Integrand size = 13, antiderivative size = 80 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {b^4 x^9}{630}-\frac {1}{70} b^3 x^8 \text {arctanh}(\tanh (a+b x))+\frac {2}{35} b^2 x^7 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{15} b x^6 \text {arctanh}(\tanh (a+b x))^3+\frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4 \] Output:
1/630*b^4*x^9-1/70*b^3*x^8*arctanh(tanh(b*x+a))+2/35*b^2*x^7*arctanh(tanh( b*x+a))^2-2/15*b*x^6*arctanh(tanh(b*x+a))^3+1/5*x^5*arctanh(tanh(b*x+a))^4
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{630} x^5 \left (b^4 x^4-9 b^3 x^3 \text {arctanh}(\tanh (a+b x))+36 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-84 b x \text {arctanh}(\tanh (a+b x))^3+126 \text {arctanh}(\tanh (a+b x))^4\right ) \] Input:
Integrate[x^4*ArcTanh[Tanh[a + b*x]]^4,x]
Output:
(x^5*(b^4*x^4 - 9*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 36*b^2*x^2*ArcTanh[Tanh [a + b*x]]^2 - 84*b*x*ArcTanh[Tanh[a + b*x]]^3 + 126*ArcTanh[Tanh[a + b*x] ]^4))/630
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.
\(\displaystyle \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{5} b \int x^5 \text {arctanh}(\tanh (a+b x))^3dx\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{5} b \left (\frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^3-\frac {1}{2} b \int x^6 \text {arctanh}(\tanh (a+b x))^2dx\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{5} b \left (\frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^3-\frac {1}{2} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{7} b \int x^7 \text {arctanh}(\tanh (a+b x))dx\right )\right )\) |
\(\Big \downarrow \) 2599 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{5} b \left (\frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^3-\frac {1}{2} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))-\frac {b \int x^8dx}{8}\right )\right )\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{5} x^5 \text {arctanh}(\tanh (a+b x))^4-\frac {4}{5} b \left (\frac {1}{6} x^6 \text {arctanh}(\tanh (a+b x))^3-\frac {1}{2} b \left (\frac {1}{7} x^7 \text {arctanh}(\tanh (a+b x))^2-\frac {2}{7} b \left (\frac {1}{8} x^8 \text {arctanh}(\tanh (a+b x))-\frac {b x^9}{72}\right )\right )\right )\) |
Input:
Int[x^4*ArcTanh[Tanh[a + b*x]]^4,x]
Output:
(x^5*ArcTanh[Tanh[a + b*x]]^4)/5 - (4*b*((x^6*ArcTanh[Tanh[a + b*x]]^3)/6 - (b*((x^7*ArcTanh[Tanh[a + b*x]]^2)/7 - (2*b*(-1/72*(b*x^9) + (x^8*ArcTan h[Tanh[a + b*x]])/8))/7))/2))/5
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 = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92
\[\frac {x^{5} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}}{5}-\frac {4 b \left (\frac {x^{6} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3}}{6}-\frac {b \left (\frac {x^{7} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2}}{7}-\frac {2 b \left (\frac {x^{8} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}{8}-\frac {x^{9} b}{72}\right )}{7}\right )}{2}\right )}{5}\]
Input:
int(x^4*arctanh(tanh(b*x+a))^4,x)
Output:
1/5*x^5*arctanh(tanh(b*x+a))^4-4/5*b*(1/6*x^6*arctanh(tanh(b*x+a))^3-1/2*b *(1/7*x^7*arctanh(tanh(b*x+a))^2-2/7*b*(1/8*x^8*arctanh(tanh(b*x+a))-1/72* x^9*b)))
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{9} \, b^{4} x^{9} + \frac {1}{2} \, a b^{3} x^{8} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {2}{3} \, a^{3} b x^{6} + \frac {1}{5} \, a^{4} x^{5} \] Input:
integrate(x^4*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")
Output:
1/9*b^4*x^9 + 1/2*a*b^3*x^8 + 6/7*a^2*b^2*x^7 + 2/3*a^3*b*x^6 + 1/5*a^4*x^ 5
Time = 1.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\begin {cases} \frac {x^{4} \operatorname {atanh}^{5}{\left (\tanh {\left (a + b x \right )} \right )}}{5 b} - \frac {2 x^{3} \operatorname {atanh}^{6}{\left (\tanh {\left (a + b x \right )} \right )}}{15 b^{2}} + \frac {2 x^{2} \operatorname {atanh}^{7}{\left (\tanh {\left (a + b x \right )} \right )}}{35 b^{3}} - \frac {x \operatorname {atanh}^{8}{\left (\tanh {\left (a + b x \right )} \right )}}{70 b^{4}} + \frac {\operatorname {atanh}^{9}{\left (\tanh {\left (a + b x \right )} \right )}}{630 b^{5}} & \text {for}\: b \neq 0 \\\frac {x^{5} \operatorname {atanh}^{4}{\left (\tanh {\left (a \right )} \right )}}{5} & \text {otherwise} \end {cases} \] Input:
integrate(x**4*atanh(tanh(b*x+a))**4,x)
Output:
Piecewise((x**4*atanh(tanh(a + b*x))**5/(5*b) - 2*x**3*atanh(tanh(a + b*x) )**6/(15*b**2) + 2*x**2*atanh(tanh(a + b*x))**7/(35*b**3) - x*atanh(tanh(a + b*x))**8/(70*b**4) + atanh(tanh(a + b*x))**9/(630*b**5), Ne(b, 0)), (x* *5*atanh(tanh(a))**4/5, True))
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=-\frac {2}{15} \, b x^{6} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac {1}{5} \, x^{5} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac {1}{630} \, {\left (36 \, b x^{7} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + {\left (b^{2} x^{9} - 9 \, b x^{8} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \] Input:
integrate(x^4*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")
Output:
-2/15*b*x^6*arctanh(tanh(b*x + a))^3 + 1/5*x^5*arctanh(tanh(b*x + a))^4 + 1/630*(36*b*x^7*arctanh(tanh(b*x + a))^2 + (b^2*x^9 - 9*b*x^8*arctanh(tanh (b*x + a)))*b)*b
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {1}{9} \, b^{4} x^{9} + \frac {1}{2} \, a b^{3} x^{8} + \frac {6}{7} \, a^{2} b^{2} x^{7} + \frac {2}{3} \, a^{3} b x^{6} + \frac {1}{5} \, a^{4} x^{5} \] Input:
integrate(x^4*arctanh(tanh(b*x+a))^4,x, algorithm="giac")
Output:
1/9*b^4*x^9 + 1/2*a*b^3*x^8 + 6/7*a^2*b^2*x^7 + 2/3*a^3*b*x^6 + 1/5*a^4*x^ 5
Time = 3.47 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^5\,\left (126\,b^4\,x^4-84\,b^3\,x^3\,\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+36\,b^2\,x^2\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2-9\,b\,x\,{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3+{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^4\right )}{630\,b^5} \] Input:
int(x^4*atanh(tanh(a + b*x))^4,x)
Output:
(atanh(tanh(a + b*x))^5*(atanh(tanh(a + b*x))^4 + 126*b^4*x^4 + 36*b^2*x^2 *atanh(tanh(a + b*x))^2 - 9*b*x*atanh(tanh(a + b*x))^3 - 84*b^3*x^3*atanh( tanh(a + b*x))))/(630*b^5)
\[ \int x^4 \text {arctanh}(\tanh (a+b x))^4 \, dx=\int \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{4} x^{4}d x \] Input:
int(x^4*atanh(tanh(b*x+a))^4,x)
Output:
int(atanh(tanh(a + b*x))**4*x**4,x)