Integrand size = 13, antiderivative size = 87 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=-6 b^3 x (b x-\text {arctanh}(\tanh (a+b x)))+3 b^2 \text {arctanh}(\tanh (a+b x))^2-\frac {2 b \text {arctanh}(\tanh (a+b x))^3}{x}-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}+6 b^2 (b x-\text {arctanh}(\tanh (a+b x)))^2 \log (x) \]
-6*b^3*x*(b*x-arctanh(tanh(b*x+a)))+3*b^2*arctanh(tanh(b*x+a))^2-2*b*arcta nh(tanh(b*x+a))^3/x-1/2*arctanh(tanh(b*x+a))^4/x^2+6*b^2*(b*x-arctanh(tanh (b*x+a)))^2*ln(x)
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=-\frac {2 b \text {arctanh}(\tanh (a+b x))^3}{x}-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}+6 b^4 x^2 \log (x)-6 b^3 x \text {arctanh}(\tanh (a+b x)) (1+2 \log (x))+3 b^2 \text {arctanh}(\tanh (a+b x))^2 (3+2 \log (x)) \]
(-2*b*ArcTanh[Tanh[a + b*x]]^3)/x - ArcTanh[Tanh[a + b*x]]^4/(2*x^2) + 6*b ^4*x^2*Log[x] - 6*b^3*x*ArcTanh[Tanh[a + b*x]]*(1 + 2*Log[x]) + 3*b^2*ArcT anh[Tanh[a + b*x]]^2*(3 + 2*Log[x])
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2599, 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.
| \(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle 2 b \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle 2 b \left (3 b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}\) |
| \(\Big \downarrow \) 2590 |
| \(\displaystyle 2 b \left (3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}\) |
| \(\Big \downarrow \) 2589 |
| \(\displaystyle 2 b \left (3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle 2 b \left (3 b \left (\frac {1}{2} \text {arctanh}(\tanh (a+b x))^2-(b x-\text {arctanh}(\tanh (a+b x))) (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{2 x^2}\) |
-1/2*ArcTanh[Tanh[a + b*x]]^4/x^2 + 2*b*(-(ArcTanh[Tanh[a + b*x]]^3/x) + 3 *b*(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])))
3.1.75.3.1 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 = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.18
| method | result | size |
| default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{2 x^{2}}+2 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\right )\right )\) | \(103\) |
| parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{2 x^{2}}+2 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{x}+3 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}-2 b \left (b \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+a \left (x \ln \left (x \right )-x \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \left (x \ln \left (x \right )-x \right )\right )\right )\right )\) | \(103\) |
| risch | \(\text {Expression too large to display}\) | \(2355\) |
-1/2*arctanh(tanh(b*x+a))^4/x^2+2*b*(-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)+(arcta nh(tanh(b*x+a))-b*x-a)*(x*ln(x)-x))))
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.54 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=\frac {b^{4} x^{4} + 8 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} \log \left (x\right ) - 8 \, a^{3} b x - a^{4}}{2 \, x^{2}} \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=\int \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]
Time = 0.60 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=-\frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} + 3 \, {\left (2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right ) + {\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \left (x\right ) - 2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \left (x\right )\right )} b\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{2 \, x^{2}} \]
-2*b*arctanh(tanh(b*x + a))^3/x + 3*(2*b*arctanh(tanh(b*x + a))^2*log(x) + (b^2*x^2 + 4*a*b*x + 2*a^2*log(x) - 2*arctanh(tanh(b*x + a))^2*log(x))*b) *b - 1/2*arctanh(tanh(b*x + a))^4/x^2
Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.49 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=\frac {1}{2} \, b^{4} x^{2} + 4 \, a b^{3} x + 6 \, a^{2} b^{2} \log \left ({\left | x \right |}\right ) - \frac {8 \, a^{3} b x + a^{4}}{2 \, x^{2}} \]
Time = 3.85 (sec) , antiderivative size = 672, normalized size of antiderivative = 7.72 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^3} \, dx=\frac {9\,b^2\,{\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 )}^4}{32\,x^2}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^4}{32\,x^2}+\frac {9\,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 )}^2}{4}-3\,b^3\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\frac {b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{4\,x}+\frac {\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 )}^3}{8\,x^2}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3\,\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^2}-\frac {b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{4\,x}+\frac {3\,b^2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (x\right )}{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 )}^2}{16\,x^2}+\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 )}^2\,\ln \left (x\right )}{2}+6\,b^4\,x^2\,\ln \left (x\right )-\frac {9\,b^2\,\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\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\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 )}^2}{4\,x}-\frac {3\,b\,{\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\,x}-3\,b^2\,\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 )+6\,b^3\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )-6\,b^3\,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 ) \]
(9*b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/4 - log((exp(2*a)*exp(2*b*x))/( exp(2*a)*exp(2*b*x) + 1))^4/(32*x^2) - log(1/(exp(2*a)*exp(2*b*x) + 1))^4/ (32*x^2) + (9*b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/ 4 - 3*b^3*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + (b*log( 1/(exp(2*a)*exp(2*b*x) + 1))^3)/(4*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))^3)/(8*x^2) + (log(1/( exp(2*a)*exp(2*b*x) + 1))^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/(8*x^2) - (b*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^ 3)/(4*x) + (3*b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log(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))^2)/(16*x^2) + (3*b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x ) + 1))^2*log(x))/2 + 6*b^4*x^2*log(x) - (9*b^2*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*l og(1/(exp(2*a)*exp(2*b*x) + 1)) + (3*b*log(1/(exp(2*a)*exp(2*b*x) + 1))*lo g((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(4*x) - (3*b*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*x) - 3*b^2*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) + 6*b^3*x*log(1/(exp(2*a)*exp(2*b *x) + 1))*log(x) - 6*b^3*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(x)