\(\int \frac {(a+b \log (c x^n)) \log (d (e+f x)^m)}{x} \, dx\) [80]

Optimal result

Integrand size = 24, antiderivative size = 100 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )}{2 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 b n}-m \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )+b m n \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{x} \, dx=-\frac {1}{2} b n \log ^2(x) \log \left (d (e+f x)^m\right )+a \log \left (-\frac {f x}{e}\right ) \log \left (d (e+f x)^m\right )+b \log (x) \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} b m n \log ^2(x) \log \left (1+\frac {f x}{e}\right )-b m \log (x) \log \left (c x^n\right ) \log \left (1+\frac {f x}{e}\right )-b m \log \left (c x^n\right ) \operatorname {PolyLog}\left (2,-\frac {f x}{e}\right )+a m \operatorname {PolyLog}\left (2,1+\frac {f x}{e}\right )+b m n \operatorname {PolyLog}\left (3,-\frac {f x}{e}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2822, 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.

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

\(\Big \downarrow \) 2822

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

\(\Big \downarrow \) 2754

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

\(\Big \downarrow \) 2821

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 2822

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

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 = 9.58 (sec) , antiderivative size = 793, normalized size of antiderivative = 7.93

method	result	size
risch	\(\text {Expression too large to display}\)	\(793\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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