3.2.67 \(\int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx\) [167]

3.2.67.1 Optimal result

Integrand size = 16, antiderivative size = 191 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=-c^2 \log (x)+c^2 \log (1-c x)-\frac {c \log (1-c x)}{x}-\frac {1}{4} c^2 \log ^2(1-c x)+\frac {\log ^2(1-c x)}{4 x^2}+\frac {1}{2} c^2 \log (c x) \log ^2(1-c x)-\frac {1}{2} c^2 \operatorname {PolyLog}(2,c x)+\frac {c \operatorname {PolyLog}(2,c x)}{2 x}+\frac {1}{2} c^2 \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{2 x^2}+c^2 \log (1-c x) \operatorname {PolyLog}(2,1-c x)-\frac {1}{2} c^2 \operatorname {PolyLog}(3,c x)-c^2 \operatorname {PolyLog}(3,1-c x) \]

output

-c^2*ln(x)+c^2*ln(-c*x+1)-c*ln(-c*x+1)/x-1/4*c^2*ln(-c*x+1)^2+1/4*ln(-c*x+ 
1)^2/x^2+1/2*c^2*ln(c*x)*ln(-c*x+1)^2-1/2*c^2*polylog(2,c*x)+1/2*c*polylog 
(2,c*x)/x+1/2*c^2*ln(-c*x+1)*polylog(2,c*x)-1/2*ln(-c*x+1)*polylog(2,c*x)/ 
x^2+c^2*ln(-c*x+1)*polylog(2,-c*x+1)-1/2*c^2*polylog(3,c*x)-c^2*polylog(3, 
-c*x+1)
 

3.2.67.2 Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.97 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\frac {1}{4} \left (-2 c^2 \log (x)-2 c^2 \log (c x)+4 c^2 \log (1-c x)-\frac {4 c \log (1-c x)}{x}+2 c^2 \log (c x) \log (1-c x)-c^2 \log ^2(1-c x)+\frac {\log ^2(1-c x)}{x^2}+2 c^2 \log (c x) \log ^2(1-c x)+\frac {2 \left (c x+\left (-1+c^2 x^2\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)}{x^2}+2 c^2 (1+2 \log (1-c x)) \operatorname {PolyLog}(2,1-c x)-2 c^2 \operatorname {PolyLog}(3,c x)-4 c^2 \operatorname {PolyLog}(3,1-c x)\right ) \]

input

Integrate[(Log[1 - c*x]*PolyLog[2, c*x])/x^3,x]
 

output

(-2*c^2*Log[x] - 2*c^2*Log[c*x] + 4*c^2*Log[1 - c*x] - (4*c*Log[1 - c*x])/ 
x + 2*c^2*Log[c*x]*Log[1 - c*x] - c^2*Log[1 - c*x]^2 + Log[1 - c*x]^2/x^2 
+ 2*c^2*Log[c*x]*Log[1 - c*x]^2 + (2*(c*x + (-1 + c^2*x^2)*Log[1 - c*x])*P 
olyLog[2, c*x])/x^2 + 2*c^2*(1 + 2*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 2*c 
^2*PolyLog[3, c*x] - 4*c^2*PolyLog[3, 1 - c*x])/4
 

3.2.67.3 Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {7157, 2009, 2845, 2857, 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.

\(\displaystyle \int \frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{x^3} \, dx\)

\(\Big \downarrow \) 7157

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2845

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

\(\Big \downarrow \) 2857

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (c \left (-c \operatorname {PolyLog}(2,c x)-\frac {1}{2} c \log ^2(1-c x)+c \log (1-c x)-\frac {\log (1-c x)}{x}-c \log (x)\right )+\frac {\log ^2(1-c x)}{2 x^2}\right )-\frac {\operatorname {PolyLog}(2,c x) \log (1-c x)}{2 x^2}-\frac {1}{2} c \left (-\frac {\operatorname {PolyLog}(2,c x)}{x}+c \operatorname {PolyLog}(3,c x)+2 c \operatorname {PolyLog}(3,1-c x)-c \operatorname {PolyLog}(2,c x) \log (1-c x)-2 c \operatorname {PolyLog}(2,1-c x) \log (1-c x)-c \log (c x) \log ^2(1-c x)-c \log (1-c x)+\frac {\log (1-c x)}{x}+c \log (x)\right )\)

input

Int[(Log[1 - c*x]*PolyLog[2, c*x])/x^3,x]
 

output

-1/2*(Log[1 - c*x]*PolyLog[2, c*x])/x^2 + (Log[1 - c*x]^2/(2*x^2) + c*(-(c 
*Log[x]) + c*Log[1 - c*x] - Log[1 - c*x]/x - (c*Log[1 - c*x]^2)/2 - c*Poly 
Log[2, c*x]))/2 - (c*(c*Log[x] - c*Log[1 - c*x] + Log[1 - c*x]/x - c*Log[c 
*x]*Log[1 - c*x]^2 - PolyLog[2, c*x]/x - c*Log[1 - c*x]*PolyLog[2, c*x] - 
2*c*Log[1 - c*x]*PolyLog[2, 1 - c*x] + c*PolyLog[3, c*x] + 2*c*PolyLog[3, 
1 - c*x]))/2
 

3.2.67.3.1 Defintions of rubi rules used

rule 2009

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

rule 2845

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^ 
n])^p/(g*(q + 1))), x] - Simp[b*e*n*(p/(g*(q + 1)))   Int[(f + g*x)^(q + 1) 
*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, 
d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && In 
tegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))
 

rule 2857

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symb 
ol] :> Int[ExpandIntegrand[Log[c*(d + e*x)], x^m/(f + g*x), x], x] /; FreeQ 
[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m]
 

rule 7157

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(x_)^(m_.)*PolyLo 
g[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> Simp[x^(m + 1)*(g + h*Log[f* 
(d + e*x)^n])*(PolyLog[2, c*(a + b*x)]/(m + 1)), x] + (Simp[b/(m + 1)   Int 
[ExpandIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], x^(m + 1) 
/(a + b*x), x], x], x] - Simp[e*h*(n/(m + 1))   Int[ExpandIntegrand[PolyLog 
[2, c*(a + b*x)], x^(m + 1)/(d + e*x), x], x], x]) /; FreeQ[{a, b, c, d, e, 
 f, g, h, n}, x] && IntegerQ[m] && NeQ[m, -1]
 

3.2.67.4 Maple [F]

\[\int \frac {\ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )}{x^{3}}d x\]

input

int(ln(-c*x+1)*polylog(2,c*x)/x^3,x)
 

output

int(ln(-c*x+1)*polylog(2,c*x)/x^3,x)
 

3.2.67.5 Fricas [F]

\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{3}} \,d x } \]

input

integrate(log(-c*x+1)*polylog(2,c*x)/x^3,x, algorithm="fricas")
 

output

integral(dilog(c*x)*log(-c*x + 1)/x^3, x)
 

3.2.67.6 Sympy [F]

\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\int \frac {\log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{3}}\, dx \]

input

integrate(ln(-c*x+1)*polylog(2,c*x)/x**3,x)
 

output

Integral(log(-c*x + 1)*polylog(2, c*x)/x**3, x)
 

3.2.67.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.85 \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} c^{2} + \frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} c^{2} - c^{2} \log \left (x\right ) - \frac {1}{2} \, c^{2} {\rm Li}_{3}(c x) - \frac {{\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )^{2} - 2 \, {\left (c x + {\left (c^{2} x^{2} - 1\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 4 \, {\left (c^{2} x^{2} - c x\right )} \log \left (-c x + 1\right )}{4 \, x^{2}} \]

input

integrate(log(-c*x+1)*polylog(2,c*x)/x^3,x, algorithm="maxima")
 

output

1/2*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log(-c*x + 1) - 2*polylo 
g(3, -c*x + 1))*c^2 + 1/2*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*c^2 - 
 c^2*log(x) - 1/2*c^2*polylog(3, c*x) - 1/4*((c^2*x^2 - 1)*log(-c*x + 1)^2 
 - 2*(c*x + (c^2*x^2 - 1)*log(-c*x + 1))*dilog(c*x) - 4*(c^2*x^2 - c*x)*lo 
g(-c*x + 1))/x^2
 

3.2.67.8 Giac [F]

\[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\int { \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{3}} \,d x } \]

input

integrate(log(-c*x+1)*polylog(2,c*x)/x^3,x, algorithm="giac")
 

output

integrate(dilog(c*x)*log(-c*x + 1)/x^3, x)
 

3.2.67.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )}{x^3} \,d x \]

input

int((log(1 - c*x)*polylog(2, c*x))/x^3,x)
 

output

int((log(1 - c*x)*polylog(2, c*x))/x^3, x)