Integrand size = 19, antiderivative size = 19 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\text {Int}\left (\frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x},x\right ) \]
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Not integrable
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx \\ \end{align*}
Not integrable
Time = 3.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx \]
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Not integrable
Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
\[\int \frac {\operatorname {arccot}\left (c +\left (i+c \right ) \tanh \left (b x +a \right )\right )}{x}d x\]
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Not integrable
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int { \frac {\operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.64 (sec) , antiderivative size = 81, normalized size of antiderivative = 4.26 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int { \frac {\operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right )}{x} \,d x } \]
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Not integrable
Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int { \frac {\operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right )}{x} \,d x } \]
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Not integrable
Time = 1.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {\cot ^{-1}(c+(i+c) \tanh (a+b x))}{x} \, dx=\int \frac {\mathrm {acot}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c+1{}\mathrm {i}\right )\right )}{x} \,d x \]
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