Integrand size = 23, antiderivative size = 198 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {p}{d x}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^2}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^2} \] Output:
p/d/x-(a+b/x)*ln(c*(a+b/x)^p)/b/d+e*ln(c*(a+b/x)^p)*ln(-b/a/x)/d^2+e*ln(c* (a+b/x)^p)*ln(e*x+d)/d^2+e*p*ln(-e*x/d)*ln(e*x+d)/d^2-e*p*ln(-e*(a*x+b)/(a *d-b*e))*ln(e*x+d)/d^2+e*p*polylog(2,1+b/a/x)/d^2-e*p*polylog(2,a*(e*x+d)/ (a*d-b*e))/d^2+e*p*polylog(2,1+e*x/d)/d^2
Time = 0.08 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {p}{d x}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^2}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {a+\frac {b}{x}}{a}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2} \] Input:
Integrate[Log[c*(a + b/x)^p]/(x^2*(d + e*x)),x]
Output:
p/(d*x) - ((a + b/x)*Log[c*(a + b/x)^p])/(b*d) + (e*Log[c*(a + b/x)^p]*Log [-(b/(a*x))])/d^2 + (e*Log[c*(a + b/x)^p]*Log[d + e*x])/d^2 + (e*p*Log[-(( e*x)/d)]*Log[d + e*x])/d^2 - (e*p*Log[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/d^2 + (e*p*PolyLog[2, (a + b/x)/a])/d^2 + (e*p*PolyLog[2, (d + e*x )/d])/d^2 - (e*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/d^2
Time = 0.89 (sec) , antiderivative size = 198, 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.
\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx\) |
\(\Big \downarrow \) 2916 |
\(\displaystyle \int \left (\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2 (d+e x)}-\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2 x}+\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2}+\frac {e \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e p \operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2}-\frac {e p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}+\frac {p}{d x}\) |
Input:
Int[Log[c*(a + b/x)^p]/(x^2*(d + e*x)),x]
Output:
p/(d*x) - ((a + b/x)*Log[c*(a + b/x)^p])/(b*d) + (e*Log[c*(a + b/x)^p]*Log [-(b/(a*x))])/d^2 + (e*Log[c*(a + b/x)^p]*Log[d + e*x])/d^2 + (e*p*Log[-(( e*x)/d)]*Log[d + e*x])/d^2 - (e*p*Log[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/d^2 + (e*p*PolyLog[2, 1 + b/(a*x)])/d^2 - (e*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/d^2 + (e*p*PolyLog[2, 1 + (e*x)/d])/d^2
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]
Time = 1.29 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.33
method | result | size |
parts | \(\frac {e \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{d x}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e \ln \left (x \right )}{d^{2}}+p b \left (\frac {e \left (-\frac {\left (\frac {\operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{a}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{a}\right ) a}{b}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b}\right )}{d^{2}}+\frac {1}{d b x}+\frac {a \ln \left (x \right )}{d \,b^{2}}-\frac {a \ln \left (a x +b \right )}{d \,b^{2}}-\frac {e \ln \left (x \right )^{2}}{2 d^{2} b}+\frac {e \operatorname {dilog}\left (\frac {a x +b}{b}\right )}{d^{2} b}+\frac {e \ln \left (x \right ) \ln \left (\frac {a x +b}{b}\right )}{d^{2} b}\right )\) | \(264\) |
Input:
int(ln(c*(a+b/x)^p)/x^2/(e*x+d),x,method=_RETURNVERBOSE)
Output:
e*ln(c*(a+b/x)^p)*ln(e*x+d)/d^2-ln(c*(a+b/x)^p)/d/x-ln(c*(a+b/x)^p)*e/d^2* ln(x)+p*b*(e/d^2*(-(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)/b*a+(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d ))/b)+1/d/b/x+1/d/b^2*a*ln(x)-1/d/b^2*a*ln(a*x+b)-1/2*e/d^2*ln(x)^2/b+e/d^ 2/b*dilog((a*x+b)/b)+e/d^2/b*ln(x)*ln((a*x+b)/b))
\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}} \,d x } \] Input:
integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="fricas")
Output:
integral(log(c*((a*x + b)/x)^p)/(e*x^3 + d*x^2), x)
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(a+b/x)**p)/x**2/(e*x+d),x)
Output:
Timed out
Time = 0.07 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {1}{2} \, b p {\left (\frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} e}{b d^{2}} - \frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} e}{b d^{2}} - \frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )\right )} e}{b d^{2}} - \frac {2 \, a \log \left (a x + b\right )}{b^{2} d} + \frac {2 \, a \log \left (x\right )}{b^{2} d} + \frac {2 \, e \log \left (e x + d\right ) \log \left (x\right ) - e \log \left (x\right )^{2}}{b d^{2}} + \frac {2}{b d x}\right )} + {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \] Input:
integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="maxima")
Output:
1/2*b*p*(2*(log(a*x/b + 1)*log(x) + dilog(-a*x/b))*e/(b*d^2) - 2*(log(e*x/ d + 1)*log(x) + dilog(-e*x/d))*e/(b*d^2) - 2*(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/(b*d^2) - 2*a*l og(a*x + b)/(b^2*d) + 2*a*log(x)/(b^2*d) + (2*e*log(e*x + d)*log(x) - e*lo g(x)^2)/(b*d^2) + 2/(b*d*x)) + (e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x ))*log((a + b/x)^p*c)
\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}} \,d x } \] Input:
integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="giac")
Output:
integrate(log((a + b/x)^p*c)/((e*x + d)*x^2), x)
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{x^2\,\left (d+e\,x\right )} \,d x \] Input:
int(log(c*(a + b/x)^p)/(x^2*(d + e*x)),x)
Output:
int(log(c*(a + b/x)^p)/(x^2*(d + e*x)), x)
\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right )}{a e \,x^{3}+a d \,x^{2}+b e \,x^{2}+b d x}d x \right ) a b d p x -2 \left (\int \frac {\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right )}{a e \,x^{3}+a d \,x^{2}+b e \,x^{2}+b d x}d x \right ) b^{2} e p x +\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right )^{2} a x -2 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) a p x -2 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) b p +2 b \,p^{2}}{2 b d p x} \] Input:
int(log(c*(a+b/x)^p)/x^2/(e*x+d),x)
Output:
(2*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*d*p*x - 2*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**2*e*p*x + log(((a*x + b)**p*c)/x**p)**2*a*x - 2*log( ((a*x + b)**p*c)/x**p)*a*p*x - 2*log(((a*x + b)**p*c)/x**p)*b*p + 2*b*p**2 )/(2*b*d*p*x)