Integrand size = 17, antiderivative size = 185 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {b n x}{2 e}+\frac {3}{16} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{4 e^2}-\frac {1}{4} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \operatorname {PolyLog}(2,e x)}{4 e^2}-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \]
1/2*b*n*x/e+3/16*b*n*x^2-1/4*x*(a+b*ln(c*x^n))/e-1/8*x^2*(a+b*ln(c*x^n))+1 /4*b*n*ln(-e*x+1)/e^2-1/4*b*n*x^2*ln(-e*x+1)-1/4*(a+b*ln(c*x^n))*ln(-e*x+1 )/e^2+1/4*x^2*(a+b*ln(c*x^n))*ln(-e*x+1)-1/4*b*n*polylog(2,e*x)/e^2-1/4*b* n*x^2*polylog(2,e*x)+1/2*x^2*(a+b*ln(c*x^n))*polylog(2,e*x)
Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.91 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (-e x (2+e x)+2 \left (-1+e^2 x^2\right ) \log (1-e x)+4 e^2 x^2 \operatorname {PolyLog}(2,e x)\right )}{8 e^2}+\frac {b n \left (8 e x+3 e^2 x^2+4 \log (1-e x)-4 e^2 x^2 \log (1-e x)+\log (x) \left (-2 e x (2+e x)+4 \left (-1+e^2 x^2\right ) \log (1-e x)\right )+\left (-4-4 e^2 x^2+8 e^2 x^2 \log (x)\right ) \operatorname {PolyLog}(2,e x)\right )}{16 e^2} \]
((a - b*n*Log[x] + b*Log[c*x^n])*(-(e*x*(2 + e*x)) + 2*(-1 + e^2*x^2)*Log[ 1 - e*x] + 4*e^2*x^2*PolyLog[2, e*x]))/(8*e^2) + (b*n*(8*e*x + 3*e^2*x^2 + 4*Log[1 - e*x] - 4*e^2*x^2*Log[1 - e*x] + Log[x]*(-2*e*x*(2 + e*x) + 4*(- 1 + e^2*x^2)*Log[1 - e*x]) + (-4 - 4*e^2*x^2 + 8*e^2*x^2*Log[x])*PolyLog[2 , e*x]))/(16*e^2)
Time = 0.51 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2832, 25, 2823, 2009, 2842, 49, 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.
| \(\displaystyle \int x \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2832 |
| \(\displaystyle -\frac {1}{2} \int -x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)dx+\frac {1}{4} b n \int -x \log (1-e x)dx+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)dx-\frac {1}{4} b n \int x \log (1-e x)dx+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle \frac {1}{2} \left (-b n \int \left (\frac {1}{2} \log (1-e x) x-\frac {x}{4}-\frac {1}{2 e}-\frac {\log (1-e x)}{2 e^2 x}\right )dx-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )\right )-\frac {1}{4} b n \int x \log (1-e x)dx+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{4} b n \int x \log (1-e x)dx+\frac {1}{2} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{2 e^2}-\frac {\log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \log (1-e x)-\frac {3 x}{4 e}-\frac {x^2}{4}\right )\right )+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 2842 |
| \(\displaystyle -\frac {1}{4} b n \left (\frac {1}{2} e \int \frac {x^2}{1-e x}dx+\frac {1}{2} x^2 \log (1-e x)\right )+\frac {1}{2} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{2 e^2}-\frac {\log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \log (1-e x)-\frac {3 x}{4 e}-\frac {x^2}{4}\right )\right )+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle -\frac {1}{4} b n \left (\frac {1}{2} e \int \left (-\frac {x}{e}-\frac {1}{e^2 (e x-1)}-\frac {1}{e^2}\right )dx+\frac {1}{2} x^2 \log (1-e x)\right )+\frac {1}{2} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{2 e^2}-\frac {\log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \log (1-e x)-\frac {3 x}{4 e}-\frac {x^2}{4}\right )\right )+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}(2,e x)}{2 e^2}-\frac {\log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \log (1-e x)-\frac {3 x}{4 e}-\frac {x^2}{4}\right )\right )+\frac {1}{2} x^2 \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n \left (\frac {1}{2} e \left (-\frac {\log (1-e x)}{e^3}-\frac {x}{e^2}-\frac {x^2}{2 e}\right )+\frac {1}{2} x^2 \log (1-e x)\right )-\frac {1}{4} b n x^2 \operatorname {PolyLog}(2,e x)\) |
-1/4*(b*n*((x^2*Log[1 - e*x])/2 + (e*(-(x/e^2) - x^2/(2*e) - Log[1 - e*x]/ e^3))/2)) - (b*n*x^2*PolyLog[2, e*x])/4 + (x^2*(a + b*Log[c*x^n])*PolyLog[ 2, e*x])/2 + (-1/2*(x*(a + b*Log[c*x^n]))/e - (x^2*(a + b*Log[c*x^n]))/4 - ((a + b*Log[c*x^n])*Log[1 - e*x])/(2*e^2) + (x^2*(a + b*Log[c*x^n])*Log[1 - e*x])/2 - b*n*((-3*x)/(4*e) - x^2/4 - Log[1 - e*x]/(4*e^2) + (x^2*Log[1 - e*x])/4 + PolyLog[2, e*x]/(2*e^2)))/2
3.3.9.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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 = 12.65 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.23
| method | result | size |
| parallelrisch | \(\frac {8 b \,\operatorname {Li}_{2}\left (e x \right ) \ln \left (c \,x^{n}\right ) x^{2} e^{2}+4 b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) x^{2} e^{2}-4 x^{2} \operatorname {Li}_{2}\left (e x \right ) b \,e^{2} n -4 x^{2} \ln \left (-e x +1\right ) b \,e^{2} n -2 b \ln \left (c \,x^{n}\right ) e^{2} x^{2}+8 x^{2} \operatorname {Li}_{2}\left (e x \right ) a \,e^{2}+4 x^{2} \ln \left (-e x +1\right ) a \,e^{2}+3 b \,e^{2} n \,x^{2}-2 a \,e^{2} x^{2}-4 b e x \ln \left (c \,x^{n}\right )+8 b e n x +4 \ln \left (x \right ) b n -4 a e x -4 b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right )-4 \,\operatorname {Li}_{2}\left (e x \right ) b n +4 \ln \left (-e x +1\right ) b n -4 b \ln \left (c \,x^{n}\right )-4 \ln \left (-e x +1\right ) a}{16 e^{2}}\) | \(228\) |
1/16*(8*b*polylog(2,e*x)*ln(c*x^n)*x^2*e^2+4*b*ln(-e*x+1)*ln(c*x^n)*x^2*e^ 2-4*x^2*polylog(2,e*x)*b*e^2*n-4*x^2*ln(-e*x+1)*b*e^2*n-2*b*ln(c*x^n)*e^2* x^2+8*x^2*polylog(2,e*x)*a*e^2+4*x^2*ln(-e*x+1)*a*e^2+3*b*e^2*n*x^2-2*a*e^ 2*x^2-4*b*e*x*ln(c*x^n)+8*b*e*n*x+4*ln(x)*b*n-4*a*e*x-4*b*ln(-e*x+1)*ln(c* x^n)-4*polylog(2,e*x)*b*n+4*ln(-e*x+1)*b*n-4*b*ln(c*x^n)-4*ln(-e*x+1)*a)/e ^2
Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.12 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {{\left (3 \, b e^{2} n - 2 \, a e^{2}\right )} x^{2} + 4 \, {\left (2 \, b e n - a e\right )} x - 4 \, {\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2} + b n\right )} {\rm Li}_2\left (e x\right ) - 4 \, {\left ({\left (b e^{2} n - a e^{2}\right )} x^{2} - b n + a\right )} \log \left (-e x + 1\right ) + 2 \, {\left (4 \, b e^{2} x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} x^{2} - 2 \, b e x + 2 \, {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + 2 \, {\left (4 \, b e^{2} n x^{2} {\rm Li}_2\left (e x\right ) - b e^{2} n x^{2} - 2 \, b e n x + 2 \, {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{16 \, e^{2}} \]
1/16*((3*b*e^2*n - 2*a*e^2)*x^2 + 4*(2*b*e*n - a*e)*x - 4*((b*e^2*n - 2*a* e^2)*x^2 + b*n)*dilog(e*x) - 4*((b*e^2*n - a*e^2)*x^2 - b*n + a)*log(-e*x + 1) + 2*(4*b*e^2*x^2*dilog(e*x) - b*e^2*x^2 - 2*b*e*x + 2*(b*e^2*x^2 - b) *log(-e*x + 1))*log(c) + 2*(4*b*e^2*n*x^2*dilog(e*x) - b*e^2*n*x^2 - 2*b*e *n*x + 2*(b*e^2*n*x^2 - b*n)*log(-e*x + 1))*log(x))/e^2
Time = 22.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.11 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\begin {cases} - \frac {a x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {a x^{2} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {a x^{2}}{8} - \frac {a x}{4 e} + \frac {a \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} + \frac {b n x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} - \frac {b n x^{2} \operatorname {Li}_{2}\left (e x\right )}{4} + \frac {3 b n x^{2}}{16} - \frac {b x^{2} \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {b x^{2} \log {\left (c x^{n} \right )} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {b x^{2} \log {\left (c x^{n} \right )}}{8} + \frac {b n x}{2 e} - \frac {b x \log {\left (c x^{n} \right )}}{4 e} - \frac {b n \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{4 e^{2}} + \frac {b \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-a*x**2*polylog(1, e*x)/4 + a*x**2*polylog(2, e*x)/2 - a*x**2/8 - a*x/(4*e) + a*polylog(1, e*x)/(4*e**2) + b*n*x**2*polylog(1, e*x)/4 - b *n*x**2*polylog(2, e*x)/4 + 3*b*n*x**2/16 - b*x**2*log(c*x**n)*polylog(1, e*x)/4 + b*x**2*log(c*x**n)*polylog(2, e*x)/2 - b*x**2*log(c*x**n)/8 + b*n *x/(2*e) - b*x*log(c*x**n)/(4*e) - b*n*polylog(1, e*x)/(4*e**2) - b*n*poly log(2, e*x)/(4*e**2) + b*log(c*x**n)*polylog(1, e*x)/(4*e**2), Ne(e, 0)), (0, True))
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x {\rm Li}_2\left (e x\right ) \,d x } \]
1/8*b*((2*(2*e^2*x^2*log(x^n) - (e^2*n - 2*e^2*log(c))*x^2)*dilog(e*x) - 2 *((e^2*n - e^2*log(c))*x^2 - n*log(x))*log(-e*x + 1) - (e^2*x^2 + 2*e*x - 2*(e^2*x^2 - 1)*log(-e*x + 1))*log(x^n))/e^2 - 8*integrate(-1/8*(e*n*x + ( 3*e^2*n - 2*e^2*log(c))*x^2 - 2*n*log(x) - 2*n)/(e^2*x - e), x)) + 1/8*(4* e^2*x^2*dilog(e*x) - e^2*x^2 - 2*e*x + 2*(e^2*x^2 - 1)*log(-e*x + 1))*a/e^ 2
\[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x {\rm Li}_2\left (e x\right ) \,d x } \]
Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int x\,\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]