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

Optimal. Leaf size=78 \[ -\frac{1}{8} x^4 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{4} x^4 \text{PolyLog}\left (3,a x^2\right )+\frac{\log \left (1-a x^2\right )}{16 a^2}+\frac{x^2}{16 a}-\frac{1}{16} x^4 \log \left (1-a x^2\right )+\frac{x^4}{32} \]

[Out]

x^2/(16*a) + x^4/32 + Log[1 - a*x^2]/(16*a^2) - (x^4*Log[1 - a*x^2])/16 - (x^4*PolyLog[2, a*x^2])/8 + (x^4*Pol
yLog[3, a*x^2])/4

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Rubi [A]  time = 0.0606042, antiderivative size = 78, 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} \[ -\frac{1}{8} x^4 \text{PolyLog}\left (2,a x^2\right )+\frac{1}{4} x^4 \text{PolyLog}\left (3,a x^2\right )+\frac{\log \left (1-a x^2\right )}{16 a^2}+\frac{x^2}{16 a}-\frac{1}{16} x^4 \log \left (1-a x^2\right )+\frac{x^4}{32} \]

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[3, a*x^2],x]

[Out]

x^2/(16*a) + x^4/32 + Log[1 - a*x^2]/(16*a^2) - (x^4*Log[1 - a*x^2])/16 - (x^4*PolyLog[2, a*x^2])/8 + (x^4*Pol
yLog[3, a*x^2])/4

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^3 \text{Li}_3\left (a x^2\right ) \, dx &=\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )-\frac{1}{2} \int x^3 \text{Li}_2\left (a x^2\right ) \, dx\\ &=-\frac{1}{8} x^4 \text{Li}_2\left (a x^2\right )+\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )-\frac{1}{4} \int x^3 \log \left (1-a x^2\right ) \, dx\\ &=-\frac{1}{8} x^4 \text{Li}_2\left (a x^2\right )+\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )-\frac{1}{8} \operatorname{Subst}\left (\int x \log (1-a x) \, dx,x,x^2\right )\\ &=-\frac{1}{16} x^4 \log \left (1-a x^2\right )-\frac{1}{8} x^4 \text{Li}_2\left (a x^2\right )+\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )-\frac{1}{16} a \operatorname{Subst}\left (\int \frac{x^2}{1-a x} \, dx,x,x^2\right )\\ &=-\frac{1}{16} x^4 \log \left (1-a x^2\right )-\frac{1}{8} x^4 \text{Li}_2\left (a x^2\right )+\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )-\frac{1}{16} a \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}-\frac{x}{a}-\frac{1}{a^2 (-1+a x)}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{16 a}+\frac{x^4}{32}+\frac{\log \left (1-a x^2\right )}{16 a^2}-\frac{1}{16} x^4 \log \left (1-a x^2\right )-\frac{1}{8} x^4 \text{Li}_2\left (a x^2\right )+\frac{1}{4} x^4 \text{Li}_3\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0144214, size = 79, normalized size = 1.01 \[ \frac{-4 a^2 x^4 \text{PolyLog}\left (2,a x^2\right )+8 a^2 x^4 \text{PolyLog}\left (3,a x^2\right )+a^2 x^4-2 a^2 x^4 \log \left (1-a x^2\right )+2 a x^2+2 \log \left (1-a x^2\right )}{32 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[3, a*x^2],x]

[Out]

(2*a*x^2 + a^2*x^4 + 2*Log[1 - a*x^2] - 2*a^2*x^4*Log[1 - a*x^2] - 4*a^2*x^4*PolyLog[2, a*x^2] + 8*a^2*x^4*Pol
yLog[3, a*x^2])/(32*a^2)

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Maple [A]  time = 0.054, size = 72, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{a}^{2}} \left ( -{\frac{{x}^{2}a \left ( 3\,a{x}^{2}+6 \right ) }{48}}-{\frac{ \left ( -3\,{a}^{2}{x}^{4}+3 \right ) \ln \left ( -a{x}^{2}+1 \right ) }{24}}+{\frac{{x}^{4}{a}^{2}{\it polylog} \left ( 2,a{x}^{2} \right ) }{4}}-{\frac{{x}^{4}{a}^{2}{\it polylog} \left ( 3,a{x}^{2} \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a^2*(-1/48*x^2*a*(3*a*x^2+6)-1/24*(-3*a^2*x^4+3)*ln(-a*x^2+1)+1/4*x^4*a^2*polylog(2,a*x^2)-1/2*x^4*a^2*po
lylog(3,a*x^2))

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Maxima [A]  time = 0.968293, size = 93, normalized size = 1.19 \begin{align*} -\frac{4 \, a^{2} x^{4}{\rm Li}_2\left (a x^{2}\right ) - 8 \, a^{2} x^{4}{\rm Li}_{3}(a x^{2}) - a^{2} x^{4} - 2 \, a x^{2} + 2 \,{\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right )}{32 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(4*a^2*x^4*dilog(a*x^2) - 8*a^2*x^4*polylog(3, a*x^2) - a^2*x^4 - 2*a*x^2 + 2*(a^2*x^4 - 1)*log(-a*x^2 +
1))/a^2

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Fricas [C]  time = 2.70568, size = 217, normalized size = 2.78 \begin{align*} -\frac{4 \, a^{2} x^{4}{\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 ) - 8 \, a^{2} x^{4}{\rm polylog}\left (3, a x^{2}\right ) - a^{2} x^{4} - 2 \, a x^{2} + 2 \,{\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right )}{32 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(4*a^2*x^4*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) - 8*a^2*x^4*polylog(3, a*x^2) - a^2*x^4
- 2*a*x^2 + 2*(a^2*x^4 - 1)*log(-a*x^2 + 1))/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*polylog(3, a*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*polylog(3, a*x^2), x)