Integrand size = 13, antiderivative size = 49 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=-b x (b x-\text {arctanh}(\tanh (a+b x)))+\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2+(b x-\text {arctanh}(\tanh (a+b x)))^2 \log (x) \]
-b*x*(b*x-arctanh(tanh(b*x+a)))+1/2*arctanh(tanh(b*x+a))^2+(b*x-arctanh(ta nh(b*x+a)))^2*ln(x)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\frac {1}{2} (a+b x)^2-(a+b x) (a+2 b x-2 \text {arctanh}(\tanh (a+b x)))+(-b x+\text {arctanh}(\tanh (a+b x)))^2 \log (b x) \]
(a + b*x)^2/2 - (a + b*x)*(a + 2*b*x - 2*ArcTanh[Tanh[a + b*x]]) + (-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*x]
Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2590, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))\) |
ArcTanh[Tanh[a + b*x]]^2/2 - (b*x - ArcTanh[Tanh[a + b*x]])*(b*x - (b*x - ArcTanh[Tanh[a + b*x]])*Log[x])
3.1.49.3.1 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[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^n/(a*n), x] - Simp[(b*u - a*v)/a Int[v^(n - 1)/u, x], x ] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && NeQ[n, 1]
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.37
method | result | size |
default | \(\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\) | \(67\) |
parts | \(\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\) | \(67\) |
risch | \(\ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right )^{2}+b^{2} \ln \left (x \right ) x^{2}-\frac {3 b^{2} x^{2}}{2}-2 b \ln \left ({\mathrm e}^{b x +a}\right ) \ln \left (x \right ) x +2 b \ln \left ({\mathrm e}^{b x +a}\right ) 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} \ln \left (x \right )}{16}-\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 (\ln \left (x \right ) \ln \left ({\mathrm e}^{b x +a}\right )-b \left (x \ln \left (x \right )-x \right )\right )}{2}\) | \(578\) |
ln(x)*arctanh(tanh(b*x+a))^2-2*b*(b*(1/2*x^2*ln(x)-1/4*x^2)+a*(x*ln(x)-x)+ (arctanh(tanh(b*x+a))-b*x-a)*(x*ln(x)-x))
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.41 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\frac {1}{2} \, b^{2} x^{2} + 2 \, a b x + a^{2} \log \left (x\right ) \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\int \frac {\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \]
Time = 0.50 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.41 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\frac {1}{2} \, b^{2} x^{2} + 2 \, a b x + a^{2} \log \left (x\right ) \]
Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.43 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\frac {1}{2} \, b^{2} x^{2} + 2 \, a b x + a^{2} \log \left ({\left | x \right |}\right ) \]
Time = 0.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.73 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x} \, dx=\ln \left (x\right )\,\left (\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4}-a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+a^2\right )+\frac {b^2\,x^2}{2}-b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right ) \]
log(x)*((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + lo g(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2/4 - a*(2*a - log((2*exp(2*a)*exp (2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2 *b*x) + a^2) + (b^2*x^2)/2 - b*x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log(( 2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)