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

Optimal. Leaf size=64 \[ \frac{1}{4} x^4 \text{PolyLog}\left (2,a x^2\right )-\frac{\log \left (1-a x^2\right )}{8 a^2}-\frac{x^2}{8 a}+\frac{1}{8} x^4 \log \left (1-a x^2\right )-\frac{x^4}{16} \]

[Out]

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

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Rubi [A]  time = 0.0495752, antiderivative size = 64, 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, 2454, 2395, 43} \[ \frac{1}{4} x^4 \text{PolyLog}\left (2,a x^2\right )-\frac{\log \left (1-a x^2\right )}{8 a^2}-\frac{x^2}{8 a}+\frac{1}{8} x^4 \log \left (1-a x^2\right )-\frac{x^4}{16} \]

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[2, a*x^2],x]

[Out]

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

Mathematica [A]  time = 0.0172903, size = 56, normalized size = 0.88 \[ \frac{4 a^2 x^4 \text{PolyLog}\left (2,a x^2\right )+2 \left (a^2 x^4-1\right ) \log \left (1-a x^2\right )-a x^2 \left (a x^2+2\right )}{16 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[2, a*x^2],x]

[Out]

(-(a*x^2*(2 + a*x^2)) + 2*(-1 + a^2*x^4)*Log[1 - a*x^2] + 4*a^2*x^4*PolyLog[2, a*x^2])/(16*a^2)

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Maple [A]  time = 0.045, size = 54, normalized size = 0.8 \begin{align*}{\frac{{x}^{4}{\it polylog} \left ( 2,a{x}^{2} \right ) }{4}}+{\frac{{x}^{4}\ln \left ( -a{x}^{2}+1 \right ) }{8}}-{\frac{{x}^{4}}{16}}-{\frac{{x}^{2}}{8\,a}}-{\frac{\ln \left ( a{x}^{2}-1 \right ) }{8\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.96931, size = 73, normalized size = 1.14 \begin{align*} \frac{4 \, a^{2} x^{4}{\rm Li}_2\left (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 )}{16 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*x^4*dilog(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 [A]  time = 2.58406, size = 120, normalized size = 1.88 \begin{align*} \frac{4 \, a^{2} x^{4}{\rm Li}_2\left (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 )}{16 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*x^4*dilog(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 [A]  time = 13.1842, size = 48, normalized size = 0.75 \begin{align*} \begin{cases} - \frac{x^{4} \operatorname{Li}_{1}\left (a x^{2}\right )}{8} + \frac{x^{4} \operatorname{Li}_{2}\left (a x^{2}\right )}{4} - \frac{x^{4}}{16} - \frac{x^{2}}{8 a} + \frac{\operatorname{Li}_{1}\left (a x^{2}\right )}{8 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**4*polylog(1, a*x**2)/8 + x**4*polylog(2, a*x**2)/4 - x**4/16 - x**2/(8*a) + polylog(1, a*x**2)/
(8*a**2), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\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^3*polylog(2,a*x^2),x, algorithm="giac")

[Out]

integrate(x^3*dilog(a*x^2), x)