3.56 \(\int \frac {\text {Li}_3(a x^q)}{x^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac {a q^3 x^{q-1} \, _2F_1\left (1,-\frac {1-q}{q};2-\frac {1}{q};a x^q\right )}{1-q}-\frac {q \text {Li}_2\left (a x^q\right )}{x}-\frac {\text {Li}_3\left (a x^q\right )}{x}+\frac {q^2 \log \left (1-a x^q\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 84, 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} \[ -\frac {q \text {PolyLog}\left (2,a x^q\right )}{x}-\frac {\text {PolyLog}\left (3,a x^q\right )}{x}-\frac {a q^3 x^{q-1} \, _2F_1\left (1,-\frac {1-q}{q};2-\frac {1}{q};a x^q\right )}{1-q}+\frac {q^2 \log \left (1-a x^q\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^q]/x^2,x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.44 \[ -\frac {G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1+\frac {1}{q} \\ 1,0,0,0,\frac {1}{q} \\\end {array}\right )}{q x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^q]/x^2,x]

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm polylog}\left (3, a x^{q}\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(polylog(3, a*x^q)/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}(a x^{q})}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(polylog(3, a*x^q)/x^2, x)

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maple [C]  time = 0.26, size = 129, normalized size = 1.54 \[ -\frac {\left (-a \right )^{\frac {1}{q}} \left (-\frac {q^{3} \left (-a \right )^{-\frac {1}{q}} \ln \left (1-a \,x^{q}\right )}{x}+\frac {q^{2} \left (-a \right )^{-\frac {1}{q}} \polylog \left (2, a \,x^{q}\right )}{x}-\frac {q \left (-a \right )^{-\frac {1}{q}} \left (1-q \right ) \polylog \left (3, a \,x^{q}\right )}{\left (-1+q \right ) x}-q^{3} x^{-1+q} a \left (-a \right )^{-\frac {1}{q}} \Phi \left (a \,x^{q}, 1, \frac {-1+q}{q}\right )\right )}{q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {polylog}\left (3,a\,x^q\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^q)/x^2,x)

[Out]

int(polylog(3, a*x^q)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a x^{q}\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, a*x**q)/x**2, x)

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