Integrand size = 13, antiderivative size = 65 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {b \log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {b \log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^2} \] Output:
1/x/(b*x-arctanh(tanh(b*x+a)))-b*ln(x)/(b*x-arctanh(tanh(b*x+a)))^2+b*ln(a rctanh(tanh(b*x+a)))/(b*x-arctanh(tanh(b*x+a)))^2
Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {-\text {arctanh}(\tanh (a+b x))+b x (1-\log (x)+\log (\text {arctanh}(\tanh (a+b x))))}{x (-b x+\text {arctanh}(\tanh (a+b x)))^2} \] Input:
Integrate[1/(x^2*ArcTanh[Tanh[a + b*x]]),x]
Output:
(-ArcTanh[Tanh[a + b*x]] + b*x*(1 - Log[x] + Log[ArcTanh[Tanh[a + b*x]]])) /(x*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25, 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.
\(\displaystyle \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2591 |
\(\displaystyle \frac {b \left (\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {b \left (\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {b \left (\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\) |
Input:
Int[1/(x^2*ArcTanh[Tanh[a + b*x]]),x]
Output:
1/(x*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(-(Log[x]/(b*x - ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tanh[a + b*x]]]/(b*x - ArcTanh[Tanh[a + b*x]])))/( b*x - ArcTanh[Tanh[a + b*x]])
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 = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98
\[-\frac {1}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x}-\frac {b \ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}+\frac {b \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2}}\]
Input:
int(1/x^2/arctanh(tanh(b*x+a)),x)
Output:
-1/(arctanh(tanh(b*x+a))-b*x)/x-1/(arctanh(tanh(b*x+a))-b*x)^2*b*ln(x)+1/( arctanh(tanh(b*x+a))-b*x)^2*b*ln(arctanh(tanh(b*x+a)))
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.40 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{2} x} \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="fricas")
Output:
(b*x*log(b*x + a) - b*x*log(x) - a)/(a^2*x)
\[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{x^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \] Input:
integrate(1/x**2/atanh(tanh(b*x+a)),x)
Output:
Integral(1/(x**2*atanh(tanh(a + b*x))), x)
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {b \log \left (b x + a\right )}{a^{2}} - \frac {b \log \left (x\right )}{a^{2}} - \frac {1}{a x} \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="maxima")
Output:
b*log(b*x + a)/a^2 - b*log(x)/a^2 - 1/(a*x)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x} \] Input:
integrate(1/x^2/arctanh(tanh(b*x+a)),x, algorithm="giac")
Output:
b*log(abs(b*x + a))/a^2 - b*log(abs(x))/a^2 - 1/(a*x)
Time = 5.93 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.23 \[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+4\,b\,x+b\,x\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}}{x\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2} \] Input:
int(1/(x^2*atanh(tanh(a + b*x))),x)
Output:
(2*log(1/(exp(2*a)*exp(2*b*x) + 1)) - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a )*exp(2*b*x) + 1)) + 4*b*x + b*x*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)/(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*b*x)^2)
\[ \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{2}}d x \] Input:
int(1/x^2/atanh(tanh(b*x+a)),x)
Output:
int(1/(atanh(tanh(a + b*x))*x**2),x)