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

Optimal. Leaf size=15 \[ \frac {\text {Li}_2\left (-\frac {1}{a+b x}\right )}{b} \]

[Out]

polylog(2,-1/(b*x+a))/b

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, 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*}

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Mathematica [B]  time = 0.01, size = 140, normalized size = 9.33 \[ -\frac {\text {Li}_2(-a-b x)}{b}+\frac {\log ^2\left (\frac {a b-(a+1) b}{b (a+b x)}\right )}{2 b}+\frac {\log \left (\frac {b (-a-b x-1)}{(-a-1) b+a b}\right ) \log \left (\frac {a b-(a+1) b}{b (a+b x)}\right )}{b}-\frac {\log \left (\frac {a+b x+1}{a+b x}\right ) \log \left (\frac {a b-(a+1) b}{b (a+b x)}\right )}{b} \]

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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fricas [A]  time = 0.66, size = 22, normalized size = 1.47 \[ \frac {{\rm Li}_2\left (-\frac {b x + a + 1}{b x + a} + 1\right )}{b} \]

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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giac [B]  time = 39.25, size = 320, normalized size = 21.33 \[ \frac {1}{2} \, {\left ({\left (a + 1\right )} b - a b\right )}^{2} {\left (\frac {\log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a \right |}}\right )}{b^{4}} - \frac {\log \left ({\left | \frac {b x + a + 1}{b x + a} - 1 \right |}\right )}{b^{4}} - \frac {1}{b^{4} {\left (\frac {b x + a + 1}{b x + a} - 1\right )}} - \frac {\log \left (\frac {1}{a - \frac {{\left (\frac {{\left (a - \frac {{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a} - b} + 1\right )} a}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a} - b}} - a - 1\right )} b}{\frac {{\left (a - \frac {{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a} - b} + 1\right )} b}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} a}{b x + a} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a} - b}} - b}} + 1\right )}{b^{4} {\left (\frac {b x + a + 1}{b x + a} - 1\right )}^{2}}\right )} \]

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]

1/2*((a + 1)*b - a*b)^2*(log(abs(b*x + a + 1)/abs(b*x + a))/b^4 - log(abs((b*x + a + 1)/(b*x + a) - 1))/b^4 -
1/(b^4*((b*x + a + 1)/(b*x + a) - 1)) - log(1/(a - ((a - ((b*x + a + 1)*a/(b*x + a) - a - 1)*b/((b*x + a + 1)*
b/(b*x + a) - b) + 1)*a/(a - ((b*x + a + 1)*a/(b*x + a) - a - 1)*b/((b*x + a + 1)*b/(b*x + a) - b)) - a - 1)*b
/((a - ((b*x + a + 1)*a/(b*x + a) - a - 1)*b/((b*x + a + 1)*b/(b*x + a) - b) + 1)*b/(a - ((b*x + a + 1)*a/(b*x
+ a) - a - 1)*b/((b*x + a + 1)*b/(b*x + a) - b)) - b)) + 1)/(b^4*((b*x + a + 1)/(b*x + a) - 1)^2))

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maple [A]  time = 0.04, size = 15, normalized size = 1.00 \[ \frac {\dilog \left (1+\frac {1}{b x +a}\right )}{b} \]

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 = 1.21, size = 61, normalized size = 4.07 \[ \frac {2 \, \log \left (b x + a + 1\right ) \log \left (b x + a\right ) - \log \left (b x + a\right )^{2}}{2 \, b} - \frac {\log \left (b x + a + 1\right ) \log \left (b x + a\right ) + {\rm Li}_2\left (-b x - a\right )}{b} \]

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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mupad [B]  time = 4.03, size = 15, normalized size = 1.00 \[ \frac {\mathrm {polylog}\left (2,-\frac {1}{a+b\,x}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(1/(a + b*x) + 1)/(a + b*x),x)

[Out]

polylog(2, -1/(a + b*x))/b

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (1 + \frac {1}{a + b x} \right )}}{a + b x}\, dx \]

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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