3.2.13 \(\int (d x)^m \operatorname {PolyLog}(4,a x^q) \, dx\) [113]

3.2.13.1 Optimal result

Integrand size = 13, antiderivative size = 154 \[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\frac {a q^4 x^{1+q} (d x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+q}{q},\frac {1+m+2 q}{q},a x^q\right )}{(1+m)^4 (1+m+q)}+\frac {q^3 (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^4}+\frac {q^2 (d x)^{1+m} \operatorname {PolyLog}\left (2,a x^q\right )}{d (1+m)^3}-\frac {q (d x)^{1+m} \operatorname {PolyLog}\left (3,a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \operatorname {PolyLog}\left (4,a x^q\right )}{d (1+m)} \]

output

a*q^4*x^(1+q)*(d*x)^m*hypergeom([1, (1+m+q)/q],[(1+m+2*q)/q],a*x^q)/(1+m)^ 
4/(1+m+q)+q^3*(d*x)^(1+m)*ln(1-a*x^q)/d/(1+m)^4+q^2*(d*x)^(1+m)*polylog(2, 
a*x^q)/d/(1+m)^3-q*(d*x)^(1+m)*polylog(3,a*x^q)/d/(1+m)^2+(d*x)^(1+m)*poly 
log(4,a*x^q)/d/(1+m)
 

3.2.13.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.34 \[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=-\frac {x (d x)^m G_{6,6}^{1,6}\left (-a x^q|\begin {array}{c} 1,1,1,1,1,1-\frac {1+m}{q} \\ 1,0,0,0,0,-\frac {1+m}{q} \\\end {array}\right )}{q} \]

input

Integrate[(d*x)^m*PolyLog[4, a*x^q],x]
 

output

-((x*(d*x)^m*MeijerG[{{1, 1, 1, 1, 1, 1 - (1 + m)/q}, {}}, {{1}, {0, 0, 0, 
 0, -((1 + m)/q)}}, -(a*x^q)])/q)
 

3.2.13.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {7145, 7145, 7145, 25, 2905, 30, 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.

\(\displaystyle \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx\)

\(\Big \downarrow \) 7145

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \int (d x)^m \operatorname {PolyLog}\left (3,a x^q\right )dx}{m+1}\)

\(\Big \downarrow \) 7145

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right )dx}{m+1}\right )}{m+1}\)

\(\Big \downarrow \) 7145

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}-\frac {q \int -(d x)^m \log \left (1-a x^q\right )dx}{m+1}\right )}{m+1}\right )}{m+1}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {q \int (d x)^m \log \left (1-a x^q\right )dx}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\right )}{m+1}\right )}{m+1}\)

\(\Big \downarrow \) 2905

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {q \left (\frac {a q \int \frac {x^{q-1} (d x)^{m+1}}{1-a x^q}dx}{d (m+1)}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\right )}{m+1}\right )}{m+1}\)

\(\Big \downarrow \) 30

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {q \left (\frac {a q x^{-m} (d x)^m \int \frac {x^{m+q}}{1-a x^q}dx}{m+1}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\right )}{m+1}\right )}{m+1}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (4,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {(d x)^{m+1} \operatorname {PolyLog}\left (3,a x^q\right )}{d (m+1)}-\frac {q \left (\frac {q \left (\frac {a q x^{q+1} (d x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {m+q+1}{q},\frac {m+2 q+1}{q},a x^q\right )}{(m+1) (m+q+1)}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\right )}{m+1}\right )}{m+1}\)

input

Int[(d*x)^m*PolyLog[4, a*x^q],x]
 

output

-((q*(-((q*((q*((a*q*x^(1 + q)*(d*x)^m*Hypergeometric2F1[1, (1 + m + q)/q, 
 (1 + m + 2*q)/q, a*x^q])/((1 + m)*(1 + m + q)) + ((d*x)^(1 + m)*Log[1 - a 
*x^q])/(d*(1 + m))))/(1 + m) + ((d*x)^(1 + m)*PolyLog[2, a*x^q])/(d*(1 + m 
))))/(1 + m)) + ((d*x)^(1 + m)*PolyLog[3, a*x^q])/(d*(1 + m))))/(1 + m)) + 
 ((d*x)^(1 + m)*PolyLog[4, a*x^q])/(d*(1 + m))
 

3.2.13.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 888

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])
 

rule 2905

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]
 

rule 7145

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]
 

3.2.13.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5.

Time = 1.24 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.41

method	result	size
meijerg	\(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {m}{q}-\frac {1}{q}} \left (-\frac {q^{4} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \ln \left (1-a \,x^{q}\right )}{\left (1+m \right )^{4}}-\frac {q^{3} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {polylog}\left (2, a \,x^{q}\right )}{\left (1+m \right )^{3}}+\frac {q^{2} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {polylog}\left (3, a \,x^{q}\right )}{\left (1+m \right )^{2}}-\frac {q \,x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {polylog}\left (4, a \,x^{q}\right )}{1+m}-\frac {q^{4} x^{1+m +q} a \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {1+m +q}{q}\right )}{\left (1+m \right )^{4}}\right )}{q}\)	\(217\)

input

int((d*x)^m*polylog(4,a*x^q),x,method=_RETURNVERBOSE)
 

output

-(d*x)^m*x^(-m)*(-a)^(-m/q-1/q)/q*(-q^4*x^(1+m)*(-a)^(m/q+1/q)/(1+m)^4*ln( 
1-a*x^q)-q^3*x^(1+m)*(-a)^(m/q+1/q)/(1+m)^3*polylog(2,a*x^q)+q^2*x^(1+m)*( 
-a)^(m/q+1/q)/(1+m)^2*polylog(3,a*x^q)-q*x^(1+m)*(-a)^(m/q+1/q)/(1+m)*poly 
log(4,a*x^q)-q^4*x^(1+m+q)*a*(-a)^(m/q+1/q)/(1+m)^4*LerchPhi(a*x^q,1,(1+m+ 
q)/q))
 

3.2.13.5 Fricas [F]

\[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_{4}(a x^{q}) \,d x } \]

input

integrate((d*x)^m*polylog(4,a*x^q),x, algorithm="fricas")
 

output

integral((d*x)^m*polylog(4, a*x^q), x)
 

3.2.13.6 Sympy [F]

\[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\int \left (d x\right )^{m} \operatorname {Li}_{4}\left (a x^{q}\right )\, dx \]

input

integrate((d*x)**m*polylog(4,a*x**q),x)
 

output

Integral((d*x)**m*polylog(4, a*x**q), x)
 

3.2.13.7 Maxima [F]

\[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_{4}(a x^{q}) \,d x } \]

input

integrate((d*x)^m*polylog(4,a*x^q),x, algorithm="maxima")
 

output

-d^m*q^4*integrate(-x^m/(m^4 + 4*m^3 + 6*m^2 - (a*m^4 + 4*a*m^3 + 6*a*m^2 
+ 4*a*m + a)*x^q + 4*m + 1), x) - (d^m*q^4*x*x^m - (d^m*m + d^m)*q^3*x*x^m 
*log(-a*x^q + 1) - (d^m*m^2 + 2*d^m*m + d^m)*q^2*x*x^m*dilog(a*x^q) + (d^m 
*m^3 + 3*d^m*m^2 + 3*d^m*m + d^m)*q*x*x^m*polylog(3, a*x^q) - (d^m*m^4 + 4 
*d^m*m^3 + 6*d^m*m^2 + 4*d^m*m + d^m)*x*x^m*polylog(4, a*x^q))/(m^5 + 5*m^ 
4 + 10*m^3 + 10*m^2 + 5*m + 1)
 

3.2.13.8 Giac [F]

\[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_{4}(a x^{q}) \,d x } \]

input

integrate((d*x)^m*polylog(4,a*x^q),x, algorithm="giac")
 

output

integrate((d*x)^m*polylog(4, a*x^q), x)
 

3.2.13.9 Mupad [F(-1)]

Timed out. \[ \int (d x)^m \operatorname {PolyLog}\left (4,a x^q\right ) \, dx=\int {\left (d\,x\right )}^m\,\mathrm {polylog}\left (4,a\,x^q\right ) \,d x \]

input

int((d*x)^m*polylog(4, a*x^q),x)
 

output

int((d*x)^m*polylog(4, a*x^q), x)