3.44 \(\int \frac {\text {Li}_3(a x^2)}{x^6} \, dx\)

Optimal. Leaf size=80 \[ \frac {8}{125} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {8 a^2}{125 x}-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}-\frac {8 a}{375 x^3}+\frac {4 \log \left (1-a x^2\right )}{125 x^5} \]

[Out]

-8/375*a/x^3-8/125*a^2/x+8/125*a^(5/2)*arctanh(x*a^(1/2))+4/125*ln(-a*x^2+1)/x^5-2/25*polylog(2,a*x^2)/x^5-1/5
*polylog(3,a*x^2)/x^5

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac {2 \text {PolyLog}\left (2,a x^2\right )}{25 x^5}-\frac {\text {PolyLog}\left (3,a x^2\right )}{5 x^5}-\frac {8 a^2}{125 x}+\frac {8}{125} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {8 a}{375 x^3}+\frac {4 \log \left (1-a x^2\right )}{125 x^5} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/x^6,x]

[Out]

(-8*a)/(375*x^3) - (8*a^2)/(125*x) + (8*a^(5/2)*ArcTanh[Sqrt[a]*x])/125 + (4*Log[1 - a*x^2])/(125*x^5) - (2*Po
lyLog[2, a*x^2])/(25*x^5) - PolyLog[3, a*x^2]/(5*x^5)

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^6} \, dx &=-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}+\frac {2}{5} \int \frac {\text {Li}_2\left (a x^2\right )}{x^6} \, dx\\ &=-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}-\frac {4}{25} \int \frac {\log \left (1-a x^2\right )}{x^6} \, dx\\ &=\frac {4 \log \left (1-a x^2\right )}{125 x^5}-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}+\frac {1}{125} (8 a) \int \frac {1}{x^4 \left (1-a x^2\right )} \, dx\\ &=-\frac {8 a}{375 x^3}+\frac {4 \log \left (1-a x^2\right )}{125 x^5}-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}+\frac {1}{125} \left (8 a^2\right ) \int \frac {1}{x^2 \left (1-a x^2\right )} \, dx\\ &=-\frac {8 a}{375 x^3}-\frac {8 a^2}{125 x}+\frac {4 \log \left (1-a x^2\right )}{125 x^5}-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}+\frac {1}{125} \left (8 a^3\right ) \int \frac {1}{1-a x^2} \, dx\\ &=-\frac {8 a}{375 x^3}-\frac {8 a^2}{125 x}+\frac {8}{125} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {4 \log \left (1-a x^2\right )}{125 x^5}-\frac {2 \text {Li}_2\left (a x^2\right )}{25 x^5}-\frac {\text {Li}_3\left (a x^2\right )}{5 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 69, normalized size = 0.86 \[ -\frac {-24 a^{5/2} x^5 \tanh ^{-1}\left (\sqrt {a} x\right )+24 a^2 x^4+30 \text {Li}_2\left (a x^2\right )+75 \text {Li}_3\left (a x^2\right )+8 a x^2-12 \log \left (1-a x^2\right )}{375 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^2]/x^6,x]

[Out]

-1/375*(8*a*x^2 + 24*a^2*x^4 - 24*a^(5/2)*x^5*ArcTanh[Sqrt[a]*x] - 12*Log[1 - a*x^2] + 30*PolyLog[2, a*x^2] +
75*PolyLog[3, a*x^2])/x^5

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fricas [C]  time = 0.81, size = 150, normalized size = 1.88 \[ \left [\frac {12 \, a^{\frac {5}{2}} x^{5} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 24 \, a^{2} x^{4} - 8 \, a x^{2} - 30 \, {\rm Li}_2\left (a x^{2}\right ) + 12 \, \log \left (-a x^{2} + 1\right ) - 75 \, {\rm polylog}\left (3, a x^{2}\right )}{375 \, x^{5}}, -\frac {24 \, \sqrt {-a} a^{2} x^{5} \arctan \left (\sqrt {-a} x\right ) + 24 \, a^{2} x^{4} + 8 \, a x^{2} + 30 \, {\rm Li}_2\left (a x^{2}\right ) - 12 \, \log \left (-a x^{2} + 1\right ) + 75 \, {\rm polylog}\left (3, a x^{2}\right )}{375 \, x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^6,x, algorithm="fricas")

[Out]

[1/375*(12*a^(5/2)*x^5*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 24*a^2*x^4 - 8*a*x^2 - 30*dilog(a*x^2) + 1
2*log(-a*x^2 + 1) - 75*polylog(3, a*x^2))/x^5, -1/375*(24*sqrt(-a)*a^2*x^5*arctan(sqrt(-a)*x) + 24*a^2*x^4 + 8
*a*x^2 + 30*dilog(a*x^2) - 12*log(-a*x^2 + 1) + 75*polylog(3, a*x^2))/x^5]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}(a x^{2})}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^6,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/x^6, x)

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maple [B]  time = 0.17, size = 138, normalized size = 1.72 \[ \frac {a^{3} \left (-\frac {16}{375 x^{3} \left (-a \right )^{\frac {3}{2}}}-\frac {16 a}{125 x \left (-a \right )^{\frac {3}{2}}}-\frac {8 x \,a^{2} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{125 \left (-a \right )^{\frac {3}{2}} \sqrt {a \,x^{2}}}+\frac {8 \ln \left (-a \,x^{2}+1\right )}{125 x^{5} \left (-a \right )^{\frac {3}{2}} a}-\frac {4 \polylog \left (2, a \,x^{2}\right )}{25 x^{5} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \polylog \left (3, a \,x^{2}\right )}{5 x^{5} \left (-a \right )^{\frac {3}{2}} a}\right )}{2 \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/x^6,x)

[Out]

1/2*a^3/(-a)^(1/2)*(-16/375/x^3/(-a)^(3/2)-16/125/x/(-a)^(3/2)*a-8/125*x/(-a)^(3/2)*a^2/(a*x^2)^(1/2)*(ln(1-(a
*x^2)^(1/2))-ln(1+(a*x^2)^(1/2)))+8/125/x^5/(-a)^(3/2)/a*ln(-a*x^2+1)-4/25/x^5/(-a)^(3/2)*polylog(2,a*x^2)/a-2
/5/x^5/(-a)^(3/2)/a*polylog(3,a*x^2))

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maxima [A]  time = 1.13, size = 74, normalized size = 0.92 \[ -\frac {4}{125} \, a^{\frac {5}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {24 \, a^{2} x^{4} + 8 \, a x^{2} + 30 \, {\rm Li}_2\left (a x^{2}\right ) - 12 \, \log \left (-a x^{2} + 1\right ) + 75 \, {\rm Li}_{3}(a x^{2})}{375 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^6,x, algorithm="maxima")

[Out]

-4/125*a^(5/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/375*(24*a^2*x^4 + 8*a*x^2 + 30*dilog(a*x^2) - 12*log(-
a*x^2 + 1) + 75*polylog(3, a*x^2))/x^5

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mupad [B]  time = 1.03, size = 70, normalized size = 0.88 \[ \frac {4\,\ln \left (1-a\,x^2\right )}{125\,x^5}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{5\,x^5}-\frac {8\,a^2\,x^2+\frac {8\,a}{3}}{125\,x^3}-\frac {2\,\mathrm {polylog}\left (2,a\,x^2\right )}{25\,x^5}-\frac {a^{5/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{125} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/x^6,x)

[Out]

(4*log(1 - a*x^2))/(125*x^5) - (2*polylog(2, a*x^2))/(25*x^5) - polylog(3, a*x^2)/(5*x^5) - ((8*a)/3 + 8*a^2*x
^2)/(125*x^3) - (a^(5/2)*atan(a^(1/2)*x*1i)*8i)/125

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/x**6,x)

[Out]

Integral(polylog(3, a*x**2)/x**6, x)

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