3.26 \(\int x^4 \text{PolyLog}(2,a x^2) \, dx\)

Optimal. Leaf size=73 \[ \frac{1}{5} x^5 \text{PolyLog}\left (2,a x^2\right )-\frac{4 x}{25 a^2}+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{25 a^{5/2}}-\frac{4 x^3}{75 a}+\frac{2}{25} x^5 \log \left (1-a x^2\right )-\frac{4 x^5}{125} \]

[Out]

(-4*x)/(25*a^2) - (4*x^3)/(75*a) - (4*x^5)/125 + (4*ArcTanh[Sqrt[a]*x])/(25*a^(5/2)) + (2*x^5*Log[1 - a*x^2])/
25 + (x^5*PolyLog[2, a*x^2])/5

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Rubi [A]  time = 0.0450937, antiderivative size = 73, normalized size of antiderivative = 1., 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, 302, 206} \[ \frac{1}{5} x^5 \text{PolyLog}\left (2,a x^2\right )-\frac{4 x}{25 a^2}+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{25 a^{5/2}}-\frac{4 x^3}{75 a}+\frac{2}{25} x^5 \log \left (1-a x^2\right )-\frac{4 x^5}{125} \]

Antiderivative was successfully verified.

[In]

Int[x^4*PolyLog[2, a*x^2],x]

[Out]

(-4*x)/(25*a^2) - (4*x^3)/(75*a) - (4*x^5)/125 + (4*ArcTanh[Sqrt[a]*x])/(25*a^(5/2)) + (2*x^5*Log[1 - a*x^2])/
25 + (x^5*PolyLog[2, a*x^2])/5

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]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

\begin{align*} \int x^4 \text{Li}_2\left (a x^2\right ) \, dx &=\frac{1}{5} x^5 \text{Li}_2\left (a x^2\right )+\frac{2}{5} \int x^4 \log \left (1-a x^2\right ) \, dx\\ &=\frac{2}{25} x^5 \log \left (1-a x^2\right )+\frac{1}{5} x^5 \text{Li}_2\left (a x^2\right )+\frac{1}{25} (4 a) \int \frac{x^6}{1-a x^2} \, dx\\ &=\frac{2}{25} x^5 \log \left (1-a x^2\right )+\frac{1}{5} x^5 \text{Li}_2\left (a x^2\right )+\frac{1}{25} (4 a) \int \left (-\frac{1}{a^3}-\frac{x^2}{a^2}-\frac{x^4}{a}+\frac{1}{a^3 \left (1-a x^2\right )}\right ) \, dx\\ &=-\frac{4 x}{25 a^2}-\frac{4 x^3}{75 a}-\frac{4 x^5}{125}+\frac{2}{25} x^5 \log \left (1-a x^2\right )+\frac{1}{5} x^5 \text{Li}_2\left (a x^2\right )+\frac{4 \int \frac{1}{1-a x^2} \, dx}{25 a^2}\\ &=-\frac{4 x}{25 a^2}-\frac{4 x^3}{75 a}-\frac{4 x^5}{125}+\frac{4 \tanh ^{-1}\left (\sqrt{a} x\right )}{25 a^{5/2}}+\frac{2}{25} x^5 \log \left (1-a x^2\right )+\frac{1}{5} x^5 \text{Li}_2\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0723123, size = 65, normalized size = 0.89 \[ \frac{1}{375} \left (75 x^5 \text{PolyLog}\left (2,a x^2\right )-\frac{60 x}{a^2}+\frac{60 \tanh ^{-1}\left (\sqrt{a} x\right )}{a^{5/2}}-\frac{20 x^3}{a}+30 x^5 \log \left (1-a x^2\right )-12 x^5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*PolyLog[2, a*x^2],x]

[Out]

((-60*x)/a^2 - (20*x^3)/a - 12*x^5 + (60*ArcTanh[Sqrt[a]*x])/a^(5/2) + 30*x^5*Log[1 - a*x^2] + 75*x^5*PolyLog[
2, a*x^2])/375

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Maple [A]  time = 0.046, size = 58, normalized size = 0.8 \begin{align*} -{\frac{4\,x}{25\,{a}^{2}}}-{\frac{4\,{x}^{3}}{75\,a}}-{\frac{4\,{x}^{5}}{125}}+{\frac{4}{25}{\it Artanh} \left ( x\sqrt{a} \right ){a}^{-{\frac{5}{2}}}}+{\frac{2\,{x}^{5}\ln \left ( -a{x}^{2}+1 \right ) }{25}}+{\frac{{x}^{5}{\it polylog} \left ( 2,a{x}^{2} \right ) }{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/25*x/a^2-4/75*x^3/a-4/125*x^5+4/25*arctanh(x*a^(1/2))/a^(5/2)+2/25*x^5*ln(-a*x^2+1)+1/5*x^5*polylog(2,a*x^2
)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69839, size = 398, normalized size = 5.45 \begin{align*} \left [\frac{75 \, a^{3} x^{5}{\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x + 30 \, \sqrt{a} \log \left (\frac{a x^{2} + 2 \, \sqrt{a} x + 1}{a x^{2} - 1}\right )}{375 \, a^{3}}, \frac{75 \, a^{3} x^{5}{\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x - 60 \, \sqrt{-a} \arctan \left (\sqrt{-a} x\right )}{375 \, a^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/375*(75*a^3*x^5*dilog(a*x^2) + 30*a^3*x^5*log(-a*x^2 + 1) - 12*a^3*x^5 - 20*a^2*x^3 - 60*a*x + 30*sqrt(a)*l
og((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)))/a^3, 1/375*(75*a^3*x^5*dilog(a*x^2) + 30*a^3*x^5*log(-a*x^2 + 1) -
12*a^3*x^5 - 20*a^2*x^3 - 60*a*x - 60*sqrt(-a)*arctan(sqrt(-a)*x))/a^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm Li}_2\left (a x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*dilog(a*x^2), x)