3.31 \(\int \frac{\log (c (a+\frac{b}{x})^p)}{x} \, dx\)

Optimal. Leaf size=40 \[ \log \left (-\frac{b}{a x}\right ) \left (-\log \left (c \left (a+\frac{b}{x}\right )^p\right )\right )-p \text{PolyLog}\left (2,\frac{b}{a x}+1\right ) \]

[Out]

-(Log[c*(a + b/x)^p]*Log[-(b/(a*x))]) - p*PolyLog[2, 1 + b/(a*x)]

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Rubi [A]  time = 0.0370085, antiderivative size = 40, 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} \[ \log \left (-\frac{b}{a x}\right ) \left (-\log \left (c \left (a+\frac{b}{x}\right )^p\right )\right )-p \text{PolyLog}\left (2,\frac{b}{a x}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x)^p]/x,x]

[Out]

-(Log[c*(a + b/x)^p]*Log[-(b/(a*x))]) - p*PolyLog[2, 1 + b/(a*x)]

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 (a+\frac{b}{x}\right )^p\right )}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{\log \left (c (a+b x)^p\right )}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\log \left (c \left (a+\frac{b}{x}\right )^p\right ) \log \left (-\frac{b}{a x}\right )+(b p) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{b x}{a}\right )}{a+b x} \, dx,x,\frac{1}{x}\right )\\ &=-\log \left (c \left (a+\frac{b}{x}\right )^p\right ) \log \left (-\frac{b}{a x}\right )-p \text{Li}_2\left (1+\frac{b}{a x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0028915, size = 41, normalized size = 1.02 \[ \log \left (-\frac{b}{a x}\right ) \left (-\log \left (c \left (a+\frac{b}{x}\right )^p\right )\right )-p \text{PolyLog}\left (2,\frac{a+\frac{b}{x}}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x)^p]/x,x]

[Out]

-(Log[c*(a + b/x)^p]*Log[-(b/(a*x))]) - p*PolyLog[2, (a + b/x)/a]

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Maple [F]  time = 0.079, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a+b/x)^p)/x,x)

[Out]

int(ln(c*(a+b/x)^p)/x,x)

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Maxima [B]  time = 1.22406, size = 112, normalized size = 2.8 \begin{align*} \frac{1}{2} \, b p{\left (\frac{2 \, \log \left (a + \frac{b}{x}\right ) \log \left (x\right )}{b} + \frac{\log \left (x\right )^{2}}{b} - \frac{2 \,{\left (\log \left (\frac{a x}{b} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{a x}{b}\right )\right )}}{b}\right )} - p \log \left (a + \frac{b}{x}\right ) \log \left (x\right ) + \log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x,x, algorithm="maxima")

[Out]

1/2*b*p*(2*log(a + b/x)*log(x)/b + log(x)^2/b - 2*(log(a*x/b + 1)*log(x) + dilog(-a*x/b))/b) - p*log(a + b/x)*
log(x) + log((a + b/x)^p*c)*log(x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (c \left (\frac{a x + b}{x}\right )^{p}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x,x, algorithm="fricas")

[Out]

integral(log(c*((a*x + b)/x)^p)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(a+b/x)**p)/x,x)

[Out]

Integral(log(c*(a + b/x)**p)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x,x, algorithm="giac")

[Out]

integrate(log((a + b/x)^p*c)/x, x)