Integrand size = 13, antiderivative size = 77 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=b x (b x-\text {arctanh}(\tanh (a+b x)))^2-\frac {1}{2} (b x-\text {arctanh}(\tanh (a+b x))) \text {arctanh}(\tanh (a+b x))^2+\frac {1}{3} \text {arctanh}(\tanh (a+b x))^3-(b x-\text {arctanh}(\tanh (a+b x)))^3 \log (x) \]
b*x*(b*x-arctanh(tanh(b*x+a)))^2-1/2*(b*x-arctanh(tanh(b*x+a)))*arctanh(ta nh(b*x+a))^2+1/3*arctanh(tanh(b*x+a))^3-(b*x-arctanh(tanh(b*x+a)))^3*ln(x)
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.35 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} (a+b x)^3+(a+b x) \left (a^2-3 a (a+b x-\text {arctanh}(\tanh (a+b x)))+3 (a+b x-\text {arctanh}(\tanh (a+b x)))^2\right )-\frac {1}{2} (a+b x)^2 (2 a+3 b x-3 \text {arctanh}(\tanh (a+b x)))+(-b x+\text {arctanh}(\tanh (a+b x)))^3 \log (b x) \]
(a + b*x)^3/3 + (a + b*x)*(a^2 - 3*a*(a + b*x - ArcTanh[Tanh[a + b*x]]) + 3*(a + b*x - ArcTanh[Tanh[a + b*x]])^2) - ((a + b*x)^2*(2*a + 3*b*x - 3*Ar cTanh[Tanh[a + b*x]]))/2 + (-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[b*x]
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2590, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {1}{3} \text {arctanh}(\tanh (a+b x))^3-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x}dx\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle \frac {1}{3} \text {arctanh}(\tanh (a+b x))^3-(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx\right )\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle \frac {1}{3} \text {arctanh}(\tanh (a+b x))^3-(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{3} \text {arctanh}(\tanh (a+b x))^3-(b x-\text {arctanh}(\tanh (a+b x))) \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))\right )\) |
ArcTanh[Tanh[a + b*x]]^3/3 - (b*x - 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]))
3.1.59.3.1 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 = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.91
method | result | size |
default | \(\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}-3 b \left (b^{2} \left (\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}\right )+2 a b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+2 b \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a^{2} \left (x \ln \left (x \right )-x \right )+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \left (x \ln \left (x \right )-x \right )\right )\) | \(147\) |
parts | \(\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}-3 b \left (b^{2} \left (\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}\right )+2 a b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+2 b \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a^{2} \left (x \ln \left (x \right )-x \right )+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \left (x \ln \left (x \right )-x \right )\right )\) | \(147\) |
risch | \(\text {Expression too large to display}\) | \(2075\) |
ln(x)*arctanh(tanh(b*x+a))^3-3*b*(b^2*(1/3*x^3*ln(x)-1/9*x^3)+2*a*b*(1/2*x ^2*ln(x)-1/4*x^2)+2*b*(arctanh(tanh(b*x+a))-b*x-a)*(1/2*x^2*ln(x)-1/4*x^2) +a^2*(x*ln(x)-x)+2*a*(arctanh(tanh(b*x+a))-b*x-a)*(x*ln(x)-x)+(arctanh(tan h(b*x+a))-b*x-a)^2*(x*ln(x)-x))
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.40 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} \log \left (x\right ) \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x}\, dx \]
Time = 0.53 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.40 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} \log \left (x\right ) \]
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.42 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\frac {1}{3} \, b^{3} x^{3} + \frac {3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3} \log \left ({\left | x \right |}\right ) \]
Time = 3.95 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.97 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x} \, dx=\frac {b^3\,x^3}{3}-\ln \left (x\right )\,\left (\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{8}-a^3-\frac {3\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4}+\frac {3\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2}\right )-\frac {3\,b^2\,x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{4}+\frac {3\,b\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4} \]
(b^3*x^3)/3 - log(x)*((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)^3/8 - a^3 - (3*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)^2)/4 + (3*a^2*(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) )/2) - (3*b^2*x^2*(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))/4 + (3*b*x*(log(2/(exp(2*a)*e xp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/4