Integrand size = 17, antiderivative size = 74 \[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=2 b n x-x \left (a+b \log \left (c x^n\right )\right )-\frac {b n (1+e x) \log (1+e x)}{e}+\frac {(1+e x) \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{e}+\frac {b n \operatorname {PolyLog}(2,-e x)}{e} \] Output:
2*b*n*x-x*(a+b*ln(c*x^n))-b*n*(e*x+1)*ln(e*x+1)/e+(e*x+1)*(a+b*ln(c*x^n))* ln(e*x+1)/e+b*n*polylog(2,-e*x)/e
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.22 \[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\frac {-a e x+2 b e n x+a \log (1+e x)-b n \log (1+e x)+a e x \log (1+e x)-b e n x \log (1+e x)+b \log \left (c x^n\right ) (-e x+(1+e x) \log (1+e x))+b n \operatorname {PolyLog}(2,-e x)}{e} \] Input:
Integrate[(a + b*Log[c*x^n])*Log[1 + e*x],x]
Output:
(-(a*e*x) + 2*b*e*n*x + a*Log[1 + e*x] - b*n*Log[1 + e*x] + a*e*x*Log[1 + e*x] - b*e*n*x*Log[1 + e*x] + b*Log[c*x^n]*(-(e*x) + (1 + e*x)*Log[1 + e*x ]) + b*n*PolyLog[2, -(e*x)])/e
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2817, 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.
\(\displaystyle \int \log (e x+1) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2817 |
\(\displaystyle -b n \int \left (\frac {(e x+1) \log (e x+1)}{e x}-1\right )dx+\frac {(e x+1) \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(e x+1) \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {\operatorname {PolyLog}(2,-e x)}{e}+\frac {(e x+1) \log (e x+1)}{e}-2 x\right )\) |
Input:
Int[(a + b*Log[c*x^n])*Log[1 + e*x],x]
Output:
-(x*(a + b*Log[c*x^n])) + ((1 + e*x)*(a + b*Log[c*x^n])*Log[1 + e*x])/e - b*n*(-2*x + ((1 + e*x)*Log[1 + e*x])/e - PolyLog[2, -(e*x)]/e)
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))^(p_.), x_Symbol] :> With[{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])^p u, x] - Simp[b*n*p Int[(a + b*Log[c*x^n])^(p - 1)/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 1]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.78 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.58
method | result | size |
risch | \(\left (b x \ln \left (e x +1\right )+\frac {b \left (-e x +\ln \left (e x +1\right )\right )}{e}\right ) \ln \left (x^{n}\right )+\frac {\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\ln \left (e x +1\right ) \left (e x +1\right )-e x -1\right )}{e}-x b n \ln \left (e x +1\right )+2 b n x -\frac {n b \ln \left (e x +1\right )}{e}+\frac {n b \operatorname {dilog}\left (e x +1\right )}{e}+\frac {2 n b}{e}\) | \(191\) |
Input:
int((a+b*ln(c*x^n))*ln(e*x+1),x,method=_RETURNVERBOSE)
Output:
(b*x*ln(e*x+1)+b*(-e*x+ln(e*x+1))/e)*ln(x^n)+(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*(ln(e*x+1)*(e *x+1)-e*x-1)-x*b*n*ln(e*x+1)+2*b*n*x-n*b/e*ln(e*x+1)+n*b/e*dilog(e*x+1)+2* n*b/e
\[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right ) \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1),x, algorithm="fricas")
Output:
integral(b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1), x)
Timed out. \[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*x**n))*ln(e*x+1),x)
Output:
Timed out
Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\frac {{\left (\log \left (e x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-e x\right )\right )} b n}{e} - \frac {{\left (b {\left (n - \log \left (c\right )\right )} - a\right )} \log \left (e x + 1\right )}{e} + \frac {{\left ({\left (2 \, e n - e \log \left (c\right )\right )} b - a e\right )} x - {\left (b n \log \left (x\right ) + {\left ({\left (e n - e \log \left (c\right )\right )} b - a e\right )} x\right )} \log \left (e x + 1\right ) - {\left (b e x - {\left (b e x + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{e} \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1),x, algorithm="maxima")
Output:
(log(e*x + 1)*log(x) + dilog(-e*x))*b*n/e - (b*(n - log(c)) - a)*log(e*x + 1)/e + (((2*e*n - e*log(c))*b - a*e)*x - (b*n*log(x) + ((e*n - e*log(c))* b - a*e)*x)*log(e*x + 1) - (b*e*x - (b*e*x + b)*log(e*x + 1))*log(x^n))/e
\[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right ) \,d x } \] Input:
integrate((a+b*log(c*x^n))*log(e*x+1),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*log(e*x + 1), x)
Timed out. \[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\int \ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:
int(log(e*x + 1)*(a + b*log(c*x^n)),x)
Output:
int(log(e*x + 1)*(a + b*log(c*x^n)), x)
\[ \int \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx=\frac {-2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{2}+x}d x \right ) b n +2 \,\mathrm {log}\left (e x +1\right ) \mathrm {log}\left (x^{n} c \right ) b e n x +2 \,\mathrm {log}\left (e x +1\right ) a e n x +2 \,\mathrm {log}\left (e x +1\right ) a n -2 \,\mathrm {log}\left (e x +1\right ) b e \,n^{2} x -2 \,\mathrm {log}\left (e x +1\right ) b \,n^{2}+\mathrm {log}\left (x^{n} c \right )^{2} b -2 \,\mathrm {log}\left (x^{n} c \right ) b e n x -2 a e n x +4 b e \,n^{2} x}{2 e n} \] Input:
int((a+b*log(c*x^n))*log(e*x+1),x)
Output:
( - 2*int(log(x**n*c)/(e*x**2 + x),x)*b*n + 2*log(e*x + 1)*log(x**n*c)*b*e *n*x + 2*log(e*x + 1)*a*e*n*x + 2*log(e*x + 1)*a*n - 2*log(e*x + 1)*b*e*n* *2*x - 2*log(e*x + 1)*b*n**2 + log(x**n*c)**2*b - 2*log(x**n*c)*b*e*n*x - 2*a*e*n*x + 4*b*e*n**2*x)/(2*e*n)