Integrand size = 13, antiderivative size = 121 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\frac {x^3 \text {arctanh}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \text {arctanh}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \text {arctanh}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \text {arctanh}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)} \] Output:
x^3*arctanh(tanh(b*x+a))^(1+n)/b/(1+n)-3*x^2*arctanh(tanh(b*x+a))^(2+n)/b^ 2/(1+n)/(2+n)+6*x*arctanh(tanh(b*x+a))^(3+n)/b^3/(1+n)/(2+n)/(3+n)-6*arcta nh(tanh(b*x+a))^(4+n)/b^4/(1+n)/(2+n)/(3+n)/(4+n)
Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\frac {\text {arctanh}(\tanh (a+b x))^{1+n} \left (b^3 \left (24+26 n+9 n^2+n^3\right ) x^3-3 b^2 \left (12+7 n+n^2\right ) x^2 \text {arctanh}(\tanh (a+b x))+6 b (4+n) x \text {arctanh}(\tanh (a+b x))^2-6 \text {arctanh}(\tanh (a+b x))^3\right )}{b^4 (1+n) (2+n) (3+n) (4+n)} \] Input:
Integrate[x^3*ArcTanh[Tanh[a + b*x]]^n,x]
Output:
(ArcTanh[Tanh[a + b*x]]^(1 + n)*(b^3*(24 + 26*n + 9*n^2 + n^3)*x^3 - 3*b^2 *(12 + 7*n + n^2)*x^2*ArcTanh[Tanh[a + b*x]] + 6*b*(4 + n)*x*ArcTanh[Tanh[ a + b*x]]^2 - 6*ArcTanh[Tanh[a + b*x]]^3))/(b^4*(1 + n)*(2 + n)*(3 + n)*(4 + n))
Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2599, 2599, 2599, 2588, 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.
| \(\displaystyle \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^3 \text {arctanh}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {3 \int x^2 \text {arctanh}(\tanh (a+b x))^{n+1}dx}{b (n+1)}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^3 \text {arctanh}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {3 \left (\frac {x^2 \text {arctanh}(\tanh (a+b x))^{n+2}}{b (n+2)}-\frac {2 \int x \text {arctanh}(\tanh (a+b x))^{n+2}dx}{b (n+2)}\right )}{b (n+1)}\) |
| \(\Big \downarrow \) 2599 |
| \(\displaystyle \frac {x^3 \text {arctanh}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {3 \left (\frac {x^2 \text {arctanh}(\tanh (a+b x))^{n+2}}{b (n+2)}-\frac {2 \left (\frac {x \text {arctanh}(\tanh (a+b x))^{n+3}}{b (n+3)}-\frac {\int \text {arctanh}(\tanh (a+b x))^{n+3}dx}{b (n+3)}\right )}{b (n+2)}\right )}{b (n+1)}\) |
| \(\Big \downarrow \) 2588 |
| \(\displaystyle \frac {x^3 \text {arctanh}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {3 \left (\frac {x^2 \text {arctanh}(\tanh (a+b x))^{n+2}}{b (n+2)}-\frac {2 \left (\frac {x \text {arctanh}(\tanh (a+b x))^{n+3}}{b (n+3)}-\frac {\int \text {arctanh}(\tanh (a+b x))^{n+3}d\text {arctanh}(\tanh (a+b x))}{b^2 (n+3)}\right )}{b (n+2)}\right )}{b (n+1)}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {x^3 \text {arctanh}(\tanh (a+b x))^{n+1}}{b (n+1)}-\frac {3 \left (\frac {x^2 \text {arctanh}(\tanh (a+b x))^{n+2}}{b (n+2)}-\frac {2 \left (\frac {x \text {arctanh}(\tanh (a+b x))^{n+3}}{b (n+3)}-\frac {\text {arctanh}(\tanh (a+b x))^{n+4}}{b^2 (n+3) (n+4)}\right )}{b (n+2)}\right )}{b (n+1)}\) |
Input:
Int[x^3*ArcTanh[Tanh[a + b*x]]^n,x]
Output:
(x^3*ArcTanh[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - (3*((x^2*ArcTanh[Tanh[a + b*x]]^(2 + n))/(b*(2 + n)) - (2*((x*ArcTanh[Tanh[a + b*x]]^(3 + n))/(b* (3 + n)) - ArcTanh[Tanh[a + b*x]]^(4 + n)/(b^2*(3 + n)*(4 + n))))/(b*(2 + n))))/(b*(1 + n))
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c Subst [Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(276\) vs. \(2(121)=242\).
Time = 1.49 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.29
| method | result | size |
| parallelrisch | \(-\frac {6 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{4}-24 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3} x b -24 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right ) x^{3} b^{3}+36 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2} x^{2} b^{2}-x^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b^{3} n^{3}-9 x^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b^{3} n^{2}-26 x^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b^{3} n +3 x^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b^{2} n^{2}+21 x^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{2} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b^{2} n -6 x \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{3} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{n} b n}{\left (n^{3}+6 n^{2}+11 n +6\right ) \left (4+n \right ) b^{4}}\) | \(277\) |
| default | \(\frac {x^{4} {\mathrm e}^{n \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}}{4+n}+\frac {n \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{3} {\mathrm e}^{n \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}}{b \left (n^{2}+7 n +12\right )}-\frac {6 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (a^{3}+3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) {\mathrm e}^{n \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {3 n \left (a^{2}+2 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) x^{2} {\mathrm e}^{n \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}}{b^{2} \left (n^{3}+9 n^{2}+26 n +24\right )}+\frac {6 n \left (a^{3}+3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) x \,{\mathrm e}^{n \ln \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )}}{b^{3} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(345\) |
| risch | \(\text {Expression too large to display}\) | \(52442\) |
Input:
int(x^3*arctanh(tanh(b*x+a))^n,x,method=_RETURNVERBOSE)
Output:
-(6*arctanh(tanh(b*x+a))^n*arctanh(tanh(b*x+a))^4-24*arctanh(tanh(b*x+a))^ n*arctanh(tanh(b*x+a))^3*x*b-24*arctanh(tanh(b*x+a))^n*arctanh(tanh(b*x+a) )*x^3*b^3+36*arctanh(tanh(b*x+a))^n*arctanh(tanh(b*x+a))^2*x^2*b^2-x^3*arc tanh(tanh(b*x+a))*arctanh(tanh(b*x+a))^n*b^3*n^3-9*x^3*arctanh(tanh(b*x+a) )*arctanh(tanh(b*x+a))^n*b^3*n^2-26*x^3*arctanh(tanh(b*x+a))*arctanh(tanh( b*x+a))^n*b^3*n+3*x^2*arctanh(tanh(b*x+a))^2*arctanh(tanh(b*x+a))^n*b^2*n^ 2+21*x^2*arctanh(tanh(b*x+a))^2*arctanh(tanh(b*x+a))^n*b^2*n-6*x*arctanh(t anh(b*x+a))^3*arctanh(tanh(b*x+a))^n*b*n)/(n^3+6*n^2+11*n+6)/(4+n)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (121) = 242\).
Time = 0.08 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.11 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\frac {{\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \cosh \left (n \log \left (b x + a\right )\right ) + {\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \sinh \left (n \log \left (b x + a\right )\right )}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="fricas")
Output:
((6*a^3*b*n*x + (b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6*b^4)*x^4 - 6*a^4 + (a* b^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^3 - 3*(a^2*b^2*n^2 + a^2*b^2*n)*x^2)* cosh(n*log(b*x + a)) + (6*a^3*b*n*x + (b^4*n^3 + 6*b^4*n^2 + 11*b^4*n + 6* b^4)*x^4 - 6*a^4 + (a*b^3*n^3 + 3*a*b^3*n^2 + 2*a*b^3*n)*x^3 - 3*(a^2*b^2* n^2 + a^2*b^2*n)*x^2)*sinh(n*log(b*x + a)))/(b^4*n^4 + 10*b^4*n^3 + 35*b^4 *n^2 + 50*b^4*n + 24*b^4)
\[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\text {Too large to display} \] Input:
integrate(x**3*atanh(tanh(b*x+a))**n,x)
Output:
Piecewise((x**4*atanh(tanh(a))**n/4, Eq(b, 0)), (-x**3/(3*b*atanh(tanh(a + b*x))**3) - x**2/(2*b**2*atanh(tanh(a + b*x))**2) - x/(b**3*atanh(tanh(a + b*x))) + log(atanh(tanh(a + b*x)))/b**4, Eq(n, -4)), (Integral(x**3/atan h(tanh(a + b*x))**3, x), Eq(n, -3)), (Integral(x**3/atanh(tanh(a + b*x))** 2, x), Eq(n, -2)), (Integral(x**3/atanh(tanh(a + b*x)), x), Eq(n, -1)), (b **3*n**3*x**3*atanh(tanh(a + b*x))*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10 *b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 9*b**3*n**2*x**3*atanh( tanh(a + b*x))*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4 *n**2 + 50*b**4*n + 24*b**4) + 26*b**3*n*x**3*atanh(tanh(a + b*x))*atanh(t anh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24 *b**4) + 24*b**3*x**3*atanh(tanh(a + b*x))*atanh(tanh(a + b*x))**n/(b**4*n **4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 3*b**2*n**2*x** 2*atanh(tanh(a + b*x))**2*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n** 3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 21*b**2*n*x**2*atanh(tanh(a + b* x))**2*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) - 36*b**2*x**2*atanh(tanh(a + b*x))**2*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 6*b*n*x*atanh(tanh(a + b*x))**3*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10* b**4*n**3 + 35*b**4*n**2 + 50*b**4*n + 24*b**4) + 24*b*x*atanh(tanh(a + b* x))**3*atanh(tanh(a + b*x))**n/(b**4*n**4 + 10*b**4*n**3 + 35*b**4*n**2...
Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.83 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="maxima")
Output:
((n^3 + 6*n^2 + 11*n + 6)*b^4*x^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*x^3 - 3*(n^2 + n)*a^2*b^2*x^2 + 6*a^3*b*n*x - 6*a^4)*(b*x + a)^n/((n^4 + 10*n^3 + 35*n ^2 + 50*n + 24)*b^4)
Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.87 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\frac {{\left (b x + a\right )}^{n} b^{4} n^{3} x^{4} + {\left (b x + a\right )}^{n} a b^{3} n^{3} x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} n^{2} x^{4} + 3 \, {\left (b x + a\right )}^{n} a b^{3} n^{2} x^{3} + 11 \, {\left (b x + a\right )}^{n} b^{4} n x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n^{2} x^{2} + 2 \, {\left (b x + a\right )}^{n} a b^{3} n x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n x^{2} + 6 \, {\left (b x + a\right )}^{n} a^{3} b n x - 6 \, {\left (b x + a\right )}^{n} a^{4}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \] Input:
integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="giac")
Output:
((b*x + a)^n*b^4*n^3*x^4 + (b*x + a)^n*a*b^3*n^3*x^3 + 6*(b*x + a)^n*b^4*n ^2*x^4 + 3*(b*x + a)^n*a*b^3*n^2*x^3 + 11*(b*x + a)^n*b^4*n*x^4 - 3*(b*x + a)^n*a^2*b^2*n^2*x^2 + 2*(b*x + a)^n*a*b^3*n*x^3 + 6*(b*x + a)^n*b^4*x^4 - 3*(b*x + a)^n*a^2*b^2*n*x^2 + 6*(b*x + a)^n*a^3*b*n*x - 6*(b*x + a)^n*a^ 4)/(b^4*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 50*b^4*n + 24*b^4)
Time = 3.56 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.45 \[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=-{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{8\,b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {3\,n\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {n\,x^3\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (n^2+3\,n+2\right )}{2\,b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \] Input:
int(x^3*atanh(tanh(a + b*x))^n,x)
Output:
-(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2* a)*exp(2*b*x) + 1))/2)^n*((3*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4)/(8*b^4*(50*n + 3 5*n^2 + 10*n^3 + n^4 + 24)) - (x^4*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^ 2 + 10*n^3 + n^4 + 24) + (3*n*x*(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)/(4*b^3*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (n*x^3*(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*n + n ^2 + 2))/(2*b*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (3*n*x^2*(n + 1)*(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)/(4*b^2*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)))
\[ \int x^3 \text {arctanh}(\tanh (a+b x))^n \, dx=\int \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{n} x^{3}d x \] Input:
int(x^3*atanh(tanh(b*x+a))^n,x)
Output:
int(atanh(tanh(a + b*x))**n*x**3,x)