Integrand size = 13, antiderivative size = 80 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {b^4}{1260 x^6}-\frac {b^3 \text {arctanh}(\tanh (a+b x))}{210 x^7}-\frac {b^2 \text {arctanh}(\tanh (a+b x))^2}{60 x^8}-\frac {2 b \text {arctanh}(\tanh (a+b x))^3}{45 x^9}-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}} \]
-1/1260*b^4/x^6-1/210*b^3*arctanh(tanh(b*x+a))/x^7-1/60*b^2*arctanh(tanh(b *x+a))^2/x^8-2/45*b*arctanh(tanh(b*x+a))^3/x^9-1/10*arctanh(tanh(b*x+a))^4 /x^10
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {b^4 x^4+6 b^3 x^3 \text {arctanh}(\tanh (a+b x))+21 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2+56 b x \text {arctanh}(\tanh (a+b x))^3+126 \text {arctanh}(\tanh (a+b x))^4}{1260 x^{10}} \]
-1/1260*(b^4*x^4 + 6*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 21*b^2*x^2*ArcTanh[T anh[a + b*x]]^2 + 56*b*x*ArcTanh[Tanh[a + b*x]]^3 + 126*ArcTanh[Tanh[a + b *x]]^4)/x^10
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.
\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{5} b \int \frac {\text {arctanh}(\tanh (a+b x))^3}{x^{10}}dx-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{5} b \left (\frac {1}{3} b \int \frac {\text {arctanh}(\tanh (a+b x))^2}{x^9}dx-\frac {\text {arctanh}(\tanh (a+b x))^3}{9 x^9}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{5} b \left (\frac {1}{3} b \left (\frac {1}{4} b \int \frac {\text {arctanh}(\tanh (a+b x))}{x^8}dx-\frac {\text {arctanh}(\tanh (a+b x))^2}{8 x^8}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{9 x^9}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}}\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {2}{5} b \left (\frac {1}{3} b \left (\frac {1}{4} b \left (\frac {1}{7} b \int \frac {1}{x^7}dx-\frac {\text {arctanh}(\tanh (a+b x))}{7 x^7}\right )-\frac {\text {arctanh}(\tanh (a+b x))^2}{8 x^8}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{9 x^9}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {2}{5} b \left (\frac {1}{3} b \left (\frac {1}{4} b \left (-\frac {\text {arctanh}(\tanh (a+b x))}{7 x^7}-\frac {b}{42 x^6}\right )-\frac {\text {arctanh}(\tanh (a+b x))^2}{8 x^8}\right )-\frac {\text {arctanh}(\tanh (a+b x))^3}{9 x^9}\right )-\frac {\text {arctanh}(\tanh (a+b x))^4}{10 x^{10}}\) |
-1/10*ArcTanh[Tanh[a + b*x]]^4/x^10 + (2*b*(-1/9*ArcTanh[Tanh[a + b*x]]^3/ x^9 + (b*(-1/8*ArcTanh[Tanh[a + b*x]]^2/x^8 + (b*(-1/42*b/x^6 - ArcTanh[Ta nh[a + b*x]]/(7*x^7)))/4))/3))/5
3.1.83.3.1 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 = 2.90 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-\frac {b^{4} x^{4}+6 b^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right ) x^{3}+21 b^{2} x^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}+56 b x \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}+126 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{1260 x^{10}}\) | \(70\) |
default | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{10 x^{10}}+\frac {2 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{9 x^{9}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{8 x^{8}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{7 x^{7}}-\frac {b}{42 x^{6}}\right )}{4}\right )}{3}\right )}{5}\) | \(74\) |
parts | \(-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{10 x^{10}}+\frac {2 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{9 x^{9}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{8 x^{8}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{7 x^{7}}-\frac {b}{42 x^{6}}\right )}{4}\right )}{3}\right )}{5}\) | \(74\) |
risch | \(\text {Expression too large to display}\) | \(22625\) |
-1/1260*(b^4*x^4+6*b^3*arctanh(tanh(b*x+a))*x^3+21*b^2*x^2*arctanh(tanh(b* x+a))^2+56*b*x*arctanh(tanh(b*x+a))^3+126*arctanh(tanh(b*x+a))^4)/x^10
Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {210 \, b^{4} x^{4} + 720 \, a b^{3} x^{3} + 945 \, a^{2} b^{2} x^{2} + 560 \, a^{3} b x + 126 \, a^{4}}{1260 \, x^{10}} \]
Time = 2.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=- \frac {b^{4}}{1260 x^{6}} - \frac {b^{3} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{210 x^{7}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{60 x^{8}} - \frac {2 b \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{45 x^{9}} - \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{10 x^{10}} \]
-b**4/(1260*x**6) - b**3*atanh(tanh(a + b*x))/(210*x**7) - b**2*atanh(tanh (a + b*x))**2/(60*x**8) - 2*b*atanh(tanh(a + b*x))**3/(45*x**9) - atanh(ta nh(a + b*x))**4/(10*x**10)
Time = 0.40 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {1}{1260} \, {\left (b {\left (\frac {b^{2}}{x^{6}} + \frac {6 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{x^{7}}\right )} + \frac {21 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x^{8}}\right )} b - \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{45 \, x^{9}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{10 \, x^{10}} \]
-1/1260*(b*(b^2/x^6 + 6*b*arctanh(tanh(b*x + a))/x^7) + 21*b*arctanh(tanh( b*x + a))^2/x^8)*b - 2/45*b*arctanh(tanh(b*x + a))^3/x^9 - 1/10*arctanh(ta nh(b*x + a))^4/x^10
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {210 \, b^{4} x^{4} + 720 \, a b^{3} x^{3} + 945 \, a^{2} b^{2} x^{2} + 560 \, a^{3} b x + 126 \, a^{4}}{1260 \, x^{10}} \]
Time = 3.70 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^4}{x^{11}} \, dx=-\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4}{10\,x^{10}}-\frac {b^4}{1260\,x^6}-\frac {b^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2}{60\,x^8}-\frac {b^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}{210\,x^7}-\frac {2\,b\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3}{45\,x^9} \]