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

Optimal. Leaf size=42 \[ -\frac{1}{10} b x^5 \tanh ^{-1}(\tanh (a+b x))+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^2+\frac{b^2 x^6}{60} \]

[Out]

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

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Rubi [A]  time = 0.0234797, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ -\frac{1}{10} b x^5 \tanh ^{-1}(\tanh (a+b x))+\frac{1}{4} x^4 \tanh ^{-1}(\tanh (a+b x))^2+\frac{b^2 x^6}{60} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0295993, size = 37, normalized size = 0.88 \[ \frac{1}{60} x^4 \left (-6 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(b^2*x^2 - 6*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/60

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Maple [A]  time = 0.036, size = 38, normalized size = 0.9 \begin{align*}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 = 1.34876, size = 49, normalized size = 1.17 \begin{align*} \frac{1}{60} \, b^{2} x^{6} - \frac{1}{10} \, b x^{5} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) + \frac{1}{4} \, x^{4} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*b^2*x^6 - 1/10*b*x^5*arctanh(tanh(b*x + a)) + 1/4*x^4*arctanh(tanh(b*x + a))^2

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Fricas [A]  time = 1.95857, size = 55, normalized size = 1.31 \begin{align*} \frac{1}{6} \, b^{2} x^{6} + \frac{2}{5} \, a b x^{5} + \frac{1}{4} \, a^{2} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.62917, size = 37, normalized size = 0.88 \begin{align*} \frac{b^{2} x^{6}}{60} - \frac{b x^{5} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{10} + \frac{x^{4} \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15066, size = 32, normalized size = 0.76 \begin{align*} \frac{1}{6} \, b^{2} x^{6} + \frac{2}{5} \, a b x^{5} + \frac{1}{4} \, a^{2} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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