Integrand size = 13, antiderivative size = 80 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {b^4 x^{10}}{1260}-\frac {1}{126} b^3 x^9 \text {arctanh}(\tanh (a+b x))+\frac {1}{28} b^2 x^8 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{21} b x^7 \text {arctanh}(\tanh (a+b x))^3+\frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4 \]
1/1260*b^4*x^10-1/126*b^3*x^9*arctanh(tanh(b*x+a))+1/28*b^2*x^8*arctanh(ta nh(b*x+a))^2-2/21*b*x^7*arctanh(tanh(b*x+a))^3+1/6*x^6*arctanh(tanh(b*x+a) )^4
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {x^6 \left (b^4 x^4-10 b^3 x^3 \text {arctanh}(\tanh (a+b x))+45 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-120 b x \text {arctanh}(\tanh (a+b x))^3+210 \text {arctanh}(\tanh (a+b x))^4\right )}{1260} \]
(x^6*(b^4*x^4 - 10*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 45*b^2*x^2*ArcTanh[Tan h[a + b*x]]^2 - 120*b*x*ArcTanh[Tanh[a + b*x]]^3 + 210*ArcTanh[Tanh[a + b* x]]^4))/1260
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11, 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, 2599, 15}
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.
\(\displaystyle \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4-\frac {2}{3} b \int x^6 \text {arctanh}(\tanh (a+b x))^3dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4-\frac {2}{3} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{7} b \int x^7 \text {arctanh}(\tanh (a+b x))^2dx\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4-\frac {2}{3} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^2-\frac {1}{4} b \int x^8 \text {arctanh}(\tanh (a+b x))dx\right )\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4-\frac {2}{3} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^2-\frac {1}{4} b \left (\frac {1}{9} x^9 \text {arctanh}(\tanh (a+b x))-\frac {b \int x^9dx}{9}\right )\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^4-\frac {2}{3} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^2-\frac {1}{4} b \left (\frac {1}{9} x^9 \text {arctanh}(\tanh (a+b x))-\frac {b x^{10}}{90}\right )\right )\right )\) |
(x^6*ArcTanh[Tanh[a + b*x]]^4)/6 - (2*b*((x^7*ArcTanh[Tanh[a + b*x]]^3)/7 - (3*b*((x^8*ArcTanh[Tanh[a + b*x]]^2)/8 - (b*(-1/90*(b*x^10) + (x^9*ArcTa nh[Tanh[a + b*x]])/9))/4))/7))/3
3.1.67.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -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 = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92
\[\frac {x^{6} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{6}-\frac {2 b \left (\frac {x^{7} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{7}-\frac {3 b \left (\frac {x^{8} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{8}-\frac {b \left (\frac {x^{9} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{9}-\frac {x^{10} b}{90}\right )}{4}\right )}{7}\right )}{3}\]
1/6*x^6*arctanh(tanh(b*x+a))^4-2/3*b*(1/7*x^7*arctanh(tanh(b*x+a))^3-3/7*b *(1/8*x^8*arctanh(tanh(b*x+a))^2-1/4*b*(1/9*x^9*arctanh(tanh(b*x+a))-1/90* x^10*b)))
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{10} \, b^{4} x^{10} + \frac {4}{9} \, a b^{3} x^{9} + \frac {3}{4} \, a^{2} b^{2} x^{8} + \frac {4}{7} \, a^{3} b x^{7} + \frac {1}{6} \, a^{4} x^{6} \]
Time = 1.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {b^{4} x^{10}}{1260} - \frac {b^{3} x^{9} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{126} + \frac {b^{2} x^{8} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{28} - \frac {2 b x^{7} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{21} + \frac {x^{6} \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{6} \]
b**4*x**10/1260 - b**3*x**9*atanh(tanh(a + b*x))/126 + b**2*x**8*atanh(tan h(a + b*x))**2/28 - 2*b*x**7*atanh(tanh(a + b*x))**3/21 + x**6*atanh(tanh( a + b*x))**4/6
Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=-\frac {2}{21} \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{6} \, x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{1260} \, {\left (45 \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{10} - 10 \, b x^{9} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]
-2/21*b*x^7*arctanh(tanh(b*x + a))^3 + 1/6*x^6*arctanh(tanh(b*x + a))^4 + 1/1260*(45*b*x^8*arctanh(tanh(b*x + a))^2 + (b^2*x^10 - 10*b*x^9*arctanh(t anh(b*x + a)))*b)*b
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{10} \, b^{4} x^{10} + \frac {4}{9} \, a b^{3} x^{9} + \frac {3}{4} \, a^{2} b^{2} x^{8} + \frac {4}{7} \, a^{3} b x^{7} + \frac {1}{6} \, a^{4} x^{6} \]
Time = 0.15 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.02 \[ \int x^5 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {x^6\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{96}+\frac {b^4\,x^{10}}{10}-\frac {b\,x^7\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{14}-\frac {2\,b^3\,x^9\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{9}+\frac {3\,b^2\,x^8\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{16} \]
(x^6*(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)/96 + (b^4*x^10)/10 - (b*x^7*(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)^3)/14 - (2*b^3*x^9*(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))/9 + (3*b^2*x^8 *(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)^2)/16