\(\int \frac {a+b \log (c (d+e x)^n)}{x (f+g x)^2} \, dx\) [253]

Optimal result

Integrand size = 25, antiderivative size = 179 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=-\frac {b e n \log (d+e x)}{f (e f-d g)}+\frac {a+b \log \left (c (d+e x)^n\right )}{f (f+g x)}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {b e n \log (f+g x)}{f (e f-d g)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^2}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^2}+\frac {b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\frac {\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}-\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )+b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \] Input:

Integrate[(a + b*Log[c*(d + e*x)^n])/(x*(f + g*x)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2863, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*Log[c*(d + e*x)^n])/(x*(f + g*x)^2),x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

Maple [C] (warning: unable to verify)

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

Time = 1.41 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.98

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f \left (g x +f \right )}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f^{2}}-\frac {b n \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {b e n \ln \left (g x +f \right )}{f \left (d g -e f \right )}+\frac {b e n \ln \left (e x +d \right )}{f \left (d g -e f \right )}+\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}+\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (g x +f \right )}{f^{2}}+\frac {1}{f \left (g x +f \right )}+\frac {\ln \left (x \right )}{f^{2}}\right )\)	\(354\)

Input:

int((a+b*ln(c*(e*x+d)^n))/x/(g*x+f)^2,x,method=_RETURNVERBOSE)
 

Output:

-b*ln((e*x+d)^n)/f^2*ln(g*x+f)+b*ln((e*x+d)^n)/f/(g*x+f)+b*ln((e*x+d)^n)/f 
^2*ln(x)-b*n/f^2*dilog((e*x+d)/d)-b*n/f^2*ln(x)*ln((e*x+d)/d)-b*e*n/f/(d*g 
-e*f)*ln(g*x+f)+b*e*n/f/(d*g-e*f)*ln(e*x+d)+b*n/f^2*dilog(((g*x+f)*e+d*g-e 
*f)/(d*g-e*f))+b*n/f^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+(1/2*I* 
b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)* 
csgn(I*c*(e*x+d)^n)*csgn(I*c)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*I*b*Pi* 
csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+b*ln(c)+a)*(-1/f^2*ln(g*x+f)+1/f/(g*x+f)+1 
/f^2*ln(x))
 

Fricas [F]

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

integrate((a+b*log(c*(e*x+d)^n))/x/(g*x+f)^2,x, algorithm="fricas")
 

Output:

integral((b*log((e*x + d)^n*c) + a)/(g^2*x^3 + 2*f*g*x^2 + f^2*x), x)
 

Sympy [F(-1)]

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

integrate((a+b*ln(c*(e*x+d)**n))/x/(g*x+f)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*log(c*(e*x+d)^n))/x/(g*x+f)^2,x, algorithm="maxima")
 

Output:

a*(1/(f*g*x + f^2) - log(g*x + f)/f^2 + log(x)/f^2) + b*integrate((log((e* 
x + d)^n) + log(c))/(g^2*x^3 + 2*f*g*x^2 + f^2*x), x)
 

Giac [F]

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

integrate((a+b*log(c*(e*x+d)^n))/x/(g*x+f)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x (f+g x)^2} \, dx=\frac {\left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{3}+2 f g \,x^{2}+f^{2} x}d x \right ) b \,f^{3}+\left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g^{2} x^{3}+2 f g \,x^{2}+f^{2} x}d x \right ) b \,f^{2} g x -\mathrm {log}\left (g x +f \right ) a f -\mathrm {log}\left (g x +f \right ) a g x +\mathrm {log}\left (x \right ) a f +\mathrm {log}\left (x \right ) a g x -a g x}{f^{2} \left (g x +f \right )} \] Input:

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

Output:

(int(log((d + e*x)**n*c)/(f**2*x + 2*f*g*x**2 + g**2*x**3),x)*b*f**3 + int 
(log((d + e*x)**n*c)/(f**2*x + 2*f*g*x**2 + g**2*x**3),x)*b*f**2*g*x - log 
(f + g*x)*a*f - log(f + g*x)*a*g*x + log(x)*a*f + log(x)*a*g*x - a*g*x)/(f 
**2*(f + g*x))