Integrand size = 19, antiderivative size = 202 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=-\frac {b e n}{2 x}+\frac {1}{4} b e^2 n \log (x)-\frac {1}{8} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac {1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{4 x^2}-\frac {1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac {1}{4} b e^2 n \operatorname {PolyLog}(2,e x)-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{2 x^2} \]
-1/2*b*e*n/x+1/4*b*e^2*n*ln(x)-1/8*b*e^2*n*ln(x)^2-1/4*e*(a+b*ln(c*x^n))/x +1/4*e^2*ln(x)*(a+b*ln(c*x^n))-1/4*b*e^2*n*ln(-e*x+1)+1/4*b*n*ln(-e*x+1)/x ^2-1/4*e^2*(a+b*ln(c*x^n))*ln(-e*x+1)+1/4*(a+b*ln(c*x^n))*ln(-e*x+1)/x^2-1 /4*b*e^2*n*polylog(2,e*x)-1/4*b*n*polylog(2,e*x)/x^2-1/2*(a+b*ln(c*x^n))*p olylog(2,e*x)/x^2
Time = 0.15 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (-e x+e^2 x^2 \log (x)+\log (1-e x)-e^2 x^2 \log (1-e x)-2 \operatorname {PolyLog}(2,e x)\right )}{4 x^2}+\frac {b n \left (-4 e x+e^2 x^2 \log ^2(x)+2 \log (1-e x)-2 e^2 x^2 \log (1-e x)-2 (-1+e x) \log (x) (-e x+(1+e x) \log (1-e x))-2 \left (1+e^2 x^2+2 \log (x)\right ) \operatorname {PolyLog}(2,e x)\right )}{8 x^2} \]
((a - b*n*Log[x] + b*Log[c*x^n])*(-(e*x) + e^2*x^2*Log[x] + Log[1 - e*x] - e^2*x^2*Log[1 - e*x] - 2*PolyLog[2, e*x]))/(4*x^2) + (b*n*(-4*e*x + e^2*x ^2*Log[x]^2 + 2*Log[1 - e*x] - 2*e^2*x^2*Log[1 - e*x] - 2*(-1 + e*x)*Log[x ]*(-(e*x) + (1 + e*x)*Log[1 - e*x]) - 2*(1 + e^2*x^2 + 2*Log[x])*PolyLog[2 , e*x]))/(8*x^2)
Time = 0.53 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2832, 25, 2823, 2009, 2842, 54, 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.
| \(\displaystyle \int \frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2832 |
| \(\displaystyle \frac {1}{2} \int -\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3}dx+\frac {1}{4} b n \int -\frac {\log (1-e x)}{x^3}dx-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3}dx-\frac {1}{4} b n \int \frac {\log (1-e x)}{x^3}dx-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle \frac {1}{2} \left (b n \int \left (-\frac {\log (x) e^2}{2 x}+\frac {\log (1-e x) e^2}{2 x}+\frac {e}{2 x^2}-\frac {\log (1-e x)}{2 x^3}\right )dx+\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\right )-\frac {1}{4} b n \int \frac {\log (1-e x)}{x^3}dx-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} b n \int \frac {\log (1-e x)}{x^3}dx+\frac {1}{2} \left (\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+b n \left (-\frac {1}{2} e^2 \operatorname {PolyLog}(2,e x)-\frac {1}{4} e^2 \log ^2(x)+\frac {1}{4} e^2 \log (x)-\frac {1}{4} e^2 \log (1-e x)+\frac {\log (1-e x)}{4 x^2}-\frac {3 e}{4 x}\right )\right )-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 2842 |
| \(\displaystyle -\frac {1}{4} b n \left (-\frac {1}{2} e \int \frac {1}{x^2 (1-e x)}dx-\frac {\log (1-e x)}{2 x^2}\right )+\frac {1}{2} \left (\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+b n \left (-\frac {1}{2} e^2 \operatorname {PolyLog}(2,e x)-\frac {1}{4} e^2 \log ^2(x)+\frac {1}{4} e^2 \log (x)-\frac {1}{4} e^2 \log (1-e x)+\frac {\log (1-e x)}{4 x^2}-\frac {3 e}{4 x}\right )\right )-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle -\frac {1}{4} b n \left (-\frac {1}{2} e \int \left (-\frac {e^2}{e x-1}+\frac {e}{x}+\frac {1}{x^2}\right )dx-\frac {\log (1-e x)}{2 x^2}\right )+\frac {1}{2} \left (\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+b n \left (-\frac {1}{2} e^2 \operatorname {PolyLog}(2,e x)-\frac {1}{4} e^2 \log ^2(x)+\frac {1}{4} e^2 \log (x)-\frac {1}{4} e^2 \log (1-e x)+\frac {\log (1-e x)}{4 x^2}-\frac {3 e}{4 x}\right )\right )-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+b n \left (-\frac {1}{2} e^2 \operatorname {PolyLog}(2,e x)-\frac {1}{4} e^2 \log ^2(x)+\frac {1}{4} e^2 \log (x)-\frac {1}{4} e^2 \log (1-e x)+\frac {\log (1-e x)}{4 x^2}-\frac {3 e}{4 x}\right )\right )-\frac {\operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {b n \operatorname {PolyLog}(2,e x)}{4 x^2}-\frac {1}{4} b n \left (-\frac {\log (1-e x)}{2 x^2}-\frac {1}{2} e \left (e \log (x)-e \log (1-e x)-\frac {1}{x}\right )\right )\) |
-1/4*(b*n*(-1/2*Log[1 - e*x]/x^2 - (e*(-x^(-1) + e*Log[x] - e*Log[1 - e*x] ))/2)) - (b*n*PolyLog[2, e*x])/(4*x^2) - ((a + b*Log[c*x^n])*PolyLog[2, e* x])/(2*x^2) + (-1/2*(e*(a + b*Log[c*x^n]))/x + (e^2*Log[x]*(a + b*Log[c*x^ n]))/2 - (e^2*(a + b*Log[c*x^n])*Log[1 - e*x])/2 + ((a + b*Log[c*x^n])*Log [1 - e*x])/(2*x^2) + b*n*((-3*e)/(4*x) + (e^2*Log[x])/4 - (e^2*Log[x]^2)/4 - (e^2*Log[1 - e*x])/4 + Log[1 - e*x]/(4*x^2) - (e^2*PolyLog[2, e*x])/2)) /2
3.3.13.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.)*PolyLog[k_, (e _.)*(x_)^(q_.)], x_Symbol] :> Simp[(-b)*n*(d*x)^(m + 1)*(PolyLog[k, e*x^q]/ (d*(m + 1)^2)), x] + (Simp[(d*x)^(m + 1)*PolyLog[k, e*x^q]*((a + b*Log[c*x^ n])/(d*(m + 1))), x] - Simp[q/(m + 1) Int[(d*x)^m*PolyLog[k - 1, e*x^q]*( a + b*Log[c*x^n]), x], x] + Simp[b*n*(q/(m + 1)^2) Int[(d*x)^m*PolyLog[k - 1, e*x^q], x], x]) /; FreeQ[{a, b, c, d, e, m, n, q}, x] && IGtQ[k, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Time = 4.71 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.33
| method | result | size |
| parallelrisch | \(\frac {4 \ln \left (x \right ) x^{2} b \,e^{2} n^{2}-2 e^{2} b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) x^{2} n -2 e^{2} b \,n^{2} \operatorname {Li}_{2}\left (e x \right ) x^{2}-2 x^{2} \ln \left (-e x +1\right ) b \,e^{2} n^{2}+2 \ln \left (x \right ) x^{2} a \,e^{2} n +e^{2} b \ln \left (c \,x^{n}\right )^{2} x^{2}-2 x^{2} \ln \left (c \,x^{n}\right ) b \,e^{2} n -2 x^{2} \ln \left (-e x +1\right ) a \,e^{2} n -4 x^{2} b \,e^{2} n^{2}-2 x^{2} a \,e^{2} n -2 x \ln \left (c \,x^{n}\right ) b e n -4 x b e \,n^{2}-2 x a e n -4 \ln \left (c \,x^{n}\right ) \operatorname {Li}_{2}\left (e x \right ) b n +2 b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) n -2 b \,n^{2} \operatorname {Li}_{2}\left (e x \right )+2 \ln \left (-e x +1\right ) b \,n^{2}-4 \,\operatorname {Li}_{2}\left (e x \right ) a n +2 \ln \left (-e x +1\right ) a n}{8 x^{2} n}\) | \(268\) |
1/8*(4*ln(x)*x^2*b*e^2*n^2-2*e^2*b*ln(-e*x+1)*ln(c*x^n)*x^2*n-2*e^2*b*n^2* polylog(2,e*x)*x^2-2*x^2*ln(-e*x+1)*b*e^2*n^2+2*ln(x)*x^2*a*e^2*n+e^2*b*ln (c*x^n)^2*x^2-2*x^2*ln(c*x^n)*b*e^2*n-2*x^2*ln(-e*x+1)*a*e^2*n-4*x^2*b*e^2 *n^2-2*x^2*a*e^2*n-2*x*ln(c*x^n)*b*e*n-4*x*b*e*n^2-2*x*a*e*n-4*ln(c*x^n)*p olylog(2,e*x)*b*n+2*b*ln(-e*x+1)*ln(c*x^n)*n-2*b*n^2*polylog(2,e*x)+2*ln(- e*x+1)*b*n^2-4*polylog(2,e*x)*a*n+2*ln(-e*x+1)*a*n)/x^2/n
Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\frac {b e^{2} n x^{2} \log \left (x\right )^{2} - 2 \, {\left (2 \, b e n + a e\right )} x - 2 \, {\left (b e^{2} n x^{2} + b n + 2 \, a\right )} {\rm Li}_2\left (e x\right ) - 2 \, {\left ({\left (b e^{2} n + a e^{2}\right )} x^{2} - b n - a\right )} \log \left (-e x + 1\right ) - 2 \, {\left (b e x + 2 \, b {\rm Li}_2\left (e x\right ) + {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + 2 \, {\left (b e^{2} x^{2} \log \left (c\right ) - b e n x + {\left (b e^{2} n + a e^{2}\right )} x^{2} - 2 \, b n {\rm Li}_2\left (e x\right ) - {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{8 \, x^{2}} \]
1/8*(b*e^2*n*x^2*log(x)^2 - 2*(2*b*e*n + a*e)*x - 2*(b*e^2*n*x^2 + b*n + 2 *a)*dilog(e*x) - 2*((b*e^2*n + a*e^2)*x^2 - b*n - a)*log(-e*x + 1) - 2*(b* e*x + 2*b*dilog(e*x) + (b*e^2*x^2 - b)*log(-e*x + 1))*log(c) + 2*(b*e^2*x^ 2*log(c) - b*e*n*x + (b*e^2*n + a*e^2)*x^2 - 2*b*n*dilog(e*x) - (b*e^2*n*x ^2 - b*n)*log(-e*x + 1))*log(x))/x^2
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \operatorname {Li}_{2}\left (e x\right )}{x^{3}}\, dx \]
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right )}{x^{3}} \,d x } \]
1/4*(e^2*log(x) - (e*x + (e^2*x^2 - 1)*log(-e*x + 1) + 2*dilog(e*x))/x^2)* a - 1/4*b*(((n + 2*log(c) + 2*log(x^n))*dilog(e*x) - (e^2*n*x^2*log(x) + n + log(c))*log(-e*x + 1) - (e^2*x^2*log(x) - e*x - (e^2*x^2 - 1)*log(-e*x + 1))*log(x^n))/x^2 + 4*integrate(-1/4*(e^2*n*x - 2*e*n - e*log(c) - (2*e^ 3*n*x^2 - e^2*n*x)*log(x))/(e*x^3 - x^2), x))
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x)}{x^3} \, dx=\int \frac {\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \]