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

Optimal. Leaf size=113 \[ -\frac{p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{e}+\frac{p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{e}+\frac{p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{e} \]

[Out]

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

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Rubi [A]  time = 0.146883, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {2462, 260, 2416, 2394, 2315, 2393, 2391} \[ -\frac{p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{e}+\frac{p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{e}+\frac{p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x)^p]/(d + e*x),x]

[Out]

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

Rule 2462

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[f +
g*x]*(a + b*Log[c*(d + e*x^n)^p]))/g, x] - Dist[(b*e*n*p)/g, Int[(x^(n - 1)*Log[f + g*x])/(d + e*x^n), x], x]
/; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]

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 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 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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0174174, size = 114, normalized size = 1.01 \[ -\frac{p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{e}+\frac{p \text{PolyLog}\left (2,\frac{d+e x}{d}\right )}{e}+\frac{\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e}-\frac{p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{e}+\frac{p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x)^p]/(d + e*x),x]

[Out]

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

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ex+d}\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)/(e*x+d),x)

[Out]

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

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Maxima [A]  time = 1.04896, size = 215, normalized size = 1.9 \begin{align*} \frac{b p{\left (\frac{\log \left (e x + d\right ) \log \left (a + \frac{b}{x}\right )}{b} - \frac{\log \left (e x + d\right ) \log \left (-\frac{a e x + a d}{a d - b e} + 1\right ) +{\rm Li}_2\left (\frac{a e x + a d}{a d - b e}\right )}{b} + \frac{\log \left (e x + d\right ) \log \left (-\frac{e x + d}{d} + 1\right ) +{\rm Li}_2\left (\frac{e x + d}{d}\right )}{b}\right )}}{e} - \frac{p \log \left (e x + d\right ) \log \left (a + \frac{b}{x}\right )}{e} + \frac{\log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(c*((a*x + b)/x)^p)/(e*x + d), 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 )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b/x)**p)/(d + e*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 )}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((a + b/x)^p*c)/(e*x + d), x)