3.173 \(\int \frac{\log (c (d+e x^n)^p)}{x} \, dx\)

Optimal. Leaf size=44 \[ \frac{p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.0397916, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2454, 2394, 2315} \[ \frac{p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*x^n)^p]/x,x]

[Out]

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 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{\log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac{(e p) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}\\ &=\frac{\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{p \text{Li}_2\left (1+\frac{e x^n}{d}\right )}{n}\\ \end{align*}

Mathematica [A]  time = 0.0052101, size = 43, normalized size = 0.98 \[ \frac{p \text{PolyLog}\left (2,\frac{d+e x^n}{d}\right )+\log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x^n)^p]/x,x]

[Out]

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

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Maple [C]  time = 0.079, size = 177, normalized size = 4. \begin{align*} \ln \left ( x \right ) \ln \left ( \left ( d+e{x}^{n} \right ) ^{p} \right ) +{\frac{i}{2}}\ln \left ( x \right ) \pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{2}}\ln \left ( x \right ) \pi \,{\it csgn} \left ( i \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{2}}\ln \left ( x \right ) \pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{3}+{\frac{i}{2}}\ln \left ( x \right ) \pi \, \left ({\it csgn} \left ( ic \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +\ln \left ( c \right ) \ln \left ( x \right ) -{\frac{p}{n}{\it dilog} \left ({\frac{d+e{x}^{n}}{d}} \right ) }-p\ln \left ( x \right ) \ln \left ({\frac{d+e{x}^{n}}{d}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d n p \int \frac{\log \left (x\right )}{e x x^{n} + d x}\,{d x} - \frac{1}{2} \, n p \log \left (x\right )^{2} + \log \left ({\left (e x^{n} + d\right )}^{p}\right ) \log \left (x\right ) + \log \left (c\right ) \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10443, size = 150, normalized size = 3.41 \begin{align*} \frac{n p \log \left (e x^{n} + d\right ) \log \left (x\right ) - n p \log \left (x\right ) \log \left (\frac{e x^{n} + d}{d}\right ) + n \log \left (c\right ) \log \left (x\right ) - p{\rm Li}_2\left (-\frac{e x^{n} + d}{d} + 1\right )}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(d + e*x**n)**p)/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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(log(c*(d+e*x^n)^p)/x,x, algorithm="giac")

[Out]

integrate(log((e*x^n + d)^p*c)/x, x)