Integrand size = 20, antiderivative size = 163 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=-\frac {3 b e n}{4 x}-\frac {1}{4} b e^2 n \log (x)+\frac {1}{4} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} b e^2 n \log (1+e x)-\frac {b n \log (1+e x)}{4 x^2}+\frac {1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac {1}{2} b e^2 n \operatorname {PolyLog}(2,-e x) \] Output:
-3/4*b*e*n/x-1/4*b*e^2*n*ln(x)+1/4*b*e^2*n*ln(x)^2-1/2*e*(a+b*ln(c*x^n))/x -1/2*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 +1/2*e^2*(a+b*ln(c*x^n))*ln(e*x+1)-1/2*(a+b*ln(c*x^n))*ln(e*x+1)/x^2+1/2*b *e^2*n*polylog(2,-e*x)
Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=-\frac {1}{4} b e^2 \log (x) \left (n+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {b \left (-e n-2 e \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{4 x}-\frac {a \log (1+e x)}{2 x^2}+\frac {1}{4} b e^2 \left (n+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log (1+e x)-\frac {b \left (n+2 n \log (x)+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log (1+e x)}{4 x^2}+\frac {1}{2} a e \left (-\frac {1}{x}-e \log (x)+e \log (1+e x)\right )+\frac {1}{2} b e n \left (-\frac {1}{x}-\frac {\log (x)}{x}-\frac {1}{2} e \log ^2(x)+e^2 \left (\frac {\log (x) \log (1+e x)}{e}+\frac {\operatorname {PolyLog}(2,-e x)}{e}\right )\right ) \] Input:
Integrate[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]
Output:
-1/4*(b*e^2*Log[x]*(n + 2*(-(n*Log[x]) + Log[c*x^n]))) + (b*(-(e*n) - 2*e* (-(n*Log[x]) + Log[c*x^n])))/(4*x) - (a*Log[1 + e*x])/(2*x^2) + (b*e^2*(n + 2*(-(n*Log[x]) + Log[c*x^n]))*Log[1 + e*x])/4 - (b*(n + 2*n*Log[x] + 2*( -(n*Log[x]) + Log[c*x^n]))*Log[1 + e*x])/(4*x^2) + (a*e*(-x^(-1) - e*Log[x ] + e*Log[1 + e*x]))/2 + (b*e*n*(-x^(-1) - Log[x]/x - (e*Log[x]^2)/2 + e^2 *((Log[x]*Log[1 + e*x])/e + PolyLog[2, -(e*x)]/e)))/2
Time = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96, 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.
| \(\displaystyle \int \frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (-\frac {\log (x) e^2}{2 x}+\frac {\log (e x+1) e^2}{2 x}-\frac {e}{2 x^2}-\frac {\log (e x+1)}{2 x^3}\right )dx-\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-b n \left (-\frac {1}{2} e^2 \operatorname {PolyLog}(2,-e x)-\frac {1}{4} e^2 \log ^2(x)+\frac {1}{4} e^2 \log (x)-\frac {1}{4} e^2 \log (e x+1)+\frac {\log (e x+1)}{4 x^2}+\frac {3 e}{4 x}\right )\) |
Input:
Int[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]
Output:
-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)
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 = 2.22 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.50
| method | result | size |
| risch | \(\left (-\frac {\ln \left (e x +1\right ) b}{2 x^{2}}-\frac {b e \left (e x \ln \left (x \right )-e \ln \left (e x +1\right ) x +1\right )}{2 x}\right ) \ln \left (x^{n}\right )+\frac {n \,e^{2} b \operatorname {dilog}\left (e x +1\right )}{2}-\frac {n \,e^{2} b \ln \left (e x \right )}{4}-\frac {3 b e n}{4 x}+\frac {b \,e^{2} n \ln \left (e x +1\right )}{4}-\frac {b n \ln \left (e x +1\right )}{4 x^{2}}+\frac {b \,e^{2} n \ln \left (x \right )^{2}}{4}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) e^{2} \left (-\frac {\ln \left (e x \right )}{2}-\frac {1}{2 e x}+\frac {\ln \left (e x +1\right ) \left (e x +1\right ) \left (e x -1\right )}{2 x^{2} e^{2}}\right )\) | \(244\) |
Input:
int((a+b*ln(c*x^n))*ln(e*x+1)/x^3,x,method=_RETURNVERBOSE)
Output:
(-1/2*ln(e*x+1)/x^2*b-1/2*b*e*(e*x*ln(x)-e*ln(e*x+1)*x+1)/x)*ln(x^n)+1/2*n *e^2*b*dilog(e*x+1)-1/4*n*e^2*b*ln(e*x)-3/4*b*e*n/x+1/4*b*e^2*n*ln(e*x+1)- 1/4*b*n*ln(e*x+1)/x^2+1/4*b*e^2*n*ln(x)^2+(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c *x^n)^2-1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c *x^n)^3+1/2*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)*e^2*(-1/2*ln(e*x)- 1/2/e/x+1/2*ln(e*x+1)*(e*x+1)*(e*x-1)/x^2/e^2)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{3}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="fricas")
Output:
integral((b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1))/x^3, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*x**n))*ln(e*x+1)/x**3,x)
Output:
Timed out
Time = 0.07 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\frac {1}{2} \, {\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b e^{2} n + \frac {1}{4} \, {\left (2 \, a e^{2} + {\left (e^{2} n + 2 \, e^{2} \log \left (c\right )\right )} b\right )} \log \left (e x + 1\right ) + \frac {b e^{2} n x^{2} \log \left (x\right )^{2} - {\left (2 \, a e^{2} + {\left (e^{2} n + 2 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} \log \left (x\right ) - {\left ({\left (3 \, e n + 2 \, e \log \left (c\right )\right )} b + 2 \, a e\right )} x - {\left (2 \, b e^{2} n x^{2} \log \left (x\right ) + b {\left (n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} \log \left (e x + 1\right ) - 2 \, {\left (b e^{2} x^{2} \log \left (x\right ) + b e x - {\left (b e^{2} x^{2} - b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{4 \, x^{2}} \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="maxima")
Output:
1/2*(log(e*x + 1)*log(x) + dilog(-e*x))*b*e^2*n + 1/4*(2*a*e^2 + (e^2*n + 2*e^2*log(c))*b)*log(e*x + 1) + 1/4*(b*e^2*n*x^2*log(x)^2 - (2*a*e^2 + (e^ 2*n + 2*e^2*log(c))*b)*x^2*log(x) - ((3*e*n + 2*e*log(c))*b + 2*a*e)*x - ( 2*b*e^2*n*x^2*log(x) + b*(n + 2*log(c)) + 2*a)*log(e*x + 1) - 2*(b*e^2*x^2 *log(x) + b*e*x - (b*e^2*x^2 - b)*log(e*x + 1))*log(x^n))/x^2
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )}{x^{3}} \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*log(e*x + 1)/x^3, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \] Input:
int((log(e*x + 1)*(a + b*log(c*x^n)))/x^3,x)
Output:
int((log(e*x + 1)*(a + b*log(c*x^n)))/x^3, x)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{4}+x^{3}}d x \right ) b \,x^{2}-4 \,\mathrm {log}\left (e x +1\right ) \mathrm {log}\left (x^{n} c \right ) b +4 \,\mathrm {log}\left (e x +1\right ) a \,e^{2} x^{2}-4 \,\mathrm {log}\left (e x +1\right ) a +2 \,\mathrm {log}\left (e x +1\right ) b \,e^{2} n \,x^{2}-2 \,\mathrm {log}\left (e x +1\right ) b n -2 \,\mathrm {log}\left (x^{n} c \right ) b -4 \,\mathrm {log}\left (x \right ) a \,e^{2} x^{2}-2 \,\mathrm {log}\left (x \right ) b \,e^{2} n \,x^{2}-4 a e x -2 b e n x -b n}{8 x^{2}} \] Input:
int((a+b*log(c*x^n))*log(e*x+1)/x^3,x)
Output:
( - 4*int(log(x**n*c)/(e*x**4 + x**3),x)*b*x**2 - 4*log(e*x + 1)*log(x**n* c)*b + 4*log(e*x + 1)*a*e**2*x**2 - 4*log(e*x + 1)*a + 2*log(e*x + 1)*b*e* *2*n*x**2 - 2*log(e*x + 1)*b*n - 2*log(x**n*c)*b - 4*log(x)*a*e**2*x**2 - 2*log(x)*b*e**2*n*x**2 - 4*a*e*x - 2*b*e*n*x - b*n)/(8*x**2)