3.88 \(\int \frac{\log (\frac{c x}{a+b x})}{a+b x} \, dx\)

Optimal. Leaf size=46 \[ -\frac{\text{PolyLog}\left (2,1-\frac{a}{a+b x}\right )}{b}-\frac{\log \left (\frac{a}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.158601, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2488, 2411, 2343, 2333, 2315} \[ -\frac{\text{PolyLog}\left (2,1-\frac{a}{a+b x}\right )}{b}-\frac{\log \left (\frac{a}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[(c*x)/(a + b*x)]/(a + b*x),x]

[Out]

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

Rule 2488

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_)),
x_Symbol] :> -Simp[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/h, x] + Dist[(p
*r*s*(b*c - a*d))/h, Int[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a
+ b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q,
0] && EqQ[b*g - a*h, 0] && IGtQ[s, 0]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

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

Mathematica [A]  time = 0.0167248, size = 84, normalized size = 1.83 \[ -\frac{\text{PolyLog}\left (2,\frac{a+b x}{a}\right )}{b}-\frac{\log \left (\frac{a}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{b}+\frac{\log ^2\left (\frac{a}{a+b x}\right )}{2 b}+\frac{\log \left (-\frac{b x}{a}\right ) \log \left (\frac{a}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(c*x)/(a + b*x)]/(a + b*x),x]

[Out]

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

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Maple [B]  time = 0.098, size = 97, normalized size = 2.1 \begin{align*} -{\frac{1}{b}{\it dilog} \left ( -{\frac{1}{c} \left ( b \left ({\frac{c}{b}}-{\frac{ac}{b \left ( bx+a \right ) }} \right ) -c \right ) } \right ) }-{\frac{1}{b}\ln \left ({\frac{c}{b}}-{\frac{ac}{b \left ( bx+a \right ) }} \right ) \ln \left ( -{\frac{1}{c} \left ( b \left ({\frac{c}{b}}-{\frac{ac}{b \left ( bx+a \right ) }} \right ) -c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-dilog(-(b*(c/b-a*c/b/(b*x+a))-c)/c)/b-ln(c/b-a*c/b/(b*x+a))*ln(-(b*(c/b-a*c/b/(b*x+a))-c)/c)/b

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Maxima [B]  time = 1.12936, size = 128, normalized size = 2.78 \begin{align*} \frac{\log \left (b x + a\right ) \log \left (\frac{c x}{b x + a}\right )}{b} - \frac{\frac{c \log \left (b x + a\right )^{2}}{b} - \frac{2 \,{\left (\log \left (\frac{b x}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a}\right )\right )} c}{b}}{2 \, c} + \frac{{\left (c \log \left (b x + a\right ) - c \log \left (x\right )\right )} \log \left (b x + a\right )}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(c*x/(b*x + a))/(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*x/(a + b*x))/(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(c*x/(b*x + a))/(b*x + a), x)