\(\int \frac {\log (c (d+e x^n)^p)}{x (f+g x^{2 n})} \, dx\) [370]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 266 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{2 n}\right )} \, dx=\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x^n\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f n}-\frac {p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f n}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f n}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{f n} \] Output:

ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/f/n-1/2*ln(c*(d+e*x^n)^p)*ln(e*((-f)^(1/2)- 
g^(1/2)*x^n)/(e*(-f)^(1/2)+d*g^(1/2)))/f/n-1/2*ln(c*(d+e*x^n)^p)*ln(e*((-f 
)^(1/2)+g^(1/2)*x^n)/(e*(-f)^(1/2)-d*g^(1/2)))/f/n-1/2*p*polylog(2,-g^(1/2 
)*(d+e*x^n)/(e*(-f)^(1/2)-d*g^(1/2)))/f/n-1/2*p*polylog(2,g^(1/2)*(d+e*x^n 
)/(e*(-f)^(1/2)+d*g^(1/2)))/f/n+p*polylog(2,1+e*x^n/d)/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 = 1.08 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/f - (Log[c*(d + e*x^n)^p]*Log[(e 
*(Sqrt[-f] - Sqrt[g]*x^n))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f) - (Log[c*(d + 
e*x^n)^p]*Log[(e*(Sqrt[-f] + Sqrt[g]*x^n))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*f 
) - (p*PolyLog[2, -((Sqrt[g]*(d + e*x^n))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*f 
) - (p*PolyLog[2, (Sqrt[g]*(d + e*x^n))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f) + 
 (p*PolyLog[2, 1 + (e*x^n)/d])/f)/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 = 6.31 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.80

method result size
risch \(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (x^{n}\right )}{n f}-\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (f +g \,x^{2 n}\right )}{2 n f}-\frac {p \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n f}-\frac {p \ln \left (x^{n}\right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )}{n f}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (f +g \,x^{2 n}\right )}{2 n f}-\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-g f}-\left (d +e \,x^{n}\right ) g +d g}{e \sqrt {-g f}+d g}\right )}{2 n f}-\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-g f}+\left (d +e \,x^{n}\right ) g -d g}{e \sqrt {-g f}-d g}\right )}{2 n f}-\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-g f}-\left (d +e \,x^{n}\right ) g +d g}{e \sqrt {-g f}+d g}\right )}{2 n f}-\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-g f}+\left (d +e \,x^{n}\right ) g -d g}{e \sqrt {-g f}-d g}\right )}{2 n f}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\ln \left (x^{n}\right )}{n f}-\frac {\ln \left (f +g \,x^{2 n}\right )}{2 n f}\right )\) \(478\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Giac [F]

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

Mupad [F(-1)]

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

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

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

Output:

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