3.2.33 \(\int \operatorname {PolyLog}(3,c (a+b x)) \, dx\) [133]

3.2.33.1 Optimal result

Integrand size = 9, antiderivative size = 84 \[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=x+\frac {(1-a c-b c x) \log (1-a c-b c x)}{b c}-\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}-x \operatorname {PolyLog}(2,c (a+b x))+\frac {a \operatorname {PolyLog}(3,c (a+b x))}{b}+x \operatorname {PolyLog}(3,c (a+b x)) \]

output

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

3.2.33.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=\frac {(a+b x) \left (1-\log (1-c (a+b x))+\frac {\log (1-c (a+b x))}{c (a+b x)}-\operatorname {PolyLog}(2,c (a+b x))+\operatorname {PolyLog}(3,c (a+b x))\right )}{b} \]

input

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

output

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

3.2.33.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.111, Rules used = {7149, 7143, 7149, 25, 2868, 2840, 2838, 2894, 2836, 2732}

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.

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

\(\Big \downarrow \) 7149

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7149

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2868

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

\(\Big \downarrow \) 2840

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 2894

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

\(\Big \downarrow \) 2836

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

\(\Big \downarrow \) 2732

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

input

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

output

(-1 + a*c + b*c*x + (1 - a*c - b*c*x)*Log[1 - a*c - b*c*x])/(b*c) - (a*Pol 
yLog[2, c*(a + b*x)])/b - x*PolyLog[2, c*(a + b*x)] + (a*PolyLog[3, c*(a + 
 b*x)])/b + x*PolyLog[3, c*(a + b*x)]
 

3.2.33.3.1 Defintions of rubi rules used

rule 25

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

rule 2732

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x 
] /; FreeQ[{c, n}, x]
 

rule 2836

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : 
> Simp[1/e   Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ 
a, b, c, d, e, n, p}, x]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 2840

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

rule 2868

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> In 
t[ExpandToSum[u, x]^q*(a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, 
 b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(BinomialMatch 
Q[u, x] && LinearMatchQ[v, x])
 

rule 2894

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*( 
a + b*Log[c*ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p}, x] && Lin 
earQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f 
_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7149

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

3.2.33.4 Maple [F]

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

input

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

output

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

3.2.33.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=\frac {b c x - {\left (b c x + a c\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right ) + {\left (b c x + a c\right )} {\rm polylog}\left (3, b c x + a c\right )}{b c} \]

input

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

output

(b*c*x - (b*c*x + a*c)*dilog(b*c*x + a*c) - (b*c*x + a*c - 1)*log(-b*c*x - 
 a*c + 1) + (b*c*x + a*c)*polylog(3, b*c*x + a*c))/(b*c)
 

3.2.33.6 Sympy [F]

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

input

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

output

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

3.2.33.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.43 \[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a}{b} + \frac {a {\rm Li}_{3}(b c x + a c)}{b} - \frac {b c x {\rm Li}_2\left (b c x + a c\right ) - b c x {\rm Li}_{3}(b c x + a c) - b c x + {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \]

input

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

output

(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a/b + a 
*polylog(3, b*c*x + a*c)/b - (b*c*x*dilog(b*c*x + a*c) - b*c*x*polylog(3, 
b*c*x + a*c) - b*c*x + (b*c*x + a*c - 1)*log(-b*c*x - a*c + 1))/(b*c)
 

3.2.33.8 Giac [F]

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

input

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

output

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

3.2.33.9 Mupad [B] (verification not implemented)

Time = 6.69 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=x-\frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}+\frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}+\frac {\ln \left (c\,\left (a+b\,x\right )-1\right )}{b\,c}-\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b} \]

input

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

output

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