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

Optimal. Leaf size=126 \[ \frac{f p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac{g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac{e^2 g p \log \left (d+e x^n\right )}{2 d^2 n}-\frac{e^2 g p \log (x)}{2 d^2}-\frac{e g p x^{-n}}{2 d n} \]

[Out]

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

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Rubi [A]  time = 0.169208, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2475, 14, 2416, 2395, 44, 2394, 2315} \[ \frac{f p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac{g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac{e^2 g p \log \left (d+e x^n\right )}{2 d^2 n}-\frac{e^2 g p \log (x)}{2 d^2}-\frac{e g p x^{-n}}{2 d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.197987, size = 104, normalized size = 0.83 \[ -\frac{-2 f \left (p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )+\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )\right )+g x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )+\frac{e g p x^{-n} \left (-e x^n \log \left (d+e x^n\right )+d+e n x^n \log (x)\right )}{d^2}}{2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 4.392, size = 448, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{e g p x^{n} + d g \log \left (c\right ) +{\left (d f n^{2} p \log \left (x\right )^{2} - 2 \, d f n \log \left (c\right ) \log \left (x\right )\right )} x^{2 \, n} -{\left (2 \, d f n x^{2 \, n} \log \left (x\right ) - d g\right )} \log \left ({\left (e x^{n} + d\right )}^{p}\right )}{2 \, d n x^{2 \, n}} + \int \frac{2 \, d^{2} f n p \log \left (x\right ) - e^{2} g p}{2 \,{\left (d e x x^{n} + d^{2} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(e*g*p*x^n + d*g*log(c) + (d*f*n^2*p*log(x)^2 - 2*d*f*n*log(c)*log(x))*x^(2*n) - (2*d*f*n*x^(2*n)*log(x)
- d*g)*log((e*x^n + d)^p))/(d*n*x^(2*n)) + integrate(1/2*(2*d^2*f*n*p*log(x) - e^2*g*p)/(d*e*x*x^n + d^2*x), x
)

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Fricas [A]  time = 2.1054, size = 352, normalized size = 2.79 \begin{align*} -\frac{2 \, d^{2} f n p x^{2 \, n} \log \left (x\right ) \log \left (\frac{e x^{n} + d}{d}\right ) + 2 \, d^{2} f p x^{2 \, n}{\rm Li}_2\left (-\frac{e x^{n} + d}{d} + 1\right ) + d e g p x^{n} + d^{2} g \log \left (c\right ) +{\left (e^{2} g n p - 2 \, d^{2} f n \log \left (c\right )\right )} x^{2 \, n} \log \left (x\right ) +{\left (d^{2} g p -{\left (2 \, d^{2} f n p \log \left (x\right ) + e^{2} g p\right )} x^{2 \, n}\right )} \log \left (e x^{n} + d\right )}{2 \, d^{2} n x^{2 \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*d^2*f*n*p*x^(2*n)*log(x)*log((e*x^n + d)/d) + 2*d^2*f*p*x^(2*n)*dilog(-(e*x^n + d)/d + 1) + d*e*g*p*x^
n + d^2*g*log(c) + (e^2*g*n*p - 2*d^2*f*n*log(c))*x^(2*n)*log(x) + (d^2*g*p - (2*d^2*f*n*p*log(x) + e^2*g*p)*x
^(2*n))*log(e*x^n + d))/(d^2*n*x^(2*n))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f + \frac{g}{x^{2 \, n}}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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