Integrand size = 30, antiderivative size = 41 \[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {a \text {arctanh}\left (\frac {f x}{e}\right )}{e f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \]
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Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2450, 214, 2449, 2352} \[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {a \text {arctanh}\left (\frac {f x}{e}\right )}{e f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 e}{e+f x}\right )}{2 e f} \]
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Rule 214
Rule 2352
Rule 2449
Rule 2450
Rubi steps \begin{align*} \text {integral}& = a \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 \\ & = \frac {a \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f}+\frac {b \text {Subst}\left (\int \frac {\log (2 e x)}{1-2 e x} \, dx,x,\frac {1}{e+f x}\right )}{f} \\ & = \frac {a \tanh ^{-1}\left (\frac {f x}{e}\right )}{e f}+\frac {b \text {Li}_2\left (1-\frac {2 e}{e+f x}\right )}{2 e f} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.00 \[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {-\left (\left (a+b \log \left (\frac {2 e}{e+f x}\right )\right ) \left (a+2 b \log \left (\frac {e-f x}{2 e}\right )+b \log \left (\frac {2 e}{e+f x}\right )\right )\right )+2 b^2 \operatorname {PolyLog}\left (2,\frac {e+f x}{2 e}\right )}{4 b e f} \]
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Time = 0.66 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07
method | result | size |
derivativedivides | \(-\frac {2 e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{4 e^{2}}-\frac {b \operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{4 e^{2}}\right )}{f}\) | \(44\) |
default | \(-\frac {2 e \left (\frac {a \ln \left (\frac {2 e}{f x +e}-1\right )}{4 e^{2}}-\frac {b \operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{4 e^{2}}\right )}{f}\) | \(44\) |
risch | \(-\frac {a \ln \left (f x -e \right )}{2 e f}+\frac {a \ln \left (f x +e \right )}{2 e f}+\frac {b \operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2 f e}\) | \(54\) |
parts | \(-\frac {a \ln \left (f x -e \right )}{2 e f}+\frac {a \ln \left (f x +e \right )}{2 e f}+\frac {b \operatorname {dilog}\left (\frac {2 e}{f x +e}\right )}{2 f e}\) | \(54\) |
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Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\frac {b {\rm Li}_2\left (-\frac {2 \, e}{f x + e} + 1\right ) + a \log \left (f x + e\right ) - a \log \left (f x - e\right )}{2 \, e f} \]
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\[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=- \int \frac {a}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {b \log {\left (2 \right )}}{- e^{2} + f^{2} x^{2}}\, dx - \int \frac {b \log {\left (\frac {e}{e + f x} \right )}}{- e^{2} + f^{2} x^{2}}\, dx \]
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\[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {b \log \left (\frac {2 \, e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}} \,d x } \]
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\[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=\int { -\frac {b \log \left (\frac {2 \, e}{f x + e}\right ) + a}{f^{2} x^{2} - e^{2}} \,d x } \]
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Time = 1.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (\frac {2 e}{e+f x}\right )}{e^2-f^2 x^2} \, dx=-\frac {a\,\ln \left (f\,x-e\right )-b\,{\mathrm {Li}}_{\mathrm {2}}\left (\frac {2\,e}{e+f\,x}\right )+a\,\ln \left (\frac {1}{e+f\,x}\right )}{2\,e\,f} \]
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