3.267 \(\int x^3 \tanh ^{-1}(\tanh (a+b x))^n \, dx\)

Optimal. Leaf size=121 \[ -\frac {6 \tanh ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

[Out]

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/(3+n)/(n^2+3*n+2)-6*arctanh(tanh(b*x+a))^(4+n)/b^4/(n^2+5*n+4)/(n^2+5*n+6)

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Rubi [A]  time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{n+2}}{b^2 (n+1) (n+2)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{n+3}}{b^3 (n+1) (n+2) (n+3)}-\frac {6 \tanh ^{-1}(\tanh (a+b x))^{n+4}}{b^4 (n+1) (n+2) (n+3) (n+4)}+\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^n,x]

[Out]

(x^3*ArcTanh[Tanh[a + b*x]]^(1 + n))/(b*(1 + n)) - (3*x^2*ArcTanh[Tanh[a + b*x]]^(2 + n))/(b^2*(1 + n)*(2 + n)
) + (6*x*ArcTanh[Tanh[a + b*x]]^(3 + n))/(b^3*(1 + n)*(2 + n)*(3 + n)) - (6*ArcTanh[Tanh[a + b*x]]^(4 + n))/(b
^4*(1 + n)*(2 + n)*(3 + n)*(4 + n))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(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] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int x^3 \tanh ^{-1}(\tanh (a+b x))^n \, dx &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{1+n} \, dx}{b (1+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 \int x \tanh ^{-1}(\tanh (a+b x))^{2+n} \, dx}{b^2 (1+n) (2+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \int \tanh ^{-1}(\tanh (a+b x))^{3+n} \, dx}{b^3 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \operatorname {Subst}\left (\int x^{3+n} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4 (1+n) (2+n) (3+n)}\\ &=\frac {x^3 \tanh ^{-1}(\tanh (a+b x))^{1+n}}{b (1+n)}-\frac {3 x^2 \tanh ^{-1}(\tanh (a+b x))^{2+n}}{b^2 (1+n) (2+n)}+\frac {6 x \tanh ^{-1}(\tanh (a+b x))^{3+n}}{b^3 (1+n) (2+n) (3+n)}-\frac {6 \tanh ^{-1}(\tanh (a+b x))^{4+n}}{b^4 (1+n) (2+n) (3+n) (4+n)}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 106, normalized size = 0.88 \[ \frac {\tanh ^{-1}(\tanh (a+b x))^{n+1} \left (-3 b^2 \left (n^2+7 n+12\right ) x^2 \tanh ^{-1}(\tanh (a+b x))+6 b (n+4) x \tanh ^{-1}(\tanh (a+b x))^2-6 \tanh ^{-1}(\tanh (a+b x))^3+b^3 \left (n^3+9 n^2+26 n+24\right ) x^3\right )}{b^4 (n+1) (n+2) (n+3) (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTanh[Tanh[a + b*x]]^n,x]

[Out]

(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))

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fricas [B]  time = 0.51, size = 255, normalized size = 2.11 \[ \frac {{\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \cosh \left (n \log \left (b x + a\right )\right ) + {\left (6 \, a^{3} b n x + {\left (b^{4} n^{3} + 6 \, b^{4} n^{2} + 11 \, b^{4} n + 6 \, b^{4}\right )} x^{4} - 6 \, a^{4} + {\left (a b^{3} n^{3} + 3 \, a b^{3} n^{2} + 2 \, a b^{3} n\right )} x^{3} - 3 \, {\left (a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x^{2}\right )} \sinh \left (n \log \left (b x + a\right )\right )}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="fricas")

[Out]

((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)

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giac [A]  time = 0.14, size = 226, normalized size = 1.87 \[ \frac {{\left (b x + a\right )}^{n} b^{4} n^{3} x^{4} + {\left (b x + a\right )}^{n} a b^{3} n^{3} x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} n^{2} x^{4} + 3 \, {\left (b x + a\right )}^{n} a b^{3} n^{2} x^{3} + 11 \, {\left (b x + a\right )}^{n} b^{4} n x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n^{2} x^{2} + 2 \, {\left (b x + a\right )}^{n} a b^{3} n x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} x^{4} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} n x^{2} + 6 \, {\left (b x + a\right )}^{n} a^{3} b n x - 6 \, {\left (b x + a\right )}^{n} a^{4}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="giac")

[Out]

((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)

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maple [B]  time = 0.11, size = 492, normalized size = 4.07 \[ \frac {x^{4} {\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{4+n}+\frac {n \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{3} {\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )}}{b \left (n^{2}+7 n +12\right )}-\frac {6 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{4}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {24 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {36 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {24 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {6 \,{\mathrm e}^{n \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right )} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {3 n \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right ) x^{2} {\mathrm e}^{n \ln \left (\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 (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) x \,{\mathrm e}^{n \ln \left (\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(tanh(b*x+a))^n,x)

[Out]

1/(4+n)*x^4*exp(n*ln(arctanh(tanh(b*x+a))))+n*(arctanh(tanh(b*x+a))-b*x)/b/(n^2+7*n+12)*x^3*exp(n*ln(arctanh(t
anh(b*x+a))))-6/b^4/(n^4+10*n^3+35*n^2+50*n+24)*exp(n*ln(arctanh(tanh(b*x+a))))*a^4-24/b^4/(n^4+10*n^3+35*n^2+
50*n+24)*exp(n*ln(arctanh(tanh(b*x+a))))*a^3*(arctanh(tanh(b*x+a))-b*x-a)-36/b^4/(n^4+10*n^3+35*n^2+50*n+24)*e
xp(n*ln(arctanh(tanh(b*x+a))))*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2-24/b^4/(n^4+10*n^3+35*n^2+50*n+24)*exp(n*ln(
arctanh(tanh(b*x+a))))*a*(arctanh(tanh(b*x+a))-b*x-a)^3-6/b^4/(n^4+10*n^3+35*n^2+50*n+24)*exp(n*ln(arctanh(tan
h(b*x+a))))*(arctanh(tanh(b*x+a))-b*x-a)^4-3*n/b^2*(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))
-b*x-a)^2)/(n^3+9*n^2+26*n+24)*x^2*exp(n*ln(arctanh(tanh(b*x+a))))+6*n*(a^3+3*a^2*(arctanh(tanh(b*x+a))-b*x-a)
+3*a*(arctanh(tanh(b*x+a))-b*x-a)^2+(arctanh(tanh(b*x+a))-b*x-a)^3)/b^3/(n^4+10*n^3+35*n^2+50*n+24)*x*exp(n*ln
(arctanh(tanh(b*x+a))))

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maxima [A]  time = 0.54, size = 101, normalized size = 0.83 \[ \frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(tanh(b*x+a))^n,x, algorithm="maxima")

[Out]

((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)

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mupad [B]  time = 1.42, size = 418, normalized size = 3.45 \[ -{\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}\right )}^n\,\left (\frac {3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}{8\,b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {3\,n\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{4\,b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {n\,x^3\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\,\left (n^2+3\,n+2\right )}{2\,b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {3\,n\,x^2\,\left (n+1\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(tanh(a + b*x))^n,x)

[Out]

-(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*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4)/(8*b^4*(50*n +
35*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*(l
og(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*(5
0*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))/(e
xp(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)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(tanh(b*x+a))**n,x)

[Out]

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/at
anh(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))*a
tanh(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**3*x**3*atanh(ta
nh(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 + 50*b**4*n + 24*b**4) - 6*atanh(tanh(a + b*x))**4*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), True))

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