\(\int \frac {x^2 \log (c (a+\frac {b}{x})^p)}{d+e x} \, dx\) [241]

Optimal result

Integrand size = 23, antiderivative size = 219 \[ \int \frac {x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b p x}{2 a e}-\frac {d x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^2}+\frac {x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}-\frac {b d p \log (b+a x)}{a e^2}-\frac {b^2 p \log (b+a x)}{2 a^2 e}+\frac {d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^3}+\frac {d^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^3}+\frac {d^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05 \[ \int \frac {x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=-\frac {b d p \log \left (a+\frac {b}{x}\right )}{a e^2}-\frac {d x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^2}+\frac {x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e}-\frac {b d p \log (x)}{a e^2}+\frac {b p \left (\frac {x}{a}-\frac {b \log (b+a x)}{a^2}\right )}{2 e}+\frac {d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^3}+\frac {d^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^3}+\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e^3}-\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 219, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*p*x)/(2*a*e) - (d*x*Log[c*(a + b/x)^p])/e^2 + (x^2*Log[c*(a + b/x)^p])/ 
(2*e) - (b*d*p*Log[b + a*x])/(a*e^2) - (b^2*p*Log[b + a*x])/(2*a^2*e) + (d 
^2*Log[c*(a + b/x)^p]*Log[d + e*x])/e^3 + (d^2*p*Log[-((e*x)/d)]*Log[d + e 
*x])/e^3 - (d^2*p*Log[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/e^3 - (d 
^2*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/e^3 + (d^2*p*PolyLog[2, 1 + (e 
*x)/d])/e^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 = 2.83 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.10

method	result	size
parts	\(\frac {x^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 e}-\frac {d x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e^{2}}+\frac {d^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{3}}+p b e \left (-\frac {d^{2} \operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{4} b}-\frac {d^{2} \ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{4} b}+\frac {d^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4} b}+\frac {d^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{4} b}+\frac {\frac {e x +d}{a}+\frac {\left (-2 d a -b e \right ) \ln \left (d a -a \left (e x +d \right )-b e \right )}{a^{2}}}{2 e^{3}}\right )\)	\(240\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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