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

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

[Out]

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

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Rubi [A]  time = 0.0314029, 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} \[ \frac{\text{PolyLog}\left (2,1-\frac{2 e}{e+f x}\right )}{2 e f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0056395, size = 27, normalized size = 1.12 \[ \frac{\text{PolyLog}\left (2,\frac{f x-e}{e+f x}\right )}{2 e f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 20, normalized size = 0.8 \begin{align*}{\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(2*e/(f*x+e))/(-f^2*x^2+e^2),x)

[Out]

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

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Maxima [B]  time = 1.19639, size = 162, normalized size = 6.75 \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{2 \, e}{f x + e}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(2*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(2*e/(f*x + e))

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Fricas [A]  time = 1.60103, size = 50, normalized size = 2.08 \begin{align*} \frac{{\rm Li}_2\left (-\frac{2 \, e}{f x + e} + 1\right )}{2 \, e f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*dilog(-2*e/(f*x + e) + 1)/(e*f)

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

[Out]

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

[Out]

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