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

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

[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))

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Rubi [A]  time = 0.0750075, antiderivative size = 121, normalized size of antiderivative = 1., 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 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]))

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 30

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

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*}

Mathematica [A]  time = 0.0767393, 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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Maple [B]  time = 0.046, size = 492, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

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 = 1.81333, size = 136, normalized size = 1.12 \begin{align*} \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}} \end{align*}

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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Fricas [B]  time = 2.16174, size = 536, normalized size = 4.43 \begin{align*} \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}} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.21447, size = 305, normalized size = 2.52 \begin{align*} \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}} \end{align*}

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)