\(\int x \operatorname {PolyLog}(2,c (a+b x)) \, dx\) [125]

Optimal result

Integrand size = 11, antiderivative size = 152 \[ \int x \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {a x}{2 b}-\frac {(1-a c) x}{4 b c}-\frac {x^2}{8}-\frac {(1-a c)^2 \log (1-a c-b c x)}{4 b^2 c^2}+\frac {1}{4} x^2 \log (1-a c-b c x)+\frac {a (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}-\frac {a^2 \operatorname {PolyLog}(2,c (a+b x))}{2 b^2}+\frac {1}{2} x^2 \operatorname {PolyLog}(2,c (a+b x)) \] Output:

1/2*a*x/b-1/4*(-a*c+1)*x/b/c-1/8*x^2-1/4*(-a*c+1)^2*ln(-b*c*x-a*c+1)/b^2/c 
^2+1/4*x^2*ln(-b*c*x-a*c+1)+1/2*a*(-b*c*x-a*c+1)*ln(-b*c*x-a*c+1)/b^2/c-1/ 
2*a^2*polylog(2,c*(b*x+a))/b^2+1/2*x^2*polylog(2,c*(b*x+a))
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.63 \[ \int x \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {-b c x (2-6 a c+b c x)+\left (-2-6 a^2 c^2+2 b^2 c^2 x^2-4 a c (-2+b c x)\right ) \log (1-a c-b c x)-4 c^2 \left (a^2-b^2 x^2\right ) \operatorname {PolyLog}(2,c (a+b x))}{8 b^2 c^2} \] Input:

Integrate[x*PolyLog[2, c*(a + b*x)],x]
 

Output:

(-(b*c*x*(2 - 6*a*c + b*c*x)) + (-2 - 6*a^2*c^2 + 2*b^2*c^2*x^2 - 4*a*c*(- 
2 + b*c*x))*Log[1 - a*c - b*c*x] - 4*c^2*(a^2 - b^2*x^2)*PolyLog[2, c*(a + 
 b*x)])/(8*b^2*c^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {7152, 2863, 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.

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

\(\Big \downarrow \) 7152

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} b \left (-\frac {a^2 \operatorname {PolyLog}(2,c (a+b x))}{b^3}-\frac {(1-a c)^2 \log (-a c-b c x+1)}{2 b^3 c^2}+\frac {a (-a c-b c x+1) \log (-a c-b c x+1)}{b^3 c}-\frac {x (1-a c)}{2 b^2 c}+\frac {a x}{b^2}+\frac {x^2 \log (-a c-b c x+1)}{2 b}-\frac {x^2}{4 b}\right )+\frac {1}{2} x^2 \operatorname {PolyLog}(2,c (a+b x))\)

Input:

Int[x*PolyLog[2, c*(a + b*x)],x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

rule 7152

Int[((d_.) + (e_.)*(x_))^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Sy 
mbol] :> Simp[(d + e*x)^(m + 1)*(PolyLog[2, c*(a + b*x)]/(e*(m + 1))), x] + 
 Simp[b/(e*(m + 1))   Int[(d + e*x)^(m + 1)*(Log[1 - a*c - b*c*x]/(a + b*x) 
), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08

method	result	size
parts	\(\frac {x^{2} \operatorname {polylog}\left (2, c \left (b x +a \right )\right )}{2}-\frac {-2 a \,c^{2} \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )-c \left (\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{2}-\frac {\left (-b c x -a c +1\right )^{2}}{4}\right )+c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+a^{2} c^{3} \operatorname {dilog}\left (-b c x -a c +1\right )}{2 b^{2} c^{3}}\)	\(164\)
derivativedivides	\(\frac {-\operatorname {polylog}\left (2, b c x +a c \right ) a c \left (b c x +a c \right )+\frac {\operatorname {polylog}\left (2, b c x +a c \right ) \left (b c x +a c \right )^{2}}{2}+a c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{4}-\frac {\left (-b c x -a c +1\right )^{2}}{8}-\frac {\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )}{2}+\frac {1}{2}-\frac {b c x}{2}-\frac {a c}{2}}{c^{2} b^{2}}\)	\(166\)
default	\(\frac {-\operatorname {polylog}\left (2, b c x +a c \right ) a c \left (b c x +a c \right )+\frac {\operatorname {polylog}\left (2, b c x +a c \right ) \left (b c x +a c \right )^{2}}{2}+a c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{4}-\frac {\left (-b c x -a c +1\right )^{2}}{8}-\frac {\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )}{2}+\frac {1}{2}-\frac {b c x}{2}-\frac {a c}{2}}{c^{2} b^{2}}\)	\(166\)
parallelrisch	\(\frac {4 x^{2} \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) b^{2} c^{2}+2 x^{2} \ln \left (1-c \left (b x +a \right )\right ) b^{2} c^{2}-2-x^{2} b^{2} c^{2}-4 x \ln \left (1-c \left (b x +a \right )\right ) a b \,c^{2}+6 a b \,c^{2} x -4 \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a^{2} c^{2}-6 \ln \left (1-c \left (b x +a \right )\right ) a^{2} c^{2}-11 a^{2} c^{2}-2 b c x +8 \ln \left (1-c \left (b x +a \right )\right ) a c +9 a c -2 \ln \left (1-c \left (b x +a \right )\right )}{8 b^{2} c^{2}}\)	\(172\)

Input:

int(x*polylog(2,c*(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.72 \[ \int x \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {b^{2} c^{2} x^{2} - 2 \, {\left (3 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \] Input:

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

Output:

-1/8*(b^2*c^2*x^2 - 2*(3*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 - a^2*c^2)*dilo 
g(b*c*x + a*c) - 2*(b^2*c^2*x^2 - 2*a*b*c^2*x - 3*a^2*c^2 + 4*a*c - 1)*log 
(-b*c*x - a*c + 1))/(b^2*c^2)
 

Sympy [A] (verification not implemented)

Time = 2.16 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01 \[ \int x \operatorname {PolyLog}(2,c (a+b x)) \, dx=\begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\\frac {x^{2} \operatorname {Li}_{2}\left (a c\right )}{2} & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {3 a^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2}} - \frac {a^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2 b^{2}} + \frac {a x \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b} + \frac {3 a x}{4 b} - \frac {a \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{2} c} - \frac {x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4} + \frac {x^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2} - \frac {x^{2}}{8} - \frac {x}{4 b c} + \frac {\operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2} c^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((0, Eq(b, 0) & Eq(c, 0)), (x**2*polylog(2, a*c)/2, Eq(b, 0)), (0 
, Eq(c, 0)), (3*a**2*polylog(1, a*c + b*c*x)/(4*b**2) - a**2*polylog(2, a* 
c + b*c*x)/(2*b**2) + a*x*polylog(1, a*c + b*c*x)/(2*b) + 3*a*x/(4*b) - a* 
polylog(1, a*c + b*c*x)/(b**2*c) - x**2*polylog(1, a*c + b*c*x)/4 + x**2*p 
olylog(2, a*c + b*c*x)/2 - x**2/8 - x/(4*b*c) + polylog(1, a*c + b*c*x)/(4 
*b**2*c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int x \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{2}}{2 \, b^{2}} + \frac {4 \, b^{2} c^{2} x^{2} {\rm Li}_2\left (b c x + a c\right ) - b^{2} c^{2} x^{2} + 2 \, {\left (3 \, a b c^{2} - b c\right )} x + 2 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \] Input:

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

Output:

1/2*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^2 
/b^2 + 1/8*(4*b^2*c^2*x^2*dilog(b*c*x + a*c) - b^2*c^2*x^2 + 2*(3*a*b*c^2 
- b*c)*x + 2*(b^2*c^2*x^2 - 2*a*b*c^2*x - 3*a^2*c^2 + 4*a*c - 1)*log(-b*c* 
x - a*c + 1))/(b^2*c^2)
 

Giac [F]

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

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

Output:

integrate(x*dilog((b*x + a)*c), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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