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

Optimal. Leaf size=419 \[ -\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x^n\right )}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f^2 n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f^2 n}+\frac {\log \left (c \left (d+e x^n\right )^p\right )}{2 f n \left (f+g x^{2 n}\right )}-\frac {d e \sqrt {g} p \tan ^{-1}\left (\frac {\sqrt {g} x^n}{\sqrt {f}}\right )}{2 f^{3/2} n \left (d^2 g+e^2 f\right )}+\frac {e^2 p \log \left (f+g x^{2 n}\right )}{4 f n \left (d^2 g+e^2 f\right )}-\frac {e^2 p \log \left (d+e x^n\right )}{2 f n \left (d^2 g+e^2 f\right )}-\frac {p \text {Li}_2\left (-\frac {\sqrt {g} \left (e x^n+d\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}-\frac {p \text {Li}_2\left (\frac {\sqrt {g} \left (e x^n+d\right )}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f^2 n}+\frac {p \text {Li}_2\left (\frac {e x^n}{d}+1\right )}{f^2 n} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.57, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {2475, 266, 44, 2416, 2394, 2315, 2413, 706, 31, 635, 205, 260, 2393, 2391} \[ -\frac {p \text {PolyLog}\left (2,-\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}-\frac {p \text {PolyLog}\left (2,\frac {\sqrt {g} \left (d+e x^n\right )}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f^2 n}+\frac {p \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{f^2 n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x^n\right )}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f^2 n}-\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}+\frac {\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{f^2 n}+\frac {\log \left (c \left (d+e x^n\right )^p\right )}{2 f n \left (f+g x^{2 n}\right )}-\frac {d e \sqrt {g} p \tan ^{-1}\left (\frac {\sqrt {g} x^n}{\sqrt {f}}\right )}{2 f^{3/2} n \left (d^2 g+e^2 f\right )}-\frac {e^2 p \log \left (d+e x^n\right )}{2 f n \left (d^2 g+e^2 f\right )}+\frac {e^2 p \log \left (f+g x^{2 n}\right )}{4 f n \left (d^2 g+e^2 f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

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 2391

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

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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 2413

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

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 {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{2 n}\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x \left (f+g x^2\right )^2} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\log \left (c (d+e x)^p\right )}{f^2 x}-\frac {g x \log \left (c (d+e x)^p\right )}{f \left (f+g x^2\right )^2}-\frac {g x \log \left (c (d+e x)^p\right )}{f^2 \left (f+g x^2\right )}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{f^2 n}-\frac {g \operatorname {Subst}\left (\int \frac {x \log \left (c (d+e x)^p\right )}{f+g x^2} \, dx,x,x^n\right )}{f^2 n}-\frac {g \operatorname {Subst}\left (\int \frac {x \log \left (c (d+e x)^p\right )}{\left (f+g x^2\right )^2} \, dx,x,x^n\right )}{f 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 {g \operatorname {Subst}\left (\int \left (-\frac {\log \left (c (d+e x)^p\right )}{2 \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {\log \left (c (d+e x)^p\right )}{2 \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}\right ) \, dx,x,x^n\right )}{f^2 n}-\frac {(e p) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{f^2 n}-\frac {(e p) \operatorname {Subst}\left (\int \frac {1}{(d+e x) \left (f+g x^2\right )} \, dx,x,x^n\right )}{2 f 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 {p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{f^2 n}+\frac {\sqrt {g} \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{\sqrt {-f}-\sqrt {g} x} \, dx,x,x^n\right )}{2 f^2 n}-\frac {\sqrt {g} \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{\sqrt {-f}+\sqrt {g} x} \, dx,x,x^n\right )}{2 f^2 n}-\frac {(e p) \operatorname {Subst}\left (\int \frac {d g-e g x}{f+g x^2} \, dx,x,x^n\right )}{2 f \left (e^2 f+d^2 g\right ) n}-\frac {\left (e^3 p\right ) \operatorname {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^n\right )}{2 f \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 {p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{f^2 n}+\frac {(e p) \operatorname {Subst}\left (\int \frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{d+e x} \, dx,x,x^n\right )}{2 f^2 n}+\frac {(e p) \operatorname {Subst}\left (\int \frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{d+e x} \, dx,x,x^n\right )}{2 f^2 n}-\frac {(d e g p) \operatorname {Subst}\left (\int \frac {1}{f+g x^2} \, dx,x,x^n\right )}{2 f \left (e^2 f+d^2 g\right ) n}+\frac {\left (e^2 g p\right ) \operatorname {Subst}\left (\int \frac {x}{f+g x^2} \, dx,x,x^n\right )}{2 f \left (e^2 f+d^2 g\right ) n}\\ &=-\frac {d e \sqrt {g} p \tan ^{-1}\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 \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{f^2 n}+\frac {p \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {g} x}{e \sqrt {-f}-d \sqrt {g}}\right )}{x} \, dx,x,d+e x^n\right )}{2 f^2 n}+\frac {p \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {g} x}{e \sqrt {-f}+d \sqrt {g}}\right )}{x} \, dx,x,d+e x^n\right )}{2 f^2 n}\\ &=-\frac {d e \sqrt {g} p \tan ^{-1}\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 \text {Li}_2\left (-\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2 n}-\frac {p \text {Li}_2\left (\frac {\sqrt {g} \left (d+e x^n\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2 n}+\frac {p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{f^2 n}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 8.11, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (c \left (d+e x^n\right )^p\right )}{x \left (f+g x^{2 n}\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{g^{2} x x^{4 \, n} + 2 \, f g x x^{2 \, n} + f^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.60, size = 1036, normalized size = 2.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________