Integrand size = 13, antiderivative size = 92 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {b}{x (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}-\frac {b^2 \log (x)}{(b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {b^2 \log (\text {arctanh}(\tanh (a+b x)))}{(b x-\text {arctanh}(\tanh (a+b x)))^3} \]
b/x/(b*x-arctanh(tanh(b*x+a)))^2+1/2/x^2/(b*x-arctanh(tanh(b*x+a)))-b^2*ln (x)/(b*x-arctanh(tanh(b*x+a)))^3+b^2*ln(arctanh(tanh(b*x+a)))/(b*x-arctanh (tanh(b*x+a)))^3
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {-4 b x \text {arctanh}(\tanh (a+b x))+\text {arctanh}(\tanh (a+b x))^2+b^2 x^2 (3-2 \log (x)+2 \log (\text {arctanh}(\tanh (a+b x))))}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))^3} \]
(-4*b*x*ArcTanh[Tanh[a + b*x]] + ArcTanh[Tanh[a + b*x]]^2 + b^2*x^2*(3 - 2 *Log[x] + 2*Log[ArcTanh[Tanh[a + b*x]]]))/(2*x^2*(b*x - ArcTanh[Tanh[a + b *x]])^3)
Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2594, 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.
\(\displaystyle \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \int \frac {1}{x^2 \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2594 |
\(\displaystyle \frac {b \left (\frac {b \int \frac {1}{x \text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2591 |
\(\displaystyle \frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\int \frac {1}{x}dx}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {b \left (\frac {b \left (\frac {b \int \frac {1}{\text {arctanh}(\tanh (a+b x))}dx}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 2588 |
\(\displaystyle \frac {b \left (\frac {b \left (\frac {\int \frac {1}{\text {arctanh}(\tanh (a+b x))}d\text {arctanh}(\tanh (a+b x))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}+\frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{2 x^2 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {1}{x (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {b \left (\frac {\log (\text {arctanh}(\tanh (a+b x)))}{b x-\text {arctanh}(\tanh (a+b x))}-\frac {\log (x)}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\right )}{b x-\text {arctanh}(\tanh (a+b x))}\) |
1/(2*x^2*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(1/(x*(b*x - ArcTanh[Tanh[a + b*x]])) + (b*(-(Log[x]/(b*x - ArcTanh[Tanh[a + b*x]])) + Log[ArcTanh[Tan h[a + b*x]]]/(b*x - ArcTanh[Tanh[a + b*x]])))/(b*x - ArcTanh[Tanh[a + b*x] ])))/(b*x - ArcTanh[Tanh[a + b*x]])
3.1.92.3.1 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 = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95
\[-\frac {1}{2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{2}}+\frac {b^{2} \ln \left (x \right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}+\frac {b}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x}-\frac {b^{2} \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{3}}\]
-1/2/(arctanh(tanh(b*x+a))-b*x)/x^2+1/(arctanh(tanh(b*x+a))-b*x)^3*b^2*ln( x)+1/(arctanh(tanh(b*x+a))-b*x)^2*b/x-1/(arctanh(tanh(b*x+a))-b*x)^3*b^2*l n(arctanh(tanh(b*x+a)))
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \, b^{2} x^{2} \log \left (b x + a\right ) - 2 \, b^{2} x^{2} \log \left (x\right ) - 2 \, a b x + a^{2}}{2 \, a^{3} x^{2}} \]
\[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=\int \frac {1}{x^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Time = 0.35 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {b^{2} \log \left (b x + a\right )}{a^{3}} + \frac {b^{2} \log \left (x\right )}{a^{3}} + \frac {2 \, b x - a}{2 \, a^{2} x^{2}} \]
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=-\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{a^{3}} + \frac {b^{2} \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {2 \, a b x - a^{2}}{2 \, a^{3} x^{2}} \]
Time = 5.81 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.11 \[ \int \frac {1}{x^3 \text {arctanh}(\tanh (a+b x))} \, dx=\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\left (2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+8\,b\,x\right )+{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2+12\,b^2\,x^2+8\,b\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+b^2\,x^2\,\mathrm {atan}\left (\frac {-\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\right )\,16{}\mathrm {i}}{x^2\,{\left (\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3} \]
(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))*(2*log(1/(exp(2*a)*exp(2*b*x) + 1)) + 8*b*x) + log((exp(2 *a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 + 12*b^2*x^2 + 8*b*x*log(1/(e xp(2*a)*exp(2*b*x) + 1)) + b^2*x^2*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)/(lo g(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))*16i)/(x^2*(log(1/(exp(2*a)*exp(2*b*x) + 1)) - log((e xp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3)