\(\int \frac {x (a+b \log (c (d+e x)^n))}{(f+g x)^2} \, dx\) [251]

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

Optimal result

Integrand size = 23, antiderivative size = 138 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=-\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+b n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((b*e*f*n*Log[d + e*x])/(g^2*(e*f - d*g))) + (f*(a + b*Log[c*(d + e*x)^n] 
))/(g^2*(f + g*x)) + (b*e*f*n*Log[f + g*x])/(g^2*(e*f - d*g)) + ((a + b*Lo 
g[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^2 + (b*n*PolyLog[2, -( 
(g*(d + e*x))/(e*f - d*g))])/g^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 = 1.56 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.25

method result size
risch \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f}{g^{2} \left (g x +f \right )}-\frac {b n \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}-\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}-\frac {b e n f \ln \left (g x +f \right )}{g^{2} \left (d g -e f \right )}+\frac {b e n f \ln \left (\left (g x +f \right ) e +d g -e f \right )}{g^{2} \left (d g -e f \right )}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (g x +f \right )}{g^{2}}+\frac {f}{g^{2} \left (g x +f \right )}\right )\) \(311\)

Input:

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

Output:

b*ln((e*x+d)^n)/g^2*ln(g*x+f)+b*ln((e*x+d)^n)/g^2*f/(g*x+f)-b*n/g^2*dilog( 
((g*x+f)*e+d*g-e*f)/(d*g-e*f))-b*n/g^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d 
*g-e*f))-b*e*n/g^2*f/(d*g-e*f)*ln(g*x+f)+b*e*n/g^2*f/(d*g-e*f)*ln((g*x+f)* 
e+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/g^2*ln(g*x 
+f)+1/g^2*f/(g*x+f))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int((log((d + e*x)**n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2*x**2 + e*f**2* 
x + 2*e*f*g*x**2 + e*g**2*x**3),x)*b*d**3*f*g**4*n + 2*int((log((d + e*x)* 
*n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2*x**2 + e*f**2*x + 2*e*f*g*x**2 + e*g 
**2*x**3),x)*b*d**3*g**5*n*x - 6*int((log((d + e*x)**n*c)*x)/(d*f**2 + 2*d 
*f*g*x + d*g**2*x**2 + e*f**2*x + 2*e*f*g*x**2 + e*g**2*x**3),x)*b*d**2*e* 
f**2*g**3*n - 6*int((log((d + e*x)**n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2*x 
**2 + e*f**2*x + 2*e*f*g*x**2 + e*g**2*x**3),x)*b*d**2*e*f*g**4*n*x + 6*in 
t((log((d + e*x)**n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2*x**2 + e*f**2*x + 2 
*e*f*g*x**2 + e*g**2*x**3),x)*b*d*e**2*f**3*g**2*n + 6*int((log((d + e*x)* 
*n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2*x**2 + e*f**2*x + 2*e*f*g*x**2 + e*g 
**2*x**3),x)*b*d*e**2*f**2*g**3*n*x - 2*int((log((d + e*x)**n*c)*x)/(d*f** 
2 + 2*d*f*g*x + d*g**2*x**2 + e*f**2*x + 2*e*f*g*x**2 + e*g**2*x**3),x)*b* 
e**3*f**4*g*n - 2*int((log((d + e*x)**n*c)*x)/(d*f**2 + 2*d*f*g*x + d*g**2 
*x**2 + e*f**2*x + 2*e*f*g*x**2 + e*g**2*x**3),x)*b*e**3*f**3*g**2*n*x + 2 
*log(d + e*x)*b*d*e*f**2*g*n**2 + 2*log(d + e*x)*b*d*e*f*g**2*n**2*x + 2*l 
og(f + g*x)*a*d**2*f*g**2*n + 2*log(f + g*x)*a*d**2*g**3*n*x - 2*log(f + g 
*x)*a*d*e*f**2*g*n - 2*log(f + g*x)*a*d*e*f*g**2*n*x - 2*log(f + g*x)*b*e* 
*2*f**3*n**2 - 2*log(f + g*x)*b*e**2*f**2*g*n**2*x + log((d + e*x)**n*c)** 
2*b*d**2*f*g**2 + log((d + e*x)**n*c)**2*b*d**2*g**3*x - log((d + e*x)**n* 
c)**2*b*d*e*f**2*g - log((d + e*x)**n*c)**2*b*d*e*f*g**2*x - 2*log((d +...