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

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))
 

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
 

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)]
 

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]
 

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)
 

Fricas [A] (verification not implemented)

Time = 0.10 (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)
 

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)
 

Maxima [A] (verification not implemented)

Time = 0.04 (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)
 

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)
 

Mupad [B] (verification not implemented)

Time = 1.26 (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
 

Reduce [F]

\[ \int \operatorname {PolyLog}(3,c (a+b x)) \, dx=\int \mathit {polylog}\left (3, b c x +a c \right )d x \] Input:

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

Output:

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