Integrand size = 13, antiderivative size = 60 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=3 b^3 x-\frac {3 b \text {arctanh}(\tanh (a+b x))^2}{2 x}-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}-3 b^2 (b x-\text {arctanh}(\tanh (a+b x))) \log (x) \]
3*b^3*x-3/2*b*arctanh(tanh(b*x+a))^2/x-1/2*arctanh(tanh(b*x+a))^3/x^2-3*b^ 2*(b*x-arctanh(tanh(b*x+a)))*ln(x)
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=b^3 x-\frac {3 b (-b x+\text {arctanh}(\tanh (a+b x)))^2}{x}-\frac {(-b x+\text {arctanh}(\tanh (a+b x)))^3}{2 x^2}+3 b^2 (-b x+\text {arctanh}(\tanh (a+b x))) \log (x) \]
b^3*x - (3*b*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)/x - (-(b*x) + ArcTanh[Ta nh[a + b*x]])^3/(2*x^2) + 3*b^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])*Log[x]
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2599, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{2} b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {3}{2} b \left (2 b \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {3}{2} b \left (2 b \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {3}{2} b \left (2 b (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\) |
-1/2*ArcTanh[Tanh[a + b*x]]^3/x^2 + (3*b*(-(ArcTanh[Tanh[a + b*x]]^2/x) + 2*b*(b*x - (b*x - ArcTanh[Tanh[a + b*x]])*Log[x])))/2
3.1.61.3.1 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 = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{2 x^{2}}+\frac {3 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\right )}{2}\) | \(59\) |
parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{2 x^{2}}+\frac {3 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\right )}{2}\) | \(59\) |
risch | \(\text {Expression too large to display}\) | \(2008\) |
-1/2*arctanh(tanh(b*x+a))^3/x^2+3/2*b*(-arctanh(tanh(b*x+a))^2/x+2*b*(ln(x )*arctanh(tanh(b*x+a))-b*(x*ln(x)-x)))
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.62 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=\frac {2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} \log \left (x\right ) - 6 \, a^{2} b x - a^{3}}{2 \, x^{2}} \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=3 \, {\left (b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac {3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{2 \, x} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{2 \, x^{2}} \]
3*(b*arctanh(tanh(b*x + a))*log(x) - (b*(x + a/b)*log(x) - b*(x + a*log(x) /b))*b)*b - 3/2*b*arctanh(tanh(b*x + a))^2/x - 1/2*arctanh(tanh(b*x + a))^ 3/x^2
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=b^{3} x + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) - \frac {6 \, a^{2} b x + a^{3}}{2 \, x^{2}} \]
Time = 3.60 (sec) , antiderivative size = 365, normalized size of antiderivative = 6.08 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3} \, dx=\frac {9\,b^2\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{4}-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{16\,x^2}-\frac {9\,b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{4}-\frac {3\,b^3\,x}{2}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{16\,x^2}-\frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{8\,x}-\frac {3\,b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )}{2}+\frac {3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{16\,x^2}-\frac {3\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{16\,x^2}-\frac {3\,b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{8\,x}+\frac {3\,b^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )}{2}-3\,b^3\,x\,\ln \left (x\right )+\frac {3\,b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{4\,x} \]
(9*b^2*log(exp(2*b*x)/(exp(2*a)*exp(2*b*x) + 1)))/4 - log((exp(2*a)*exp(2* b*x))/(exp(2*a)*exp(2*b*x) + 1))^3/(16*x^2) - (9*b^2*log(1/(exp(2*a)*exp(2 *b*x) + 1)))/4 - (3*b^3*x)/2 + log(1/(exp(2*a)*exp(2*b*x) + 1))^3/(16*x^2) - (3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/(8*x) - (3*b^2*log(1/(exp(2*a) *exp(2*b*x) + 1))*log(x))/2 + (3*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)/(16*x^2) - (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) ))/(16*x^2) - (3*b*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2) /(8*x) + (3*b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(x ))/2 - 3*b^3*x*log(x) + (3*b*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)))/(4*x)