3.158 \(\int x \text{PolyLog}(n,d (F^{c (a+b x)})^p) \, dx\)

Optimal. Leaf size=65 \[ \frac{x \text{PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\text{PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)} \]

[Out]

(x*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - PolyLog[2 + n, d*(F^(c*(a + b*x)))^p]/(b^2*c^2*p^2*
Log[F]^2)

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Rubi [A]  time = 0.0331011, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {6609, 2282, 6589} \[ \frac{x \text{PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\text{PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]

[Out]

(x*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - PolyLog[2 + n, d*(F^(c*(a + b*x)))^p]/(b^2*c^2*p^2*
Log[F]^2)

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x \text{Li}_n\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx &=\frac{x \text{Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\int \text{Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx}{b c p \log (F)}\\ &=\frac{x \text{Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_{1+n}\left (d x^p\right )}{x} \, dx,x,F^{c (a+b x)}\right )}{b^2 c^2 p \log ^2(F)}\\ &=\frac{x \text{Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\text{Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}\\ \end{align*}

Mathematica [A]  time = 0.0052208, size = 65, normalized size = 1. \[ \frac{x \text{PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac{\text{PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]

[Out]

(x*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - PolyLog[2 + n, d*(F^(c*(a + b*x)))^p]/(b^2*c^2*p^2*
Log[F]^2)

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int x{\it polylog} \left ( n,d \left ({F}^{c \left ( bx+a \right ) } \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(n,d*(F^(c*(b*x+a)))^p),x)

[Out]

int(x*polylog(n,d*(F^(c*(b*x+a)))^p),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="maxima")

[Out]

integrate(x*polylog(n, (F^((b*x + a)*c))^p*d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x{\rm polylog}\left (n,{\left (F^{b c x + a c}\right )}^{p} d\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="fricas")

[Out]

integral(x*polylog(n, (F^(b*c*x + a*c))^p*d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{Li}_{n}\left (d \left (F^{a c} F^{b c x}\right )^{p}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(n,d*(F**(c*(b*x+a)))**p),x)

[Out]

Integral(x*polylog(n, d*(F**(a*c)*F**(b*c*x))**p), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="giac")

[Out]

integrate(x*polylog(n, (F^((b*x + a)*c))^p*d), x)