Integrand size = 13, antiderivative size = 80 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {b^4 x^{11}}{2310}-\frac {1}{210} b^3 x^{10} \text {arctanh}(\tanh (a+b x))+\frac {1}{42} b^2 x^9 \text {arctanh}(\tanh (a+b x))^2-\frac {1}{14} b x^8 \text {arctanh}(\tanh (a+b x))^3+\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4 \]
1/2310*b^4*x^11-1/210*b^3*x^10*arctanh(tanh(b*x+a))+1/42*b^2*x^9*arctanh(t anh(b*x+a))^2-1/14*b*x^8*arctanh(tanh(b*x+a))^3+1/7*x^7*arctanh(tanh(b*x+a ))^4
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {x^7 \left (b^4 x^4-11 b^3 x^3 \text {arctanh}(\tanh (a+b x))+55 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-165 b x \text {arctanh}(\tanh (a+b x))^3+330 \text {arctanh}(\tanh (a+b x))^4\right )}{2310} \]
(x^7*(b^4*x^4 - 11*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 55*b^2*x^2*ArcTanh[Tan h[a + b*x]]^2 - 165*b*x*ArcTanh[Tanh[a + b*x]]^3 + 330*ArcTanh[Tanh[a + b* x]]^4))/2310
Time = 0.29 (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.
\(\displaystyle \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{7} b \int x^7 \text {arctanh}(\tanh (a+b x))^3dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{8} b \int x^8 \text {arctanh}(\tanh (a+b x))^2dx\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{8} b \left (\frac {1}{9} x^9 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{9} b \int x^9 \text {arctanh}(\tanh (a+b x))dx\right )\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{8} b \left (\frac {1}{9} x^9 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{9} b \left (\frac {1}{10} x^{10} \text {arctanh}(\tanh (a+b x))-\frac {b \int x^{10}dx}{10}\right )\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))^3-\frac {3}{8} b \left (\frac {1}{9} x^9 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{9} b \left (\frac {1}{10} x^{10} \text {arctanh}(\tanh (a+b x))-\frac {b x^{11}}{110}\right )\right )\right )\) |
(x^7*ArcTanh[Tanh[a + b*x]]^4)/7 - (4*b*((x^8*ArcTanh[Tanh[a + b*x]]^3)/8 - (3*b*((x^9*ArcTanh[Tanh[a + b*x]]^2)/9 - (2*b*(-1/110*(b*x^11) + (x^10*A rcTanh[Tanh[a + b*x]])/10))/9))/8))/7
3.1.66.3.1 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
\[\frac {x^{7} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{7}-\frac {4 b \left (\frac {x^{8} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{8}-\frac {3 b \left (\frac {x^{9} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{9}-\frac {2 b \left (\frac {x^{10} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{10}-\frac {x^{11} b}{110}\right )}{9}\right )}{8}\right )}{7}\]
1/7*x^7*arctanh(tanh(b*x+a))^4-4/7*b*(1/8*x^8*arctanh(tanh(b*x+a))^3-3/8*b *(1/9*x^9*arctanh(tanh(b*x+a))^2-2/9*b*(1/10*x^10*arctanh(tanh(b*x+a))-1/1 10*x^11*b)))
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \]
Time = 1.65 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {b^{4} x^{11}}{2310} - \frac {b^{3} x^{10} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{210} + \frac {b^{2} x^{9} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{42} - \frac {b x^{8} \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{14} + \frac {x^{7} \operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{7} \]
b**4*x**11/2310 - b**3*x**10*atanh(tanh(a + b*x))/210 + b**2*x**9*atanh(ta nh(a + b*x))**2/42 - b*x**8*atanh(tanh(a + b*x))**3/14 + x**7*atanh(tanh(a + b*x))**4/7
Time = 0.40 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=-\frac {1}{14} \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{7} \, x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{2310} \, {\left (55 \, b x^{9} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{11} - 11 \, b x^{10} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \]
-1/14*b*x^8*arctanh(tanh(b*x + a))^3 + 1/7*x^7*arctanh(tanh(b*x + a))^4 + 1/2310*(55*b*x^9*arctanh(tanh(b*x + a))^2 + (b^2*x^11 - 11*b*x^10*arctanh( tanh(b*x + a)))*b)*b
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{11} \, b^{4} x^{11} + \frac {2}{5} \, a b^{3} x^{10} + \frac {2}{3} \, a^{2} b^{2} x^{9} + \frac {1}{2} \, a^{3} b x^{8} + \frac {1}{7} \, a^{4} x^{7} \]
Time = 3.67 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.02 \[ \int x^6 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {x^7\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{112}+\frac {b^4\,x^{11}}{11}-\frac {b\,x^8\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{16}-\frac {b^3\,x^{10}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{5}+\frac {b^2\,x^9\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{6} \]
(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)^4)/112 + (b^4*x^11)/11 - (b*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)^3)/16 - (b^3*x^10*(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))/5 + (b^2*x^9*( log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e xp(2*b*x) + 1)) + 2*b*x)^2)/6