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

Optimal. Leaf size=105 \[ -\frac{\text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )}{d}+\frac{\text{PolyLog}\left (2,\frac{d x}{c}+1\right )}{d}+\frac{\log \left (\frac{a}{x}+b\right ) \log (c+d x)}{d}-\frac{\log (c+d x) \log \left (-\frac{d (a+b x)}{b c-a d}\right )}{d}+\frac{\log \left (-\frac{d x}{c}\right ) \log (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.167946, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {2465, 2462, 260, 2416, 2394, 2315, 2393, 2391} \[ -\frac{\text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )}{d}+\frac{\text{PolyLog}\left (2,\frac{d x}{c}+1\right )}{d}+\frac{\log \left (\frac{a}{x}+b\right ) \log (c+d x)}{d}-\frac{\log (c+d x) \log \left (-\frac{d (a+b x)}{b c-a d}\right )}{d}+\frac{\log \left (-\frac{d x}{c}\right ) \log (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Log[(a + b*x)/x]/(c + d*x),x]

[Out]

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

Rule 2465

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.)*(u_)^(r_.), x_Symbol] :> Int[ExpandToSum[u, x]^r*(a + b*Log[c*
ExpandToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, p, q, r}, x] && LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMa
tchQ[u, x] && BinomialMatchQ[v, x])

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

Mathematica [A]  time = 0.0286185, size = 80, normalized size = 0.76 \[ \frac{-\text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )+\text{PolyLog}\left (2,\frac{d x}{c}+1\right )+\log (c+d x) \left (-\log \left (\frac{d (a+b x)}{a d-b c}\right )+\log \left (\frac{a}{x}+b\right )+\log \left (-\frac{d x}{c}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.411, size = 114, normalized size = 1.1 \begin{align*} -{\frac{1}{d}\ln \left ( b+{\frac{a}{x}} \right ) \ln \left ( -{\frac{a}{bx}} \right ) }-{\frac{1}{d}{\it dilog} \left ( -{\frac{a}{bx}} \right ) }+{\frac{1}{d}{\it dilog} \left ({\frac{1}{ad-bc} \left ( c \left ( b+{\frac{a}{x}} \right ) +ad-bc \right ) } \right ) }+{\frac{1}{d}\ln \left ( b+{\frac{a}{x}} \right ) \ln \left ({\frac{1}{ad-bc} \left ( c \left ( b+{\frac{a}{x}} \right ) +ad-bc \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*ln(b+a/x)*ln(-a/b/x)-1/d*dilog(-a/b/x)+1/d*dilog((c*(b+a/x)+a*d-b*c)/(a*d-b*c))+1/d*ln(b+a/x)*ln((c*(b+a/
x)+a*d-b*c)/(a*d-b*c))

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Maxima [A]  time = 1.0664, size = 167, normalized size = 1.59 \begin{align*} -\frac{{\left (\log \left (b x + a\right ) - \log \left (x\right )\right )} \log \left (d x + c\right )}{d} + \frac{\log \left (d x + c\right ) \log \left (\frac{b x + a}{x}\right )}{d} - \frac{\log \left (\frac{d x}{c} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{d x}{c}\right )}{d} + \frac{\log \left (b x + a\right ) \log \left (\frac{b d x + a d}{b c - a d} + 1\right ) +{\rm Li}_2\left (-\frac{b d x + a d}{b c - a d}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a/x + b)/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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