\(\int \frac {\log (c (a+\frac {b}{x})^p)}{x^3 (d+e x)} \, dx\) [246]

Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^3} \] Output:

1/4*p/d/x^2-1/2*a*p/b/d/x-e*p/d^2/x+1/2*a^2*p*ln(a+b/x)/b^2/d+e*(a+b/x)*ln 
(c*(a+b/x)^p)/b/d^2-1/2*ln(c*(a+b/x)^p)/d/x^2-e^2*ln(c*(a+b/x)^p)*ln(-b/a/ 
x)/d^3-e^2*ln(c*(a+b/x)^p)*ln(e*x+d)/d^3-e^2*p*ln(-e*x/d)*ln(e*x+d)/d^3+e^ 
2*p*ln(-e*(a*x+b)/(a*d-b*e))*ln(e*x+d)/d^3-e^2*p*polylog(2,1+b/a/x)/d^3+e^ 
2*p*polylog(2,a*(e*x+d)/(a*d-b*e))/d^3-e^2*p*polylog(2,1+e*x/d)/d^3
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.92 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=-\frac {-\frac {d^2 p}{x^2}+\frac {2 a d^2 p}{b x}+\frac {4 d e p}{x}-\frac {2 a^2 d^2 p \log \left (a+\frac {b}{x}\right )}{b^2}-\frac {4 d e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b}+\frac {2 d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2}+4 e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )+4 e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)+4 e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)-4 e^2 p \log \left (\frac {e (b+a x)}{-a d+b e}\right ) \log (d+e x)+4 e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )-4 e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )+4 e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{4 d^3} \] Input:

Integrate[Log[c*(a + b/x)^p]/(x^3*(d + e*x)),x]
 

Output:

-1/4*(-((d^2*p)/x^2) + (2*a*d^2*p)/(b*x) + (4*d*e*p)/x - (2*a^2*d^2*p*Log[ 
a + b/x])/b^2 - (4*d*e*(a + b/x)*Log[c*(a + b/x)^p])/b + (2*d^2*Log[c*(a + 
 b/x)^p])/x^2 + 4*e^2*Log[c*(a + b/x)^p]*Log[-(b/(a*x))] + 4*e^2*Log[c*(a 
+ b/x)^p]*Log[d + e*x] + 4*e^2*p*Log[-((e*x)/d)]*Log[d + e*x] - 4*e^2*p*Lo 
g[(e*(b + a*x))/(-(a*d) + b*e)]*Log[d + e*x] + 4*e^2*p*PolyLog[2, 1 + b/(a 
*x)] - 4*e^2*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)] + 4*e^2*p*PolyLog[2, 
1 + (e*x)/d])/d^3
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 287, 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 = {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.

\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx\)

\(\Big \downarrow \) 2916

\(\displaystyle \int \left (-\frac {e^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 (d+e x)}+\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 x}-\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2 x^2}+\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}-\frac {e^2 \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^3}-\frac {a p}{2 b d x}-\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}-\frac {e p}{d^2 x}+\frac {p}{4 d x^2}\)

Input:

Int[Log[c*(a + b/x)^p]/(x^3*(d + e*x)),x]
 

Output:

p/(4*d*x^2) - (a*p)/(2*b*d*x) - (e*p)/(d^2*x) + (a^2*p*Log[a + b/x])/(2*b^ 
2*d) + (e*(a + b/x)*Log[c*(a + b/x)^p])/(b*d^2) - Log[c*(a + b/x)^p]/(2*d* 
x^2) - (e^2*Log[c*(a + b/x)^p]*Log[-(b/(a*x))])/d^3 - (e^2*Log[c*(a + b/x) 
^p]*Log[d + e*x])/d^3 - (e^2*p*Log[-((e*x)/d)]*Log[d + e*x])/d^3 + (e^2*p* 
Log[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/d^3 - (e^2*p*PolyLog[2, 1 
+ b/(a*x)])/d^3 + (e^2*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/d^3 - (e^2 
*p*PolyLog[2, 1 + (e*x)/d])/d^3
 

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 = 1.62 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.20

method	result	size
parts	\(-\frac {e^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e}{d^{2} x}+\frac {p b \left (-\frac {-\frac {d}{2 b \,x^{2}}-\frac {-d a -2 b e}{b^{2} x}+\frac {\left (d a +2 b e \right ) a \ln \left (x \right )}{b^{3}}-\frac {\left (d a +2 b e \right ) a \ln \left (a x +b \right )}{b^{3}}}{d^{2}}+\frac {e^{2} \ln \left (x \right )^{2}}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (\frac {a x +b}{b}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (x \right ) \ln \left (\frac {a x +b}{b}\right )}{d^{3} b}+\frac {2 e^{2} \operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{d^{3} b}+\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{3} b}\right )}{2}\)	\(345\)

Input:

int(ln(c*(a+b/x)^p)/x^3/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

integrate(log(c*(a+b/x)^p)/x^3/(e*x+d),x, algorithm="fricas")
 

Output:

integral(log(c*((a*x + b)/x)^p)/(e*x^4 + d*x^3), x)
 

Sympy [F]

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

integrate(ln(c*(a+b/x)**p)/x**3/(e*x+d),x)
 

Output:

Integral(log(c*(a + b/x)**p)/(x**3*(d + e*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {1}{4} \, {\left (4 \, e {\left (\frac {a \log \left (a x + b\right )}{b^{2} d^{2}} - \frac {a \log \left (x\right )}{b^{2} d^{2}} - \frac {1}{b d^{2} x}\right )} - \frac {4 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )\right )} e^{2}}{b d^{3}} + \frac {2 \, a^{2} \log \left (a x + b\right )}{b^{3} d} - \frac {2 \, a^{2} \log \left (x\right )}{b^{3} d} - \frac {2 \, {\left (2 \, e^{2} \log \left (e x + d\right ) \log \left (x\right ) - e^{2} \log \left (x\right )^{2}\right )}}{b d^{3}} - \frac {2 \, a x - b}{b^{2} d x^{2}}\right )} b p - \frac {1}{2} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \] Input:

integrate(log(c*(a+b/x)^p)/x^3/(e*x+d),x, algorithm="maxima")
 

Output:

1/4*(4*e*(a*log(a*x + b)/(b^2*d^2) - a*log(x)/(b^2*d^2) - 1/(b*d^2*x)) - 4 
*(log(a*x/b + 1)*log(x) + dilog(-a*x/b))*e^2/(b*d^3) + 4*(log(e*x/d + 1)*l 
og(x) + dilog(-e*x/d))*e^2/(b*d^3) + 4*(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)))*e^2/(b*d^3) + 2*a^2*log 
(a*x + b)/(b^3*d) - 2*a^2*log(x)/(b^3*d) - 2*(2*e^2*log(e*x + d)*log(x) - 
e^2*log(x)^2)/(b*d^3) - (2*a*x - b)/(b^2*d*x^2))*b*p - 1/2*(2*e^2*log(e*x 
+ d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2))*log((a + b/x)^p*c)
 

Giac [F]

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

integrate(log(c*(a+b/x)^p)/x^3/(e*x+d),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \] Input:

int(log(c*(a + b/x)^p)/(x^3*(d + e*x)),x)
 

Output:

int(log(c*(a + b/x)^p)/(x^3*(d + e*x)), x)
 

Reduce [F]

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

int(log(c*(a+b/x)^p)/x^3/(e*x+d),x)
 

Output:

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