\(\int \frac {\log (c (d+e x^n)^p)}{x (f+g x^{2 n})^2} \, dx\) [374]

Optimal result

Integrand size = 27, antiderivative size = 419 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{2 n}\right )^2} \, dx=-\frac {d e \sqrt {g} p \arctan \left (\frac {\sqrt {g} x^n}{\sqrt {f}}\right )}{2 f^{3/2} \left (e^2 f+d^2 g\right ) n}-\frac {e^2 p \log \left (d+e x^n\right )}{2 f \left (e^2 f+d^2 g\right ) n}+\frac {\log \left (c \left (d+e x^n\right )^p\right )}{2 f n \left (f+g x^{2 n}\right )}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f^2 n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x^n\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2 n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}+\frac {e^2 p \log \left (f+g x^{2 n}\right )}{4 f \left (e^2 f+d^2 g\right ) n}-\frac {p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2 n}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{f^2 n} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.94, 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.

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {d e \sqrt {g} p \arctan \left (\frac {\sqrt {g} x^n}{\sqrt {f}}\right )}{2 f^{3/2} \left (d^2 g+e^2 f\right )}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x^n\right )}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f^2}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f^2}+\frac {\log \left (c \left (d+e x^n\right )^p\right )}{2 f \left (f+g x^{2 n}\right )}+\frac {e^2 p \log \left (f+g x^{2 n}\right )}{4 f \left (d^2 g+e^2 f\right )}-\frac {e^2 p \log \left (d+e x^n\right )}{2 f \left (d^2 g+e^2 f\right )}-\frac {p \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} \left (e x^n+d\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {g} \left (e x^n+d\right )}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f^2}+\frac {p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{f^2}}{n}\)

Input:

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

Output:

(-1/2*(d*e*Sqrt[g]*p*ArcTan[(Sqrt[g]*x^n)/Sqrt[f]])/(f^(3/2)*(e^2*f + d^2* 
g)) - (e^2*p*Log[d + e*x^n])/(2*f*(e^2*f + d^2*g)) + Log[c*(d + e*x^n)^p]/ 
(2*f*(f + g*x^(2*n))) + (Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/f^2 - (Lo 
g[c*(d + e*x^n)^p]*Log[(e*(Sqrt[-f] - Sqrt[g]*x^n))/(e*Sqrt[-f] + d*Sqrt[g 
])])/(2*f^2) - (Log[c*(d + e*x^n)^p]*Log[(e*(Sqrt[-f] + Sqrt[g]*x^n))/(e*S 
qrt[-f] - d*Sqrt[g])])/(2*f^2) + (e^2*p*Log[f + g*x^(2*n)])/(4*f*(e^2*f + 
d^2*g)) - (p*PolyLog[2, -((Sqrt[g]*(d + e*x^n))/(e*Sqrt[-f] - d*Sqrt[g]))] 
)/(2*f^2) - (p*PolyLog[2, (Sqrt[g]*(d + e*x^n))/(e*Sqrt[-f] + d*Sqrt[g])]) 
/(2*f^2) + (p*PolyLog[2, 1 + (e*x^n)/d])/f^2)/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 = 22.05 (sec) , antiderivative size = 635, normalized size of antiderivative = 1.52

method	result	size
risch	\(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (x^{n}\right )}{n \,f^{2}}-\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (f +g \,x^{2 n}\right )}{2 n \,f^{2}}+\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n f \left (f +g \,x^{2 n}\right )}-\frac {e^{2} p \ln \left (d +e \,x^{n}\right )}{2 f \left (d^{2} g +f \,e^{2}\right ) n}+\frac {e^{2} p \ln \left (f +g \,x^{2 n}\right )}{4 f \left (d^{2} g +f \,e^{2}\right ) n}-\frac {e p g d \arctan \left (\frac {x^{n} g}{\sqrt {g f}}\right )}{2 n f \left (d^{2} g +f \,e^{2}\right ) \sqrt {g f}}-\frac {p \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n \,f^{2}}-\frac {p \ln \left (x^{n}\right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )}{n \,f^{2}}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (f +g \,x^{2 n}\right )}{2 n \,f^{2}}-\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^{2}}-\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^{2}}-\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^{2}}-\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^{2}}+\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^{2}}-\frac {\ln \left (f +g \,x^{2 n}\right )}{2 n \,f^{2}}+\frac {1}{2 n f \left (f +g \,x^{2 n}\right )}\right )\)	\(635\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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