3.280 \(\int \frac{\log (\frac{e}{e+f x})}{e^2-f^2 x^2} \, dx\)

Optimal. Leaf size=42 \[ \frac{\text{PolyLog}\left (2,1-\frac{2 e}{e+f x}\right )}{2 e f}-\frac{\log (2) \tanh ^{-1}\left (\frac{f x}{e}\right )}{e f} \]

[Out]

-((ArcTanh[(f*x)/e]*Log[2])/(e*f)) + PolyLog[2, 1 - (2*e)/(e + f*x)]/(2*e*f)

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Rubi [A]  time = 0.0544162, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2403, 208, 2402, 2315} \[ \frac{\text{PolyLog}\left (2,1-\frac{2 e}{e+f x}\right )}{2 e f}-\frac{\log (2) \tanh ^{-1}\left (\frac{f x}{e}\right )}{e f} \]

Antiderivative was successfully verified.

[In]

Int[Log[e/(e + f*x)]/(e^2 - f^2*x^2),x]

[Out]

-((ArcTanh[(f*x)/e]*Log[2])/(e*f)) + PolyLog[2, 1 - (2*e)/(e + f*x)]/(2*e*f)

Rule 2403

Int[((a_.) + Log[(c_.)/((d_) + (e_.)*(x_))]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[a + b*Log[c/(2*d)]
, Int[1/(f + g*x^2), x], x] + Dist[b, Int[Log[(2*d)/(d + e*x)]/(f + g*x^2), x], x] /; FreeQ[{a, b, c, d, e, f,
g}, x] && EqQ[e^2*f + d^2*g, 0] && GtQ[c/(2*d), 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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{e}{e+f x}\right )}{e^2-f^2 x^2} \, dx &=-\left (\log (2) \int \frac{1}{e^2-f^2 x^2} \, dx\right )+\int \frac{\log \left (\frac{2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx\\ &=-\frac{\tanh ^{-1}\left (\frac{f x}{e}\right ) \log (2)}{e f}+\frac{\operatorname{Subst}\left (\int \frac{\log (2 e x)}{1-2 e x} \, dx,x,\frac{1}{e+f x}\right )}{f}\\ &=-\frac{\tanh ^{-1}\left (\frac{f x}{e}\right ) \log (2)}{e f}+\frac{\text{Li}_2\left (1-\frac{2 e}{e+f x}\right )}{2 e f}\\ \end{align*}

Mathematica [A]  time = 0.0158733, size = 81, normalized size = 1.93 \[ \frac{\text{PolyLog}\left (2,\frac{e+f x}{2 e}\right )}{2 e f}-\frac{\log ^2\left (\frac{e}{e+f x}\right )}{4 e f}-\frac{\log \left (\frac{e-f x}{2 e}\right ) \log \left (\frac{e}{e+f x}\right )}{2 e f} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[e/(e + f*x)]/(e^2 - f^2*x^2),x]

[Out]

-(Log[(e - f*x)/(2*e)]*Log[e/(e + f*x)])/(2*e*f) - Log[e/(e + f*x)]^2/(4*e*f) + PolyLog[2, (e + f*x)/(2*e)]/(2
*e*f)

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Maple [B]  time = 0.062, size = 84, normalized size = 2. \begin{align*} -{\frac{1}{2\,fe}\ln \left ( 1-2\,{\frac{e}{fx+e}} \right ) \ln \left ({\frac{e}{fx+e}} \right ) }+{\frac{1}{2\,fe}\ln \left ( 1-2\,{\frac{e}{fx+e}} \right ) \ln \left ( 2\,{\frac{e}{fx+e}} \right ) }+{\frac{1}{2\,fe}{\it dilog} \left ( 2\,{\frac{e}{fx+e}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(e/(f*x+e))/(-f^2*x^2+e^2),x)

[Out]

-1/2/f/e*ln(1-2*e/(f*x+e))*ln(e/(f*x+e))+1/2/f/e*ln(1-2*e/(f*x+e))*ln(2*e/(f*x+e))+1/2/f/e*dilog(2*e/(f*x+e))

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Maxima [B]  time = 1.09472, size = 161, normalized size = 3.83 \begin{align*} \frac{1}{4} \, f{\left (\frac{\log \left (f x + e\right )^{2} - 2 \, \log \left (f x + e\right ) \log \left (f x - e\right )}{e f^{2}} + \frac{2 \,{\left (\log \left (f x + e\right ) \log \left (-\frac{f x + e}{2 \, e} + 1\right ) +{\rm Li}_2\left (\frac{f x + e}{2 \, e}\right )\right )}}{e f^{2}}\right )} + \frac{1}{2} \,{\left (\frac{\log \left (f x + e\right )}{e f} - \frac{\log \left (f x - e\right )}{e f}\right )} \log \left (\frac{e}{f x + e}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*f*((log(f*x + e)^2 - 2*log(f*x + e)*log(f*x - e))/(e*f^2) + 2*(log(f*x + e)*log(-1/2*(f*x + e)/e + 1) + di
log(1/2*(f*x + e)/e))/(e*f^2)) + 1/2*(log(f*x + e)/(e*f) - log(f*x - e)/(e*f))*log(e/(f*x + e))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\log \left (\frac{e}{f x + e}\right )}{f^{2} x^{2} - e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-log(e/(f*x + e))/(f^2*x^2 - e^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(e/(f*x+e))/(-f**2*x**2+e**2),x)

[Out]

-Integral(log(e/(e + f*x))/(-e**2 + f**2*x**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-log(e/(f*x + e))/(f^2*x^2 - e^2), x)