Integrand size = 13, antiderivative size = 39 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=2 b^2 x-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}-2 b (b x-\text {arctanh}(\tanh (a+b x))) \log (x) \]
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}-2 b^2 x \log (x)+2 b \text {arctanh}(\tanh (a+b x)) (1+\log (x)) \]
Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2599, 2589, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 2 b \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle 2 b \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle 2 b (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\) |
3.1.50.3.1 Defintions of rubi rules used
Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Simp[(b*u - a*v)/a Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]
Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 ))) Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} , x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 ] && !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ [n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]) || (ILt Q[m, 0] && !IntegerQ[n]))
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\) | \(41\) |
parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\) | \(41\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{b x +a}\right )^{2}}{x}-2 \ln \left (x \right ) x \,b^{2}+2 \ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right ) b +2 b^{2} x +\frac {\pi ^{2} {\left (\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 )-\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}+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\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}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right )}^{2}}{16 x}-\frac {i \pi \left (\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 )-\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}+\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \,\operatorname {csgn}\left (i {\mathrm e}^{b x +a}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\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}+\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}\right ) \left (-\frac {\ln \left ({\mathrm e}^{b x +a}\right )}{x}+b \ln \left (x \right )\right )}{2}\) | \(561\) |
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.62 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=\frac {b^{2} x^{2} + 2 \, a b x \log \left (x\right ) - a^{2}}{x} \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=\int \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx \]
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - 2 \, {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x} \]
2*b*arctanh(tanh(b*x + a))*log(x) - 2*(b*(x + a/b)*log(x) - b*(x + a*log(x )/b))*b - arctanh(tanh(b*x + a))^2/x
Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=b^{2} x + 2 \, a b \log \left ({\left | x \right |}\right ) - \frac {a^{2}}{x} \]
Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 5.08 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2} \, dx=b\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4\,x}-b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4\,x}+b\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )-2\,b^2\,x\,\ln \left (x\right )-b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )+\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2\,x} \]
b*log(exp(2*b*x)/(exp(2*a)*exp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(e xp(2*a)*exp(2*b*x) + 1))^2/(4*x) - b*log(1/(exp(2*a)*exp(2*b*x) + 1)) - lo g(1/(exp(2*a)*exp(2*b*x) + 1))^2/(4*x) + b*log((exp(2*a)*exp(2*b*x))/(exp( 2*a)*exp(2*b*x) + 1))*log(x) - 2*b^2*x*log(x) - b*log(1/(exp(2*a)*exp(2*b* x) + 1))*log(x) + (log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b* x))/(exp(2*a)*exp(2*b*x) + 1)))/(2*x)