Integrand size = 27, antiderivative size = 24 \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {\operatorname {PolyLog}\left (2,1-\frac {2 a}{a+b x}\right )}{2 a b} \]
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Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2458, 2378, 2370, 2352} \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {\operatorname {PolyLog}\left (2,1-\frac {2 a}{a+b x}\right )}{2 a b} \]
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Rule 2352
Rule 2370
Rule 2378
Rule 2458
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\log \left (\frac {2 a}{x}\right )}{(2 a-x) x} \, dx,x,a+b x\right )}{b} \\ & = -\frac {\text {Subst}\left (\int \frac {\log (2 a x)}{\left (2 a-\frac {1}{x}\right ) x} \, dx,x,\frac {1}{a+b x}\right )}{b} \\ & = -\frac {\text {Subst}\left (\int \frac {\log (2 a x)}{-1+2 a x} \, dx,x,\frac {1}{a+b x}\right )}{b} \\ & = \frac {\text {Li}_2\left (1-\frac {2 a}{a+b x}\right )}{2 a b} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {-a+b x}{a+b x}\right )}{2 a b} \]
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Time = 0.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (\frac {2 a}{b x +a}\right )}{2 b a}\) | \(20\) |
default | \(\frac {\operatorname {dilog}\left (\frac {2 a}{b x +a}\right )}{2 b a}\) | \(20\) |
risch | \(\frac {\operatorname {dilog}\left (\frac {2 a}{b x +a}\right )}{2 b a}\) | \(20\) |
parts | \(\frac {\ln \left (\frac {2 a}{b x +a}\right ) \ln \left (b x +a \right )}{2 a b}-\frac {\ln \left (\frac {2 a}{b x +a}\right ) \ln \left (-b x +a \right )}{2 a b}+\frac {b \left (\frac {\ln \left (b x +a \right )^{2}}{2 a \,b^{2}}+\frac {-\operatorname {dilog}\left (-\frac {-b x -a}{2 a}\right )-\ln \left (-b x +a \right ) \ln \left (-\frac {-b x -a}{2 a}\right )}{a \,b^{2}}\right )}{2}\) | \(120\) |
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none
Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {{\rm Li}_2\left (-\frac {2 \, a}{b x + a} + 1\right )}{2 \, a b} \]
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\[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=- \int \frac {\log {\left (2 \right )}}{- a^{2} + b^{2} x^{2}}\, dx - \int \frac {\log {\left (\frac {a}{a + b x} \right )}}{- a^{2} + b^{2} x^{2}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (21) = 42\).
Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 5.00 \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {1}{4} \, b {\left (\frac {\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (b x - a\right )}{a b^{2}} + \frac {2 \, {\left (\log \left (b x + a\right ) \log \left (-\frac {b x + a}{2 \, a} + 1\right ) + {\rm Li}_2\left (\frac {b x + a}{2 \, a}\right )\right )}}{a b^{2}}\right )} + \frac {1}{2} \, {\left (\frac {\log \left (b x + a\right )}{a b} - \frac {\log \left (b x - a\right )}{a b}\right )} \log \left (\frac {2 \, a}{b x + a}\right ) \]
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\[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\int { -\frac {\log \left (\frac {2 \, a}{b x + a}\right )}{{\left (b x + a\right )} {\left (b x - a\right )}} \,d x } \]
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Time = 1.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\log \left (\frac {2 a}{a+b x}\right )}{(a-b x) (a+b x)} \, dx=\frac {{\mathrm {Li}}_{\mathrm {2}}\left (\frac {2\,a}{a+b\,x}\right )}{2\,a\,b} \]
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