^2))/(9*x^3) - polylog(3, a*x^2)/(3*x^3) - (8*a)/(27*x) - (a^(3
/2)*atan(a^(1/2)*x*1i)*8i)/27

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/x**4,x)

[Out]

Integral(polylog(3, a*x**2)/x**4, x)

________________________________________________________________________________________