Integrand size = 13, antiderivative size = 68 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=-3 b^2 x (b x-\text {arctanh}(\tanh (a+b x)))+\frac {3}{2} b \text {arctanh}(\tanh (a+b x))^2-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}+3 b (b x-\text {arctanh}(\tanh (a+b x)))^2 \log (x) \]
-3*b^2*x*(b*x-arctanh(tanh(b*x+a)))+3/2*b*arctanh(tanh(b*x+a))^2-arctanh(t anh(b*x+a))^3/x+3*b*(b*x-arctanh(tanh(b*x+a)))^2*ln(x)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}-6 b^2 x \text {arctanh}(\tanh (a+b x)) \log (x)+3 b \text {arctanh}(\tanh (a+b x))^2 (1+\log (x))+\frac {3}{2} b^3 x^2 (-1+2 \log (x)) \]
-(ArcTanh[Tanh[a + b*x]]^3/x) - 6*b^2*x*ArcTanh[Tanh[a + b*x]]*Log[x] + 3* b*ArcTanh[Tanh[a + b*x]]^2*(1 + Log[x]) + (3*b^3*x^2*(-1 + 2*Log[x]))/2
Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2599, 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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle 3 b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\) |
\(\Big \downarrow \) 2590 |
\(\displaystyle 3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\) |
\(\Big \downarrow \) 2589 |
\(\displaystyle 3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle 3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\) |
-(ArcTanh[Tanh[a + b*x]]^3/x) + 3*b*(ArcTanh[Tanh[a + b*x]]^2/2 - (b*x - A rcTanh[Tanh[a + b*x]])*(b*x - (b*x - ArcTanh[Tanh[a + b*x]])*Log[x]))
3.1.60.3.1 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]
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 = 0.15 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.25
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\right )\) | \(85\) |
parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\right )\) | \(85\) |
risch | \(\text {Expression too large to display}\) | \(2029\) |
-arctanh(tanh(b*x+a))^3/x+3*b*(ln(x)*arctanh(tanh(b*x+a))^2-2*b*(b*(1/2*x^ 2*ln(x)-1/4*x^2)+a*(x*ln(x)-x)+(arctanh(tanh(b*x+a))-b*x-a)*(x*ln(x)-x)))
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.53 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=\frac {b^{3} x^{3} + 6 \, a b^{2} x^{2} + 6 \, a^{2} b x \log \left (x\right ) - 2 \, a^{3}}{2 \, x} \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=\int \frac {\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{2}}\, dx \]
Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=3 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) + \frac {3}{2} \, {\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right ) - 2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} \]
3*b*arctanh(tanh(b*x + a))^2*log(x) + 3/2*(b^2*x^2 + 4*a*b*x + 2*a^2*log(x ) - 2*arctanh(tanh(b*x + a))^2*log(x))*b - arctanh(tanh(b*x + a))^3/x
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.49 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=\frac {1}{2} \, b^{3} x^{2} + 3 \, a b^{2} x + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac {a^{3}}{x} \]
Time = 0.21 (sec) , antiderivative size = 415, normalized size of antiderivative = 6.10 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2} \, dx=\frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{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}{8\,x}+\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}{4}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{8\,x}-\frac {3\,b^3\,x^2}{2}+\frac {3\,b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (x\right )}{4}+\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}{8\,x}-\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 )}{8\,x}+\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\,\ln \left (x\right )}{4}-\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 )}{2}+3\,b^3\,x^2\,\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 )\,\ln \left (x\right )}{2}+3\,b^2\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )-3\,b^2\,x\,\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 ) \]
(3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/4 - log((exp(2*a)*exp(2*b*x))/(ex p(2*a)*exp(2*b*x) + 1))^3/(8*x) + (3*b*log((exp(2*a)*exp(2*b*x))/(exp(2*a) *exp(2*b*x) + 1))^2)/4 + log(1/(exp(2*a)*exp(2*b*x) + 1))^3/(8*x) - (3*b^3 *x^2)/2 + (3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log(x))/4 + (3*log(1/(ex p(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1 ))^2)/(8*x) - (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)))/(8*x) + (3*b*log((exp(2*a)*exp(2*b*x))/(ex p(2*a)*exp(2*b*x) + 1))^2*log(x))/4 - (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)))/2 + 3*b^3*x^2*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))*log(x))/2 + 3*b^2*x*log(1/(exp(2*a)*exp(2*b*x) + 1)) *log(x) - 3*b^2*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log (x)