\(\int \frac {(f+g x^{-n}) \log (c (d+e x^n)^p)}{x} \, dx\) [363]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\left (f+g x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {e g p \log (x)}{d}-\frac {e g p \log \left (d+e x^n\right )}{d n}-\frac {g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \] Output:

e*g*p*ln(x)/d-e*g*p*ln(d+e*x^n)/d/n-g*ln(c*(d+e*x^n)^p)/n/(x^n)+f*ln(-e*x^ 
n/d)*ln(c*(d+e*x^n)^p)/n+f*p*polylog(2,1+e*x^n/d)/n
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.93 \[ \int \frac {\left (f+g x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {e g n p \log (x)-e g p \log \left (d+e x^n\right )-d g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )+d f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+d f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{d n} \] Input:

Integrate[((f + g/x^n)*Log[c*(d + e*x^n)^p])/x,x]
 

Output:

(e*g*n*p*Log[x] - e*g*p*Log[d + e*x^n] - (d*g*Log[c*(d + e*x^n)^p])/x^n + 
d*f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + d*f*p*PolyLog[2, 1 + (e*x^n)/ 
d])/(d*n)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2005, 2925, 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.

\(\displaystyle \int \frac {\left (f+g x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx\)

\(\Big \downarrow \) 2005

\(\displaystyle \int x^{-n-1} \left (f x^n+g\right ) \log \left (c \left (d+e x^n\right )^p\right )dx\)

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

Int[((f + g/x^n)*Log[c*(d + e*x^n)^p])/x,x]
 

Output:

((e*g*p*Log[x^n])/d - (e*g*p*Log[d + e*x^n])/d - (g*Log[c*(d + e*x^n)^p])/ 
x^n + f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + f*p*PolyLog[2, 1 + (e*x^n 
)/d])/n
 

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 2925
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Si 
mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], 
x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer 
Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 
] || IGtQ[q, 0])
 
Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 6.84 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.49

method result size
risch \(\frac {\left (f \ln \left (x \right ) n \,x^{n}-g \right ) x^{-n} \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{n}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {g \,x^{-n}}{n}+\frac {\ln \left (x^{n}\right ) f}{n}\right )-\frac {p f \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p f \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )-\frac {e g p \ln \left (d +e \,x^{n}\right )}{d n}+\frac {e p g \ln \left (x^{n}\right )}{n d}\) \(242\)

Input:

int((f+g/(x^n))*ln(c*(d+e*x^n)^p)/x,x,method=_RETURNVERBOSE)
 

Output:

(f*ln(x)*n*x^n-g)/n/(x^n)*ln((d+e*x^n)^p)+(1/2*I*Pi*csgn(I*(d+e*x^n)^p)*cs 
gn(I*c*(d+e*x^n)^p)^2-1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*c 
sgn(I*c)-1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^3+1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^2 
*csgn(I*c)+ln(c))*(-1/n*g/(x^n)+1/n*ln(x^n)*f)-p/n*f*dilog((d+e*x^n)/d)-p* 
f*ln(x)*ln((d+e*x^n)/d)-e*g*p*ln(d+e*x^n)/d/n+e*p/n*g/d*ln(x^n)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18 \[ \int \frac {\left (f+g x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {d f n p x^{n} \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + d f p x^{n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + d g \log \left (c\right ) - {\left (e g n p + d f n \log \left (c\right )\right )} x^{n} \log \left (x\right ) + {\left (d g p - {\left (d f n p \log \left (x\right ) - e g p\right )} x^{n}\right )} \log \left (e x^{n} + d\right )}{d n x^{n}} \] Input:

integrate((f+g/(x^n))*log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")
 

Output:

-(d*f*n*p*x^n*log(x)*log((e*x^n + d)/d) + d*f*p*x^n*dilog(-(e*x^n + d)/d + 
 1) + d*g*log(c) - (e*g*n*p + d*f*n*log(c))*x^n*log(x) + (d*g*p - (d*f*n*p 
*log(x) - e*g*p)*x^n)*log(e*x^n + d))/(d*n*x^n)
 

Sympy [F]

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

integrate((f+g/(x**n))*ln(c*(d+e*x**n)**p)/x,x)
 

Output:

Integral((f*x**n + g)*log(c*(d + e*x**n)**p)/(x*x**n), x)
 

Maxima [F]

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

integrate((f+g/(x^n))*log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")
 

Output:

-1/2*((f*n^2*p*log(x)^2 - 2*f*n*log(c)*log(x))*x^n - 2*(f*n*x^n*log(x) - g 
)*log((e*x^n + d)^p) + 2*g*log(c))/(n*x^n) + integrate((d*f*n*p*log(x) + e 
*g*p)/(e*x*x^n + d*x), x)
 

Giac [F]

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

integrate((f+g/(x^n))*log(c*(d+e*x^n)^p)/x,x, algorithm="giac")
 

Output:

integrate((f + g/x^n)*log((e*x^n + d)^p*c)/x, x)
 

Mupad [F(-1)]

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

int((log(c*(d + e*x^n)^p)*(f + g/x^n))/x,x)
 

Output:

int((log(c*(d + e*x^n)^p)*(f + g/x^n))/x, x)
 

Reduce [F]

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

int((f+g/(x^n))*log(c*(d+e*x^n)^p)/x,x)
 

Output:

( - 2*x**n*int(log((x**n*e + d)**p*c)/(x**(2*n)*e*x + x**n*d*x),x)*d**3*f* 
n*p - 2*x**n*log(x**n*e + d)*d*e*f*p**2 - 2*x**n*log(x**n*e + d)*e**2*g*p* 
*2 + x**n*log((x**n*e + d)**p*c)**2*d*e*f + 2*x**n*log(x)*d*e*f*n*p**2 + 2 
*x**n*log(x)*e**2*g*n*p**2 - 2*log((x**n*e + d)**p*c)*d**2*f*p - 2*log((x* 
*n*e + d)**p*c)*d*e*g*p)/(2*x**n*d*e*n*p)