Integrand size = 27, antiderivative size = 377 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=-\frac {d e \sqrt {g} p \arctan \left (\frac {\sqrt {f} x^n}{\sqrt {g}}\right )}{2 f^{3/2} \left (d^2 f+e^2 g\right ) n}-\frac {e^2 g p \log \left (d+e x^n\right )}{2 f^2 \left (d^2 f+e^2 g\right ) n}+\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{2 f^2 n \left (g+f x^{2 n}\right )}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (\sqrt {g}+\sqrt {-f} x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {e^2 g p \log \left (g+f x^{2 n}\right )}{4 f^2 \left (d^2 f+e^2 g\right ) n}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n} \]
-1/2*e^2*g*p*ln(d+e*x^n)/f^2/(d^2*f+e^2*g)/n+1/2*g*ln(c*(d+e*x^n)^p)/f^2/n /(g+f*x^(2*n))+1/4*e^2*g*p*ln(g+f*x^(2*n))/f^2/(d^2*f+e^2*g)/n+1/2*ln(c*(d +e*x^n)^p)*ln(-e*(x^n*(-f)^(1/2)+g^(1/2))/(d*(-f)^(1/2)-e*g^(1/2)))/f^2/n+ 1/2*ln(c*(d+e*x^n)^p)*ln(e*(-x^n*(-f)^(1/2)+g^(1/2))/(d*(-f)^(1/2)+e*g^(1/ 2)))/f^2/n+1/2*p*polylog(2,(d+e*x^n)*(-f)^(1/2)/(d*(-f)^(1/2)-e*g^(1/2)))/ f^2/n+1/2*p*polylog(2,(d+e*x^n)*(-f)^(1/2)/(d*(-f)^(1/2)+e*g^(1/2)))/f^2/n -1/2*d*e*p*arctan(x^n*f^(1/2)/g^(1/2))*g^(1/2)/f^(3/2)/(d^2*f+e^2*g)/n
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx \]
Time = 0.76 (sec) , antiderivative size = 357, normalized size of antiderivative = 0.95, 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.
| \(\displaystyle \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {x^{4 n-1} \log \left (c \left (d+e x^n\right )^p\right )}{\left (f x^{2 n}+g\right )^2}dx\) |
| \(\Big \downarrow \) 2925 |
| \(\displaystyle \frac {\int \frac {x^{3 n} \log \left (c \left (e x^n+d\right )^p\right )}{\left (f x^{2 n}+g\right )^2}dx^n}{n}\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \frac {\int \left (\frac {x^n \log \left (c \left (e x^n+d\right )^p\right )}{f \left (f x^{2 n}+g\right )}-\frac {g x^n \log \left (c \left (e x^n+d\right )^p\right )}{f \left (f x^{2 n}+g\right )^2}\right )dx^n}{n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {d e \sqrt {g} p \arctan \left (\frac {\sqrt {f} x^n}{\sqrt {g}}\right )}{2 f^{3/2} \left (d^2 f+e^2 g\right )}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (\sqrt {-f} x^n+\sqrt {g}\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2}+\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{2 f^2 \left (f x^{2 n}+g\right )}+\frac {e^2 g p \log \left (f x^{2 n}+g\right )}{4 f^2 \left (d^2 f+e^2 g\right )}-\frac {e^2 g p \log \left (d+e x^n\right )}{2 f^2 \left (d^2 f+e^2 g\right )}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (e x^n+d\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt {-f} \left (e x^n+d\right )}{\sqrt {-f} d+e \sqrt {g}}\right )}{2 f^2}}{n}\) |
(-1/2*(d*e*Sqrt[g]*p*ArcTan[(Sqrt[f]*x^n)/Sqrt[g]])/(f^(3/2)*(d^2*f + e^2* g)) - (e^2*g*p*Log[d + e*x^n])/(2*f^2*(d^2*f + e^2*g)) + (g*Log[c*(d + e*x ^n)^p])/(2*f^2*(g + f*x^(2*n))) + (Log[c*(d + e*x^n)^p]*Log[(e*(Sqrt[g] - Sqrt[-f]*x^n))/(d*Sqrt[-f] + e*Sqrt[g])])/(2*f^2) + (Log[c*(d + e*x^n)^p]* Log[-((e*(Sqrt[g] + Sqrt[-f]*x^n))/(d*Sqrt[-f] - e*Sqrt[g]))])/(2*f^2) + ( e^2*g*p*Log[g + f*x^(2*n)])/(4*f^2*(d^2*f + e^2*g)) + (p*PolyLog[2, (Sqrt[ -f]*(d + e*x^n))/(d*Sqrt[-f] - e*Sqrt[g])])/(2*f^2) + (p*PolyLog[2, (Sqrt[ -f]*(d + e*x^n))/(d*Sqrt[-f] + e*Sqrt[g])])/(2*f^2))/n
3.4.77.3.1 Defintions of rubi rules used
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]
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]
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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 59.38 (sec) , antiderivative size = 561, normalized size of antiderivative = 1.49
| method | result | size |
| risch | \(\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) \ln \left (g +f \,x^{2 n}\right )}{2 n \,f^{2}}+\frac {\ln \left (\left (d +e \,x^{n}\right )^{p}\right ) g}{2 n \,f^{2} \left (g +f \,x^{2 n}\right )}-\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (g +f \,x^{2 n}\right )}{2 n \,f^{2}}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-f g}-f \left (d +e \,x^{n}\right )+d f}{e \sqrt {-f g}+d f}\right )}{2 n \,f^{2}}+\frac {p \ln \left (d +e \,x^{n}\right ) \ln \left (\frac {e \sqrt {-f g}+f \left (d +e \,x^{n}\right )-d f}{e \sqrt {-f g}-d f}\right )}{2 n \,f^{2}}+\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-f g}-f \left (d +e \,x^{n}\right )+d f}{e \sqrt {-f g}+d f}\right )}{2 n \,f^{2}}+\frac {p \operatorname {dilog}\left (\frac {e \sqrt {-f g}+f \left (d +e \,x^{n}\right )-d f}{e \sqrt {-f g}-d f}\right )}{2 n \,f^{2}}-\frac {e^{2} g p \ln \left (d +e \,x^{n}\right )}{2 f^{2} \left (d^{2} f +e^{2} g \right ) n}+\frac {e^{2} g p \ln \left (g +f \,x^{2 n}\right )}{4 f^{2} \left (d^{2} f +e^{2} g \right ) n}-\frac {p e g d \arctan \left (\frac {x^{n} f}{\sqrt {f g}}\right )}{2 n f \left (d^{2} f +e^{2} g \right ) \sqrt {f g}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\ln \left (g +f \,x^{2 n}\right )}{2 n \,f^{2}}+\frac {g}{2 n \,f^{2} \left (g +f \,x^{2 n}\right )}\right )\) | \(561\) |
1/2/n*ln((d+e*x^n)^p)/f^2*ln(g+f*(x^n)^2)+1/2/n*ln((d+e*x^n)^p)*g/f^2/(g+f *(x^n)^2)-1/2/n*p/f^2*ln(d+e*x^n)*ln(g+f*(x^n)^2)+1/2/n*p/f^2*ln(d+e*x^n)* ln((e*(-f*g)^(1/2)-f*(d+e*x^n)+d*f)/(e*(-f*g)^(1/2)+d*f))+1/2/n*p/f^2*ln(d +e*x^n)*ln((e*(-f*g)^(1/2)+f*(d+e*x^n)-d*f)/(e*(-f*g)^(1/2)-d*f))+1/2/n*p/ f^2*dilog((e*(-f*g)^(1/2)-f*(d+e*x^n)+d*f)/(e*(-f*g)^(1/2)+d*f))+1/2/n*p/f ^2*dilog((e*(-f*g)^(1/2)+f*(d+e*x^n)-d*f)/(e*(-f*g)^(1/2)-d*f))-1/2*e^2*g* p*ln(d+e*x^n)/f^2/(d^2*f+e^2*g)/n+1/4/n*p*e^2*g/f^2/(d^2*f+e^2*g)*ln(g+f*( x^n)^2)-1/2/n*p*e*g/f/(d^2*f+e^2*g)*d/(f*g)^(1/2)*arctan(x^n*f/(f*g)^(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))*(1/2/n/f^2*ln(g+f*(x^n )^2)+1/2/n*g/f^2/(g+f*(x^n)^2))
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} x} \,d x } \]
\[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{-2 n}\right )^2} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}{x\,{\left (f+\frac {g}{x^{2\,n}}\right )}^2} \,d x \]