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

Optimal. Leaf size=59 \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

[Out]

(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(7/2))/(105*b^3)

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Rubi [A]  time = 0.0297581, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2157, 30} \[ \frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (8*x*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2) + (16*ArcTanh[Tanh[a
+ b*x]]^(7/2))/(105*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 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{4 \int x \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{8 \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{15 b^2}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{8 \operatorname{Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{15 b^3}\\ &=\frac{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac{8 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac{16 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{105 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0274241, size = 49, normalized size = 0.83 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-28 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+35 b^2 x^2\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(35*b^2*x^2 - 28*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(10
5*b^3)

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Maple [A]  time = 0.046, size = 69, normalized size = 1.2 \begin{align*} 2\,{\frac{1/7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}+1/5\, \left ( -2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}+1/3\, \left ( bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(tanh(b*x+a))^(1/2),x)

[Out]

2/b^3*(1/7*arctanh(tanh(b*x+a))^(7/2)+1/5*(-2*arctanh(tanh(b*x+a))+2*b*x)*arctanh(tanh(b*x+a))^(5/2)+1/3*(b*x-
arctanh(tanh(b*x+a)))^2*arctanh(tanh(b*x+a))^(3/2))

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Maxima [A]  time = 1.79401, size = 57, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)/b^3

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Fricas [A]  time = 1.542, size = 97, normalized size = 1.64 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*b^3*x^3 + 3*a*b^2*x^2 - 4*a^2*b*x + 8*a^3)*sqrt(b*x + a)/b^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(tanh(b*x+a))**(1/2),x)

[Out]

Integral(x**2*sqrt(atanh(tanh(a + b*x))), x)

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Giac [A]  time = 1.17487, size = 50, normalized size = 0.85 \begin{align*} \frac{2 \,{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2}\right )}}{105 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

2/105*(15*(b*x + a)^(7/2) - 42*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2)/b^3