3.56 \(\int x^2 \tanh ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=53 \[ \frac{\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

[Out]

(x^2*ArcTanh[Tanh[a + b*x]]^4)/(4*b) - (x*ArcTanh[Tanh[a + b*x]]^5)/(10*b^2) + ArcTanh[Tanh[a + b*x]]^6/(60*b^
3)

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Rubi [A]  time = 0.0286999, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac{\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3}-\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(x^2*ArcTanh[Tanh[a + b*x]]^4)/(4*b) - (x*ArcTanh[Tanh[a + b*x]]^5)/(10*b^2) + ArcTanh[Tanh[a + b*x]]^6/(60*b^
3)

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^2 \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{\int x \tanh ^{-1}(\tanh (a+b x))^4 \, dx}{2 b}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\int \tanh ^{-1}(\tanh (a+b x))^5 \, dx}{10 b^2}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\operatorname{Subst}\left (\int x^5 \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{10 b^3}\\ &=\frac{x^2 \tanh ^{-1}(\tanh (a+b x))^4}{4 b}-\frac{x \tanh ^{-1}(\tanh (a+b x))^5}{10 b^2}+\frac{\tanh ^{-1}(\tanh (a+b x))^6}{60 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0220387, size = 54, normalized size = 1.02 \[ -\frac{1}{60} x^3 \left (-6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+15 b x \tanh ^{-1}(\tanh (a+b x))^2-20 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

-(x^3*(b^3*x^3 - 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 15*b*x*ArcTanh[Tanh[a + b*x]]^2 - 20*ArcTanh[Tanh[a + b*x]
]^3))/60

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Maple [A]  time = 0.036, size = 56, normalized size = 1.1 \begin{align*}{\frac{{x}^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{3}}-b \left ({\frac{{x}^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{4}}-{\frac{b}{2} \left ({\frac{{x}^{5}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{5}}-{\frac{{x}^{6}b}{30}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(tanh(b*x+a))^3,x)

[Out]

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 [A]  time = 1.56134, size = 73, normalized size = 1.38 \begin{align*} -\frac{1}{4} \, b x^{4} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac{1}{3} \, x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac{1}{60} \,{\left (b^{2} x^{6} - 6 \, b x^{5} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-1/4*b*x^4*arctanh(tanh(b*x + a))^2 + 1/3*x^3*arctanh(tanh(b*x + a))^3 - 1/60*(b^2*x^6 - 6*b*x^5*arctanh(tanh(
b*x + a)))*b

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Fricas [A]  time = 1.4896, size = 80, normalized size = 1.51 \begin{align*} \frac{1}{6} \, b^{3} x^{6} + \frac{3}{5} \, a b^{2} x^{5} + \frac{3}{4} \, a^{2} b x^{4} + \frac{1}{3} \, a^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.11896, size = 56, normalized size = 1.06 \begin{align*} - \frac{b^{3} x^{6}}{60} + \frac{b^{2} x^{5} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{10} - \frac{b x^{4} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{4} + \frac{x^{3} \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(tanh(b*x+a))**3,x)

[Out]

-b**3*x**6/60 + b**2*x**5*atanh(tanh(a + b*x))/10 - b*x**4*atanh(tanh(a + b*x))**2/4 + x**3*atanh(tanh(a + b*x
))**3/3

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Giac [A]  time = 1.15411, size = 47, normalized size = 0.89 \begin{align*} \frac{1}{6} \, b^{3} x^{6} + \frac{3}{5} \, a b^{2} x^{5} + \frac{3}{4} \, a^{2} b x^{4} + \frac{1}{3} \, a^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

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