3.1.46 \(\int x \text {PolyLog}(2,a x^q) \, dx\) [46]

Optimal. Leaf size=71 \[ \frac {a q^2 x^{2+q} \, _2F_1\left (1,\frac {2+q}{q};2 \left (1+\frac {1}{q}\right );a x^q\right )}{4 (2+q)}+\frac {1}{4} q x^2 \log \left (1-a x^q\right )+\frac {1}{2} x^2 \text {PolyLog}\left (2,a x^q\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6726, 2505, 371} \begin {gather*} \frac {a q^2 x^{q+2} \, _2F_1\left (1,\frac {q+2}{q};2 \left (1+\frac {1}{q}\right );a x^q\right )}{4 (q+2)}+\frac {1}{2} x^2 \text {Li}_2\left (a x^q\right )+\frac {1}{4} q x^2 \log \left (1-a x^q\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[2, a*x^q],x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2505

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

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Mathematica [A]
time = 0.03, size = 69, normalized size = 0.97 \begin {gather*} \frac {q x^2 \left (a q x^q \, _2F_1\left (1,\frac {2+q}{q};2+\frac {2}{q};a x^q\right )+(2+q) \log \left (1-a x^q\right )\right )}{4 (2+q)}+\frac {1}{2} x^2 \text {PolyLog}\left (2,a x^q\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[2, a*x^q],x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.11, size = 108, normalized size = 1.52

method	result	size
meijerg	\(-\frac {\left (-a \right )^{-\frac {2}{q}} \left (-\frac {q^{2} x^{2} \left (-a \right )^{\frac {2}{q}} \ln \left (1-a \,x^{q}\right )}{4}-\frac {q \,x^{2} \left (-a \right )^{\frac {2}{q}} \left (1+\frac {q}{2}\right ) \polylog \left (2, a \,x^{q}\right )}{2+q}-\frac {q^{2} x^{2+q} a \left (-a \right )^{\frac {2}{q}} \Phi \left (a \,x^{q}, 1, \frac {2+q}{q}\right )}{4}\right )}{q}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [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*polylog(2,a*x^q),x, algorithm="fricas")

[Out]

integral(x*dilog(a*x^q), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {Li}_{2}\left (a x^{q}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*polylog(2, a*x**q), x)

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

[Out]

integrate(x*dilog(a*x^q), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {polylog}\left (2,a\,x^q\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(2, a*x^q),x)

[Out]

int(x*polylog(2, a*x^q), x)

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