\(\int \frac {x \log (c (a+\frac {b}{x})^p)}{d+e x} \, dx\) [242]

Optimal result

Integrand size = 21, antiderivative size = 151 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}+\frac {b p \log (b+a x)}{a e}-\frac {d \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^2}-\frac {d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^2}-\frac {d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \] Output:

x*ln(c*(a+b/x)^p)/e+b*p*ln(a*x+b)/a/e-d*ln(c*(a+b/x)^p)*ln(e*x+d)/e^2-d*p* 
ln(-e*x/d)*ln(e*x+d)/e^2+d*p*ln(-e*(a*x+b)/(a*d-b*e))*ln(e*x+d)/e^2+d*p*po 
lylog(2,a*(e*x+d)/(a*d-b*e))/e^2-d*p*polylog(2,1+e*x/d)/e^2
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b p \log \left (a+\frac {b}{x}\right )}{a e}+\frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}+\frac {b p \log (x)}{a e}-\frac {d \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^2}-\frac {d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^2}-\frac {d p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2916, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2916

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

Maple [A] (verified)

Time = 2.63 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.25

method	result	size
parts	\(\frac {x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e}-\frac {d \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}+p b e \left (\frac {\ln \left (d a -a \left (e x +d \right )-b e \right )}{e^{2} a}+\frac {d \operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{3} b}+\frac {d \ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{3} b}-\frac {d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3} b}-\frac {d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{3} b}\right )\)	\(188\)

Input:

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

Output:

x*ln(c*(a+b/x)^p)/e-d*ln(c*(a+b/x)^p)*ln(e*x+d)/e^2+p*b*e*(1/e^2*ln(d*a-a* 
(e*x+d)-b*e)/a+1/e^3*d/b*dilog((-d*a+a*(e*x+d)+b*e)/(-a*d+b*e))+1/e^3*d/b* 
ln(e*x+d)*ln((-d*a+a*(e*x+d)+b*e)/(-a*d+b*e))-1/e^3*d/b*ln(e*x+d)*ln(-e*x/ 
d)-1/e^3*d/b*dilog(-e*x/d))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((log(((a*x + b)**p*c)/x**p)*x)/(d + e*x),x)