3.53 \(\int x \text {Li}_3(a x^q) \, dx\)

Optimal. Leaf size=88 \[ -\frac {a q^3 x^{q+2} \, _2F_1\left (1,\frac {q+2}{q};2 \left (1+\frac {1}{q}\right );a x^q\right )}{8 (q+2)}-\frac {1}{4} q x^2 \text {Li}_2\left (a x^q\right )+\frac {1}{2} x^2 \text {Li}_3\left (a x^q\right )-\frac {1}{8} q^2 x^2 \log \left (1-a x^q\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6591, 2455, 364} \[ -\frac {1}{4} q x^2 \text {PolyLog}\left (2,a x^q\right )+\frac {1}{2} x^2 \text {PolyLog}\left (3,a x^q\right )-\frac {a q^3 x^{q+2} \, _2F_1\left (1,\frac {q+2}{q};2 \left (1+\frac {1}{q}\right );a x^q\right )}{8 (q+2)}-\frac {1}{8} q^2 x^2 \log \left (1-a x^q\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[3, a*x^q],x]

[Out]

-(a*q^3*x^(2 + q)*Hypergeometric2F1[1, (2 + q)/q, 2*(1 + q^(-1)), a*x^q])/(8*(2 + q)) - (q^2*x^2*Log[1 - a*x^q
])/8 - (q*x^2*PolyLog[2, a*x^q])/4 + (x^2*PolyLog[3, a*x^q])/2

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 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 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]

Rubi steps

\begin {align*} \int x \text {Li}_3\left (a x^q\right ) \, dx &=\frac {1}{2} x^2 \text {Li}_3\left (a x^q\right )-\frac {1}{2} q \int x \text {Li}_2\left (a x^q\right ) \, dx\\ &=-\frac {1}{4} q x^2 \text {Li}_2\left (a x^q\right )+\frac {1}{2} x^2 \text {Li}_3\left (a x^q\right )-\frac {1}{4} q^2 \int x \log \left (1-a x^q\right ) \, dx\\ &=-\frac {1}{8} q^2 x^2 \log \left (1-a x^q\right )-\frac {1}{4} q x^2 \text {Li}_2\left (a x^q\right )+\frac {1}{2} x^2 \text {Li}_3\left (a x^q\right )-\frac {1}{8} \left (a q^3\right ) \int \frac {x^{1+q}}{1-a x^q} \, dx\\ &=-\frac {a q^3 x^{2+q} \, _2F_1\left (1,\frac {2+q}{q};2 \left (1+\frac {1}{q}\right );a x^q\right )}{8 (2+q)}-\frac {1}{8} q^2 x^2 \log \left (1-a x^q\right )-\frac {1}{4} q x^2 \text {Li}_2\left (a x^q\right )+\frac {1}{2} x^2 \text {Li}_3\left (a x^q\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.47 \[ -\frac {x^2 G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,\frac {q-2}{q} \\ 1,0,0,0,-\frac {2}{q} \\\end {array}\right )}{q} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*PolyLog[3, a*x^q],x]

[Out]

-((x^2*MeijerG[{{1, 1, 1, 1, (-2 + q)/q}, {}}, {{1}, {0, 0, 0, -2/q}}, -(a*x^q)])/q)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x {\rm polylog}\left (3, a x^{q}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*polylog(3, a*x^q), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm Li}_{3}(a x^{q})\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*polylog(3, a*x^q), x)

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maple [C]  time = 0.28, size = 132, normalized size = 1.50 \[ -\frac {\left (-a \right )^{-\frac {2}{q}} \left (\frac {q^{3} x^{2} \left (-a \right )^{\frac {2}{q}} \ln \left (1-a \,x^{q}\right )}{8}+\frac {q^{2} x^{2} \left (-a \right )^{\frac {2}{q}} \polylog \left (2, a \,x^{q}\right )}{4}-\frac {q \,x^{2} \left (-a \right )^{\frac {2}{q}} \left (1+\frac {q}{2}\right ) \polylog \left (3, a \,x^{q}\right )}{2+q}+\frac {q^{3} x^{2+q} a \left (-a \right )^{\frac {2}{q}} \Phi \left (a \,x^{q}, 1, \frac {2+q}{q}\right )}{8}\right )}{q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, q^{3} x^{2} - \frac {1}{8} \, q^{2} x^{2} \log \left (-a x^{q} + 1\right ) - \frac {1}{4} \, q x^{2} {\rm Li}_2\left (a x^{q}\right ) + q^{3} \int \frac {x}{8 \, {\left (a x^{q} - 1\right )}}\,{d x} + \frac {1}{2} \, x^{2} {\rm Li}_{3}(a x^{q}) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*q^3*x^2 - 1/8*q^2*x^2*log(-a*x^q + 1) - 1/4*q*x^2*dilog(a*x^q) + q^3*integrate(1/8*x/(a*x^q - 1), x) + 1/
2*x^2*polylog(3, a*x^q)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {polylog}\left (3,a\,x^q\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(3, a*x^q),x)

[Out]

int(x*polylog(3, a*x^q), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {Li}_{3}\left (a x^{q}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*polylog(3, a*x**q), x)

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