\(\int x^2 (a+b \log (c x^n)) \operatorname {PolyLog}(2,e x) \, dx\) [214]

Optimal result

Integrand size = 19, antiderivative size = 217 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {5 b n x}{27 e^2}+\frac {7 b n x^2}{108 e}+\frac {1}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n \log (1-e x)}{27 e^3}-\frac {2}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \operatorname {PolyLog}(2,e x)}{9 e^3}-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \] Output:

5/27*b*n*x/e^2+7/108*b*n*x^2/e+1/27*b*n*x^3-1/9*x*(a+b*ln(c*x^n))/e^2-1/18 
*x^2*(a+b*ln(c*x^n))/e-1/27*x^3*(a+b*ln(c*x^n))+2/27*b*n*ln(-e*x+1)/e^3-2/ 
27*b*n*x^3*ln(-e*x+1)-1/9*(a+b*ln(c*x^n))*ln(-e*x+1)/e^3+1/9*x^3*(a+b*ln(c 
*x^n))*ln(-e*x+1)-1/9*b*n*polylog(2,e*x)/e^3-1/9*b*n*x^3*polylog(2,e*x)+1/ 
3*x^3*(a+b*ln(c*x^n))*polylog(2,e*x)
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (-e x \left (6+3 e x+2 e^2 x^2\right )+6 \left (-1+e^3 x^3\right ) \log (1-e x)+18 e^3 x^3 \operatorname {PolyLog}(2,e x)\right )}{54 e^3}+\frac {b n \left (20 e x+7 e^2 x^2+4 e^3 x^3+8 \log (1-e x)-8 e^3 x^3 \log (1-e x)+2 \log (x) \left (-e x \left (6+3 e x+2 e^2 x^2\right )+6 \left (-1+e^3 x^3\right ) \log (1-e x)\right )+12 \left (-1-e^3 x^3+3 e^3 x^3 \log (x)\right ) \operatorname {PolyLog}(2,e x)\right )}{108 e^3} \] Input:

Integrate[x^2*(a + b*Log[c*x^n])*PolyLog[2, e*x],x]
 

Output:

((a - b*n*Log[x] + b*Log[c*x^n])*(-(e*x*(6 + 3*e*x + 2*e^2*x^2)) + 6*(-1 + 
 e^3*x^3)*Log[1 - e*x] + 18*e^3*x^3*PolyLog[2, e*x]))/(54*e^3) + (b*n*(20* 
e*x + 7*e^2*x^2 + 4*e^3*x^3 + 8*Log[1 - e*x] - 8*e^3*x^3*Log[1 - e*x] + 2* 
Log[x]*(-(e*x*(6 + 3*e*x + 2*e^2*x^2)) + 6*(-1 + e^3*x^3)*Log[1 - e*x]) + 
12*(-1 - e^3*x^3 + 3*e^3*x^3*Log[x])*PolyLog[2, e*x]))/(108*e^3)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2832, 25, 2823, 2009, 2842, 49, 2009}

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.

\(\displaystyle \int x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2832

\(\displaystyle -\frac {1}{3} \int -x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)dx+\frac {1}{9} b n \int -x^2 \log (1-e x)dx+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)dx-\frac {1}{9} b n \int x^2 \log (1-e x)dx+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 2823

\(\displaystyle \frac {1}{3} \left (-b n \int \left (\frac {1}{3} \log (1-e x) x^2-\frac {x^2}{9}-\frac {x}{6 e}-\frac {1}{3 e^2}-\frac {\log (1-e x)}{3 e^3 x}\right )dx-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{9} b n \int x^2 \log (1-e x)dx+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{9} b n \int x^2 \log (1-e x)dx+\frac {1}{3} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{3 e^3}-\frac {\log (1-e x)}{9 e^3}-\frac {4 x}{9 e^2}+\frac {1}{9} x^3 \log (1-e x)-\frac {5 x^2}{36 e}-\frac {2 x^3}{27}\right )\right )+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 2842

\(\displaystyle -\frac {1}{9} b n \left (\frac {1}{3} e \int \frac {x^3}{1-e x}dx+\frac {1}{3} x^3 \log (1-e x)\right )+\frac {1}{3} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{3 e^3}-\frac {\log (1-e x)}{9 e^3}-\frac {4 x}{9 e^2}+\frac {1}{9} x^3 \log (1-e x)-\frac {5 x^2}{36 e}-\frac {2 x^3}{27}\right )\right )+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 49

\(\displaystyle -\frac {1}{9} b n \left (\frac {1}{3} e \int \left (-\frac {x^2}{e}-\frac {x}{e^2}-\frac {1}{e^3 (e x-1)}-\frac {1}{e^3}\right )dx+\frac {1}{3} x^3 \log (1-e x)\right )+\frac {1}{3} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{3 e^3}-\frac {\log (1-e x)}{9 e^3}-\frac {4 x}{9 e^2}+\frac {1}{9} x^3 \log (1-e x)-\frac {5 x^2}{36 e}-\frac {2 x^3}{27}\right )\right )+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{3 e^3}-\frac {\log (1-e x)}{9 e^3}-\frac {4 x}{9 e^2}+\frac {1}{9} x^3 \log (1-e x)-\frac {5 x^2}{36 e}-\frac {2 x^3}{27}\right )\right )+\frac {1}{3} x^3 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n \left (\frac {1}{3} e \left (-\frac {\log (1-e x)}{e^4}-\frac {x}{e^3}-\frac {x^2}{2 e^2}-\frac {x^3}{3 e}\right )+\frac {1}{3} x^3 \log (1-e x)\right )-\frac {1}{9} b n x^3 \operatorname {PolyLog}(2,e x)\)

Input:

Int[x^2*(a + b*Log[c*x^n])*PolyLog[2, e*x],x]
 

Output:

-1/9*(b*n*((x^3*Log[1 - e*x])/3 + (e*(-(x/e^3) - x^2/(2*e^2) - x^3/(3*e) - 
 Log[1 - e*x]/e^4))/3)) - (b*n*x^3*PolyLog[2, e*x])/9 + (x^3*(a + b*Log[c* 
x^n])*PolyLog[2, e*x])/3 + (-1/3*(x*(a + b*Log[c*x^n]))/e^2 - (x^2*(a + b* 
Log[c*x^n]))/(6*e) - (x^3*(a + b*Log[c*x^n]))/9 - ((a + b*Log[c*x^n])*Log[ 
1 - e*x])/(3*e^3) + (x^3*(a + b*Log[c*x^n])*Log[1 - e*x])/3 - b*n*((-4*x)/ 
(9*e^2) - (5*x^2)/(36*e) - (2*x^3)/27 - Log[1 - e*x]/(9*e^3) + (x^3*Log[1 
- e*x])/9 + PolyLog[2, e*x]/(3*e^3)))/3
 

Defintions of rubi rules used

rule 25

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

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2823

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* 
(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])   u, x] - Simp[b*n   Int[1/x 
 u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q 
+ 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
 

rule 2832

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.)*PolyLog[k_, (e 
_.)*(x_)^(q_.)], x_Symbol] :> Simp[(-b)*n*(d*x)^(m + 1)*(PolyLog[k, e*x^q]/ 
(d*(m + 1)^2)), x] + (Simp[(d*x)^(m + 1)*PolyLog[k, e*x^q]*((a + b*Log[c*x^ 
n])/(d*(m + 1))), x] - Simp[q/(m + 1)   Int[(d*x)^m*PolyLog[k - 1, e*x^q]*( 
a + b*Log[c*x^n]), x], x] + Simp[b*n*(q/(m + 1)^2)   Int[(d*x)^m*PolyLog[k 
- 1, e*x^q], x], x]) /; FreeQ[{a, b, c, d, e, m, n, q}, x] && IGtQ[k, 0]
 

rule 2842

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ 
))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( 
g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1)))   Int[(f + g*x)^(q + 1)/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && 
NeQ[q, -1]
 

Maple [A] (verified)

Time = 21.11 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.24

method	result	size
parallelrisch	\(\frac {-12 a -6 a \,e^{2} x^{2}+20 b e n x -12 b \ln \left (c \,x^{n}\right )+20 n b -4 a \,e^{3} x^{3}+12 \ln \left (x \right ) n b +36 b \operatorname {polylog}\left (2, e x \right ) \ln \left (c \,x^{n}\right ) x^{3} e^{3}+12 b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) x^{3} e^{3}-12 x^{3} \operatorname {polylog}\left (2, e x \right ) b \,e^{3} n -8 x^{3} \ln \left (-e x +1\right ) b \,e^{3} n -12 \ln \left (-e x +1\right ) a +7 b \,e^{2} n \,x^{2}-12 a e x -12 b e x \ln \left (c \,x^{n}\right )-6 b \ln \left (c \,x^{n}\right ) x^{2} e^{2}-4 b \ln \left (c \,x^{n}\right ) x^{3} e^{3}+36 x^{3} \operatorname {polylog}\left (2, e x \right ) a \,e^{3}+12 x^{3} \ln \left (-e x +1\right ) a \,e^{3}+8 \ln \left (-e x +1\right ) b n -12 \operatorname {polylog}\left (2, e x \right ) b n -12 b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right )+4 b \,e^{3} n \,x^{3}}{108 e^{3}}\)	\(269\)

Input:

int(x^2*(a+b*ln(c*x^n))*polylog(2,e*x),x,method=_RETURNVERBOSE)
 

Output:

1/108*(-12*a-6*a*e^2*x^2+20*b*e*n*x-12*b*ln(c*x^n)+20*n*b-4*a*e^3*x^3+12*l 
n(x)*n*b+36*b*polylog(2,e*x)*ln(c*x^n)*x^3*e^3+12*b*ln(-e*x+1)*ln(c*x^n)*x 
^3*e^3-12*x^3*polylog(2,e*x)*b*e^3*n-8*x^3*ln(-e*x+1)*b*e^3*n-12*ln(-e*x+1 
)*a+7*b*e^2*n*x^2-12*a*e*x-12*b*e*x*ln(c*x^n)-6*b*ln(c*x^n)*x^2*e^2-4*b*ln 
(c*x^n)*x^3*e^3+36*x^3*polylog(2,e*x)*a*e^3+12*x^3*ln(-e*x+1)*a*e^3+8*ln(- 
e*x+1)*b*n-12*polylog(2,e*x)*b*n-12*b*ln(-e*x+1)*ln(c*x^n)+4*b*e^3*n*x^3)/ 
e^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.14 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {4 \, {\left (b e^{3} n - a e^{3}\right )} x^{3} + {\left (7 \, b e^{2} n - 6 \, a e^{2}\right )} x^{2} + 4 \, {\left (5 \, b e n - 3 \, a e\right )} x - 12 \, {\left ({\left (b e^{3} n - 3 \, a e^{3}\right )} x^{3} + b n\right )} {\rm Li}_2\left (e x\right ) - 4 \, {\left ({\left (2 \, b e^{3} n - 3 \, a e^{3}\right )} x^{3} - 2 \, b n + 3 \, a\right )} \log \left (-e x + 1\right ) + 2 \, {\left (18 \, b e^{3} x^{3} {\rm Li}_2\left (e x\right ) - 2 \, b e^{3} x^{3} - 3 \, b e^{2} x^{2} - 6 \, b e x + 6 \, {\left (b e^{3} x^{3} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + 2 \, {\left (18 \, b e^{3} n x^{3} {\rm Li}_2\left (e x\right ) - 2 \, b e^{3} n x^{3} - 3 \, b e^{2} n x^{2} - 6 \, b e n x + 6 \, {\left (b e^{3} n x^{3} - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{108 \, e^{3}} \] Input:

integrate(x^2*(a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/108*(4*(b*e^3*n - a*e^3)*x^3 + (7*b*e^2*n - 6*a*e^2)*x^2 + 4*(5*b*e*n - 
3*a*e)*x - 12*((b*e^3*n - 3*a*e^3)*x^3 + b*n)*dilog(e*x) - 4*((2*b*e^3*n - 
 3*a*e^3)*x^3 - 2*b*n + 3*a)*log(-e*x + 1) + 2*(18*b*e^3*x^3*dilog(e*x) - 
2*b*e^3*x^3 - 3*b*e^2*x^2 - 6*b*e*x + 6*(b*e^3*x^3 - b)*log(-e*x + 1))*log 
(c) + 2*(18*b*e^3*n*x^3*dilog(e*x) - 2*b*e^3*n*x^3 - 3*b*e^2*n*x^2 - 6*b*e 
*n*x + 6*(b*e^3*n*x^3 - b*n)*log(-e*x + 1))*log(x))/e^3
 

Sympy [A] (verification not implemented)

Time = 59.21 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.15 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\begin {cases} - \frac {a x^{3} \operatorname {Li}_{1}\left (e x\right )}{9} + \frac {a x^{3} \operatorname {Li}_{2}\left (e x\right )}{3} - \frac {a x^{3}}{27} - \frac {a x^{2}}{18 e} - \frac {a x}{9 e^{2}} + \frac {a \operatorname {Li}_{1}\left (e x\right )}{9 e^{3}} + \frac {2 b n x^{3} \operatorname {Li}_{1}\left (e x\right )}{27} - \frac {b n x^{3} \operatorname {Li}_{2}\left (e x\right )}{9} + \frac {b n x^{3}}{27} - \frac {b x^{3} \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{9} + \frac {b x^{3} \log {\left (c x^{n} \right )} \operatorname {Li}_{2}\left (e x\right )}{3} - \frac {b x^{3} \log {\left (c x^{n} \right )}}{27} + \frac {7 b n x^{2}}{108 e} - \frac {b x^{2} \log {\left (c x^{n} \right )}}{18 e} + \frac {5 b n x}{27 e^{2}} - \frac {b x \log {\left (c x^{n} \right )}}{9 e^{2}} - \frac {2 b n \operatorname {Li}_{1}\left (e x\right )}{27 e^{3}} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{9 e^{3}} + \frac {b \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{9 e^{3}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

integrate(x**2*(a+b*ln(c*x**n))*polylog(2,e*x),x)
 

Output:

Piecewise((-a*x**3*polylog(1, e*x)/9 + a*x**3*polylog(2, e*x)/3 - a*x**3/2 
7 - a*x**2/(18*e) - a*x/(9*e**2) + a*polylog(1, e*x)/(9*e**3) + 2*b*n*x**3 
*polylog(1, e*x)/27 - b*n*x**3*polylog(2, e*x)/9 + b*n*x**3/27 - b*x**3*lo 
g(c*x**n)*polylog(1, e*x)/9 + b*x**3*log(c*x**n)*polylog(2, e*x)/3 - b*x** 
3*log(c*x**n)/27 + 7*b*n*x**2/(108*e) - b*x**2*log(c*x**n)/(18*e) + 5*b*n* 
x/(27*e**2) - b*x*log(c*x**n)/(9*e**2) - 2*b*n*polylog(1, e*x)/(27*e**3) - 
 b*n*polylog(2, e*x)/(9*e**3) + b*log(c*x**n)*polylog(1, e*x)/(9*e**3), Ne 
(e, 0)), (0, True))
 

Maxima [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} {\rm Li}_2\left (e x\right ) \,d x } \] Input:

integrate(x^2*(a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="maxima")
 

Output:

1/54*b*((6*(3*e^3*x^3*log(x^n) - (e^3*n - 3*e^3*log(c))*x^3)*dilog(e*x) - 
2*((2*e^3*n - 3*e^3*log(c))*x^3 - 3*n*log(x))*log(-e*x + 1) - (2*e^3*x^3 + 
 3*e^2*x^2 + 6*e*x - 6*(e^3*x^3 - 1)*log(-e*x + 1))*log(x^n))/e^3 - 54*int 
egrate(-1/54*(e^2*n*x^2 + 6*(e^3*n - e^3*log(c))*x^3 + 3*e*n*x - 6*n*log(x 
) - 6*n)/(e^3*x - e^2), x)) + 1/54*(18*e^3*x^3*dilog(e*x) - 2*e^3*x^3 - 3* 
e^2*x^2 - 6*e*x + 6*(e^3*x^3 - 1)*log(-e*x + 1))*a/e^3
 

Giac [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} {\rm Li}_2\left (e x\right ) \,d x } \] Input:

integrate(x^2*(a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*x^2*dilog(e*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int x^2\,\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:

int(x^2*polylog(2, e*x)*(a + b*log(c*x^n)),x)
 

Output:

int(x^2*polylog(2, e*x)*(a + b*log(c*x^n)), x)
 

Reduce [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\left (\int \mathrm {log}\left (x^{n} c \right ) \mathit {polylog}\left (2, e x \right ) x^{2}d x \right ) b +\left (\int \mathit {polylog}\left (2, e x \right ) x^{2}d x \right ) a \] Input:

int(x^2*(a+b*log(c*x^n))*polylog(2,e*x),x)
 

Output:

int(log(x**n*c)*polylog(2,e*x)*x**2,x)*b + int(polylog(2,e*x)*x**2,x)*a