\(\int \frac {\log (c (a+\frac {b}{x})^p)}{d+e x} \, dx\) [201]

Optimal result

Integrand size = 20, antiderivative size = 113 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}+\frac {p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2912, 2005, 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.

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

\(\Big \downarrow \) 2912

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

\(\Big \downarrow \) 2005

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2005

Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m 
+ n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg 
Q[n]
 

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]
 

rule 2912

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_. 
)*(x_)), x_Symbol] :> Simp[Log[f + g*x]*((a + b*Log[c*(d + e*x^n)^p])/g), x 
] - Simp[b*e*n*(p/g)   Int[x^(n - 1)*(Log[f + g*x]/(d + e*x^n)), x], x] /; 
FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]
 

Maple [A] (verified)

Time = 2.89 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21

method	result	size
parts	\(\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e}+p b \left (-\frac {\left (\frac {\operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{a}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{a}\right ) a}{b e}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b e}\right )\)	\(137\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.41 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b p {\left (\frac {\log \left (e x + d\right ) \log \left (a + \frac {b}{x}\right )}{b} - \frac {\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )}{b} + \frac {\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )}{b}\right )}}{e} - \frac {p \log \left (e x + d\right ) \log \left (a + \frac {b}{x}\right )}{e} + \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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