\(\int \frac {x (a+b \log (c x^n))}{(d+e x)^2} \, dx\) [41]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac {\left (a+b n+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \] Output:

-x*(a+b*ln(c*x^n))/e/(e*x+d)+(a+b*n+b*ln(c*x^n))*ln(1+e*x/d)/e^2+b*n*polyl 
og(2,-e*x/d)/e^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {\frac {d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-b n (\log (x)-\log (d+e x))+\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2784, 2754, 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.

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

\(\Big \downarrow \) 2784

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

\(\Big \downarrow \) 2754

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

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 2784

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

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]
 

Maple [C] (warning: unable to verify)

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

Time = 0.46 (sec) , antiderivative size = 205, normalized size of antiderivative = 3.15

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) d}{e^{2} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{2}}-\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{2}}-\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{2}}+\frac {b n \ln \left (e x +d \right )}{e^{2}}-\frac {b n \ln \left (e x \right )}{e^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {d}{e^{2} \left (e x +d \right )}+\frac {\ln \left (e x +d \right )}{e^{2}}\right )\)	\(205\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(a + b*log(c*x**n))/(d + e*x)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {-2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{3}+2 d e \,x^{2}+d^{2} x}d x \right ) b \,d^{3} n -2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{3}+2 d e \,x^{2}+d^{2} x}d x \right ) b \,d^{2} e n x +2 \,\mathrm {log}\left (e x +d \right ) a d n +2 \,\mathrm {log}\left (e x +d \right ) a e n x +4 \,\mathrm {log}\left (e x +d \right ) b d \,n^{2}+4 \,\mathrm {log}\left (e x +d \right ) b e \,n^{2} x +\mathrm {log}\left (x^{n} c \right )^{2} b d +\mathrm {log}\left (x^{n} c \right )^{2} b e x -4 \,\mathrm {log}\left (x^{n} c \right ) b e n x -2 a e n x}{2 e^{2} n \left (e x +d \right )} \] Input:

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

Output:

( - 2*int(log(x**n*c)/(d**2*x + 2*d*e*x**2 + e**2*x**3),x)*b*d**3*n - 2*in 
t(log(x**n*c)/(d**2*x + 2*d*e*x**2 + e**2*x**3),x)*b*d**2*e*n*x + 2*log(d 
+ e*x)*a*d*n + 2*log(d + e*x)*a*e*n*x + 4*log(d + e*x)*b*d*n**2 + 4*log(d 
+ e*x)*b*e*n**2*x + log(x**n*c)**2*b*d + log(x**n*c)**2*b*e*x - 4*log(x**n 
*c)*b*e*n*x - 2*a*e*n*x)/(2*e**2*n*(d + e*x))