Integrand size = 13, antiderivative size = 77 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=4 b^4 x-\frac {2 b^2 \text {arctanh}(\tanh (a+b x))^2}{x}-\frac {2 b \text {arctanh}(\tanh (a+b x))^3}{3 x^2}-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}-4 b^3 (b x-\text {arctanh}(\tanh (a+b x))) \log (x) \]
4*b^4*x-2*b^2*arctanh(tanh(b*x+a))^2/x-2/3*b*arctanh(tanh(b*x+a))^3/x^2-1/ 3*arctanh(tanh(b*x+a))^4/x^3-4*b^3*(b*x-arctanh(tanh(b*x+a)))*ln(x)
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=-\frac {6 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2+2 b x \text {arctanh}(\tanh (a+b x))^3+\text {arctanh}(\tanh (a+b x))^4+2 b^4 x^4 (5+6 \log (x))-2 b^3 x^3 \text {arctanh}(\tanh (a+b x)) (11+6 \log (x))}{3 x^3} \]
-1/3*(6*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 2*b*x*ArcTanh[Tanh[a + b*x]]^3 + ArcTanh[Tanh[a + b*x]]^4 + 2*b^4*x^4*(5 + 6*Log[x]) - 2*b^3*x^3*ArcTanh[ Tanh[a + b*x]]*(11 + 6*Log[x]))/x^3
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2599, 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.
| \(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {4}{3} b \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^3}dx-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {4}{3} b \left (\frac {3}{2} b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^2}dx-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {4}{3} b \left (\frac {3}{2} b \left (2 b \int \frac {\text {arctanh}(\tanh (a+b x))}{x}dx-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}\) |
| \(\Big \downarrow \) 2589 |
| \(\displaystyle \frac {4}{3} b \left (\frac {3}{2} b \left (2 b \left (b x-(b x-\text {arctanh}(\tanh (a+b x))) \int \frac {1}{x}dx\right )-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle \frac {4}{3} b \left (\frac {3}{2} b \left (2 b (b x-\log (x) (b x-\text {arctanh}(\tanh (a+b x))))-\frac {\text {arctanh}(\tanh (a+b x))^2}{x}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{2 x^2}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{3 x^3}\) |
-1/3*ArcTanh[Tanh[a + b*x]]^4/x^3 + (4*b*(-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
3.1.76.3.1 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 = 0.46 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{3 x^{3}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{2 x^{2}}+\frac {3 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\right )}{2}\right )}{3}\) | \(77\) |
| parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{3 x^{3}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{2 x^{2}}+\frac {3 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\right )}{2}\right )}{3}\) | \(77\) |
| risch | \(\text {Expression too large to display}\) | \(2334\) |
-1/3*arctanh(tanh(b*x+a))^4/x^3+4/3*b*(-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 = 48, normalized size of antiderivative = 0.62 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=\frac {3 \, b^{4} x^{4} + 12 \, a b^{3} x^{3} \log \left (x\right ) - 18 \, a^{2} b^{2} x^{2} - 6 \, a^{3} b x - a^{4}}{3 \, x^{3}} \]
\[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=\int \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{4}}\, dx \]
Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=2 \, {\left (2 \, {\left (b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x}\right )} b - \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{2}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{3 \, x^{3}} \]
2*(2*(b*arctanh(tanh(b*x + a))*log(x) - (b*(x + a/b)*log(x) - b*(x + a*log (x)/b))*b)*b - b*arctanh(tanh(b*x + a))^2/x)*b - 2/3*b*arctanh(tanh(b*x + a))^3/x^2 - 1/3*arctanh(tanh(b*x + a))^4/x^3
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=b^{4} x + 4 \, a b^{3} \log \left ({\left | x \right |}\right ) - \frac {18 \, a^{2} b^{2} x^{2} + 6 \, a^{3} b x + a^{4}}{3 \, x^{3}} \]
Time = 3.77 (sec) , antiderivative size = 571, normalized size of antiderivative = 7.42 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^4} \, dx=\frac {11\,b^3\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\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}{48\,x^3}-\frac {11\,b^3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\frac {10\,b^4\,x}{3}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^4}{48\,x^3}+\frac {b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{12\,x^2}-2\,b^3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )+\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}{12\,x^3}+\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 )}{12\,x^3}-\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}{12\,x^2}+2\,b^3\,\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 )-4\,b^4\,x\,\ln \left (x\right )-\frac {b^2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{2\,x}-\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 )}^2}{8\,x^3}-\frac {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}{2\,x}+\frac {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 )}{x}+\frac {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^2}-\frac {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^2} \]
(11*b^3*log(exp(2*b*x)/(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))^4/(48*x^3) - (11*b^3*log(1/(exp(2*a)*exp (2*b*x) + 1)))/3 - (10*b^4*x)/3 - log(1/(exp(2*a)*exp(2*b*x) + 1))^4/(48*x ^3) + (b*log(1/(exp(2*a)*exp(2*b*x) + 1))^3)/(12*x^2) - 2*b^3*log(1/(exp(2 *a)*exp(2*b*x) + 1))*log(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)/(12*x^3) + (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)))/( 12*x^3) - (b*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3)/(12*x ^2) + 2*b^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(x) - 4*b^4*x*log(x) - (b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/(2*x) - (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)/(8*x^3) - (b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(2*x) + (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)))/x + (b*log(1/(exp(2*a)*exp(2*b*x) + 1))*l og((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(4*x^2) - (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^2)