Integrand size = 19, antiderivative size = 135 \[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\frac {x^3 \operatorname {PolyLog}\left (1+n,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \operatorname {PolyLog}\left (2+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \operatorname {PolyLog}\left (3+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \operatorname {PolyLog}\left (4+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)} \] Output:
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
Time = 0.01 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00 \[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\frac {x^3 \operatorname {PolyLog}\left (1+n,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 x^2 \operatorname {PolyLog}\left (2+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}+\frac {6 x \operatorname {PolyLog}\left (3+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^3 c^3 p^3 \log ^3(F)}-\frac {6 \operatorname {PolyLog}\left (4+n,d \left (F^{c (a+b x)}\right )^p\right )}{b^4 c^4 p^4 \log ^4(F)} \] Input:
Integrate[x^3*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]
Output:
(x^3*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - (3*x^2*PolyLo g[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)
Time = 0.69 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {7163, 7163, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {x^3 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \int x^2 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )dx}{b c p \log (F)}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {x^3 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {2 \int x \operatorname {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )dx}{b c p \log (F)}\right )}{b c p \log (F)}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {x^3 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {\int \operatorname {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )dx}{b c p \log (F)}\right )}{b c p \log (F)}\right )}{b c p \log (F)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x^3 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {\int F^{-c (a+b x)} \operatorname {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )dF^{c (a+b x)}}{b^2 c^2 p \log ^2(F)}\right )}{b c p \log (F)}\right )}{b c p \log (F)}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {x^3 \operatorname {PolyLog}\left (n+1,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (n+2,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (n+3,d \left (F^{c (a+b x)}\right )^p\right )}{b c p \log (F)}-\frac {\operatorname {PolyLog}\left (n+4,d \left (F^{c (a+b x)}\right )^p\right )}{b^2 c^2 p^2 \log ^2(F)}\right )}{b c p \log (F)}\right )}{b c p \log (F)}\) |
Input:
Int[x^3*PolyLog[n, d*(F^(c*(a + b*x)))^p],x]
Output:
(x^3*PolyLog[1 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - (3*((x^2*Poly Log[2 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - (2*((x*PolyLog[3 + n, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]) - PolyLog[4 + n, d*(F^(c*(a + b*x)) )^p]/(b^2*c^2*p^2*Log[F]^2)))/(b*c*p*Log[F])))/(b*c*p*Log[F])
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
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] - Simp[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]
\[\int x^{3} \operatorname {polylog}\left (n , d \left (F^{c \left (b x +a \right )}\right )^{p}\right )d x\]
Input:
int(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x)
Output:
int(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x)
\[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int { x^{3} {\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d) \,d x } \] Input:
integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="fricas")
Output:
integral(x^3*polylog(n, (F^(b*c*x + a*c))^p*d), x)
\[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int x^{3} \operatorname {Li}_{n}\left (d \left (F^{a c + b c x}\right )^{p}\right )\, dx \] Input:
integrate(x**3*polylog(n,d*(F**(c*(b*x+a)))**p),x)
Output:
Integral(x**3*polylog(n, d*(F**(a*c + b*c*x))**p), x)
\[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int { x^{3} {\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d) \,d x } \] Input:
integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="maxima")
Output:
integrate(x^3*polylog(n, F^((b*x + a)*c*p)*d), x)
\[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int { x^{3} {\rm Li}_{n}({\left (F^{{\left (b x + a\right )} c}\right )}^{p} d) \,d x } \] Input:
integrate(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x, algorithm="giac")
Output:
integrate(x^3*polylog(n, (F^((b*x + a)*c))^p*d), x)
Timed out. \[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int x^3\,\mathrm {polylog}\left (n,d\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^p\right ) \,d x \] Input:
int(x^3*polylog(n, d*(F^(c*(a + b*x)))^p),x)
Output:
int(x^3*polylog(n, d*(F^(c*(a + b*x)))^p), x)
\[ \int x^3 \operatorname {PolyLog}\left (n,d \left (F^{c (a+b x)}\right )^p\right ) \, dx=\int \mathit {polylog}\left (n , f^{b c p x +a c p} d \right ) x^{3}d x \] Input:
int(x^3*polylog(n,d*(F^(c*(b*x+a)))^p),x)
Output:
int(polylog(n,f**(a*c*p + b*c*p*x)*d)*x**3,x)