3.2.36 \(\int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx\) [136]

3.2.36.1 Optimal result

Integrand size = 13, antiderivative size = 629 \[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=-\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac {b^2 \log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac {b^2 \left (\log \left (1+\frac {b x}{a}\right )+\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac {b^2 \left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )^2}{4 a^2}-\frac {b^2 \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{2 a^2}-\frac {b^2 \operatorname {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b \operatorname {PolyLog}(2,c (a+b x))}{2 a x}-\frac {b^2 \log (x) \operatorname {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \operatorname {PolyLog}\left (2,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \operatorname {PolyLog}\left (2,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac {b^2 \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \operatorname {PolyLog}(2,1-c (a+b x))}{2 a^2}+\frac {b^2 \operatorname {PolyLog}\left (3,-\frac {b x}{a}\right )}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \operatorname {PolyLog}(3,c (a+b x))}{2 a^2}-\frac {b^2 \operatorname {PolyLog}\left (3,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \operatorname {PolyLog}\left (3,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \operatorname {PolyLog}(3,1-c (a+b x))}{2 a^2} \]

output

-1/2*b^2*ln(b*c*x/(-a*c+1))*ln(-b*c*x-a*c+1)/a^2-1/2*b^2*ln(x)*ln(1+b*x/a) 
*ln(1-c*(b*x+a))/a^2-1/4*b^2*(ln(1+b*x/a)+ln((-a*c+1)/(1-c*(b*x+a)))-ln((- 
a*c+1)*(b*x+a)/a/(1-c*(b*x+a))))*ln(-a*(1-c*(b*x+a))/b/x)^2/a^2-1/4*b^2*(l 
n(c*(b*x+a))-ln(1+b*x/a))*(ln(x)+ln(-a*(1-c*(b*x+a))/b/x))^2/a^2-1/2*b^2*( 
ln(1-c*(b*x+a))-ln(-a*(1-c*(b*x+a))/b/x))*polylog(2,-b*x/a)/a^2-1/2*b^2*po 
lylog(2,c*(b*x+a))/a^2-1/2*b*polylog(2,c*(b*x+a))/a/x-1/2*b^2*ln(x)*polylo 
g(2,c*(b*x+a))/a^2-1/2*b^2*polylog(2,1-b*c*x/(-a*c+1))/a^2-1/2*b^2*ln(-a*( 
1-c*(b*x+a))/b/x)*polylog(2,-b*x/a/(1-c*(b*x+a)))/a^2+1/2*b^2*ln(-a*(1-c*( 
b*x+a))/b/x)*polylog(2,-b*c*x/(1-c*(b*x+a)))/a^2-1/2*b^2*(ln(x)+ln(-a*(1-c 
*(b*x+a))/b/x))*polylog(2,1-c*(b*x+a))/a^2+1/2*b^2*polylog(3,-b*x/a)/a^2+1 
/2*(b^2-a^2/x^2)*polylog(3,c*(b*x+a))/a^2-1/2*b^2*polylog(3,-b*x/a/(1-c*(b 
*x+a)))/a^2+1/2*b^2*polylog(3,-b*c*x/(1-c*(b*x+a)))/a^2+1/2*b^2*polylog(3, 
1-c*(b*x+a))/a^2
 

3.2.36.2 Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 573, normalized size of antiderivative = 0.91 \[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\frac {-\operatorname {PolyLog}(3,c (a+b x))+\frac {b x \left (-((a+b x \log (x)-b x \log (a+b x)) \operatorname {PolyLog}(2,c (a+b x)))+b x \left (\log (c (a+b x)) \log (1-a c-b c x)-\log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-a c-b c x)+\frac {1}{2} \left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \log (1-a c-b c x) (-2 \log (x)+\log (1-a c-b c x))-\left (\log (c (a+b x))-\log \left (1+\frac {b x}{a}\right )\right ) \log (1-a c-b c x) \log \left (\frac {a (-1+a c+b c x)}{b x}\right )-\frac {1}{2} \left (\log \left (\frac {1-a c}{b c x}\right )-\log \left (\frac {(1-a c) (a+b x)}{b x}\right )+\log \left (1+\frac {b x}{a}\right )\right ) \log ^2\left (\frac {a (-1+a c+b c x)}{b x}\right )-\log (x) \left (\log (1-a c-b c x)-\log \left (1+\frac {b c x}{-1+a c}\right )\right )-\left (\log (1-a c-b c x)-\log \left (\frac {a (-1+a c+b c x)}{b x}\right )\right ) \operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )+\operatorname {PolyLog}\left (2,\frac {b c x}{1-a c}\right )-\log (a+b x) \operatorname {PolyLog}(2,c (a+b x))+\operatorname {PolyLog}(2,1-a c-b c x)-\left (\log (x)+\log \left (\frac {a (-1+a c+b c x)}{b x}\right )\right ) \operatorname {PolyLog}(2,1-a c-b c x)+\log \left (\frac {a (-1+a c+b c x)}{b x}\right ) \left (\operatorname {PolyLog}\left (2,\frac {a (-1+a c+b c x)}{b x}\right )-\operatorname {PolyLog}\left (2,\frac {-1+a c+b c x}{b c x}\right )\right )+\operatorname {PolyLog}\left (3,-\frac {b x}{a}\right )+\operatorname {PolyLog}(3,c (a+b x))+\operatorname {PolyLog}(3,1-a c-b c x)-\operatorname {PolyLog}\left (3,\frac {a (-1+a c+b c x)}{b x}\right )+\operatorname {PolyLog}\left (3,\frac {-1+a c+b c x}{b c x}\right )\right )\right )}{a^2}}{2 x^2} \]

input

Integrate[PolyLog[3, c*(a + b*x)]/x^3,x]
 

output

(-PolyLog[3, c*(a + b*x)] + (b*x*(-((a + b*x*Log[x] - b*x*Log[a + b*x])*Po 
lyLog[2, c*(a + b*x)]) + b*x*(Log[c*(a + b*x)]*Log[1 - a*c - b*c*x] - Log[ 
x]*Log[1 + (b*x)/a]*Log[1 - a*c - b*c*x] + ((Log[c*(a + b*x)] - Log[1 + (b 
*x)/a])*Log[1 - a*c - b*c*x]*(-2*Log[x] + Log[1 - a*c - b*c*x]))/2 - (Log[ 
c*(a + b*x)] - Log[1 + (b*x)/a])*Log[1 - a*c - b*c*x]*Log[(a*(-1 + a*c + b 
*c*x))/(b*x)] - ((Log[(1 - a*c)/(b*c*x)] - Log[((1 - a*c)*(a + b*x))/(b*x) 
] + Log[1 + (b*x)/a])*Log[(a*(-1 + a*c + b*c*x))/(b*x)]^2)/2 - Log[x]*(Log 
[1 - a*c - b*c*x] - Log[1 + (b*c*x)/(-1 + a*c)]) - (Log[1 - a*c - b*c*x] - 
 Log[(a*(-1 + a*c + b*c*x))/(b*x)])*PolyLog[2, -((b*x)/a)] + PolyLog[2, (b 
*c*x)/(1 - a*c)] - Log[a + b*x]*PolyLog[2, c*(a + b*x)] + PolyLog[2, 1 - a 
*c - b*c*x] - (Log[x] + Log[(a*(-1 + a*c + b*c*x))/(b*x)])*PolyLog[2, 1 - 
a*c - b*c*x] + Log[(a*(-1 + a*c + b*c*x))/(b*x)]*(PolyLog[2, (a*(-1 + a*c 
+ b*c*x))/(b*x)] - PolyLog[2, (-1 + a*c + b*c*x)/(b*c*x)]) + PolyLog[3, -( 
(b*x)/a)] + PolyLog[3, c*(a + b*x)] + PolyLog[3, 1 - a*c - b*c*x] - PolyLo 
g[3, (a*(-1 + a*c + b*c*x))/(b*x)] + PolyLog[3, (-1 + a*c + b*c*x)/(b*c*x) 
])))/a^2)/(2*x^2)
 

3.2.36.3 Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 601, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {7153, 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.

\(\displaystyle \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx\)

\(\Big \downarrow \) 7153

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

\(\Big \downarrow \) 2009

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

input

Int[PolyLog[3, c*(a + b*x)]/x^3,x]
 

output

((b^2 - a^2/x^2)*PolyLog[3, c*(a + b*x)])/(2*a^2) - (b^3*((Log[(b*c*x)/(1 
- a*c)]*Log[1 - a*c - b*c*x])/(a^2*b) + (Log[x]*Log[1 + (b*x)/a]*Log[1 - c 
*(a + b*x)])/(a^2*b) + ((Log[1 + (b*x)/a] + Log[(1 - a*c)/(1 - c*(a + b*x) 
)] - Log[((1 - a*c)*(a + b*x))/(a*(1 - c*(a + b*x)))])*Log[-((a*(1 - c*(a 
+ b*x)))/(b*x))]^2)/(2*a^2*b) + ((Log[c*(a + b*x)] - Log[1 + (b*x)/a])*(Lo 
g[x] + Log[-((a*(1 - c*(a + b*x)))/(b*x))])^2)/(2*a^2*b) + ((Log[1 - c*(a 
+ b*x)] - Log[-((a*(1 - c*(a + b*x)))/(b*x))])*PolyLog[2, -((b*x)/a)])/(a^ 
2*b) + PolyLog[2, c*(a + b*x)]/(a^2*b) + PolyLog[2, c*(a + b*x)]/(a*b^2*x) 
 + (Log[x]*PolyLog[2, c*(a + b*x)])/(a^2*b) + PolyLog[2, 1 - (b*c*x)/(1 - 
a*c)]/(a^2*b) + (Log[-((a*(1 - c*(a + b*x)))/(b*x))]*PolyLog[2, -((b*x)/(a 
*(1 - c*(a + b*x))))])/(a^2*b) - (Log[-((a*(1 - c*(a + b*x)))/(b*x))]*Poly 
Log[2, -((b*c*x)/(1 - c*(a + b*x)))])/(a^2*b) + ((Log[x] + Log[-((a*(1 - c 
*(a + b*x)))/(b*x))])*PolyLog[2, 1 - c*(a + b*x)])/(a^2*b) - PolyLog[3, -( 
(b*x)/a)]/(a^2*b) + PolyLog[3, -((b*x)/(a*(1 - c*(a + b*x))))]/(a^2*b) - P 
olyLog[3, -((b*c*x)/(1 - c*(a + b*x)))]/(a^2*b) - PolyLog[3, 1 - c*(a + b* 
x)]/(a^2*b)))/2
 

3.2.36.3.1 Defintions of rubi rules used

rule 2009

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

rule 7153

Int[(x_)^(m_.)*PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)], x_Symbol] :> 
Simp[(-(a^(m + 1) - b^(m + 1)*x^(m + 1)))*(PolyLog[n, c*(a + b*x)^p]/((m + 
1)*b^(m + 1))), x] + Simp[p/((m + 1)*b^m)   Int[ExpandIntegrand[PolyLog[n - 
 1, c*(a + b*x)^p], (a^(m + 1) - b^(m + 1)*x^(m + 1))/(a + b*x), x], x], x] 
 /; FreeQ[{a, b, c, p}, x] && GtQ[n, 0] && IntegerQ[m] && NeQ[m, -1]
 

3.2.36.4 Maple [F]

\[\int \frac {\operatorname {polylog}\left (3, c \left (b x +a \right )\right )}{x^{3}}d x\]

input

int(polylog(3,c*(b*x+a))/x^3,x)
 

output

int(polylog(3,c*(b*x+a))/x^3,x)
 

3.2.36.5 Fricas [F]

\[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\int { \frac {{\rm Li}_{3}({\left (b x + a\right )} c)}{x^{3}} \,d x } \]

input

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="fricas")
 

output

integral(polylog(3, b*c*x + a*c)/x^3, x)
 

3.2.36.6 Sympy [F]

\[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\int \frac {\operatorname {Li}_{3}\left (a c + b c x\right )}{x^{3}}\, dx \]

input

integrate(polylog(3,c*(b*x+a))/x**3,x)
 

output

Integral(polylog(3, a*c + b*c*x)/x**3, x)
 

3.2.36.7 Maxima [F]

\[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\int { \frac {{\rm Li}_{3}({\left (b x + a\right )} c)}{x^{3}} \,d x } \]

input

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="maxima")
 

output

b*integrate(1/2*dilog(b*c*x + a*c)/(b*x^3 + a*x^2), x) - 1/2*polylog(3, b* 
c*x + a*c)/x^2
 

3.2.36.8 Giac [F]

\[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\int { \frac {{\rm Li}_{3}({\left (b x + a\right )} c)}{x^{3}} \,d x } \]

input

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="giac")
 

output

integrate(polylog(3, (b*x + a)*c)/x^3, x)
 

3.2.36.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {PolyLog}(3,c (a+b x))}{x^3} \, dx=\int \frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )}{x^3} \,d x \]

input

int(polylog(3, c*(a + b*x))/x^3,x)
 

output

int(polylog(3, c*(a + b*x))/x^3, x)