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

Optimal. Leaf size=29 \[ \frac{\text{PolyLog}\left (2,-\frac{b x}{a}\right )}{b}+\frac{\log (x) \log \left (\frac{b x}{a}+1\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0202085, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2317, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{b x}{a}\right )}{b}+\frac{\log (x) \log \left (\frac{b x}{a}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(a + b*x),x]

[Out]

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

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 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 (x)}{a+b x} \, dx &=\frac{\log (x) \log \left (1+\frac{b x}{a}\right )}{b}-\frac{\int \frac{\log \left (1+\frac{b x}{a}\right )}{x} \, dx}{b}\\ &=\frac{\log (x) \log \left (1+\frac{b x}{a}\right )}{b}+\frac{\text{Li}_2\left (-\frac{b x}{a}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.0024513, size = 30, normalized size = 1.03 \[ \frac{\text{PolyLog}\left (2,-\frac{b x}{a}\right )}{b}+\frac{\log (x) \log \left (\frac{a+b x}{a}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(a + b*x),x]

[Out]

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

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Maple [A]  time = 0.007, size = 32, normalized size = 1.1 \begin{align*}{\frac{1}{b}{\it dilog} \left ({\frac{bx+a}{a}} \right ) }+{\frac{\ln \left ( x \right ) }{b}\ln \left ({\frac{bx+a}{a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.958278, size = 34, normalized size = 1.17 \begin{align*} \frac{\log \left (\frac{b x}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x}{a}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(x)/(b*x + a), x)

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Sympy [C]  time = 7.30554, size = 151, normalized size = 5.21 \begin{align*} \begin{cases} \frac{\log{\left (\frac{a}{b} \right )} \log{\left (\frac{a}{b} + x \right )}}{b} + \frac{i \pi \log{\left (\frac{a}{b} + x \right )}}{b} - \frac{\operatorname{Li}_{2}\left (\frac{b \left (\frac{a}{b} + x\right )}{a}\right )}{b} & \text{for}\: \left |{\frac{a}{b} + x}\right | < 1 \\- \frac{\log{\left (\frac{a}{b} \right )} \log{\left (\frac{1}{\frac{a}{b} + x} \right )}}{b} - \frac{i \pi \log{\left (\frac{1}{\frac{a}{b} + x} \right )}}{b} - \frac{\operatorname{Li}_{2}\left (\frac{b \left (\frac{a}{b} + x\right )}{a}\right )}{b} & \text{for}\: \frac{1}{\left |{\frac{a}{b} + x}\right |} < 1 \\- \frac{{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{\frac{a}{b} + x} \right )} \log{\left (\frac{a}{b} \right )}}{b} - \frac{i \pi{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{\frac{a}{b} + x} \right )}}{b} + \frac{{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{\frac{a}{b} + x} \right )} \log{\left (\frac{a}{b} \right )}}{b} + \frac{i \pi{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{\frac{a}{b} + x} \right )}}{b} - \frac{\operatorname{Li}_{2}\left (\frac{b \left (\frac{a}{b} + x\right )}{a}\right )}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(a/b)*log(a/b + x)/b + I*pi*log(a/b + x)/b - polylog(2, b*(a/b + x)/a)/b, Abs(a/b + x) < 1), (-l
og(a/b)*log(1/(a/b + x))/b - I*pi*log(1/(a/b + x))/b - polylog(2, b*(a/b + x)/a)/b, 1/Abs(a/b + x) < 1), (-mei
jerg(((), (1, 1)), ((0, 0), ()), a/b + x)*log(a/b)/b - I*pi*meijerg(((), (1, 1)), ((0, 0), ()), a/b + x)/b + m
eijerg(((1, 1), ()), ((), (0, 0)), a/b + x)*log(a/b)/b + I*pi*meijerg(((1, 1), ()), ((), (0, 0)), a/b + x)/b -
polylog(2, b*(a/b + x)/a)/b, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(x)/(b*x + a), x)