3.337 \(\int \frac{\log (\frac{2 a}{a+b x})}{a^2-b^2 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac{\text{PolyLog}\left (2,1-\frac{2 a}{a+b x}\right )}{2 a b} \]

[Out]

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

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Rubi [A]  time = 0.0309399, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2402, 2315} \[ \frac{\text{PolyLog}\left (2,1-\frac{2 a}{a+b x}\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*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]

Rubi steps

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

Mathematica [A]  time = 0.0059873, size = 27, normalized size = 1.12 \[ \frac{\text{PolyLog}\left (2,\frac{b x-a}{a+b x}\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 20, normalized size = 0.8 \begin{align*}{\frac{1}{2\,ab}{\it dilog} \left ( 2\,{\frac{a}{bx+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(2*a/(b*x+a))/(-b^2*x^2+a^2),x)

[Out]

1/2/b/a*dilog(2*a/(b*x+a))

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Maxima [B]  time = 1.03673, size = 162, normalized size = 6.75 \begin{align*} \frac{1}{4} \, b{\left (\frac{\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (b x - a\right )}{a b^{2}} + \frac{2 \,{\left (\log \left (b x + a\right ) \log \left (-\frac{b x + a}{2 \, a} + 1\right ) +{\rm Li}_2\left (\frac{b x + a}{2 \, a}\right )\right )}}{a b^{2}}\right )} + \frac{1}{2} \,{\left (\frac{\log \left (b x + a\right )}{a b} - \frac{\log \left (b x - a\right )}{a b}\right )} \log \left (\frac{2 \, a}{b x + a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(2*a/(b*x+a))/(-b^2*x^2+a^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67994, size = 50, normalized size = 2.08 \begin{align*} \frac{{\rm Li}_2\left (-\frac{2 \, a}{b x + a} + 1\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(2*a/(b*x+a))/(-b^2*x^2+a^2),x, algorithm="fricas")

[Out]

1/2*dilog(-2*a/(b*x + a) + 1)/(a*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\log{\left (2 \right )}}{- a^{2} + b^{2} x^{2}}\, dx - \int \frac{\log{\left (\frac{a}{a + b x} \right )}}{- a^{2} + b^{2} x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(2*a/(b*x+a))/(-b**2*x**2+a**2),x)

[Out]

-Integral(log(2)/(-a**2 + b**2*x**2), x) - Integral(log(a/(a + b*x))/(-a**2 + b**2*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\log \left (\frac{2 \, a}{b x + a}\right )}{b^{2} x^{2} - a^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(2*a/(b*x+a))/(-b^2*x^2+a^2),x, algorithm="giac")

[Out]

integrate(-log(2*a/(b*x + a))/(b^2*x^2 - a^2), x)