3.3.79 \(\int \frac {\log (\frac {2 e}{e+f x})}{e^2-f^2 x^2} \, dx\) [279]

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 = 0.02, 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 = {2449, 2352} \begin {gather*} \frac {\text {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \end {gather*}

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 {\text {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 \begin {gather*} \frac {\text {Li}_2\left (\frac {-e+f x}{e+f x}\right )}{2 e f} \end {gather*}

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 = 0.72, size = 20, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\dilog \left (\frac {2 e}{f x +e}\right )}{2 f e}\)	\(20\)
default	\(\frac {\dilog \left (\frac {2 e}{f x +e}\right )}{2 f e}\)	\(20\)
risch	\(\frac {\dilog \left (\frac {2 e}{f x +e}\right )}{2 f e}\)	\(20\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (22) = 44\).
time = 0.27, size = 124, normalized size = 5.17 \begin {gather*} \frac {1}{4} \, f {\left (\frac {{\left (\log \left (f x + e\right )^{2} - 2 \, \log \left (f x + e\right ) \log \left (f x - e\right )\right )} e^{\left (-1\right )}}{f^{2}} + \frac {2 \, {\left (\log \left (f x + e\right ) \log \left (-\frac {1}{2} \, {\left (f x + e\right )} e^{\left (-1\right )} + 1\right ) + {\rm Li}_2\left (\frac {1}{2} \, {\left (f x + e\right )} e^{\left (-1\right )}\right )\right )} e^{\left (-1\right )}}{f^{2}}\right )} + \frac {1}{2} \, {\left (\frac {e^{\left (-1\right )} \log \left (f x + e\right )}{f} - \frac {e^{\left (-1\right )} \log \left (f x - e\right )}{f}\right )} \log \left (\frac {2 \, e}{f x + e}\right ) \end {gather*}

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

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Fricas [A]
time = 0.35, size = 22, normalized size = 0.92 \begin {gather*} \frac {{\rm Li}_2\left (-\frac {2 \, e}{f x + e} + 1\right ) e^{\left (-1\right )}}{2 \, f} \end {gather*}

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^(-1)/f

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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