\(\int \frac {\log (\frac {e}{e+f x})}{e^2-f^2 x^2} \, dx\) [280]

Optimal result

Integrand size = 25, antiderivative size = 42 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=-\frac {\text {arctanh}\left (\frac {f x}{e}\right ) \log (2)}{e f}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2450, 214, 2449, 2352} \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {\operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f}-\frac {\log (2) \text {arctanh}\left (\frac {f x}{e}\right )}{e f} \]

[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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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]

Rule 2450

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]

Rubi steps \begin{align*} \text {integral}& = -\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 {\text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.93 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=-\frac {\log \left (\frac {e-f x}{2 e}\right ) \log \left (\frac {e}{e+f x}\right )}{2 e f}-\frac {\log ^2\left (\frac {e}{e+f x}\right )}{4 e f}+\frac {\operatorname {PolyLog}\left (2,\frac {e+f x}{2 e}\right )}{2 e f} \]

[In]

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

[Out]

-1/2*(Log[(e - f*x)/(2*e)]*Log[e/(e + f*x)])/(e*f) - Log[e/(e + f*x)]^2/(4*e*f) + PolyLog[2, (e + f*x)/(2*e)]/
(2*e*f)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.48

method	result	size
derivativedivides	\(-\frac {\frac {\left (\ln \left (\frac {e}{f x +e}\right )-\ln \left (\frac {2 e}{f x +e}\right )\right ) \ln \left (1-\frac {2 e}{f x +e}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2}}{f e}\)	\(62\)
default	\(-\frac {\frac {\left (\ln \left (\frac {e}{f x +e}\right )-\ln \left (\frac {2 e}{f x +e}\right )\right ) \ln \left (1-\frac {2 e}{f x +e}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2}}{f e}\)	\(62\)
risch	\(-\frac {\ln \left (1-\frac {2 e}{f x +e}\right ) \ln \left (\frac {e}{f x +e}\right )}{2 e f}+\frac {\ln \left (1-\frac {2 e}{f x +e}\right ) \ln \left (\frac {2 e}{f x +e}\right )}{2 e f}+\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2 f e}\)	\(84\)
parts	\(\frac {\ln \left (\frac {e}{f x +e}\right ) \ln \left (f x +e \right )}{2 e f}-\frac {\ln \left (\frac {e}{f x +e}\right ) \ln \left (-f x +e \right )}{2 e f}+\frac {f \left (\frac {\ln \left (f x +e \right )^{2}}{2 e \,f^{2}}+\frac {-\operatorname {dilog}\left (-\frac {-f x -e}{2 e}\right )-\ln \left (-f x +e \right ) \ln \left (-\frac {-f x -e}{2 e}\right )}{e \,f^{2}}\right )}{2}\)	\(118\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {\log \left (\frac {e}{f x + e}\right )}{f^{2} x^{2} - e^{2}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=- \int \frac {\log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (39) = 78\).

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.83 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {1}{4} \, f {\left (\frac {\log \left (f x + e\right )^{2} - 2 \, \log \left (f x + e\right ) \log \left (f x - e\right )}{e f^{2}} + \frac {2 \, {\left (\log \left (f x + e\right ) \log \left (-\frac {f x + e}{2 \, e} + 1\right ) + {\rm Li}_2\left (\frac {f x + e}{2 \, e}\right )\right )}}{e f^{2}}\right )} + \frac {1}{2} \, {\left (\frac {\log \left (f x + e\right )}{e f} - \frac {\log \left (f x - e\right )}{e f}\right )} \log \left (\frac {e}{f x + e}\right ) \]

[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))

Giac [F]

\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {\log \left (\frac {e}{f x + e}\right )}{f^{2} x^{2} - e^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int \frac {\ln \left (\frac {e}{e+f\,x}\right )}{e^2-f^2\,x^2} \,d x \]

[In]

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

[Out]

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