\(\int \frac {(f+g x^{2 n}) \log (c (d+e x^n)^p)}{x} \, dx\) [361]

Optimal result

Integrand size = 25, antiderivative size = 124 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {d g p x^n}{2 e n}-\frac {g p x^{2 n}}{4 n}-\frac {d^2 g p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \] Output:

1/2*d*g*p*x^n/e/n-1/4*g*p*x^(2*n)/n-1/2*d^2*g*p*ln(d+e*x^n)/e^2/n+1/2*g*x^ 
(2*n)*ln(c*(d+e*x^n)^p)/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 = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {-e g p x^n \left (-2 d+e x^n\right )-2 d^2 g p \log \left (d+e x^n\right )+2 e^2 \left (g x^{2 n}+2 f \log \left (-\frac {e x^n}{d}\right )\right ) \log \left (c \left (d+e x^n\right )^p\right )+4 e^2 f p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{4 e^2 n} \] Input:

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

Output:

(-(e*g*p*x^n*(-2*d + e*x^n)) - 2*d^2*g*p*Log[d + e*x^n] + 2*e^2*(g*x^(2*n) 
 + 2*f*Log[-((e*x^n)/d)])*Log[c*(d + e*x^n)^p] + 4*e^2*f*p*PolyLog[2, 1 + 
(e*x^n)/d])/(4*e^2*n)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {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.

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((d*g*p*x^n)/(2*e) - (g*p*x^(2*n))/4 - (d^2*g*p*Log[d + e*x^n])/(2*e^2) + 
(g*x^(2*n)*Log[c*(d + e*x^n)^p])/2 + 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 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 = 6.44 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.01

method	result	size
risch	\(\frac {\left (2 f \ln \left (x \right ) n +g \,x^{2 n}\right ) \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (f \ln \left (x \right )+\frac {g \,x^{2 n}}{2 n}\right )-\frac {g p \,x^{2 n}}{4 n}+\frac {d g p \,x^{n}}{2 e n}-\frac {d^{2} g p \ln \left (d +e \,x^{n}\right )}{2 e^{2} n}-\frac {p f \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p f \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\)	\(249\)

Input:

int((f+g*x^(2*n))*ln(c*(d+e*x^n)^p)/x,x,method=_RETURNVERBOSE)
 

Output:

1/2*(2*f*ln(x)*n+g*(x^n)^2)/n*ln((d+e*x^n)^p)+(1/2*I*Pi*csgn(I*(d+e*x^n)^p 
)*csgn(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)*csgn(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))*(f*ln(x)+1/2*g*(x^n)^2/n)-1/4*p/n*g*(x^n)^2+1/2*d*g* 
p*x^n/e/n-1/2*d^2*g*p*ln(d+e*x^n)/e^2/n-p/n*f*dilog((d+e*x^n)/d)-p*f*ln(x) 
*ln((d+e*x^n)/d)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {\left (f+g x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {4 \, e^{2} f n p \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) - 4 \, e^{2} f n \log \left (c\right ) \log \left (x\right ) - 2 \, d e g p x^{n} + 4 \, e^{2} f p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + {\left (e^{2} g p - 2 \, e^{2} g \log \left (c\right )\right )} x^{2 \, n} - 2 \, {\left (2 \, e^{2} f n p \log \left (x\right ) + e^{2} g p x^{2 \, n} - d^{2} g p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \] Input:

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")
 

Output:

-1/4*(4*e^2*f*n*p*log(x)*log((e*x^n + d)/d) - 4*e^2*f*n*log(c)*log(x) - 2* 
d*e*g*p*x^n + 4*e^2*f*p*dilog(-(e*x^n + d)/d + 1) + (e^2*g*p - 2*e^2*g*log 
(c))*x^(2*n) - 2*(2*e^2*f*n*p*log(x) + e^2*g*p*x^(2*n) - d^2*g*p)*log(e*x^ 
n + d))/(e^2*n)
 

Sympy [F]

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

integrate((f+g*x**(2*n))*ln(c*(d+e*x**n)**p)/x,x)
 

Output:

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

Maxima [F]

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

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")
 

Output:

-1/4*(2*e^2*f*n^2*p*log(x)^2 - 2*d*e*g*p*x^n + (e^2*g*p - 2*e^2*g*log(c))* 
x^(2*n) - 2*(2*e^2*f*n*log(x) + e^2*g*x^(2*n))*log((e*x^n + d)^p) + 2*(d^2 
*g*n*p - 2*e^2*f*n*log(c))*log(x))/(e^2*n) + integrate(1/2*(2*d*e^2*f*n*p* 
log(x) + d^3*g*p)/(e^3*x*x^n + d*e^2*x), x)
 

Giac [F]

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

integrate((f+g*x^(2*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*x**(2*n)*log((x**n*e + d)**p*c)*e**2*g*p - x**(2*n)*e**2*g*p**2 + 2*x** 
n*d*e*g*p**2 + 4*int(log((x**n*e + d)**p*c)/(x**n*e*x + d*x),x)*d*e**2*f*n 
*p + 2*log((x**n*e + d)**p*c)**2*e**2*f - 2*log((x**n*e + d)**p*c)*d**2*g* 
p)/(4*e**2*n*p)