3.2.89 \(\int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx\) [189]

3.2.89.1 Optimal result

Integrand size = 21, antiderivative size = 331 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=-a c^2 \log (x)+a c^2 \log (1-c x)-\frac {a c \log (1-c x)}{x}-\frac {1}{4} a c^2 \log ^2(1-c x)+\frac {a \log ^2(1-c x)}{4 x^2}+\frac {b (1-c x) \log ^2(1-c x)}{x}-\frac {b^2 \log (c x) \log ^2(1-c x)}{2 a}+\frac {(b+a c)^2 \log (c x) \log ^2(1-c x)}{2 a}-2 b c \operatorname {PolyLog}(2,c x)-\frac {1}{2} a c^2 \operatorname {PolyLog}(2,c x)+\frac {a c \operatorname {PolyLog}(2,c x)}{2 x}+\frac {(b+a c)^2 \log (1-c x) \operatorname {PolyLog}(2,c x)}{2 a}-\frac {(a+b x)^2 \log (1-c x) \operatorname {PolyLog}(2,c x)}{2 a x^2}-\frac {b^2 \log (1-c x) \operatorname {PolyLog}(2,1-c x)}{a}+\frac {(b+a c)^2 \log (1-c x) \operatorname {PolyLog}(2,1-c x)}{a}-\frac {1}{2} c (2 b+a c) \operatorname {PolyLog}(3,c x)+\frac {b^2 \operatorname {PolyLog}(3,1-c x)}{a}-\frac {(b+a c)^2 \operatorname {PolyLog}(3,1-c x)}{a} \]

output

-a*c^2*ln(x)+a*c^2*ln(-c*x+1)-a*c*ln(-c*x+1)/x-1/4*a*c^2*ln(-c*x+1)^2+1/4* 
a*ln(-c*x+1)^2/x^2+b*(-c*x+1)*ln(-c*x+1)^2/x-1/2*b^2*ln(c*x)*ln(-c*x+1)^2/ 
a+1/2*(a*c+b)^2*ln(c*x)*ln(-c*x+1)^2/a-2*b*c*polylog(2,c*x)-1/2*a*c^2*poly 
log(2,c*x)+1/2*a*c*polylog(2,c*x)/x+1/2*(a*c+b)^2*ln(-c*x+1)*polylog(2,c*x 
)/a-1/2*(b*x+a)^2*ln(-c*x+1)*polylog(2,c*x)/a/x^2-b^2*ln(-c*x+1)*polylog(2 
,-c*x+1)/a+(a*c+b)^2*ln(-c*x+1)*polylog(2,-c*x+1)/a-1/2*c*(a*c+2*b)*polylo 
g(3,c*x)+b^2*polylog(3,-c*x+1)/a-(a*c+b)^2*polylog(3,-c*x+1)/a
 

3.2.89.2 Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=\frac {1}{4} \left (-4 a c^2 \log (c x)+4 a c^2 \log (1-c x)-\frac {4 a c \log (1-c x)}{x}+8 b c \log (c x) \log (1-c x)+2 a c^2 \log (c x) \log (1-c x)-4 b c \log ^2(1-c x)-a c^2 \log ^2(1-c x)+\frac {a \log ^2(1-c x)}{x^2}+\frac {4 b \log ^2(1-c x)}{x}+4 b c \log (c x) \log ^2(1-c x)+2 a c^2 \log (c x) \log ^2(1-c x)+\frac {2 (a c x+(-1+c x) (a+2 b x+a c x) \log (1-c x)) \operatorname {PolyLog}(2,c x)}{x^2}+2 c (4 b+a c+2 (2 b+a c) \log (1-c x)) \operatorname {PolyLog}(2,1-c x)-4 b c \operatorname {PolyLog}(3,c x)-2 a c^2 \operatorname {PolyLog}(3,c x)-8 b c \operatorname {PolyLog}(3,1-c x)-4 a c^2 \operatorname {PolyLog}(3,1-c x)\right ) \]

input

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

output

(-4*a*c^2*Log[c*x] + 4*a*c^2*Log[1 - c*x] - (4*a*c*Log[1 - c*x])/x + 8*b*c 
*Log[c*x]*Log[1 - c*x] + 2*a*c^2*Log[c*x]*Log[1 - c*x] - 4*b*c*Log[1 - c*x 
]^2 - a*c^2*Log[1 - c*x]^2 + (a*Log[1 - c*x]^2)/x^2 + (4*b*Log[1 - c*x]^2) 
/x + 4*b*c*Log[c*x]*Log[1 - c*x]^2 + 2*a*c^2*Log[c*x]*Log[1 - c*x]^2 + (2* 
(a*c*x + (-1 + c*x)*(a + 2*b*x + a*c*x)*Log[1 - c*x])*PolyLog[2, c*x])/x^2 
 + 2*c*(4*b + a*c + 2*(2*b + a*c)*Log[1 - c*x])*PolyLog[2, 1 - c*x] - 4*b* 
c*PolyLog[3, c*x] - 2*a*c^2*PolyLog[3, c*x] - 8*b*c*PolyLog[3, 1 - c*x] - 
4*a*c^2*PolyLog[3, 1 - c*x])/4
 

3.2.89.3 Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {7160, 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.

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

\(\Big \downarrow \) 7160

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.2.89.3.1 Defintions of rubi rules used

rule 2009

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

rule 7160

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

3.2.89.4 Maple [F]

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

input

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

output

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

3.2.89.5 Fricas [F]

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

input

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

output

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

3.2.89.6 Sympy [F]

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

input

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

output

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

3.2.89.7 Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.64 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^3} \, dx=-a c^{2} \log \left (x\right ) + \frac {1}{2} \, {\left (a c^{2} + 2 \, b c\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} + \frac {1}{2} \, {\left (a c^{2} + 4 \, b c\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} - \frac {1}{2} \, {\left (a c^{2} + 2 \, b c\right )} {\rm Li}_{3}(c x) - \frac {{\left ({\left (a c^{2} + 4 \, b c\right )} x^{2} - 4 \, b x - a\right )} \log \left (-c x + 1\right )^{2} - 2 \, {\left (a c x + {\left ({\left (a c^{2} + 2 \, b c\right )} x^{2} - 2 \, b x - a\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) - 4 \, {\left (a c^{2} x^{2} - a c x\right )} \log \left (-c x + 1\right )}{4 \, x^{2}} \]

input

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

output

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

3.2.89.8 Giac [F]

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

input

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

output

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

3.2.89.9 Mupad [F(-1)]

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

input

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

output

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