3.156 \(\int x^3 \text {Li}_n(d (F^{c (a+b x)})^p) \, dx\)

Optimal. Leaf size=135 \[ -\frac {6 \text {Li}_{n+4}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)}+\frac {6 x \text {Li}_{n+3}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {3 x^2 \text {Li}_{n+2}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {x^3 \text {Li}_{n+1}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)} \]

[Out]

x^3*polylog(1+n,d*(F^(c*(b*x+a)))^p)/b/c/p/ln(F)-3*x^2*polylog(2+n,d*(F^(c*(b*x+a)))^p)/b^2/c^2/p^2/ln(F)^2+6*
x*polylog(3+n,d*(F^(c*(b*x+a)))^p)/b^3/c^3/p^3/ln(F)^3-6*polylog(4+n,d*(F^(c*(b*x+a)))^p)/b^4/c^4/p^4/ln(F)^4

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6609, 2282, 6589} \[ -\frac {3 x^2 \text {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \text {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \text {PolyLog}\left (n+4,d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)}+\frac {x^3 \text {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]

[Out]

(x^3*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - (3*x^2*PolyLog[2 + n, d*(F^(c*(a + b*x)))^p])/(b^
2*c^2*p^2*Log[F]^2) + (6*x*PolyLog[3 + n, d*(F^(c*(a + b*x)))^p])/(b^3*c^3*p^3*Log[F]^3) - (6*PolyLog[4 + n, d
*(F^(c*(a + b*x)))^p])/(b^4*c^4*p^4*Log[F]^4)

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]

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]

Rubi steps

\begin {align*} \int x^3 \text {Li}_n\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx &=\frac {x^3 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \int x^2 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx}{b c p \log (F)}\\ &=\frac {x^3 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \text {Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 \int x \text {Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx}{b^2 c^2 p^2 \log ^2(F)}\\ &=\frac {x^3 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \text {Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \text {Li}_{3+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \int \text {Li}_{3+n}\left (d \left (F^{c (a+b x)}\right )^p\right ) \, dx}{b^3 c^3 p^3 \log ^3(F)}\\ &=\frac {x^3 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \text {Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \text {Li}_{3+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_{3+n}\left (d x^p\right )}{x} \, dx,x,F^{c (a+b x)}\right )}{b^4 c^4 p^3 \log ^4(F)}\\ &=\frac {x^3 \text {Li}_{1+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \text {Li}_{2+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \text {Li}_{3+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \text {Li}_{4+n}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 135, normalized size = 1.00 \[ -\frac {6 \text {Li}_{n+4}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)}+\frac {6 x \text {Li}_{n+3}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {3 x^2 \text {Li}_{n+2}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {x^3 \text {Li}_{n+1}\left (d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]

[Out]

(x^3*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - (3*x^2*PolyLog[2 + n, d*(F^(c*(a + b*x)))^p])/(b^
2*c^2*p^2*Log[F]^2) + (6*x*PolyLog[3 + n, d*(F^(c*(a + b*x)))^p])/(b^3*c^3*p^3*Log[F]^3) - (6*PolyLog[4 + n, d
*(F^(c*(a + b*x)))^p])/(b^4*c^4*p^4*Log[F]^4)

________________________________________________________________________________________

fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} {\rm polylog}\left (n, {\left (F^{b c x + a c}\right )}^{p} d\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="fricas")

[Out]

integral(x^3*polylog(n, (F^(b*c*x + a*c))^p*d), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="giac")

[Out]

integrate(x^3*polylog(n, (F^((b*x + a)*c))^p*d), x)

________________________________________________________________________________________

maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int x^{3} \polylog \left (n , d \left (F^{c \left (b x +a \right )}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x)

[Out]

int(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="maxima")

[Out]

integrate(x^3*polylog(n, (F^((b*x + a)*c))^p*d), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {polylog}\left (n,d\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^p\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(n, d*(F^(c*(a + b*x)))^p),x)

[Out]

int(x^3*polylog(n, d*(F^(c*(a + b*x)))^p), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {Li}_{n}\left (d \left (F^{a c} F^{b c x}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*polylog(n,d*(F**(c*(b*x+a)))**p),x)

[Out]

Integral(x**3*polylog(n, d*(F**(a*c)*F**(b*c*x))**p), x)

________________________________________________________________________________________