Integrand size = 11, antiderivative size = 93 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=-\frac {a q^3 x^{-3+q} \operatorname {Hypergeometric2F1}\left (1,-\frac {3-q}{q},2-\frac {3}{q},a x^q\right )}{27 (3-q)}+\frac {q^2 \log \left (1-a x^q\right )}{27 x^3}-\frac {q \operatorname {PolyLog}\left (2,a x^q\right )}{9 x^3}-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3} \] Output:
-1/27*a*q^3*x^(-3+q)*hypergeom([1, -(3-q)/q],[2-3/q],a*x^q)/(3-q)+1/27*q^2 *ln(1-a*x^q)/x^3-1/9*q*polylog(2,a*x^q)/x^3-1/3*polylog(3,a*x^q)/x^3
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.44 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=-\frac {G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,\frac {3+q}{q} \\ 1,0,0,0,\frac {3}{q} \\\end {array}\right )}{q x^3} \] Input:
Integrate[PolyLog[3, a*x^q]/x^4,x]
Output:
-(MeijerG[{{1, 1, 1, 1, (3 + q)/q}, {}}, {{1}, {0, 0, 0, 3/q}}, -(a*x^q)]/ (q*x^3))
Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {7145, 7145, 25, 2905, 888}
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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{3} q \int \frac {\operatorname {PolyLog}\left (2,a x^q\right )}{x^4}dx-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{3} q \left (\frac {1}{3} q \int -\frac {\log \left (1-a x^q\right )}{x^4}dx-\frac {\operatorname {PolyLog}\left (2,a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} q \left (-\frac {1}{3} q \int \frac {\log \left (1-a x^q\right )}{x^4}dx-\frac {\operatorname {PolyLog}\left (2,a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {1}{3} q \left (-\frac {1}{3} q \left (-\frac {1}{3} a q \int \frac {x^{q-4}}{1-a x^q}dx-\frac {\log \left (1-a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (2,a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {1}{3} q \left (-\frac {1}{3} q \left (\frac {a q x^{q-3} \operatorname {Hypergeometric2F1}\left (1,-\frac {3-q}{q},2-\frac {3}{q},a x^q\right )}{3 (3-q)}-\frac {\log \left (1-a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (2,a x^q\right )}{3 x^3}\right )-\frac {\operatorname {PolyLog}\left (3,a x^q\right )}{3 x^3}\) |
Input:
Int[PolyLog[3, a*x^q]/x^4,x]
Output:
(q*(-1/3*(q*((a*q*x^(-3 + q)*Hypergeometric2F1[1, -((3 - q)/q), 2 - 3/q, a *x^q])/(3*(3 - q)) - Log[1 - a*x^q]/(3*x^3))) - PolyLog[2, a*x^q]/(3*x^3)) )/3 - PolyLog[3, a*x^q]/(3*x^3)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
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] - Simp[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]
Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbo l] :> Simp[(d*x)^(m + 1)*(PolyLog[n, a*(b*x^p)^q]/(d*(m + 1))), x] - Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 5.
Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.42
method | result | size |
meijerg | \(-\frac {\left (-a \right )^{\frac {3}{q}} \left (-\frac {q^{3} \left (-a \right )^{-\frac {3}{q}} \ln \left (1-a \,x^{q}\right )}{27 x^{3}}+\frac {q^{2} \left (-a \right )^{-\frac {3}{q}} \operatorname {polylog}\left (2, a \,x^{q}\right )}{9 x^{3}}-\frac {q \left (-a \right )^{-\frac {3}{q}} \left (1-\frac {q}{3}\right ) \operatorname {polylog}\left (3, a \,x^{q}\right )}{\left (-3+q \right ) x^{3}}-\frac {q^{3} x^{-3+q} a \left (-a \right )^{-\frac {3}{q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {-3+q}{q}\right )}{27}\right )}{q}\) | \(132\) |
Input:
int(polylog(3,a*x^q)/x^4,x,method=_RETURNVERBOSE)
Output:
-(-a)^(3/q)/q*(-1/27*q^3/x^3*(-a)^(-3/q)*ln(1-a*x^q)+1/9*q^2/x^3*(-a)^(-3/ q)*polylog(2,a*x^q)-q/(-3+q)/x^3*(-a)^(-3/q)*(1-1/3*q)*polylog(3,a*x^q)-1/ 27*q^3*x^(-3+q)*a*(-a)^(-3/q)*LerchPhi(a*x^q,1,(-3+q)/q))
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{x^{4}} \,d x } \] Input:
integrate(polylog(3,a*x^q)/x^4,x, algorithm="fricas")
Output:
integral(polylog(3, a*x^q)/x^4, x)
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{q}\right )}{x^{4}}\, dx \] Input:
integrate(polylog(3,a*x**q)/x**4,x)
Output:
Integral(polylog(3, a*x**q)/x**4, x)
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{x^{4}} \,d x } \] Input:
integrate(polylog(3,a*x^q)/x^4,x, algorithm="maxima")
Output:
-q^3*integrate(1/27/(a*x^4*x^q - x^4), x) + 1/81*(q^3 + 3*q^2*log(-a*x^q + 1) - 9*q*dilog(a*x^q) - 27*polylog(3, a*x^q))/x^3
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{x^{4}} \,d x } \] Input:
integrate(polylog(3,a*x^q)/x^4,x, algorithm="giac")
Output:
integrate(polylog(3, a*x^q)/x^4, x)
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int \frac {\mathrm {polylog}\left (3,a\,x^q\right )}{x^4} \,d x \] Input:
int(polylog(3, a*x^q)/x^4,x)
Output:
int(polylog(3, a*x^q)/x^4, x)
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{x^4} \, dx=\int \frac {\mathit {polylog}\left (3, x^{q} a \right )}{x^{4}}d x \] Input:
int(polylog(3,a*x^q)/x^4,x)
Output:
int(polylog(3,x**q*a)/x**4,x)