Integrand size = 25, antiderivative size = 42 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=-\frac {\text {arctanh}\left (\frac {f x}{e}\right ) \log (2)}{e f}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \] Output:
-arctanh(f*x/e)*ln(2)/e/f+1/2*polylog(2,1-2*e/(f*x+e))/e/f
Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.93 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=-\frac {\log \left (\frac {e-f x}{2 e}\right ) \log \left (\frac {e}{e+f x}\right )}{2 e f}-\frac {\log ^2\left (\frac {e}{e+f x}\right )}{4 e f}+\frac {\operatorname {PolyLog}\left (2,\frac {e+f x}{2 e}\right )}{2 e f} \] Input:
Integrate[Log[e/(e + f*x)]/(e^2 - f^2*x^2),x]
Output:
-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)
Time = 0.46 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2850, 221, 2849, 2752}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx\) |
\(\Big \downarrow \) 2850 |
\(\displaystyle \int \frac {\log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2}dx-\log (2) \int \frac {1}{e^2-f^2 x^2}dx\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \int \frac {\log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2}dx-\frac {\log (2) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle \frac {\int \frac {\log \left (\frac {2 e}{e+f x}\right )}{1-\frac {2 e}{e+f x}}d\frac {1}{e+f x}}{f}-\frac {\log (2) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f}-\frac {\log (2) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}\) |
Input:
Int[Log[e/(e + f*x)]/(e^2 - f^2*x^2),x]
Output:
-((ArcTanh[(f*x)/e]*Log[2])/(e*f)) + PolyLog[2, 1 - (2*e)/(e + f*x)]/(2*e* f)
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]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-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]
Int[((a_.) + Log[(c_.)/((d_) + (e_.)*(x_))]*(b_.))/((f_) + (g_.)*(x_)^2), x _Symbol] :> Simp[(a + b*Log[c/(2*d)]) Int[1/(f + g*x^2), x], x] + Simp[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]
Time = 1.85 (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 | \(-\frac {\ln \left (\frac {e}{f x +e}\right ) \ln \left (-f x +e \right )}{2 e f}+\frac {\ln \left (\frac {e}{f x +e}\right ) \ln \left (f x +e \right )}{2 e f}+\frac {f \left (\frac {\ln \left (f x +e \right )^{2}}{2 e \,f^{2}}+\frac {-\operatorname {dilog}\left (-\frac {-f x -e}{2 e}\right )-\ln \left (-f x +e \right ) \ln \left (-\frac {-f x -e}{2 e}\right )}{e \,f^{2}}\right )}{2}\) | \(118\) |
Input:
int(ln(e/(f*x+e))/(-f^2*x^2+e^2),x,method=_RETURNVERBOSE)
Output:
-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)))
\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {\log \left (\frac {e}{f x + e}\right )}{f^{2} x^{2} - e^{2}} \,d x } \] Input:
integrate(log(e/(f*x+e))/(-f^2*x^2+e^2),x, algorithm="fricas")
Output:
integral(-log(e/(f*x + e))/(f^2*x^2 - e^2), x)
\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=- \int \frac {\log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \] Input:
integrate(ln(e/(f*x+e))/(-f**2*x**2+e**2),x)
Output:
-Integral(log(e/(e + f*x))/(-e**2 + f**2*x**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (39) = 78\).
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.83 \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {1}{4} \, f {\left (\frac {\log \left (f x + e\right )^{2} - 2 \, \log \left (f x + e\right ) \log \left (f x - e\right )}{e f^{2}} + \frac {2 \, {\left (\log \left (f x + e\right ) \log \left (-\frac {f x + e}{2 \, e} + 1\right ) + {\rm Li}_2\left (\frac {f x + e}{2 \, e}\right )\right )}}{e f^{2}}\right )} + \frac {1}{2} \, {\left (\frac {\log \left (f x + e\right )}{e f} - \frac {\log \left (f x - e\right )}{e f}\right )} \log \left (\frac {e}{f x + e}\right ) \] Input:
integrate(log(e/(f*x+e))/(-f^2*x^2+e^2),x, algorithm="maxima")
Output:
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) + dilog(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))
\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {\log \left (\frac {e}{f x + e}\right )}{f^{2} x^{2} - e^{2}} \,d x } \] Input:
integrate(log(e/(f*x+e))/(-f^2*x^2+e^2),x, algorithm="giac")
Output:
integrate(-log(e/(f*x + e))/(f^2*x^2 - e^2), x)
Timed out. \[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int \frac {\ln \left (\frac {e}{e+f\,x}\right )}{e^2-f^2\,x^2} \,d x \] Input:
int(log(e/(e + f*x))/(e^2 - f^2*x^2),x)
Output:
int(log(e/(e + f*x))/(e^2 - f^2*x^2), x)
\[ \int \frac {\log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int \frac {\mathrm {log}\left (\frac {e}{f x +e}\right )}{-f^{2} x^{2}+e^{2}}d x \] Input:
int(log(e/(f*x+e))/(-f^2*x^2+e^2),x)
Output:
int(log(e/(e + f*x))/(e**2 - f**2*x**2),x)