\(\int \frac {(f+g x^{-2 n})^2 \log (c (d+e x^n)^p)}{x} \, dx\) [369]

Optimal result

Integrand size = 27, antiderivative size = 257 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {e g^2 p x^{-3 n}}{12 d n}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}-\frac {e f g p x^{-n}}{d n}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}-\frac {e^2 f g p \log (x)}{d^2}-\frac {e^4 g^2 p \log (x)}{4 d^4}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]

[Out]

-1/12*e*g^2*p/d/n/(x^(3*n))+1/8*e^2*g^2*p/d^2/n/(x^(2*n))-e*f*g*p/d/n/(x^n)-1/4*e^3*g^2*p/d^3/n/(x^n)-e^2*f*g*
p*ln(x)/d^2-1/4*e^4*g^2*p*ln(x)/d^4+e^2*f*g*p*ln(d+e*x^n)/d^2/n+1/4*e^4*g^2*p*ln(d+e*x^n)/d^4/n-1/4*g^2*ln(c*(
d+e*x^n)^p)/n/(x^(4*n))-f*g*ln(c*(d+e*x^n)^p)/n/(x^(2*n))+f^2*ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/n+f^2*p*polylog(2
,1+e*x^n/d)/n

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2525, 269, 272, 45, 2463, 2442, 46, 2441, 2352} \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {e^4 g^2 p \log (x)}{4 d^4}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}-\frac {e^2 f g p \log (x)}{d^2}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}+\frac {f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}-\frac {e f g p x^{-n}}{d n}-\frac {e g^2 p x^{-3 n}}{12 d n} \]

[In]

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

[Out]

-1/12*(e*g^2*p)/(d*n*x^(3*n)) + (e^2*g^2*p)/(8*d^2*n*x^(2*n)) - (e*f*g*p)/(d*n*x^n) - (e^3*g^2*p)/(4*d^3*n*x^n
) - (e^2*f*g*p*Log[x])/d^2 - (e^4*g^2*p*Log[x])/(4*d^4) + (e^2*f*g*p*Log[d + e*x^n])/(d^2*n) + (e^4*g^2*p*Log[
d + e*x^n])/(4*d^4*n) - (g^2*Log[c*(d + e*x^n)^p])/(4*n*x^(4*n)) - (f*g*Log[c*(d + e*x^n)^p])/(n*x^(2*n)) + (f
^2*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (f^2*p*PolyLog[2, 1 + (e*x^n)/d])/n

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2463

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 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(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] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (f+\frac {g}{x^2}\right )^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {g^2 \log \left (c (d+e x)^p\right )}{x^5}+\frac {2 f g \log \left (c (d+e x)^p\right )}{x^3}+\frac {f^2 \log \left (c (d+e x)^p\right )}{x}\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {f^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac {(2 f g) \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^3} \, dx,x,x^n\right )}{n}+\frac {g^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^5} \, dx,x,x^n\right )}{n} \\ & = -\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {\left (e f^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}+\frac {(e f g p) \text {Subst}\left (\int \frac {1}{x^2 (d+e x)} \, dx,x,x^n\right )}{n}+\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \frac {1}{x^4 (d+e x)} \, dx,x,x^n\right )}{4 n} \\ & = -\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n}+\frac {(e f g p) \text {Subst}\left (\int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx,x,x^n\right )}{n}+\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \left (\frac {1}{d x^4}-\frac {e}{d^2 x^3}+\frac {e^2}{d^3 x^2}-\frac {e^3}{d^4 x}+\frac {e^4}{d^4 (d+e x)}\right ) \, dx,x,x^n\right )}{4 n} \\ & = -\frac {e g^2 p x^{-3 n}}{12 d n}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}-\frac {e f g p x^{-n}}{d n}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}-\frac {e^2 f g p \log (x)}{d^2}-\frac {e^4 g^2 p \log (x)}{4 d^4}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.73 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {\frac {24 e f g p \left (d x^{-n}+e n \log (x)-e \log \left (d+e x^n\right )\right )}{d^2}+\frac {e g^2 p \left (d x^{-3 n} \left (2 d^2-3 d e x^n+6 e^2 x^{2 n}\right )+6 e^3 n \log (x)-6 e^3 \log \left (d+e x^n\right )\right )}{d^4}+6 g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )+24 f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )-24 f^2 \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )\right )}{24 n} \]

[In]

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

[Out]

-1/24*((24*e*f*g*p*(d/x^n + e*n*Log[x] - e*Log[d + e*x^n]))/d^2 + (e*g^2*p*((d*(2*d^2 - 3*d*e*x^n + 6*e^2*x^(2
*n)))/x^(3*n) + 6*e^3*n*Log[x] - 6*e^3*Log[d + e*x^n]))/d^4 + (6*g^2*Log[c*(d + e*x^n)^p])/x^(4*n) + (24*f*g*L
og[c*(d + e*x^n)^p])/x^(2*n) - 24*f^2*(Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + p*PolyLog[2, 1 + (e*x^n)/d]))/
n

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 6.13 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.54

method	result	size
risch	\(\frac {\left (4 f^{2} \ln \left (x \right ) n \,x^{4 n}-4 f g \,x^{2 n}-g^{2}\right ) x^{-4 n} \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{4 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (f^{2} \ln \left (x^{n}\right )-f g \,x^{-2 n}-\frac {g^{2} x^{-4 n}}{4}\right )}{n}+\frac {e^{4} g^{2} p \ln \left (d +e \,x^{n}\right )}{4 d^{4} n}-\frac {p e \,g^{2} x^{-3 n}}{12 n d}-\frac {e^{3} g^{2} p \,x^{-n}}{4 d^{3} n}+\frac {p \,e^{2} g^{2} x^{-2 n}}{8 n \,d^{2}}-\frac {p \,e^{4} g^{2} \ln \left (x^{n}\right )}{4 n \,d^{4}}+\frac {e^{2} f g p \ln \left (d +e \,x^{n}\right )}{d^{2} n}-\frac {e f g p \,x^{-n}}{d n}-\frac {p \,e^{2} f g \ln \left (x^{n}\right )}{n \,d^{2}}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\)	\(396\)

[In]

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

[Out]

1/4*(4*f^2*ln(x)*n*(x^n)^4-4*f*g*(x^n)^2-g^2)/n/(x^n)^4*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))/n*(f^2*ln(x^n)-f*g/(x^n)^2-1/4*g^2/(x^n)^4)+1/4*e^4*g^2*p*ln
(d+e*x^n)/d^4/n-1/12*p*e/n*g^2/d/(x^n)^3-1/4*e^3*g^2*p/d^3/n/(x^n)+1/8*p*e^2/n*g^2/d^2/(x^n)^2-1/4*p*e^4/n*g^2
/d^4*ln(x^n)+e^2*f*g*p*ln(d+e*x^n)/d^2/n-e*f*g*p/d/n/(x^n)-p*e^2/n*f*g/d^2*ln(x^n)-p/n*f^2*dilog((d+e*x^n)/d)-
p*f^2*ln(x)*ln((d+e*x^n)/d)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.03 \[ \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {24 \, d^{4} f^{2} n p x^{4 \, n} \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + 24 \, d^{4} f^{2} p x^{4 \, n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + 2 \, d^{3} e g^{2} p x^{n} + 6 \, d^{4} g^{2} \log \left (c\right ) + 6 \, {\left (4 \, d^{3} e f g + d e^{3} g^{2}\right )} p x^{3 \, n} - 6 \, {\left (4 \, d^{4} f^{2} n \log \left (c\right ) - {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} n p\right )} x^{4 \, n} \log \left (x\right ) - 3 \, {\left (d^{2} e^{2} g^{2} p - 8 \, d^{4} f g \log \left (c\right )\right )} x^{2 \, n} + 6 \, {\left (4 \, d^{4} f g p x^{2 \, n} + d^{4} g^{2} p - {\left (4 \, d^{4} f^{2} n p \log \left (x\right ) + {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} p\right )} x^{4 \, n}\right )} \log \left (e x^{n} + d\right )}{24 \, d^{4} n x^{4 \, n}} \]

[In]

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

[Out]

-1/24*(24*d^4*f^2*n*p*x^(4*n)*log(x)*log((e*x^n + d)/d) + 24*d^4*f^2*p*x^(4*n)*dilog(-(e*x^n + d)/d + 1) + 2*d
^3*e*g^2*p*x^n + 6*d^4*g^2*log(c) + 6*(4*d^3*e*f*g + d*e^3*g^2)*p*x^(3*n) - 6*(4*d^4*f^2*n*log(c) - (4*d^2*e^2
*f*g + e^4*g^2)*n*p)*x^(4*n)*log(x) - 3*(d^2*e^2*g^2*p - 8*d^4*f*g*log(c))*x^(2*n) + 6*(4*d^4*f*g*p*x^(2*n) +
d^4*g^2*p - (4*d^4*f^2*n*p*log(x) + (4*d^2*e^2*f*g + e^4*g^2)*p)*x^(4*n))*log(e*x^n + d))/(d^4*n*x^(4*n))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-1/24*(2*d^2*e*g^2*p*x^n + 6*d^3*g^2*log(c) + 12*(d^3*f^2*n^2*p*log(x)^2 - 2*d^3*f^2*n*log(c)*log(x))*x^(4*n)
+ 6*(4*d^2*e*f*g*p + e^3*g^2*p)*x^(3*n) - 3*(d*e^2*g^2*p - 8*d^3*f*g*log(c))*x^(2*n) - 6*(4*d^3*f^2*n*x^(4*n)*
log(x) - 4*d^3*f*g*x^(2*n) - d^3*g^2)*log((e*x^n + d)^p))/(d^3*n*x^(4*n)) + integrate(1/4*(4*d^4*f^2*n*p*log(x
) - 4*d^2*e^2*f*g*p - e^4*g^2*p)/(d^3*e*x*x^n + d^4*x), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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