Integrand size = 13, antiderivative size = 70 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}+\frac {\log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^2}-\frac {\log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^2} \]
-1/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))+ln(x)/(b*x-arctanh(tanh (b*x+a)))^2-ln(arctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^2
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {-b x+\text {arctanh}(\tanh (a+b x)) (1+\log (b x)-\log (\text {arctanh}(\tanh (a+b x))))}{\text {arctanh}(\tanh (a+b x)) (-b x+\text {arctanh}(\tanh (a+b x)))^2} \]
(-(b*x) + ArcTanh[Tanh[a + b*x]]*(1 + Log[b*x] - Log[ArcTanh[Tanh[a + b*x] ]]))/(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.26, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2594, 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.
\(\displaystyle \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle -\frac {\int \frac {1}{x \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2591 |
\(\displaystyle -\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle -\frac {\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle -\frac {\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle -\frac {1}{(b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))}-\frac {\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}}{b x-\text {arctanh}(\tanh (a+b x))}\) |
-(1/((b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]])) - (-(Log[x]/( b*x - ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b*x - ArcTan h[Tanh[a + b*x]]))/(b*x - ArcTanh[Tanh[a + b*x]])
3.1.99.3.1 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]
Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[ D[v, x]]}, Simp[v^(n + 1)/((n + 1)*(b*u - a*v)), x] - Simp[a*((n + 1)/((n + 1)*(b*u - a*v))) Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Piecew iseLinearQ[u, v, x] && LtQ[n, -1]
Time = 270.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {\ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}+\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}+\frac {\ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}\) | \(67\) |
risch | \(\text {Expression too large to display}\) | \(1382\) |
-1/(arctanh(tanh(b*x+a))-b*x)^2*ln(arctanh(tanh(b*x+a)))+1/(arctanh(tanh(b *x+a))-b*x)/arctanh(tanh(b*x+a))+1/(arctanh(tanh(b*x+a))-b*x)^2*ln(x)
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {{\left (b x + a\right )} \log \left (b x + a\right ) - {\left (b x + a\right )} \log \left (x\right ) - a}{a^{2} b x + a^{3}} \]
\[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.51 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {1}{a b x + a^{2}} - \frac {\log \left (b x + a\right )}{a^{2}} + \frac {\log \left (x\right )}{a^{2}} \]
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=-\frac {\log \left ({\left | b x + a \right |}\right )}{a^{2}} + \frac {\log \left ({\left | x \right |}\right )}{a^{2}} + \frac {1}{{\left (b x + a\right )} a} \]
Time = 6.03 (sec) , antiderivative size = 359, normalized size of antiderivative = 5.13 \[ \int \frac {1}{x \text {arctanh}(\tanh (a+b x))^2} \, dx=\frac {8\,b\,x-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\left (-4+\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}\right )+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\left (-4+\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}\right )}{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2} \]
(8*b*x - log(1/(exp(2*a)*exp(2*b*x) + 1))*(atan((log((exp(2*a)*exp(2*b*x)) /(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x *2i)/(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*b*x))*8i - 4) + log((exp(2*a)*exp(2*b*x))/(exp(2*a )*exp(2*b*x) + 1))*(atan((log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*1i - log(1/(exp(2*a)*exp(2*b*x) + 1))*1i + b*x*2i)/(log(1/(exp(2*a)*e xp(2*b*x) + 1)) - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2 *b*x))*8i - 4))/((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)))*(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*b*x)^2)