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

Optimal. Leaf size=257 \[ \frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {e^4 g^2 p \log (x)}{4 d^4}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}-\frac {e^2 f g p \log (x)}{d^2}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}+\frac {f^2 p \text {Li}_2\left (\frac {e x^n}{d}+1\right )}{n}-\frac {e f g p x^{-n}}{d n}-\frac {e g^2 p x^{-3 n}}{12 d n} \]

[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]  time = 0.32, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2475, 263, 266, 43, 2416, 2395, 44, 2394, 2315} \[ \frac {f^2 p \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}+\frac {e^2 f g p \log \left (d+e x^n\right )}{d^2 n}-\frac {e^2 f g p \log (x)}{d^2}+\frac {e^2 g^2 p x^{-2 n}}{8 d^2 n}-\frac {e^3 g^2 p x^{-n}}{4 d^3 n}+\frac {e^4 g^2 p \log \left (d+e x^n\right )}{4 d^4 n}-\frac {e^4 g^2 p \log (x)}{4 d^4}-\frac {e f g p x^{-n}}{d n}-\frac {e g^2 p x^{-3 n}}{12 d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*g^2*p)/(12*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 43

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 44

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] && L
tQ[m + n + 2, 0])

Rule 263

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 266

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 2315

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

Rule 2394

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 2395

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 2416

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 2475

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*} \int \frac {\left (f+g x^{-2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac {\operatorname {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 {\operatorname {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 \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac {(2 f g) \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^3} \, dx,x,x^n\right )}{n}+\frac {g^2 \operatorname {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 ) \operatorname {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) \operatorname {Subst}\left (\int \frac {1}{x^2 (d+e x)} \, dx,x,x^n\right )}{n}+\frac {\left (e g^2 p\right ) \operatorname {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) \operatorname {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 ) \operatorname {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]  time = 0.58, size = 188, normalized size = 0.73 \[ -\frac {-24 f^2 \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \text {Li}_2\left (\frac {e x^n}{d}+1\right )\right )+24 f g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )+6 g^2 x^{-4 n} \log \left (c \left (d+e x^n\right )^p\right )+\frac {24 e f g p \left (-e \log \left (d+e x^n\right )+d x^{-n}+e n \log (x)\right )}{d^2}+\frac {e g^2 p \left (d x^{-3 n} \left (2 d^2-3 d e x^n+6 e^2 x^{2 n}\right )-6 e^3 \log \left (d+e x^n\right )+6 e^3 n \log (x)\right )}{d^4}}{24 n} \]

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 265, normalized size = 1.03 \[ -\frac {24 \, d^{4} f^{2} n p x^{4 \, n} \log \relax (x) \log \left (\frac {e x^{n} + d}{d}\right ) + 24 \, d^{4} f^{2} p x^{4 \, n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + 2 \, d^{3} e g^{2} p x^{n} + 6 \, d^{4} g^{2} \log \relax (c) + 6 \, {\left (4 \, d^{3} e f g + d e^{3} g^{2}\right )} p x^{3 \, n} - 6 \, {\left (4 \, d^{4} f^{2} n \log \relax (c) - {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} n p\right )} x^{4 \, n} \log \relax (x) - 3 \, {\left (d^{2} e^{2} g^{2} p - 8 \, d^{4} f g \log \relax (c)\right )} x^{2 \, n} + 6 \, {\left (4 \, d^{4} f g p x^{2 \, n} + d^{4} g^{2} p - {\left (4 \, d^{4} f^{2} n p \log \relax (x) + {\left (4 \, d^{2} e^{2} f g + e^{4} g^{2}\right )} p\right )} x^{4 \, n}\right )} \log \left (e x^{n} + d\right )}{24 \, d^{4} n x^{4 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f + \frac {g}{x^{2 \, n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [C]  time = 3.64, size = 755, normalized size = 2.94 \[ -\frac {e \,g^{2} p \,x^{-3 n}}{12 d n}-\frac {f^{2} p \dilog \left (\frac {e \,x^{n}+d}{d}\right )}{n}+\frac {f^{2} \ln \relax (c ) \ln \left (x^{n}\right )}{n}+\frac {\left (4 f^{2} n \,x^{4 n} \ln \relax (x )-4 f g \,x^{2 n}-g^{2}\right ) x^{-4 n} \ln \left (\left (e \,x^{n}+d \right )^{p}\right )}{4 n}-\frac {g^{2} x^{-4 n} \ln \relax (c )}{4 n}-\frac {e^{4} g^{2} p \ln \left (x^{n}\right )}{4 d^{4} n}-\frac {i \pi \,g^{2} x^{-4 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{8 n}-\frac {i \pi \,g^{2} x^{-4 n} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{8 n}-f^{2} p \ln \relax (x ) \ln \left (\frac {e \,x^{n}+d}{d}\right )+\frac {e^{2} g^{2} p \,x^{-2 n}}{8 d^{2} n}-\frac {e^{3} g^{2} p \,x^{-n}}{4 d^{3} n}-\frac {e^{2} f g p \ln \left (x^{n}\right )}{d^{2} n}+\frac {e^{4} g^{2} p \ln \left (e \,x^{n}+d \right )}{4 d^{4} n}+\frac {i \pi f g \,x^{-2 n} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{2 n}-\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right ) \ln \left (x^{n}\right )}{2 n}-\frac {i \pi f g \,x^{-2 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{2 n}-\frac {i \pi f g \,x^{-2 n} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{2 n}+\frac {i \pi \,g^{2} x^{-4 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )}{8 n}-\frac {f g \,x^{-2 n} \ln \relax (c )}{n}-\frac {e f g p \,x^{-n}}{d n}+\frac {i \pi \,g^{2} x^{-4 n} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{8 n}-\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3} \ln \left (x^{n}\right )}{2 n}+\frac {e^{2} f g p \ln \left (e \,x^{n}+d \right )}{d^{2} n}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \left (x^{n}\right )}{2 n}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \left (x^{n}\right )}{2 n}+\frac {i \pi f g \,x^{-2 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )}{2 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, d^{2} e g^{2} p x^{n} + 6 \, d^{3} g^{2} \log \relax (c) + 12 \, {\left (d^{3} f^{2} n^{2} p \log \relax (x)^{2} - 2 \, d^{3} f^{2} n \log \relax (c) \log \relax (x)\right )} x^{4 \, n} + 6 \, {\left (4 \, d^{2} e f g p + e^{3} g^{2} p\right )} x^{3 \, n} - 3 \, {\left (d e^{2} g^{2} p - 8 \, d^{3} f g \log \relax (c)\right )} x^{2 \, n} - 6 \, {\left (4 \, d^{3} f^{2} n x^{4 \, n} \log \relax (x) - 4 \, d^{3} f g x^{2 \, n} - d^{3} g^{2}\right )} \log \left ({\left (e x^{n} + d\right )}^{p}\right )}{24 \, d^{3} n x^{4 \, n}} + \int \frac {4 \, d^{4} f^{2} n p \log \relax (x) - 4 \, d^{2} e^{2} f g p - e^{4} g^{2} p}{4 \, {\left (d^{3} e x x^{n} + d^{4} x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+\frac {g}{x^{2\,n}}\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________