Optimal. Leaf size=92 \[ -\coth ^{-1}(a+b x) \log \left (\frac {2}{1+a+b x}\right )+\coth ^{-1}(a+b x) \log \left (\frac {2 b x}{(1-a) (1+a+b x)}\right )+\frac {1}{2} \text {PolyLog}\left (2,1-\frac {2}{1+a+b x}\right )-\frac {1}{2} \text {PolyLog}\left (2,1-\frac {2 b x}{(1-a) (1+a+b x)}\right ) \]
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Rubi [A]
time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6247, 6058,
2449, 2352, 2497} \begin {gather*} \frac {1}{2} \text {Li}_2\left (1-\frac {2}{a+b x+1}\right )-\frac {1}{2} \text {Li}_2\left (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) \end {gather*}
Antiderivative was successfully verified.
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Rule 2352
Rule 2449
Rule 2497
Rule 6058
Rule 6247
Rubi steps
\begin {align*} \int \frac {\coth ^{-1}(a+b x)}{x} \, dx &=\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} \text {Li}_2\left (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} \text {Li}_2\left (1-\frac {2}{1+a+b x}\right )-\frac {1}{2} \text {Li}_2\left (1-\frac {2 b x}{(1-a) (1+a+b x)}\right )\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.11, size = 259, normalized size = 2.82 \begin {gather*} \left (\coth ^{-1}(a+b x)-\tanh ^{-1}(a+b x)\right ) \log (x)+\tanh ^{-1}(a+b x) \left (-\log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+\log \left (-i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )\right )+\frac {1}{8} \left (4 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )^2-\left (\pi -2 i \tanh ^{-1}(a+b x)\right )^2-8 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right ) \log \left (1-e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )-4 i \left (\pi -2 i \tanh ^{-1}(a+b x)\right ) \log \left (1+e^{2 \tanh ^{-1}(a+b x)}\right )+4 \left (i \pi +2 \tanh ^{-1}(a+b x)\right ) \log \left (\frac {2}{\sqrt {1-(a+b x)^2}}\right )+8 \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right ) \log \left (-2 i \sinh \left (\tanh ^{-1}(a)-\tanh ^{-1}(a+b x)\right )\right )-4 \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(a)-2 \tanh ^{-1}(a+b x)}\right )-4 \text {PolyLog}\left (2,-e^{2 \tanh ^{-1}(a+b x)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.28, size = 104, normalized size = 1.13
method | result | size |
risch | \(\frac {\dilog \left (\frac {b x}{-a -1}\right )}{2}+\frac {\ln \left (b x +a +1\right ) \ln \left (\frac {b x}{-a -1}\right )}{2}-\frac {\dilog \left (\frac {b x}{1-a}\right )}{2}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{1-a}\right )}{2}\) | \(68\) |
derivativedivides | \(\ln \left (-b x \right ) \mathrm {arccoth}\left (b x +a \right )+\frac {\dilog \left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}-\frac {\dilog \left (\frac {-b x -a -1}{-a -1}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-a -1}\right )}{2}\) | \(104\) |
default | \(\ln \left (-b x \right ) \mathrm {arccoth}\left (b x +a \right )+\frac {\dilog \left (\frac {-b x -a +1}{1-a}\right )}{2}+\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a +1}{1-a}\right )}{2}-\frac {\dilog \left (\frac {-b x -a -1}{-a -1}\right )}{2}-\frac {\ln \left (-b x \right ) \ln \left (\frac {-b x -a -1}{-a -1}\right )}{2}\) | \(104\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 128, normalized size = 1.39 \begin {gather*} -\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 ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acoth}{\left (a + b x \right )}}{x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acoth}\left (a+b\,x\right )}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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