3.107 \(\int \frac{\log (1+\frac{1}{a+b x})}{a+b x} \, dx\)

Optimal. Leaf size=15 \[ \frac{\text{PolyLog}\left (2,-\frac{1}{a+b x}\right )}{b} \]

[Out]

PolyLog[2, -(a + b*x)^(-1)]/b

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Rubi [A]  time = 0.0140898, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {2447} \[ \frac{\text{PolyLog}\left (2,-\frac{1}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + (a + b*x)^(-1)]/(a + b*x),x]

[Out]

PolyLog[2, -(a + b*x)^(-1)]/b

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [B]  time = 0.0144037, size = 140, normalized size = 9.33 \[ -\frac{\text{PolyLog}(2,-a-b x)}{b}+\frac{\log ^2\left (\frac{a b-(a+1) b}{b (a+b x)}\right )}{2 b}+\frac{\log \left (\frac{b (-a-b x-1)}{(-a-1) b+a b}\right ) \log \left (\frac{a b-(a+1) b}{b (a+b x)}\right )}{b}-\frac{\log \left (\frac{a+b x+1}{a+b x}\right ) \log \left (\frac{a b-(a+1) b}{b (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 + (a + b*x)^(-1)]/(a + b*x),x]

[Out]

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

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Maple [A]  time = 0.045, size = 15, normalized size = 1. \begin{align*}{\frac{{\it dilog} \left ( 1+ \left ( bx+a \right ) ^{-1} \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(1+1/(b*x+a))/(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.21988, size = 82, normalized size = 5.47 \begin{align*} \frac{2 \, \log \left (b x + a + 1\right ) \log \left (b x + a\right ) - \log \left (b x + a\right )^{2}}{2 \, b} - \frac{\log \left (b x + a + 1\right ) \log \left (b x + a\right ) +{\rm Li}_2\left (-b x - a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+1/(b*x+a))/(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.952507, size = 53, normalized size = 3.53 \begin{align*} \frac{{\rm Li}_2\left (-\frac{b x + a + 1}{b x + a} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+1/(b*x+a))/(b*x+a),x, algorithm="fricas")

[Out]

dilog(-(b*x + a + 1)/(b*x + a) + 1)/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (1 + \frac{1}{a + b x} \right )}}{a + b x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(1+1/(b*x+a))/(b*x+a),x)

[Out]

Integral(log(1 + 1/(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{1}{b x + a} + 1\right )}{b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+1/(b*x+a))/(b*x+a),x, algorithm="giac")

[Out]

integrate(log(1/(b*x + a) + 1)/(b*x + a), x)