Integrand size = 13, antiderivative size = 276 \[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=-\frac {b^2 c}{6 a (1-a c) x}+\frac {b^3 c^2 \log (x)}{6 a (1-a c)^2}-\frac {b^3 c \log (x)}{3 a^2 (1-a c)}-\frac {b^3 c^2 \log (1-a c-b c x)}{6 a (1-a c)^2}+\frac {b^3 c \log (1-a c-b c x)}{3 a^2 (1-a c)}+\frac {b \log (1-a c-b c x)}{6 a x^2}-\frac {b^2 \log (1-a c-b c x)}{3 a^2 x}-\frac {b^3 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{3 a^3}-\frac {b^3 \operatorname {PolyLog}(2,c (a+b x))}{3 a^3}-\frac {\operatorname {PolyLog}(2,c (a+b x))}{3 x^3}-\frac {b^3 \operatorname {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{3 a^3} \]
-1/6*b^2*c/a/(-a*c+1)/x+1/6*b^3*c^2*ln(x)/a/(-a*c+1)^2-1/3*b^3*c*ln(x)/a^2 /(-a*c+1)-1/6*b^3*c^2*ln(-b*c*x-a*c+1)/a/(-a*c+1)^2+1/3*b^3*c*ln(-b*c*x-a* c+1)/a^2/(-a*c+1)+1/6*b*ln(-b*c*x-a*c+1)/a/x^2-1/3*b^2*ln(-b*c*x-a*c+1)/a^ 2/x-1/3*b^3*ln(b*c*x/(-a*c+1))*ln(-b*c*x-a*c+1)/a^3-1/3*b^3*polylog(2,c*(b *x+a))/a^3-1/3*polylog(2,c*(b*x+a))/x^3-1/3*b^3*polylog(2,1-b*c*x/(-a*c+1) )/a^3
Time = 0.24 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.80 \[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=-\frac {b \left (\frac {2 a b^2 c \log (x)}{1-a c}+\frac {2 a b^2 c \log (1-a c-b c x)}{-1+a c}-\frac {a^2 \log (1-a c-b c x)}{x^2}+\frac {2 a b \log (1-a c-b c x)}{x}+2 b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)-\frac {a^2 b c (-1+a c+b c x \log (x)-b c x \log (1-a c-b c x))}{(-1+a c)^2 x}+2 b^2 \operatorname {PolyLog}(2,c (a+b x))+2 b^2 \operatorname {PolyLog}\left (2,\frac {-1+a c+b c x}{-1+a c}\right )\right )}{6 a^3}-\frac {\operatorname {PolyLog}(2,a c+b c x)}{3 x^3} \]
-1/6*(b*((2*a*b^2*c*Log[x])/(1 - a*c) + (2*a*b^2*c*Log[1 - a*c - b*c*x])/( -1 + a*c) - (a^2*Log[1 - a*c - b*c*x])/x^2 + (2*a*b*Log[1 - a*c - b*c*x])/ x + 2*b^2*Log[(b*c*x)/(1 - a*c)]*Log[1 - a*c - b*c*x] - (a^2*b*c*(-1 + a*c + b*c*x*Log[x] - b*c*x*Log[1 - a*c - b*c*x]))/((-1 + a*c)^2*x) + 2*b^2*Po lyLog[2, c*(a + b*x)] + 2*b^2*PolyLog[2, (-1 + a*c + b*c*x)/(-1 + a*c)]))/ a^3 - PolyLog[2, a*c + b*c*x]/(3*x^3)
Time = 0.56 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx\) |
| \(\Big \downarrow \) 7152 |
| \(\displaystyle -\frac {1}{3} b \int \frac {\log (-a c-b x c+1)}{x^3 (a+b x)}dx-\frac {\operatorname {PolyLog}(2,c (a+b x))}{3 x^3}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle -\frac {1}{3} b \int \left (-\frac {\log (-a c-b x c+1) b^3}{a^3 (a+b x)}+\frac {\log (-a c-b x c+1) b^2}{a^3 x}-\frac {\log (-a c-b x c+1) b}{a^2 x^2}+\frac {\log (-a c-b x c+1)}{a x^3}\right )dx-\frac {\operatorname {PolyLog}(2,c (a+b x))}{3 x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{3} b \left (\frac {b^2 \operatorname {PolyLog}(2,c (a+b x))}{a^3}+\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{a^3}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (-a c-b c x+1)}{a^3}+\frac {b^2 c \log (x)}{a^2 (1-a c)}-\frac {b^2 c \log (-a c-b c x+1)}{a^2 (1-a c)}+\frac {b \log (-a c-b c x+1)}{a^2 x}-\frac {b^2 c^2 \log (x)}{2 a (1-a c)^2}+\frac {b^2 c^2 \log (-a c-b c x+1)}{2 a (1-a c)^2}-\frac {\log (-a c-b c x+1)}{2 a x^2}+\frac {b c}{2 a x (1-a c)}\right )-\frac {\operatorname {PolyLog}(2,c (a+b x))}{3 x^3}\) |
-1/3*PolyLog[2, c*(a + b*x)]/x^3 - (b*((b*c)/(2*a*(1 - a*c)*x) - (b^2*c^2* Log[x])/(2*a*(1 - a*c)^2) + (b^2*c*Log[x])/(a^2*(1 - a*c)) + (b^2*c^2*Log[ 1 - a*c - b*c*x])/(2*a*(1 - a*c)^2) - (b^2*c*Log[1 - a*c - b*c*x])/(a^2*(1 - a*c)) - Log[1 - a*c - b*c*x]/(2*a*x^2) + (b*Log[1 - a*c - b*c*x])/(a^2* x) + (b^2*Log[(b*c*x)/(1 - a*c)]*Log[1 - a*c - b*c*x])/a^3 + (b^2*PolyLog[ 2, c*(a + b*x)])/a^3 + (b^2*PolyLog[2, 1 - (b*c*x)/(1 - a*c)])/a^3))/3
3.2.30.3.1 Defintions of rubi rules used
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]
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]
Time = 3.18 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.88
| method | result | size |
| parts | \(-\frac {\operatorname {polylog}\left (2, c \left (b x +a \right )\right )}{3 x^{3}}-\frac {b^{3} c^{2} \left (\frac {\operatorname {dilog}\left (-b c x -a c +1\right )}{a^{3} c^{2}}+\frac {-\frac {\ln \left (-b c x \right )}{a c -1}-\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x -a c +1\right )}{\left (a c -1\right ) b c x}}{a^{2} c}+\frac {-\frac {\frac {a c -1}{b c x}+\ln \left (-b c x \right )}{2 \left (a c -1\right )^{2}}+\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x +a c -1\right ) \left (-b c x -a c +1\right )}{2 b^{2} c^{2} x^{2} \left (a c -1\right )^{2}}}{a}+\frac {\operatorname {dilog}\left (-\frac {b c x}{a c -1}\right )+\ln \left (-b c x -a c +1\right ) \ln \left (-\frac {b c x}{a c -1}\right )}{a^{3} c^{2}}\right )}{3}\) | \(242\) |
| derivativedivides | \(b^{3} c^{3} \left (-\frac {\operatorname {polylog}\left (2, b c x +a c \right )}{3 b^{3} c^{3} x^{3}}-\frac {-\frac {\ln \left (-b c x \right )}{a c -1}-\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x -a c +1\right )}{\left (a c -1\right ) b c x}}{3 a^{2} c^{2}}-\frac {\operatorname {dilog}\left (-\frac {b c x}{a c -1}\right )+\ln \left (-b c x -a c +1\right ) \ln \left (-\frac {b c x}{a c -1}\right )}{3 a^{3} c^{3}}-\frac {-\frac {\frac {a c -1}{b c x}+\ln \left (-b c x \right )}{2 \left (a c -1\right )^{2}}+\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x +a c -1\right ) \left (-b c x -a c +1\right )}{2 b^{2} c^{2} x^{2} \left (a c -1\right )^{2}}}{3 a c}-\frac {\operatorname {dilog}\left (-b c x -a c +1\right )}{3 a^{3} c^{3}}\right )\) | \(254\) |
| default | \(b^{3} c^{3} \left (-\frac {\operatorname {polylog}\left (2, b c x +a c \right )}{3 b^{3} c^{3} x^{3}}-\frac {-\frac {\ln \left (-b c x \right )}{a c -1}-\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x -a c +1\right )}{\left (a c -1\right ) b c x}}{3 a^{2} c^{2}}-\frac {\operatorname {dilog}\left (-\frac {b c x}{a c -1}\right )+\ln \left (-b c x -a c +1\right ) \ln \left (-\frac {b c x}{a c -1}\right )}{3 a^{3} c^{3}}-\frac {-\frac {\frac {a c -1}{b c x}+\ln \left (-b c x \right )}{2 \left (a c -1\right )^{2}}+\frac {\ln \left (-b c x -a c +1\right ) \left (-b c x +a c -1\right ) \left (-b c x -a c +1\right )}{2 b^{2} c^{2} x^{2} \left (a c -1\right )^{2}}}{3 a c}-\frac {\operatorname {dilog}\left (-b c x -a c +1\right )}{3 a^{3} c^{3}}\right )\) | \(254\) |
-1/3*polylog(2,c*(b*x+a))/x^3-1/3*b^3*c^2*(1/a^3/c^2*dilog(-b*c*x-a*c+1)+1 /a^2/c*(-1/(a*c-1)*ln(-b*c*x)-ln(-b*c*x-a*c+1)*(-b*c*x-a*c+1)/(a*c-1)/b/c/ x)+1/a*(-1/2/(a*c-1)^2*((a*c-1)/b/c/x+ln(-b*c*x))+1/2*ln(-b*c*x-a*c+1)*(-b *c*x+a*c-1)*(-b*c*x-a*c+1)/b^2/c^2/x^2/(a*c-1)^2)+1/a^3/c^2*(dilog(-b*c*x/ (a*c-1))+ln(-b*c*x-a*c+1)*ln(-b*c*x/(a*c-1))))
\[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=\int { \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x^{4}} \,d x } \]
\[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=\int \frac {\operatorname {Li}_{2}\left (a c + b c x\right )}{x^{4}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.09 \[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} b^{3}}{3 \, a^{3}} - \frac {{\left (\log \left (-b c x - a c + 1\right ) \log \left (-\frac {b c x + a c - 1}{a c - 1} + 1\right ) + {\rm Li}_2\left (\frac {b c x + a c - 1}{a c - 1}\right )\right )} b^{3}}{3 \, a^{3}} + \frac {{\left (3 \, a b^{3} c^{2} - 2 \, b^{3} c\right )} \log \left (x\right )}{6 \, {\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )}} + \frac {{\left (a^{2} b^{2} c^{2} - a b^{2} c\right )} x^{2} - 2 \, {\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left ({\left (3 \, a b^{3} c^{2} - 2 \, b^{3} c\right )} x^{3} + 2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a b^{2} c + b^{2}\right )} x^{2} - {\left (a^{3} b c^{2} - 2 \, a^{2} b c + a b\right )} x\right )} \log \left (-b c x - a c + 1\right )}{6 \, {\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )} x^{3}} \]
1/3*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*b^3 /a^3 - 1/3*(log(-b*c*x - a*c + 1)*log(-(b*c*x + a*c - 1)/(a*c - 1) + 1) + dilog((b*c*x + a*c - 1)/(a*c - 1)))*b^3/a^3 + 1/6*(3*a*b^3*c^2 - 2*b^3*c)* log(x)/(a^4*c^2 - 2*a^3*c + a^2) + 1/6*((a^2*b^2*c^2 - a*b^2*c)*x^2 - 2*(a ^4*c^2 - 2*a^3*c + a^2)*dilog(b*c*x + a*c) - ((3*a*b^3*c^2 - 2*b^3*c)*x^3 + 2*(a^2*b^2*c^2 - 2*a*b^2*c + b^2)*x^2 - (a^3*b*c^2 - 2*a^2*b*c + a*b)*x) *log(-b*c*x - a*c + 1))/((a^4*c^2 - 2*a^3*c + a^2)*x^3)
\[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=\int { \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {PolyLog}(2,c (a+b x))}{x^4} \, dx=\int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{x^4} \,d x \]