Integrand size = 20, antiderivative size = 195 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=-\frac {5 b e n}{36 x^2}+\frac {4 b e^2 n}{9 x}+\frac {1}{9} b e^3 n \log (x)-\frac {1}{6} b e^3 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b e^3 n \log (1+e x)-\frac {b n \log (1+e x)}{9 x^3}-\frac {1}{3} e^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 x^3}-\frac {1}{3} b e^3 n \operatorname {PolyLog}(2,-e x) \]
-5/36*b*e*n/x^2+4/9*b*e^2*n/x+1/9*b*e^3*n*ln(x)-1/6*b*e^3*n*ln(x)^2-1/6*e* (a+b*ln(c*x^n))/x^2+1/3*e^2*(a+b*ln(c*x^n))/x+1/3*e^3*ln(x)*(a+b*ln(c*x^n) )-1/9*b*e^3*n*ln(e*x+1)-1/9*b*n*ln(e*x+1)/x^3-1/3*e^3*(a+b*ln(c*x^n))*ln(e *x+1)-1/3*(a+b*ln(c*x^n))*ln(e*x+1)/x^3-1/3*b*e^3*n*polylog(2,-e*x)
Time = 0.07 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=-\frac {6 a e x+5 b e n x-12 a e^2 x^2-16 b e^2 n x^2+6 b e^3 n x^3 \log ^2(x)+6 b e x \log \left (c x^n\right )-12 b e^2 x^2 \log \left (c x^n\right )-4 e^3 x^3 \log (x) \left (3 a+b n+3 b \log \left (c x^n\right )\right )+12 a \log (1+e x)+4 b n \log (1+e x)+12 a e^3 x^3 \log (1+e x)+4 b e^3 n x^3 \log (1+e x)+12 b \log \left (c x^n\right ) \log (1+e x)+12 b e^3 x^3 \log \left (c x^n\right ) \log (1+e x)+12 b e^3 n x^3 \operatorname {PolyLog}(2,-e x)}{36 x^3} \]
-1/36*(6*a*e*x + 5*b*e*n*x - 12*a*e^2*x^2 - 16*b*e^2*n*x^2 + 6*b*e^3*n*x^3 *Log[x]^2 + 6*b*e*x*Log[c*x^n] - 12*b*e^2*x^2*Log[c*x^n] - 4*e^3*x^3*Log[x ]*(3*a + b*n + 3*b*Log[c*x^n]) + 12*a*Log[1 + e*x] + 4*b*n*Log[1 + e*x] + 12*a*e^3*x^3*Log[1 + e*x] + 4*b*e^3*n*x^3*Log[1 + e*x] + 12*b*Log[c*x^n]*L og[1 + e*x] + 12*b*e^3*x^3*Log[c*x^n]*Log[1 + e*x] + 12*b*e^3*n*x^3*PolyLo g[2, -(e*x)])/x^3
Time = 0.33 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2823, 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.
| \(\displaystyle \int \frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {\log (x) e^3}{3 x}-\frac {\log (e x+1) e^3}{3 x}+\frac {e^2}{3 x^2}-\frac {e}{6 x^3}-\frac {\log (e x+1)}{3 x^4}\right )dx+\frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{3} e^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{3} e^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}-b n \left (\frac {1}{3} e^3 \operatorname {PolyLog}(2,-e x)+\frac {1}{6} e^3 \log ^2(x)-\frac {1}{9} e^3 \log (x)+\frac {1}{9} e^3 \log (e x+1)-\frac {4 e^2}{9 x}+\frac {\log (e x+1)}{9 x^3}+\frac {5 e}{36 x^2}\right )\) |
-1/6*(e*(a + b*Log[c*x^n]))/x^2 + (e^2*(a + b*Log[c*x^n]))/(3*x) + (e^3*Lo g[x]*(a + b*Log[c*x^n]))/3 - (e^3*(a + b*Log[c*x^n])*Log[1 + e*x])/3 - ((a + b*Log[c*x^n])*Log[1 + e*x])/(3*x^3) - b*n*((5*e)/(36*x^2) - (4*e^2)/(9* x) - (e^3*Log[x])/9 + (e^3*Log[x]^2)/6 + (e^3*Log[1 + e*x])/9 + Log[1 + e* x]/(9*x^3) + (e^3*PolyLog[2, -(e*x)])/3)
3.1.9.3.1 Defintions of rubi rules used
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.23 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.45
| method | result | size |
| risch | \(\left (-\frac {\ln \left (e x +1\right ) b}{3 x^{3}}-\frac {b e \left (2 e^{2} \ln \left (e x +1\right ) x^{2}-2 e^{2} \ln \left (x \right ) x^{2}-2 e x +1\right )}{6 x^{2}}\right ) \ln \left (x^{n}\right )+\frac {4 b \,e^{2} n}{9 x}-\frac {5 b e n}{36 x^{2}}-\frac {b \,e^{3} n \ln \left (x \right )^{2}}{6}+\frac {n \,e^{3} b \ln \left (e x \right )}{9}-\frac {b \,e^{3} n \ln \left (e x +1\right )}{9}-\frac {b n \ln \left (e x +1\right )}{9 x^{3}}-\frac {e^{3} b n \operatorname {dilog}\left (e x +1\right )}{3}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) e^{3} \left (\frac {\ln \left (e x \right )}{3}-\frac {1}{6 x^{2} e^{2}}+\frac {1}{3 e x}-\frac {\ln \left (e x +1\right ) \left (e x +1\right ) \left (\left (e x +1\right )^{2}-3 e x \right )}{3 x^{3} e^{3}}\right )\) | \(282\) |
(-1/3*ln(e*x+1)/x^3*b-1/6*b*e*(2*e^2*ln(e*x+1)*x^2-2*e^2*ln(x)*x^2-2*e*x+1 )/x^2)*ln(x^n)+4/9*b*e^2*n/x-5/36*b*e*n/x^2-1/6*b*e^3*n*ln(x)^2+1/9*n*e^3* b*ln(e*x)-1/9*b*e^3*n*ln(e*x+1)-1/9*b*n*ln(e*x+1)/x^3-1/3*e^3*b*n*dilog(e* x+1)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c) *csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I* c*x^n)^3+b*ln(c)+a)*e^3*(1/3*ln(e*x)-1/6/x^2/e^2+1/3/e/x-1/3*ln(e*x+1)*(e* x+1)*((e*x+1)^2-3*e*x)/x^3/e^3)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=-\frac {1}{3} \, {\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b e^{3} n - \frac {1}{9} \, {\left (3 \, a e^{3} + {\left (e^{3} n + 3 \, e^{3} \log \left (c\right )\right )} b\right )} \log \left (e x + 1\right ) - \frac {6 \, b e^{3} n x^{3} \log \left (x\right )^{2} - 4 \, {\left (3 \, a e^{3} + {\left (e^{3} n + 3 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3} \log \left (x\right ) - 4 \, {\left (3 \, a e^{2} + {\left (4 \, e^{2} n + 3 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} + {\left ({\left (5 \, e n + 6 \, e \log \left (c\right )\right )} b + 6 \, a e\right )} x - 4 \, {\left (3 \, b e^{3} n x^{3} \log \left (x\right ) - b {\left (n + 3 \, \log \left (c\right )\right )} - 3 \, a\right )} \log \left (e x + 1\right ) - 6 \, {\left (2 \, b e^{3} x^{3} \log \left (x\right ) + 2 \, b e^{2} x^{2} - b e x - 2 \, {\left (b e^{3} x^{3} + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{36 \, x^{3}} \]
-1/3*(log(e*x + 1)*log(x) + dilog(-e*x))*b*e^3*n - 1/9*(3*a*e^3 + (e^3*n + 3*e^3*log(c))*b)*log(e*x + 1) - 1/36*(6*b*e^3*n*x^3*log(x)^2 - 4*(3*a*e^3 + (e^3*n + 3*e^3*log(c))*b)*x^3*log(x) - 4*(3*a*e^2 + (4*e^2*n + 3*e^2*lo g(c))*b)*x^2 + ((5*e*n + 6*e*log(c))*b + 6*a*e)*x - 4*(3*b*e^3*n*x^3*log(x ) - b*(n + 3*log(c)) - 3*a)*log(e*x + 1) - 6*(2*b*e^3*x^3*log(x) + 2*b*e^2 *x^2 - b*e*x - 2*(b*e^3*x^3 + b)*log(e*x + 1))*log(x^n))/x^3
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^4} \, dx=\int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \]