Integrand size = 10, antiderivative size = 92 \[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=-\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )+\coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6247, 6058, 2449, 2352, 2497} \[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{a+b x+1}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (a+b x+1)}\right )+\log \left (\frac {2}{a+b x+1}\right ) \left (-\coth ^{-1}(a+b x)\right )+\log \left (\frac {2 b x}{(1-a) (a+b x+1)}\right ) \coth ^{-1}(a+b x) \]
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Rule 2352
Rule 2449
Rule 2497
Rule 6058
Rule 6247
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\coth ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b} \\ & = -\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )+\coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )+\text {Subst}\left (\int \frac {\log \left (\frac {2}{1+x}\right )}{1-x^2} \, dx,x,a+b x\right )-\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (-\frac {a}{b}+\frac {x}{b}\right )}{\left (\frac {1}{b}-\frac {a}{b}\right ) (1+x)}\right )}{1-x^2} \, dx,x,a+b x\right ) \\ & = -\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )+\coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right )+\text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+a+b x}\right ) \\ & = -\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )+\coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right ) \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.82 \[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\left (\coth ^{-1}(a+b x)-\text {arctanh}(a+b x)\right ) \log (x)+\text {arctanh}(a+b x) \left (-\log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+\log (-i \sinh (\text {arctanh}(a)-\text {arctanh}(a+b x)))\right )+\frac {1}{8} \left (4 (\text {arctanh}(a)-\text {arctanh}(a+b x))^2-(\pi -2 i \text {arctanh}(a+b x))^2-8 (\text {arctanh}(a)-\text {arctanh}(a+b x)) \log \left (1-e^{2 \text {arctanh}(a)-2 \text {arctanh}(a+b x)}\right )-4 i (\pi -2 i \text {arctanh}(a+b x)) \log \left (1+e^{2 \text {arctanh}(a+b x)}\right )+4 (i \pi +2 \text {arctanh}(a+b x)) \log \left (\frac {2}{\sqrt {1-(a+b x)^2}}\right )+8 (\text {arctanh}(a)-\text {arctanh}(a+b x)) \log (-2 i \sinh (\text {arctanh}(a)-\text {arctanh}(a+b x)))-4 \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}(a)-2 \text {arctanh}(a+b x)}\right )-4 \operatorname {PolyLog}\left (2,-e^{2 \text {arctanh}(a+b x)}\right )\right ) \]
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Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {\operatorname {dilog}\left (\frac {x b}{-1-a}\right )}{2}+\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {x b}{-1-a}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {b x}{1-a}\right )}{2}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{1-a}\right )}{2}\) | \(68\) |
parts | \(\ln \left (x \right ) \operatorname {arccoth}\left (b x +a \right )+b \left (\frac {\operatorname {dilog}\left (\frac {b x +a -1}{-1+a}\right )}{2 b}+\frac {\ln \left (x \right ) \ln \left (\frac {b x +a -1}{-1+a}\right )}{2 b}-\frac {\operatorname {dilog}\left (\frac {b x +a +1}{1+a}\right )}{2 b}-\frac {\ln \left (x \right ) \ln \left (\frac {b x +a +1}{1+a}\right )}{2 b}\right )\) | \(90\) |
derivativedivides | \(\ln \left (-b x \right ) \operatorname {arccoth}\left (b x +a \right )-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-a}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-1-a}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}\) | \(104\) |
default | \(\ln \left (-b x \right ) \operatorname {arccoth}\left (b x +a \right )-\frac {\operatorname {dilog}\left (\frac {-b x -a -1}{-1-a}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-1-a}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}\) | \(104\) |
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\[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcoth}\left (b x + a\right )}{x} \,d x } \]
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\[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\int \frac {\operatorname {acoth}{\left (a + b x \right )}}{x}\, dx \]
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none
Time = 0.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.39 \[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=-\frac {1}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{b} - \frac {\log \left (b x + a - 1\right )}{b}\right )} \log \left (x\right ) + \frac {1}{2} \, b {\left (\frac {\log \left (b x + a + 1\right ) \log \left (-\frac {b x + a + 1}{a + 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + a + 1}{a + 1}\right )}{b} - \frac {\log \left (b x + a - 1\right ) \log \left (-\frac {b x + a - 1}{a - 1} + 1\right ) + {\rm Li}_2\left (\frac {b x + a - 1}{a - 1}\right )}{b}\right )} + \operatorname {arcoth}\left (b x + a\right ) \log \left (x\right ) \]
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\[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\int { \frac {\operatorname {arcoth}\left (b x + a\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\coth ^{-1}(a+b x)}{x} \, dx=\int \frac {\mathrm {acoth}\left (a+b\,x\right )}{x} \,d x \]
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