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

Optimal. Leaf size=159 \[ \frac{p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{d}-\frac{p \text{PolyLog}\left (2,\frac{b}{a x}+1\right )}{d}-\frac{p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{d}-\frac{\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{d}-\frac{\log \left (-\frac{b}{a x}\right ) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{d}+\frac{p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{d}-\frac{p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.246079, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2466, 2454, 2394, 2315, 2462, 260, 2416, 2393, 2391} \[ \frac{p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{d}-\frac{p \text{PolyLog}\left (2,\frac{b}{a x}+1\right )}{d}-\frac{p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{d}-\frac{\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{d}-\frac{\log \left (-\frac{b}{a x}\right ) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{d}+\frac{p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{d}-\frac{p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2466

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_))^(r_.), x_S
ymbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e,
f, g, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]

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]

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

Mathematica [A]  time = 0.0606942, size = 139, normalized size = 0.87 \[ -\frac{-p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )+p \text{PolyLog}\left (2,\frac{b}{a x}+1\right )+p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )+\log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\log \left (-\frac{b}{a x}\right ) \log \left (c \left (a+\frac{b}{x}\right )^p\right )-p \log (d+e x) \log \left (\frac{e (a x+b)}{b e-a d}\right )+p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.744, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ( ex+d \right ) }\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/(e*x+d),x)

[Out]

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

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Maxima [A]  time = 1.21021, size = 242, normalized size = 1.52 \begin{align*} -\frac{1}{2} \, b p{\left (\frac{2 \, \log \left (e x + d\right ) \log \left (x\right ) - \log \left (x\right )^{2}}{b d} + \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 d} - \frac{2 \,{\left (\log \left (\frac{e x}{d} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{e x}{d}\right )\right )}}{b d} - \frac{2 \,{\left (\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 )\right )}}{b d}\right )} -{\left (\frac{\log \left (e x + d\right )}{d} - \frac{\log \left (x\right )}{d}\right )} \log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(c*((a*x + b)/x)^p)/(e*x^2 + d*x), x)

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

[Out]

Timed out

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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 )}{{\left (e x + d\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/(e*x+d),x, algorithm="giac")

[Out]

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