Integrand size = 21, antiderivative size = 460 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {7 a c^2}{36 x}-\frac {1}{2} b c^2 \log (x)-\frac {5}{12} a c^3 \log (x)-\frac {1}{6} c^2 (3 b+2 a c) \log (x)+\frac {1}{2} b c^2 \log (1-c x)+\frac {5}{12} a c^3 \log (1-c x)+\frac {1}{6} c^2 (3 b+2 a c) \log (1-c x)-\frac {7 a c \log (1-c x)}{36 x^2}-\frac {b c \log (1-c x)}{2 x}-\frac {2 a c^2 \log (1-c x)}{9 x}-\frac {c (3 b+2 a c) \log (1-c x)}{6 x}-\frac {1}{4} b c^2 \log ^2(1-c x)-\frac {1}{9} a c^3 \log ^2(1-c x)+\frac {a \log ^2(1-c x)}{9 x^3}+\frac {b \log ^2(1-c x)}{4 x^2}+\frac {1}{6} c^2 (3 b+2 a c) \log (c x) \log ^2(1-c x)-\frac {1}{2} b c^2 \operatorname {PolyLog}(2,c x)-\frac {2}{9} a c^3 \operatorname {PolyLog}(2,c x)+\frac {a c \operatorname {PolyLog}(2,c x)}{6 x^2}+\frac {c (3 b+2 a c) \operatorname {PolyLog}(2,c x)}{6 x}+\frac {1}{6} c^2 (3 b+2 a c) \log (1-c x) \operatorname {PolyLog}(2,c x)-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}\right ) \log (1-c x) \operatorname {PolyLog}(2,c x)+\frac {1}{3} c^2 (3 b+2 a c) \log (1-c x) \operatorname {PolyLog}(2,1-c x)-\frac {1}{6} c^2 (3 b+2 a c) \operatorname {PolyLog}(3,c x)-\frac {1}{3} c^2 (3 b+2 a c) \operatorname {PolyLog}(3,1-c x) \]
1/6*a*c*polylog(2,c*x)/x^2+1/6*c*(2*a*c+3*b)*polylog(2,c*x)/x-7/36*a*c*ln( -c*x+1)/x^2-1/2*b*c*ln(-c*x+1)/x-2/9*a*c^2*ln(-c*x+1)/x-1/6*c*(2*a*c+3*b)* ln(-c*x+1)/x+1/6*c^2*(2*a*c+3*b)*ln(c*x)*ln(-c*x+1)^2+1/6*c^2*(2*a*c+3*b)* ln(-c*x+1)*polylog(2,c*x)+1/3*c^2*(2*a*c+3*b)*ln(-c*x+1)*polylog(2,-c*x+1) +7/36*a*c^2/x-1/2*b*c^2*polylog(2,c*x)-2/9*a*c^3*polylog(2,c*x)-1/6*c^2*(2 *a*c+3*b)*polylog(3,c*x)-1/3*c^2*(2*a*c+3*b)*polylog(3,-c*x+1)-1/2*b*c^2*l n(x)-5/12*a*c^3*ln(x)-1/6*c^2*(2*a*c+3*b)*ln(x)+1/2*b*c^2*ln(-c*x+1)+5/12* a*c^3*ln(-c*x+1)+1/6*c^2*(2*a*c+3*b)*ln(-c*x+1)-1/4*b*c^2*ln(-c*x+1)^2-1/9 *a*c^3*ln(-c*x+1)^2+1/9*a*ln(-c*x+1)^2/x^3+1/4*b*ln(-c*x+1)^2/x^2-1/6*(2*a /x^3+3*b/x^2)*ln(-c*x+1)*polylog(2,c*x)
Time = 1.13 (sec) , antiderivative size = 389, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {1}{36} \left (-7 a c^3+\frac {7 a c^2}{x}-36 b c^2 \log (c x)-27 a c^3 \log (c x)+36 b c^2 \log (1-c x)+27 a c^3 \log (1-c x)-\frac {7 a c \log (1-c x)}{x^2}-\frac {36 b c \log (1-c x)}{x}-\frac {20 a c^2 \log (1-c x)}{x}+18 b c^2 \log (c x) \log (1-c x)+8 a c^3 \log (c x) \log (1-c x)-9 b c^2 \log ^2(1-c x)-4 a c^3 \log ^2(1-c x)+\frac {4 a \log ^2(1-c x)}{x^3}+\frac {9 b \log ^2(1-c x)}{x^2}+18 b c^2 \log (c x) \log ^2(1-c x)+12 a c^3 \log (c x) \log ^2(1-c x)+\frac {6 \left (c x (a+3 b x+2 a c x)+\left (-2 a-3 b x+3 b c^2 x^3+2 a c^3 x^3\right ) \log (1-c x)\right ) \operatorname {PolyLog}(2,c x)}{x^3}+2 c^2 (9 b+4 a c+6 (3 b+2 a c) \log (1-c x)) \operatorname {PolyLog}(2,1-c x)-18 b c^2 \operatorname {PolyLog}(3,c x)-12 a c^3 \operatorname {PolyLog}(3,c x)-36 b c^2 \operatorname {PolyLog}(3,1-c x)-24 a c^3 \operatorname {PolyLog}(3,1-c x)\right ) \]
(-7*a*c^3 + (7*a*c^2)/x - 36*b*c^2*Log[c*x] - 27*a*c^3*Log[c*x] + 36*b*c^2 *Log[1 - c*x] + 27*a*c^3*Log[1 - c*x] - (7*a*c*Log[1 - c*x])/x^2 - (36*b*c *Log[1 - c*x])/x - (20*a*c^2*Log[1 - c*x])/x + 18*b*c^2*Log[c*x]*Log[1 - c *x] + 8*a*c^3*Log[c*x]*Log[1 - c*x] - 9*b*c^2*Log[1 - c*x]^2 - 4*a*c^3*Log [1 - c*x]^2 + (4*a*Log[1 - c*x]^2)/x^3 + (9*b*Log[1 - c*x]^2)/x^2 + 18*b*c ^2*Log[c*x]*Log[1 - c*x]^2 + 12*a*c^3*Log[c*x]*Log[1 - c*x]^2 + (6*(c*x*(a + 3*b*x + 2*a*c*x) + (-2*a - 3*b*x + 3*b*c^2*x^3 + 2*a*c^3*x^3)*Log[1 - c *x])*PolyLog[2, c*x])/x^3 + 2*c^2*(9*b + 4*a*c + 6*(3*b + 2*a*c)*Log[1 - c *x])*PolyLog[2, 1 - c*x] - 18*b*c^2*PolyLog[3, c*x] - 12*a*c^3*PolyLog[3, c*x] - 36*b*c^2*PolyLog[3, 1 - c*x] - 24*a*c^3*PolyLog[3, 1 - c*x])/36
Time = 0.95 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.08, 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.
| \(\displaystyle \int \frac {(a+b x) \operatorname {PolyLog}(2,c x) \log (1-c x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 7160 |
| \(\displaystyle c \int \left (-\frac {(3 b+2 a c) \operatorname {PolyLog}(2,c x) c^2}{6 (1-c x)}-\frac {(3 b+2 a c) \operatorname {PolyLog}(2,c x) c}{6 x}-\frac {(3 b+2 a c) \operatorname {PolyLog}(2,c x)}{6 x^2}-\frac {a \operatorname {PolyLog}(2,c x)}{3 x^3}\right )dx+\int \left (-\frac {b \log ^2(1-c x)}{2 x^3}-\frac {a \log ^2(1-c x)}{3 x^4}\right )dx-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}\right ) \operatorname {PolyLog}(2,c x) \log (1-c x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle c \left (-\frac {1}{6} c (2 a c+3 b) \operatorname {PolyLog}(3,c x)-\frac {1}{3} c (2 a c+3 b) \operatorname {PolyLog}(3,1-c x)+\frac {(2 a c+3 b) \operatorname {PolyLog}(2,c x)}{6 x}+\frac {1}{6} c (2 a c+3 b) \operatorname {PolyLog}(2,c x) \log (1-c x)+\frac {1}{3} c (2 a c+3 b) \operatorname {PolyLog}(2,1-c x) \log (1-c x)+\frac {1}{6} c (2 a c+3 b) \log (c x) \log ^2(1-c x)-\frac {1}{6} c \log (x) (2 a c+3 b)+\frac {1}{6} c (2 a c+3 b) \log (1-c x)-\frac {(2 a c+3 b) \log (1-c x)}{6 x}-\frac {1}{12} a c^2 \log (x)+\frac {1}{12} a c^2 \log (1-c x)+\frac {a \operatorname {PolyLog}(2,c x)}{6 x^2}-\frac {a \log (1-c x)}{12 x^2}+\frac {a c}{12 x}\right )-\frac {1}{6} \left (\frac {2 a}{x^3}+\frac {3 b}{x^2}\right ) \operatorname {PolyLog}(2,c x) \log (1-c x)-\frac {2}{9} a c^3 \operatorname {PolyLog}(2,c x)-\frac {1}{9} a c^3 \log ^2(1-c x)-\frac {1}{3} a c^3 \log (x)+\frac {1}{3} a c^3 \log (1-c x)+\frac {a c^2}{9 x}-\frac {2 a c^2 \log (1-c x)}{9 x}+\frac {a \log ^2(1-c x)}{9 x^3}-\frac {a c \log (1-c x)}{9 x^2}-\frac {1}{2} b c^2 \operatorname {PolyLog}(2,c x)-\frac {1}{4} b c^2 \log ^2(1-c x)-\frac {1}{2} b c^2 \log (x)+\frac {1}{2} b c^2 \log (1-c x)+\frac {b \log ^2(1-c x)}{4 x^2}-\frac {b c \log (1-c x)}{2 x}\) |
(a*c^2)/(9*x) - (b*c^2*Log[x])/2 - (a*c^3*Log[x])/3 + (b*c^2*Log[1 - c*x]) /2 + (a*c^3*Log[1 - c*x])/3 - (a*c*Log[1 - c*x])/(9*x^2) - (b*c*Log[1 - c* x])/(2*x) - (2*a*c^2*Log[1 - c*x])/(9*x) - (b*c^2*Log[1 - c*x]^2)/4 - (a*c ^3*Log[1 - c*x]^2)/9 + (a*Log[1 - c*x]^2)/(9*x^3) + (b*Log[1 - c*x]^2)/(4* x^2) - (b*c^2*PolyLog[2, c*x])/2 - (2*a*c^3*PolyLog[2, c*x])/9 - (((2*a)/x ^3 + (3*b)/x^2)*Log[1 - c*x]*PolyLog[2, c*x])/6 + c*((a*c)/(12*x) - (a*c^2 *Log[x])/12 - (c*(3*b + 2*a*c)*Log[x])/6 + (a*c^2*Log[1 - c*x])/12 + (c*(3 *b + 2*a*c)*Log[1 - c*x])/6 - (a*Log[1 - c*x])/(12*x^2) - ((3*b + 2*a*c)*L og[1 - c*x])/(6*x) + (c*(3*b + 2*a*c)*Log[c*x]*Log[1 - c*x]^2)/6 + (a*Poly Log[2, c*x])/(6*x^2) + ((3*b + 2*a*c)*PolyLog[2, c*x])/(6*x) + (c*(3*b + 2 *a*c)*Log[1 - c*x]*PolyLog[2, c*x])/6 + (c*(3*b + 2*a*c)*Log[1 - c*x]*Poly Log[2, 1 - c*x])/3 - (c*(3*b + 2*a*c)*PolyLog[3, c*x])/6 - (c*(3*b + 2*a*c )*PolyLog[3, 1 - c*x])/3)
3.2.90.3.1 Defintions of rubi rules used
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]
\[\int \frac {\left (b x +a \right ) \ln \left (-c x +1\right ) \operatorname {polylog}\left (2, c x \right )}{x^{4}}d x\]
\[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int { \frac {{\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{4}} \,d x } \]
\[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int \frac {\left (a + b x\right ) \log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x^{4}}\, dx \]
Time = 0.25 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.62 \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\frac {1}{6} \, {\left (2 \, a c^{3} + 3 \, b c^{2}\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} + \frac {1}{18} \, {\left (4 \, a c^{3} + 9 \, b c^{2}\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} - \frac {1}{4} \, {\left (3 \, a c^{3} + 4 \, b c^{2}\right )} \log \left (x\right ) - \frac {1}{6} \, {\left (2 \, a c^{3} + 3 \, b c^{2}\right )} {\rm Li}_{3}(c x) + \frac {7 \, a c^{2} x^{2} - {\left ({\left (4 \, a c^{3} + 9 \, b c^{2}\right )} x^{3} - 9 \, b x - 4 \, a\right )} \log \left (-c x + 1\right )^{2} + 6 \, {\left (a c x + {\left (2 \, a c^{2} + 3 \, b c\right )} x^{2} + {\left ({\left (2 \, a c^{3} + 3 \, b c^{2}\right )} x^{3} - 3 \, b x - 2 \, a\right )} \log \left (-c x + 1\right )\right )} {\rm Li}_2\left (c x\right ) + {\left (9 \, {\left (3 \, a c^{3} + 4 \, b c^{2}\right )} x^{3} - 7 \, a c x - 4 \, {\left (5 \, a c^{2} + 9 \, b c\right )} x^{2}\right )} \log \left (-c x + 1\right )}{36 \, x^{3}} \]
1/6*(2*a*c^3 + 3*b*c^2)*(log(c*x)*log(-c*x + 1)^2 + 2*dilog(-c*x + 1)*log( -c*x + 1) - 2*polylog(3, -c*x + 1)) + 1/18*(4*a*c^3 + 9*b*c^2)*(log(c*x)*l og(-c*x + 1) + dilog(-c*x + 1)) - 1/4*(3*a*c^3 + 4*b*c^2)*log(x) - 1/6*(2* a*c^3 + 3*b*c^2)*polylog(3, c*x) + 1/36*(7*a*c^2*x^2 - ((4*a*c^3 + 9*b*c^2 )*x^3 - 9*b*x - 4*a)*log(-c*x + 1)^2 + 6*(a*c*x + (2*a*c^2 + 3*b*c)*x^2 + ((2*a*c^3 + 3*b*c^2)*x^3 - 3*b*x - 2*a)*log(-c*x + 1))*dilog(c*x) + (9*(3* a*c^3 + 4*b*c^2)*x^3 - 7*a*c*x - 4*(5*a*c^2 + 9*b*c)*x^2)*log(-c*x + 1))/x ^3
\[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int { \frac {{\left (b x + a\right )} {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {(a+b x) \log (1-c x) \operatorname {PolyLog}(2,c x)}{x^4} \, dx=\int \frac {\ln \left (1-c\,x\right )\,\mathrm {polylog}\left (2,c\,x\right )\,\left (a+b\,x\right )}{x^4} \,d x \]