3.50 \(\int \frac{\text{PolyLog}(2,a x^q)}{x^3} \, dx\)

Optimal. Leaf size=78 \[ -\frac{a q^2 x^{q-2} \text{Hypergeometric2F1}\left (1,-\frac{2-q}{q},2 \left (1-\frac{1}{q}\right ),a x^q\right )}{4 (2-q)}-\frac{\text{PolyLog}\left (2,a x^q\right )}{2 x^2}+\frac{q \log \left (1-a x^q\right )}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0418889, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6591, 2455, 364} \[ -\frac{\text{PolyLog}\left (2,a x^q\right )}{2 x^2}-\frac{a q^2 x^{q-2} \, _2F_1\left (1,-\frac{2-q}{q};2 \left (1-\frac{1}{q}\right );a x^q\right )}{4 (2-q)}+\frac{q \log \left (1-a x^q\right )}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^q]/x^3,x]

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0521594, size = 61, normalized size = 0.78 \[ \frac{q \left (\frac{a q x^q \text{Hypergeometric2F1}\left (1,\frac{q-2}{q},2-\frac{2}{q},a x^q\right )}{q-2}+\log \left (1-a x^q\right )\right )-2 \text{PolyLog}\left (2,a x^q\right )}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^q]/x^3,x]

[Out]

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

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Maple [C]  time = 0.209, size = 108, normalized size = 1.4 \begin{align*} -{\frac{1}{q} \left ( -a \right ) ^{2\,{q}^{-1}} \left ( -{\frac{{q}^{2}\ln \left ( 1-a{x}^{q} \right ) }{4\,{x}^{2}} \left ( -a \right ) ^{-2\,{q}^{-1}}}-{\frac{q{\it polylog} \left ( 2,a{x}^{q} \right ) }{ \left ( -2+q \right ){x}^{2}} \left ( -a \right ) ^{-2\,{q}^{-1}} \left ( 1-{\frac{q}{2}} \right ) }-{\frac{{q}^{2}{x}^{-2+q}a}{4} \left ( -a \right ) ^{-2\,{q}^{-1}}{\it LerchPhi} \left ( a{x}^{q},1,{\frac{-2+q}{q}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -q^{2} \int \frac{1}{4 \,{\left (a x^{3} x^{q} - x^{3}\right )}}\,{d x} + \frac{q^{2} + 2 \, q \log \left (-a x^{q} + 1\right ) - 4 \,{\rm Li}_2\left (a x^{q}\right )}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm Li}_2\left (a x^{q}\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(dilog(a*x^q)/x^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (a x^{q}\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^q)/x^3, x)