\(\int \frac {a+b \log (c (d+e x)^n)}{f+g x} \, dx\) [220]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 63 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g} \] Output:

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx=\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g}+\frac {b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2841, 2840, 2838}

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.

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

\(\Big \downarrow \) 2841

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

\(\Big \downarrow \) 2840

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2838
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 2840
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_ 
Symbol] :> Simp[1/g   Subst[Int[(a + b*Log[1 + c*e*(x/g)])/x, x], x, f + g* 
x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g + c 
*(e*f - d*g), 0]
 

rule 2841
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_ 
)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x 
)^n])/g), x] - Simp[b*e*(n/g)   Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]
 
Maple [C] (warning: unable to verify)

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

Time = 1.05 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.44

method result size
risch \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g}-\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g}-\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g}+\frac {\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \ln \left (g x +f \right )}{g}\) \(217\)

Input:

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

Output:

b*ln((e*x+d)^n)*ln(g*x+f)/g-b/g*n*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-b/g 
*n*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)*ln(g*x+f)/g
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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