3.71 \(\int x \tanh ^{-1}(\tanh (a+b x))^4 \, dx\)

Optimal. Leaf size=34 \[ \frac{x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2} \]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - ArcTanh[Tanh[a + b*x]]^6/(30*b^2)

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Rubi [A]  time = 0.0139257, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ \frac{x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

(x*ArcTanh[Tanh[a + b*x]]^5)/(5*b) - ArcTanh[Tanh[a + b*x]]^6/(30*b^2)

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 \tanh ^{-1}(\tanh (a+b x))^4 \, dx &=\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{\int \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{5 b}\\ &=\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{\operatorname{Subst}\left (\int x^5 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{5 b^2}\\ &=\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{5 b}-\frac{\tanh ^{-1}(\tanh (a+b x))^6}{30 b^2}\\ \end{align*}

Mathematica [B]  time = 0.0841169, size = 125, normalized size = 3.68 \[ -\frac{(a+b x) \left (-20 \left (2 a^2+a b x-b^2 x^2\right ) \tanh ^{-1}(\tanh (a+b x))^3+(5 a-b x) (a+b x)^4-6 (4 a-b x) (a+b x)^3 \tanh ^{-1}(\tanh (a+b x))+15 (3 a-b x) (a+b x)^2 \tanh ^{-1}(\tanh (a+b x))^2+15 (a-b x) \tanh ^{-1}(\tanh (a+b x))^4\right )}{30 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTanh[Tanh[a + b*x]]^4,x]

[Out]

-((a + b*x)*((5*a - b*x)*(a + b*x)^4 - 6*(4*a - b*x)*(a + b*x)^3*ArcTanh[Tanh[a + b*x]] + 15*(3*a - b*x)*(a +
b*x)^2*ArcTanh[Tanh[a + b*x]]^2 - 20*(2*a^2 + a*b*x - b^2*x^2)*ArcTanh[Tanh[a + b*x]]^3 + 15*(a - b*x)*ArcTanh
[Tanh[a + b*x]]^4))/(30*b^2)

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Maple [B]  time = 0.04, size = 74, normalized size = 2.2 \begin{align*}{\frac{{x}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{4}}{2}}-2\,b \left ( 1/3\,{x}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}-b \left ( 1/4\,{x}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}-1/2\,b \left ( 1/5\,{x}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -1/30\,{x}^{6}b \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a))^4,x)

[Out]

1/2*x^2*arctanh(tanh(b*x+a))^4-2*b*(1/3*x^3*arctanh(tanh(b*x+a))^3-b*(1/4*x^4*arctanh(tanh(b*x+a))^2-1/2*b*(1/
5*x^5*arctanh(tanh(b*x+a))-1/30*x^6*b)))

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Maxima [B]  time = 1.74207, size = 97, normalized size = 2.85 \begin{align*} -\frac{2}{3} \, b x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{4} + \frac{1}{30} \,{\left (15 \, b x^{4} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} +{\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^4,x, algorithm="maxima")

[Out]

-2/3*b*x^3*arctanh(tanh(b*x + a))^3 + 1/2*x^2*arctanh(tanh(b*x + a))^4 + 1/30*(15*b*x^4*arctanh(tanh(b*x + a))
^2 + (b^2*x^6 - 6*b*x^5*arctanh(tanh(b*x + a)))*b)*b

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Fricas [A]  time = 1.49553, size = 104, normalized size = 3.06 \begin{align*} \frac{1}{6} \, b^{4} x^{6} + \frac{4}{5} \, a b^{3} x^{5} + \frac{3}{2} \, a^{2} b^{2} x^{4} + \frac{4}{3} \, a^{3} b x^{3} + \frac{1}{2} \, a^{4} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^4,x, algorithm="fricas")

[Out]

1/6*b^4*x^6 + 4/5*a*b^3*x^5 + 3/2*a^2*b^2*x^4 + 4/3*a^3*b*x^3 + 1/2*a^4*x^2

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Sympy [B]  time = 2.47037, size = 76, normalized size = 2.24 \begin{align*} \frac{b^{4} x^{6}}{30} - \frac{b^{3} x^{5} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{5} + \frac{b^{2} x^{4} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{2} - \frac{2 b x^{3} \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3} + \frac{x^{2} \operatorname{atanh}^{4}{\left (\tanh{\left (a + b x \right )} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**4,x)

[Out]

b**4*x**6/30 - b**3*x**5*atanh(tanh(a + b*x))/5 + b**2*x**4*atanh(tanh(a + b*x))**2/2 - 2*b*x**3*atanh(tanh(a
+ b*x))**3/3 + x**2*atanh(tanh(a + b*x))**4/2

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Giac [A]  time = 1.10129, size = 62, normalized size = 1.82 \begin{align*} \frac{1}{6} \, b^{4} x^{6} + \frac{4}{5} \, a b^{3} x^{5} + \frac{3}{2} \, a^{2} b^{2} x^{4} + \frac{4}{3} \, a^{3} b x^{3} + \frac{1}{2} \, a^{4} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(tanh(b*x+a))^4,x, algorithm="giac")

[Out]

1/6*b^4*x^6 + 4/5*a*b^3*x^5 + 3/2*a^2*b^2*x^4 + 4/3*a^3*b*x^3 + 1/2*a^4*x^2