\(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x} \, dx\) [362]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b n \log ^2\left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 m}-b n \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+b m n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right ) \] Output:

1/2*ln(f*x^m)^2*(a+b*ln(c*(e*x+d)^n))/m-1/2*b*n*ln(f*x^m)^2*ln(1+e*x/d)/m- 
b*n*ln(f*x^m)*polylog(2,-e*x/d)+b*m*n*polylog(3,-e*x/d)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.45 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {1}{2} \left (\frac {a \log ^2\left (f x^m\right )}{m}-b m \log ^2(x) \log \left (c (d+e x)^n\right )+2 b \log (x) \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b m n \log ^2(x) \log \left (1+\frac {e x}{d}\right )-2 b n \log (x) \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )-2 b n \log \left (f x^m\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+2 b m n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right ) \] Input:

Integrate[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]))/x,x]
 

Output:

((a*Log[f*x^m]^2)/m - b*m*Log[x]^2*Log[c*(d + e*x)^n] + 2*b*Log[x]*Log[f*x 
^m]*Log[c*(d + e*x)^n] + b*m*n*Log[x]^2*Log[1 + (e*x)/d] - 2*b*n*Log[x]*Lo 
g[f*x^m]*Log[1 + (e*x)/d] - 2*b*n*Log[f*x^m]*PolyLog[2, -((e*x)/d)] + 2*b* 
m*n*PolyLog[3, -((e*x)/d)])/2
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2872, 2754, 2821, 7143}

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.

\(\displaystyle \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx\)

\(\Big \downarrow \) 2872

\(\displaystyle \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b e n \int \frac {\log ^2\left (f x^m\right )}{d+e x}dx}{2 m}\)

\(\Big \downarrow \) 2754

\(\displaystyle \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b e n \left (\frac {\log \left (\frac {e x}{d}+1\right ) \log ^2\left (f x^m\right )}{e}-\frac {2 m \int \frac {\log \left (f x^m\right ) \log \left (\frac {e x}{d}+1\right )}{x}dx}{e}\right )}{2 m}\)

\(\Big \downarrow \) 2821

\(\displaystyle \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b e n \left (\frac {\log \left (\frac {e x}{d}+1\right ) \log ^2\left (f x^m\right )}{e}-\frac {2 m \left (m \int \frac {\operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{x}dx-\operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \log \left (f x^m\right )\right )}{e}\right )}{2 m}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\log ^2\left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 m}-\frac {b e n \left (\frac {\log \left (\frac {e x}{d}+1\right ) \log ^2\left (f x^m\right )}{e}-\frac {2 m \left (m \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )-\operatorname {PolyLog}\left (2,-\frac {e x}{d}\right ) \log \left (f x^m\right )\right )}{e}\right )}{2 m}\)

Input:

Int[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]))/x,x]
 

Output:

(Log[f*x^m]^2*(a + b*Log[c*(d + e*x)^n]))/(2*m) - (b*e*n*((Log[f*x^m]^2*Lo 
g[1 + (e*x)/d])/e - (2*m*(-(Log[f*x^m]*PolyLog[2, -((e*x)/d)]) + m*PolyLog 
[3, -((e*x)/d)]))/e))/(2*m)
 

Defintions of rubi rules used

rule 2754

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symb 
ol] :> Simp[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^p/e), x] - Simp[b*n*(p/e) 
  Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, 
b, c, d, e, n}, x] && IGtQ[p, 0]
 

rule 2821

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b 
_.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c 
*x^n])^p/m), x] + Simp[b*n*(p/m)   Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c 
*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 
0] && EqQ[d*e, 1]
 

rule 2872

Int[(Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b 
_.)))/(x_), x_Symbol] :> Simp[Log[f*x^m]^2*((a + b*Log[c*(d + e*x)^n])/(2*m 
)), x] - Simp[b*e*(n/(2*m))   Int[Log[f*x^m]^2/(d + e*x), x], x] /; FreeQ[{ 
a, b, c, d, e, f, m, n}, x]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 9.75 (sec) , antiderivative size = 756, normalized size of antiderivative = 8.59

method	result	size
risch	\(\text {Expression too large to display}\)	\(756\)

Input:

int(ln(f*x^m)*(a+b*ln(c*(e*x+d)^n))/x,x,method=_RETURNVERBOSE)
 

Output:

(b*ln(x)*ln(x^m)-1/2*b*m*ln(x)^2-1/2*I*ln(x)*Pi*b*csgn(I*f)*csgn(I*x^m)*cs 
gn(I*f*x^m)+1/2*I*ln(x)*Pi*b*csgn(I*f)*csgn(I*f*x^m)^2+1/2*I*ln(x)*Pi*b*cs 
gn(I*x^m)*csgn(I*f*x^m)^2-1/2*I*ln(x)*Pi*b*csgn(I*f*x^m)^3+ln(x)*ln(f)*b)* 
ln((e*x+d)^n)+(1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b* 
Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-1/4*I*b*Pi*csgn(I*c*(e* 
x+d)^n)^3+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+1/2*b*ln(c)+1/2*a)*(I 
*Pi*csgn(I*f)*csgn(I*f*x^m)^2*ln(x)+I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2*ln(x) 
+2*ln(x)*ln(f)-I*Pi*csgn(I*f*x^m)^3*ln(x)-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn( 
I*f*x^m)*ln(x)+1/m*ln(x^m)^2)-1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*x^m)*cs 
gn(I*f*x^m)^2+1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f 
*x^m)+1/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f*x^m)^3-1/2*I*n*b*ln(x)*ln((e* 
x+d)/d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/2*I*n*b*ln(x)*ln((e*x+d)/d)*Pi*cs 
gn(I*f)*csgn(I*f*x^m)^2+1/2*I*n*b*ln(x)*ln((e*x+d)/d)*Pi*csgn(I*f*x^m)^3-1 
/2*I*n*b*dilog((e*x+d)/d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/2*I*n*b*ln(x)*ln( 
(e*x+d)/d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/2*n*b*m*ln(x)^2*ln(1+e 
*x/d)+n*b*ln(x)^2*ln((e*x+d)/d)*m-n*b*ln(x)*ln((e*x+d)/d)*ln(x^m)-n*b*ln(x 
)*ln((e*x+d)/d)*ln(f)-n*b*m*ln(x)*polylog(2,-e*x/d)+n*b*dilog((e*x+d)/d)*m 
*ln(x)-n*b*dilog((e*x+d)/d)*ln(x^m)-n*b*dilog((e*x+d)/d)*ln(f)+b*m*n*polyl 
og(3,-e*x/d)
 

Fricas [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x} \,d x } \] Input:

integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x,x, algorithm="fricas")
 

Output:

integral((b*log((e*x + d)^n*c)*log(f*x^m) + a*log(f*x^m))/x, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate(ln(f*x**m)*(a+b*ln(c*(e*x+d)**n))/x,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x} \,d x } \] Input:

integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x,x, algorithm="maxima")
 

Output:

-1/2*(b*m*log(x)^2 - 2*b*log(f)*log(x) - 2*b*log(x)*log(x^m))*log((e*x + d 
)^n) - integrate(-1/2*(b*e*m*n*x*log(x)^2 - 2*b*e*n*x*log(f)*log(x) + 2*b* 
d*log(c)*log(f) + 2*a*d*log(f) + 2*(b*e*log(c)*log(f) + a*e*log(f))*x - 2* 
(b*e*n*x*log(x) - b*d*log(c) - a*d - (b*e*log(c) + a*e)*x)*log(x^m))/(e*x^ 
2 + d*x), x)
 

Giac [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x} \,d x } \] Input:

integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x,x, algorithm="giac")
 

Output:

integrate((b*log((e*x + d)^n*c) + a)*log(f*x^m)/x, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x} \,d x \] Input:

int((log(f*x^m)*(a + b*log(c*(d + e*x)^n)))/x,x)
 

Output:

int((log(f*x^m)*(a + b*log(c*(d + e*x)^n)))/x, x)
 

Reduce [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) \mathrm {log}\left (x^{m} f \right )}{x}d x \right ) b m +\mathrm {log}\left (x^{m} f \right )^{2} a}{2 m} \] Input:

int(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x,x)
 

Output:

(2*int((log((d + e*x)**n*c)*log(x**m*f))/x,x)*b*m + log(x**m*f)**2*a)/(2*m 
)