Integrand size = 29, antiderivative size = 47 \[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {\text {arctanh}\left (\frac {f x}{e}\right ) (a-b \log (2))}{e f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \]
Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.70 \[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {-\left (\left (a+b \log \left (\frac {e}{e+f x}\right )\right ) \left (a+2 b \log \left (\frac {e-f x}{2 e}\right )+b \log \left (\frac {e}{e+f x}\right )\right )\right )+2 b^2 \operatorname {PolyLog}\left (2,\frac {e+f x}{2 e}\right )}{4 b e f} \]
(-((a + b*Log[e/(e + f*x)])*(a + 2*b*Log[(e - f*x)/(2*e)] + b*Log[e/(e + f *x)])) + 2*b^2*PolyLog[2, (e + f*x)/(2*e)])/(4*b*e*f)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.
\(\displaystyle \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx\) |
\(\Big \downarrow \) 2850 |
\(\displaystyle (a-b \log (2)) \int \frac {1}{e^2-f^2 x^2}dx+b \int \frac {\log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2}dx\) |
\(\Big \downarrow \) 221 |
\(\displaystyle b \int \frac {\log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2}dx+\frac {(a-b \log (2)) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle \frac {b \int \frac {\log \left (\frac {2 e}{e+f x}\right )}{1-\frac {2 e}{e+f x}}d\frac {1}{e+f x}}{f}+\frac {(a-b \log (2)) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {(a-b \log (2)) \text {arctanh}\left (\frac {f x}{e}\right )}{e f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f}\) |
3.3.82.3.1 Defintions of rubi rules used
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 = 0.65 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.81
method | result | size |
derivativedivides | \(-\frac {e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{2 e^{2}}+\frac {b \left (\frac {\left (\ln \left (\frac {e}{f x +e}\right )-\ln \left (\frac {2 e}{f x +e}\right )\right ) \ln \left (1-\frac {2 e}{f x +e}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2}\right )}{e^{2}}\right )}{f}\) | \(85\) |
default | \(-\frac {e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{2 e^{2}}+\frac {b \left (\frac {\left (\ln \left (\frac {e}{f x +e}\right )-\ln \left (\frac {2 e}{f x +e}\right )\right ) \ln \left (1-\frac {2 e}{f x +e}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2}\right )}{e^{2}}\right )}{f}\) | \(85\) |
parts | \(-\frac {a \ln \left (f x -e \right )}{2 e f}+\frac {a \ln \left (f x +e \right )}{2 e f}-\frac {b \left (\frac {\left (\ln \left (\frac {e}{f x +e}\right )-\ln \left (\frac {2 e}{f x +e}\right )\right ) \ln \left (1-\frac {2 e}{f x +e}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2}\right )}{f e}\) | \(96\) |
risch | \(-\frac {a \ln \left (f x -e \right )}{2 e f}+\frac {a \ln \left (f x +e \right )}{2 e f}-\frac {b \ln \left (1-\frac {2 e}{f x +e}\right ) \ln \left (\frac {e}{f x +e}\right )}{2 e f}+\frac {b \ln \left (1-\frac {2 e}{f x +e}\right ) \ln \left (\frac {2 e}{f x +e}\right )}{2 e f}+\frac {b \operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2 f e}\) | \(119\) |
-1/f*e*(1/2/e^2*a*ln(2*e/(f*x+e)-1)+1/e^2*b*(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))))
\[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {b \log \left (\frac {e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}} \,d x } \]
\[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=- \int \frac {a}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {b \log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \]
\[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {b \log \left (\frac {e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}} \,d x } \]
1/2*a*(log(f*x + e)/(e*f) - log(f*x - e)/(e*f)) + b*integrate((log(f*x + e ) - log(e))/(f^2*x^2 - e^2), x)
\[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {b \log \left (\frac {e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (\frac {e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int \frac {a+b\,\ln \left (\frac {e}{e+f\,x}\right )}{e^2-f^2\,x^2} \,d x \]