3.32 \(\int x^5 \text{PolyLog}(3,a x^2) \, dx\)

Optimal. Leaf size=88 \[ -\frac{1}{18} x^6 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{6} x^6 \text{PolyLog}\left (3,a x^2\right )+\frac{x^2}{54 a^2}+\frac{\log \left (1-a x^2\right )}{54 a^3}+\frac{x^4}{108 a}-\frac{1}{54} x^6 \log \left (1-a x^2\right )+\frac{x^6}{162} \]

[Out]

x^2/(54*a^2) + x^4/(108*a) + x^6/162 + Log[1 - a*x^2]/(54*a^3) - (x^6*Log[1 - a*x^2])/54 - (x^6*PolyLog[2, a*x
^2])/18 + (x^6*PolyLog[3, a*x^2])/6

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Rubi [A]  time = 0.075215, antiderivative size = 88, normalized size of antiderivative = 1., 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, 2454, 2395, 43} \[ -\frac{1}{18} x^6 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{6} x^6 \text{PolyLog}\left (3,a x^2\right )+\frac{x^2}{54 a^2}+\frac{\log \left (1-a x^2\right )}{54 a^3}+\frac{x^4}{108 a}-\frac{1}{54} x^6 \log \left (1-a x^2\right )+\frac{x^6}{162} \]

Antiderivative was successfully verified.

[In]

Int[x^5*PolyLog[3, a*x^2],x]

[Out]

x^2/(54*a^2) + x^4/(108*a) + x^6/162 + Log[1 - a*x^2]/(54*a^3) - (x^6*Log[1 - a*x^2])/54 - (x^6*PolyLog[2, a*x
^2])/18 + (x^6*PolyLog[3, a*x^2])/6

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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^5 \text{Li}_3\left (a x^2\right ) \, dx &=\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )-\frac{1}{3} \int x^5 \text{Li}_2\left (a x^2\right ) \, dx\\ &=-\frac{1}{18} x^6 \text{Li}_2\left (a x^2\right )+\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )-\frac{1}{9} \int x^5 \log \left (1-a x^2\right ) \, dx\\ &=-\frac{1}{18} x^6 \text{Li}_2\left (a x^2\right )+\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )-\frac{1}{18} \operatorname{Subst}\left (\int x^2 \log (1-a x) \, dx,x,x^2\right )\\ &=-\frac{1}{54} x^6 \log \left (1-a x^2\right )-\frac{1}{18} x^6 \text{Li}_2\left (a x^2\right )+\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )-\frac{1}{54} a \operatorname{Subst}\left (\int \frac{x^3}{1-a x} \, dx,x,x^2\right )\\ &=-\frac{1}{54} x^6 \log \left (1-a x^2\right )-\frac{1}{18} x^6 \text{Li}_2\left (a x^2\right )+\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )-\frac{1}{54} a \operatorname{Subst}\left (\int \left (-\frac{1}{a^3}-\frac{x}{a^2}-\frac{x^2}{a}-\frac{1}{a^3 (-1+a x)}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{54 a^2}+\frac{x^4}{108 a}+\frac{x^6}{162}+\frac{\log \left (1-a x^2\right )}{54 a^3}-\frac{1}{54} x^6 \log \left (1-a x^2\right )-\frac{1}{18} x^6 \text{Li}_2\left (a x^2\right )+\frac{1}{6} x^6 \text{Li}_3\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0180466, size = 88, normalized size = 1. \[ \frac{-18 a^3 x^6 \text{PolyLog}\left (2,a x^2\right )+54 a^3 x^6 \text{PolyLog}\left (3,a x^2\right )+2 a^3 x^6+3 a^2 x^4-6 a^3 x^6 \log \left (1-a x^2\right )+6 a x^2+6 \log \left (1-a x^2\right )}{324 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*PolyLog[3, a*x^2],x]

[Out]

(6*a*x^2 + 3*a^2*x^4 + 2*a^3*x^6 + 6*Log[1 - a*x^2] - 6*a^3*x^6*Log[1 - a*x^2] - 18*a^3*x^6*PolyLog[2, a*x^2]
+ 54*a^3*x^6*PolyLog[3, a*x^2])/(324*a^3)

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Maple [A]  time = 0.058, size = 80, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{a}^{3}} \left ({\frac{{x}^{2}a \left ( 4\,{a}^{2}{x}^{4}+6\,a{x}^{2}+12 \right ) }{324}}+{\frac{ \left ( -4\,{x}^{6}{a}^{3}+4 \right ) \ln \left ( -a{x}^{2}+1 \right ) }{108}}-{\frac{{x}^{6}{a}^{3}{\it polylog} \left ( 2,a{x}^{2} \right ) }{9}}+{\frac{{x}^{6}{a}^{3}{\it polylog} \left ( 3,a{x}^{2} \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*polylog(3,a*x^2),x)

[Out]

1/2/a^3*(1/324*x^2*a*(4*a^2*x^4+6*a*x^2+12)+1/108*(-4*a^3*x^6+4)*ln(-a*x^2+1)-1/9*x^6*a^3*polylog(2,a*x^2)+1/3
*x^6*a^3*polylog(3,a*x^2))

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Maxima [A]  time = 0.97977, size = 104, normalized size = 1.18 \begin{align*} -\frac{18 \, a^{3} x^{6}{\rm Li}_2\left (a x^{2}\right ) - 54 \, a^{3} x^{6}{\rm Li}_{3}(a x^{2}) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \,{\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{324 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*polylog(3,a*x^2),x, algorithm="maxima")

[Out]

-1/324*(18*a^3*x^6*dilog(a*x^2) - 54*a^3*x^6*polylog(3, a*x^2) - 2*a^3*x^6 - 3*a^2*x^4 - 6*a*x^2 + 6*(a^3*x^6
- 1)*log(-a*x^2 + 1))/a^3

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Fricas [C]  time = 2.71681, size = 240, normalized size = 2.73 \begin{align*} -\frac{18 \, a^{3} x^{6}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x^{2} + 1\right )}{a}, -\frac{2 \, \log \left (-a x^{2} + 1\right )}{x}\right ) - 54 \, a^{3} x^{6}{\rm polylog}\left (3, a x^{2}\right ) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \,{\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{324 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*polylog(3,a*x^2),x, algorithm="fricas")

[Out]

-1/324*(18*a^3*x^6*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) - 54*a^3*x^6*polylog(3, a*x^2) - 2*a^
3*x^6 - 3*a^2*x^4 - 6*a*x^2 + 6*(a^3*x^6 - 1)*log(-a*x^2 + 1))/a^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \operatorname{Li}_{3}\left (a x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*polylog(3,a*x**2),x)

[Out]

Integral(x**5*polylog(3, a*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5}{\rm Li}_{3}(a x^{2})\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*polylog(3,a*x^2),x, algorithm="giac")

[Out]

integrate(x^5*polylog(3, a*x^2), x)