3.57 \(\int \frac {\text {Li}_3(a x^q)}{x^3} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a q^3 x^{q-2} \, _2F_1\left (1,-\frac {2-q}{q};2 \left (1-\frac {1}{q}\right );a x^q\right )}{8 (2-q)}-\frac {q \text {Li}_2\left (a x^q\right )}{4 x^2}-\frac {\text {Li}_3\left (a x^q\right )}{2 x^2}+\frac {q^2 \log \left (1-a x^q\right )}{8 x^2} \]

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

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Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6591, 2455, 364} \[ -\frac {q \text {PolyLog}\left (2,a x^q\right )}{4 x^2}-\frac {\text {PolyLog}\left (3,a x^q\right )}{2 x^2}-\frac {a q^3 x^{q-2} \, _2F_1\left (1,-\frac {2-q}{q};2 \left (1-\frac {1}{q}\right );a x^q\right )}{8 (2-q)}+\frac {q^2 \log \left (1-a x^q\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^q]/x^3,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*Log[1 - a*x^q
])/(8*x^2) - (q*PolyLog[2, a*x^q])/(4*x^2) - PolyLog[3, a*x^q]/(2*x^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 \frac {\text {Li}_3\left (a x^q\right )}{x^3} \, dx &=-\frac {\text {Li}_3\left (a x^q\right )}{2 x^2}+\frac {1}{2} q \int \frac {\text {Li}_2\left (a x^q\right )}{x^3} \, dx\\ &=-\frac {q \text {Li}_2\left (a x^q\right )}{4 x^2}-\frac {\text {Li}_3\left (a x^q\right )}{2 x^2}-\frac {1}{4} q^2 \int \frac {\log \left (1-a x^q\right )}{x^3} \, dx\\ &=\frac {q^2 \log \left (1-a x^q\right )}{8 x^2}-\frac {q \text {Li}_2\left (a x^q\right )}{4 x^2}-\frac {\text {Li}_3\left (a x^q\right )}{2 x^2}+\frac {1}{8} \left (a q^3\right ) \int \frac {x^{-3+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 {q^2 \log \left (1-a x^q\right )}{8 x^2}-\frac {q \text {Li}_2\left (a x^q\right )}{4 x^2}-\frac {\text {Li}_3\left (a x^q\right )}{2 x^2}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.43 \[ -\frac {G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,\frac {q+2}{q} \\ 1,0,0,0,\frac {2}{q} \\\end {array}\right )}{q x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^q]/x^3,x]

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm polylog}\left (3, a x^{q}\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(polylog(3, a*x^q)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(polylog(3, a*x^q)/x^3, x)

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maple [C]  time = 0.26, size = 132, normalized size = 1.39 \[ -\frac {\left (-a \right )^{\frac {2}{q}} \left (-\frac {q^{3} \left (-a \right )^{-\frac {2}{q}} \ln \left (1-a \,x^{q}\right )}{8 x^{2}}+\frac {q^{2} \left (-a \right )^{-\frac {2}{q}} \polylog \left (2, a \,x^{q}\right )}{4 x^{2}}-\frac {q \left (-a \right )^{-\frac {2}{q}} \left (1-\frac {q}{2}\right ) \polylog \left (3, a \,x^{q}\right )}{\left (-2+q \right ) x^{2}}-\frac {q^{3} x^{-2+q} a \left (-a \right )^{-\frac {2}{q}} \Phi \left (a \,x^{q}, 1, \frac {-2+q}{q}\right )}{8}\right )}{q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^q)/x^3,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 \[ -q^{3} \int \frac {1}{8 \, {\left (a x^{3} x^{q} - x^{3}\right )}}\,{d x} + \frac {q^{3} + 2 \, q^{2} \log \left (-a x^{q} + 1\right ) - 4 \, q {\rm Li}_2\left (a x^{q}\right ) - 8 \, {\rm Li}_{3}(a x^{q})}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^q)/x^3,x)

[Out]

int(polylog(3, a*x^q)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, a*x**q)/x**3, x)

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