3.111 \(\int (d x)^m \text{PolyLog}(2,a x^q) \, dx\)

Optimal. Leaf size=101 \[ \frac{a q^2 x^{q+1} (d x)^m \text{Hypergeometric2F1}\left (1,\frac{m+q+1}{q},\frac{m+2 q+1}{q},a x^q\right )}{(m+1)^2 (m+q+1)}+\frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^q\right )}{d (m+1)}+\frac{q (d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)^2} \]

[Out]

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

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Rubi [A]  time = 0.0597507, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6591, 2455, 20, 364} \[ \frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^q\right )}{d (m+1)}+\frac{a q^2 x^{q+1} (d x)^m \, _2F_1\left (1,\frac{m+q+1}{q};\frac{m+2 q+1}{q};a x^q\right )}{(m+1)^2 (m+q+1)}+\frac{q (d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[2, a*x^q],x]

[Out]

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

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 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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

Mathematica [A]  time = 0.0642053, size = 80, normalized size = 0.79 \[ \frac{x (d x)^m \left (a q^2 x^q \text{Hypergeometric2F1}\left (1,\frac{m+q+1}{q},\frac{m+2 q+1}{q},a x^q\right )+(m+q+1) \left ((m+1) \text{PolyLog}\left (2,a x^q\right )+q \log \left (1-a x^q\right )\right )\right )}{(m+1)^2 (m+q+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*PolyLog[2, a*x^q],x]

[Out]

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

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Maple [C]  time = 0.293, size = 148, normalized size = 1.5 \begin{align*} -{\frac{ \left ( dx \right ) ^{m}{x}^{-m}}{q} \left ( -a \right ) ^{-{\frac{m}{q}}-{q}^{-1}} \left ( -{\frac{{q}^{2}{x}^{1+m}\ln \left ( 1-a{x}^{q} \right ) }{ \left ( 1+m \right ) ^{2}} \left ( -a \right ) ^{{\frac{m}{q}}+{q}^{-1}}}-{\frac{q{x}^{1+m}{\it polylog} \left ( 2,a{x}^{q} \right ) }{1+m} \left ( -a \right ) ^{{\frac{m}{q}}+{q}^{-1}}}-{\frac{{q}^{2}{x}^{1+m+q}a}{ \left ( 1+m \right ) ^{2}} \left ( -a \right ) ^{{\frac{m}{q}}+{q}^{-1}}{\it LerchPhi} \left ( a{x}^{q},1,{\frac{1+m+q}{q}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(2,a*x^q),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(2,a*x^q),x, algorithm="fricas")

[Out]

integral((d*x)^m*dilog(a*x^q), 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((d*x)**m*polylog(2,a*x**q),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m*dilog(a*x^q), x)