\(\int \frac {\log (c (d+e x^n)^p)}{x (f+g x^{-2 n})} \, dx\) [373]

Optimal result

Integrand size = 27, antiderivative size = 221 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )} \, dx=\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f n}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (\sqrt {g}+\sqrt {-f} x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f n}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f n}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f n} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.

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

\(\Big \downarrow \) 2005

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((Log[c*(d + e*x^n)^p]*Log[(e*(Sqrt[g] - Sqrt[-f]*x^n))/(d*Sqrt[-f] + e*Sq 
rt[g])])/(2*f) + (Log[c*(d + e*x^n)^p]*Log[-((e*(Sqrt[g] + Sqrt[-f]*x^n))/ 
(d*Sqrt[-f] - e*Sqrt[g]))])/(2*f) + (p*PolyLog[2, (Sqrt[-f]*(d + e*x^n))/( 
d*Sqrt[-f] - e*Sqrt[g])])/(2*f) + (p*PolyLog[2, (Sqrt[-f]*(d + e*x^n))/(d* 
Sqrt[-f] + e*Sqrt[g])])/(2*f))/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 = 18.71 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.80

method	result	size
risch	\(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (g +f \,x^{2 n}\right )}{2 n f}-\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (g +f \,x^{2 n}\right )}{2 n f}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-g f}-f \left (d +e \,x^{n}\right )+d f}{e \sqrt {-g f}+d f}\right )}{2 n f}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-g f}+f \left (d +e \,x^{n}\right )-d f}{e \sqrt {-g f}-d f}\right )}{2 n f}+\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-g f}-f \left (d +e \,x^{n}\right )+d f}{e \sqrt {-g f}+d f}\right )}{2 n f}+\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-g f}+f \left (d +e \,x^{n}\right )-d f}{e \sqrt {-g f}-d f}\right )}{2 n f}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (g +f \,x^{2 n}\right )}{2 n f}\)	\(398\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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