Integrand size = 13, antiderivative size = 44 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))} \]
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=\frac {-\log (x)+\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))} \]
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2591, 14, 2588, 14}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2591 |
\(\displaystyle \frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\) |
-(Log[x]/(b*x - ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b* x - ArcTanh[Tanh[a + b*x]])
3.1.90.3.1 Defintions of rubi rules used
Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c Subst [Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, x]
Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D [v, x]]}, Simp[b/(b*u - a*v) Int[1/v, x], x] - Simp[a/(b*u - a*v) Int[1 /u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
Time = 4.53 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {\ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}+\frac {\ln \left (x \right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\) | \(43\) |
risch | \(\frac {4 i \ln \left (x \right )}{\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-4 i b x +4 i \ln \left ({\mathrm e}^{b x +a}\right )}-\frac {4 i \ln \left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+4 i \left (\ln \left ({\mathrm e}^{b x +a}\right )-b x -a \right )+4 i a +4 i b x \right )}{\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}-4 i b x +4 i \ln \left ({\mathrm e}^{b x +a}\right )}\) | \(831\) |
Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.36 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {\log \left (b x + a\right ) - \log \left (x\right )}{a} \]
\[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{x \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {\log \left (b x + a\right )}{a} + \frac {\log \left (x\right )}{a} \]
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {\log \left ({\left | b x + a \right |}\right )}{a} + \frac {\log \left ({\left | x \right |}\right )}{a} \]
Time = 6.01 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.43 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {4\,\mathrm {atanh}\left (\frac {4\,b\,x}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}-1\right )}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x} \]