3.1.20 \(\int x^3 \text {PolyLog}(2,a x^2) \, dx\) [20]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 = {6726, 2504, 2442, 45} \begin {gather*} -\frac {\log \left (1-a x^2\right )}{8 a^2}+\frac {1}{4} x^4 \text {Li}_2\left (a x^2\right )-\frac {x^2}{8 a}+\frac {1}{8} x^4 \log \left (1-a x^2\right )-\frac {x^4}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[2, a*x^2],x]

[Out]

-1/8*x^2/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 45

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])

Rule 2442

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 2504

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 6726

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 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} \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.01, size = 56, normalized size = 0.88 \begin {gather*} \frac {-a x^2 \left (2+a x^2\right )+2 \left (-1+a^2 x^4\right ) \log \left (1-a x^2\right )+4 a^2 x^4 \text {PolyLog}\left (2,a x^2\right )}{16 a^2} \end {gather*}

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 = 0.05, size = 60, normalized size = 0.94

method	result	size
meijerg	\(-\frac {\frac {a \,x^{2} \left (3 a \,x^{2}+6\right )}{24}+\frac {\left (-3 a^{2} x^{4}+3\right ) \ln \left (-a \,x^{2}+1\right )}{12}-\frac {a^{2} x^{4} \polylog \left (2, a \,x^{2}\right )}{2}}{2 a^{2}}\)	\(57\)
default	\(\frac {x^{4} \polylog \left (2, a \,x^{2}\right )}{4}+\frac {x^{4} \ln \left (-a \,x^{2}+1\right )}{8}+\frac {a \left (-\frac {\frac {1}{2} a \,x^{4}+x^{2}}{2 a^{2}}-\frac {\ln \left (a \,x^{2}-1\right )}{2 a^{3}}\right )}{4}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(2,a*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 54, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.35, size = 54, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 1.79, size = 48, normalized size = 0.75 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.27, size = 53, normalized size = 0.83 \begin {gather*} \frac {x^4\,\mathrm {polylog}\left (2,a\,x^2\right )}{4}-\frac {\ln \left (a\,x^2-1\right )}{8\,a^2}+\frac {x^4\,\ln \left (1-a\,x^2\right )}{8}-\frac {x^4}{16}-\frac {x^2}{8\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(2, a*x^2),x)

[Out]

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

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