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

Optimal. Leaf size=24 \[ \frac {\text {Li}_2\left (1-\frac {2 e}{e+f x}\right )}{2 e f} \]

[Out]

1/2*polylog(2,1-2*e/(f*x+e))/e/f

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.12 \[ \frac {\text {Li}_2\left (\frac {f x-e}{e+f x}\right )}{2 e f} \]

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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fricas [A]  time = 0.41, size = 21, normalized size = 0.88 \[ \frac {{\rm Li}_2\left (-\frac {2 \, e}{f x + e} + 1\right )}{2 \, e f} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\log \left (\frac {2 \, e}{f x + e}\right )}{f^{2} x^{2} - e^{2}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 20, normalized size = 0.83 \[ \frac {\dilog \left (\frac {2 e}{f x +e}\right )}{2 e f} \]

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 = 0.64, size = 120, normalized size = 5.00 \[ \frac {1}{4} \, f {\left (\frac {\log \left (f x + e\right )^{2} - 2 \, \log \left (f x + e\right ) \log \left (f x - e\right )}{e f^{2}} + \frac {2 \, {\left (\log \left (f x + e\right ) \log \left (-\frac {f x + e}{2 \, e} + 1\right ) + {\rm Li}_2\left (\frac {f x + e}{2 \, e}\right )\right )}}{e f^{2}}\right )} + \frac {1}{2} \, {\left (\frac {\log \left (f x + e\right )}{e f} - \frac {\log \left (f x - e\right )}{e f}\right )} \log \left (\frac {2 \, e}{f x + e}\right ) \]

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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mupad [B]  time = 0.29, size = 19, normalized size = 0.79 \[ \frac {{\mathrm {Li}}_{\mathrm {2}}\left (\frac {2\,e}{e+f\,x}\right )}{2\,e\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

dilog((2*e)/(e + f*x))/(2*e*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\log {\relax (2 )}}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {\log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \]

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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