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

Optimal. Leaf size=377 \[ \frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (\sqrt {-f} x^n+\sqrt {g}\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{2 f^2 n \left (f x^{2 n}+g\right )}-\frac {d e \sqrt {g} p \tan ^{-1}\left (\frac {\sqrt {f} x^n}{\sqrt {g}}\right )}{2 f^{3/2} n \left (d^2 f+e^2 g\right )}+\frac {e^2 g p \log \left (f x^{2 n}+g\right )}{4 f^2 n \left (d^2 f+e^2 g\right )}-\frac {e^2 g p \log \left (d+e x^n\right )}{2 f^2 n \left (d^2 f+e^2 g\right )}+\frac {p \text {Li}_2\left (\frac {\sqrt {-f} \left (e x^n+d\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {p \text {Li}_2\left (\frac {\sqrt {-f} \left (e x^n+d\right )}{\sqrt {-f} d+e \sqrt {g}}\right )}{2 f^2 n} \]

[Out]

-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

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Rubi [A]  time = 0.61, antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {2475, 263, 266, 43, 2416, 2413, 706, 31, 635, 205, 260, 2394, 2393, 2391} \[ \frac {p \text {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {p \text {PolyLog}\left (2,\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n}+\frac {g \log \left (c \left (d+e x^n\right )^p\right )}{2 f^2 n \left (f x^{2 n}+g\right )}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n}+\frac {\log \left (c \left (d+e x^n\right )^p\right ) \log \left (-\frac {e \left (\sqrt {-f} x^n+\sqrt {g}\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}-\frac {e^2 g p \log \left (d+e x^n\right )}{2 f^2 n \left (d^2 f+e^2 g\right )}+\frac {e^2 g p \log \left (f x^{2 n}+g\right )}{4 f^2 n \left (d^2 f+e^2 g\right )}-\frac {d e \sqrt {g} p \tan ^{-1}\left (\frac {\sqrt {f} x^n}{\sqrt {g}}\right )}{2 f^{3/2} n \left (d^2 f+e^2 g\right )} \]

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[f]*x^n)/Sqrt[g]])/(2*f^(3/2)*(d^2*f + e^2*g)*n) - (e^2*g*p*Log[d + e*x^n])/(2*f^2
*(d^2*f + e^2*g)*n) + (g*Log[c*(d + e*x^n)^p])/(2*f^2*n*(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*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*n) + (e^2*g*p*Log[g + f*x^(2*n)])/(4*f^2*(d^2*f + e^2*g)*n) + (p*Poly
Log[2, (Sqrt[-f]*(d + e*x^n))/(d*Sqrt[-f] - e*Sqrt[g])])/(2*f^2*n) + (p*PolyLog[2, (Sqrt[-f]*(d + e*x^n))/(d*S
qrt[-f] + e*Sqrt[g])])/(2*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 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 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 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 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 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 )}{\left (f+\frac {g}{x^2}\right )^2 x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {g x \log \left (c (d+e x)^p\right )}{f \left (g+f x^2\right )^2}+\frac {x \log \left (c (d+e x)^p\right )}{f \left (g+f x^2\right )}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \log \left (c (d+e x)^p\right )}{g+f x^2} \, dx,x,x^n\right )}{f n}-\frac {g \operatorname {Subst}\left (\int \frac {x \log \left (c (d+e x)^p\right )}{\left (g+f x^2\right )^2} \, dx,x,x^n\right )}{f 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 {\operatorname {Subst}\left (\int \left (-\frac {\sqrt {-f} \log \left (c (d+e x)^p\right )}{2 f \left (\sqrt {g}-\sqrt {-f} x\right )}+\frac {\sqrt {-f} \log \left (c (d+e x)^p\right )}{2 f \left (\sqrt {g}+\sqrt {-f} x\right )}\right ) \, dx,x,x^n\right )}{f n}-\frac {(e g p) \operatorname {Subst}\left (\int \frac {1}{(d+e x) \left (g+f x^2\right )} \, dx,x,x^n\right )}{2 f^2 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 {\operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{\sqrt {g}-\sqrt {-f} x} \, dx,x,x^n\right )}{2 (-f)^{3/2} n}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{\sqrt {g}+\sqrt {-f} x} \, dx,x,x^n\right )}{2 (-f)^{3/2} n}-\frac {(e g p) \operatorname {Subst}\left (\int \frac {d f-e f x}{g+f x^2} \, dx,x,x^n\right )}{2 f^2 \left (d^2 f+e^2 g\right ) n}-\frac {\left (e^3 g p\right ) \operatorname {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^n\right )}{2 f^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 p) \operatorname {Subst}\left (\int \frac {\log \left (\frac {e \left (\sqrt {g}-\sqrt {-f} x\right )}{d \sqrt {-f}+e \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 {g}+\sqrt {-f} x\right )}{-d \sqrt {-f}+e \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}{g+f x^2} \, dx,x,x^n\right )}{2 f \left (d^2 f+e^2 g\right ) n}+\frac {\left (e^2 g p\right ) \operatorname {Subst}\left (\int \frac {x}{g+f x^2} \, dx,x,x^n\right )}{2 f \left (d^2 f+e^2 g\right ) n}\\ &=-\frac {d e \sqrt {g} p \tan ^{-1}\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 {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {-f} x}{-d \sqrt {-f}+e \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 {-f} x}{d \sqrt {-f}+e \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 {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 \text {Li}_2\left (\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}-e \sqrt {g}}\right )}{2 f^2 n}+\frac {p \text {Li}_2\left (\frac {\sqrt {-f} \left (d+e x^n\right )}{d \sqrt {-f}+e \sqrt {g}}\right )}{2 f^2 n}\\ \end {align*}

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Mathematica [F]  time = 1.26, 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]

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{f^{2} x + \frac {2 \, f g x x^{2 \, n}}{x^{4 \, n}} + \frac {g^{2} x}{x^{4 \, n}}}, x\right ) \]

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)/(f^2*x + 2*f*g*x*x^(2*n)/x^(4*n) + g^2*x/x^(4*n)), x)

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

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)/((f + g/x^(2*n))^2*x), x)

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maple [C]  time = 0.71, size = 810, normalized size = 2.15 \[ -\frac {d e g p \arctan \left (\frac {f \,x^{n}}{\sqrt {f g}}\right )}{2 \left (d^{2} f +e^{2} g \right ) \sqrt {f g}\, f n}-\frac {e^{2} g p \ln \left (e \,x^{n}+d \right )}{2 \left (d^{2} f +e^{2} g \right ) f^{2} n}+\frac {e^{2} g p \ln \left (f \,x^{2 n}+g \right )}{4 \left (d^{2} f +e^{2} g \right ) f^{2} n}-\frac {i \pi g \,\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 )}{4 \left (f \,x^{2 n}+g \right ) f^{2} n}+\frac {i \pi g \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{4 \left (f \,x^{2 n}+g \right ) f^{2} n}+\frac {i \pi g \,\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}}{4 \left (f \,x^{2 n}+g \right ) f^{2} n}-\frac {i \pi g \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{4 \left (f \,x^{2 n}+g \right ) f^{2} n}-\frac {i \pi \,\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 (f \,x^{2 n}+g \right )}{4 f^{2} n}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \left (f \,x^{2 n}+g \right )}{4 f^{2} n}+\frac {i \pi \,\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 (f \,x^{2 n}+g \right )}{4 f^{2} n}-\frac {i \pi \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3} \ln \left (f \,x^{2 n}+g \right )}{4 f^{2} n}+\frac {p \ln \left (\frac {d f +\sqrt {-f g}\, e -\left (e \,x^{n}+d \right ) f}{d f +\sqrt {-f g}\, e}\right ) \ln \left (e \,x^{n}+d \right )}{2 f^{2} n}+\frac {p \ln \left (\frac {-d f +\sqrt {-f g}\, e +\left (e \,x^{n}+d \right ) f}{-d f +\sqrt {-f g}\, e}\right ) \ln \left (e \,x^{n}+d \right )}{2 f^{2} n}-\frac {p \ln \left (e \,x^{n}+d \right ) \ln \left (f \,x^{2 n}+g \right )}{2 f^{2} n}+\frac {g \ln \relax (c )}{2 \left (f \,x^{2 n}+g \right ) f^{2} n}+\frac {g \ln \left (\left (e \,x^{n}+d \right )^{p}\right )}{2 \left (f \,x^{2 n}+g \right ) f^{2} n}+\frac {p \dilog \left (\frac {d f +\sqrt {-f g}\, e -\left (e \,x^{n}+d \right ) f}{d f +\sqrt {-f g}\, e}\right )}{2 f^{2} n}+\frac {p \dilog \left (\frac {-d f +\sqrt {-f g}\, e +\left (e \,x^{n}+d \right ) f}{-d f +\sqrt {-f g}\, e}\right )}{2 f^{2} n}+\frac {\ln \relax (c ) \ln \left (f \,x^{2 n}+g \right )}{2 f^{2} n}+\frac {\ln \left (\left (e \,x^{n}+d \right )^{p}\right ) \ln \left (f \,x^{2 n}+g \right )}{2 f^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)/((f + g/x^(2*n))^2*x), x)

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

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)

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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

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