3.160 \(\int \frac{x^2}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

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

[Out]

(-2*x^2)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (8*x)/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*Sqrt[ArcTanh[Ta
nh[a + b*x]]])/(3*b^3)

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Rubi [A]  time = 0.0296099, 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} \[ -\frac{8 x}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 b^3}-\frac{2 x^2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.036097, size = 48, normalized size = 0.81 \[ -\frac{2 \left (4 b x \tanh ^{-1}(\tanh (a+b x))-8 \tanh ^{-1}(\tanh (a+b x))^2+b^2 x^2\right )}{3 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*(b^2*x^2 + 4*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a + b*x]]^2))/(3*b^3*ArcTanh[Tanh[a + b*x]]^(3/2)
)

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Maple [A]  time = 0.043, size = 91, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{{b}^{3}} \left ( \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-{\frac{-2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +2\,bx}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-1/3\,{\frac{{a}^{2}+2\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) + \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.79707, size = 57, normalized size = 0.97 \begin{align*} \frac{2 \,{\left (3 \, b^{3} x^{3} + 15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 8 \, a^{3}\right )}}{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/3*(3*b^3*x^3 + 15*a*b^2*x^2 + 20*a^2*b*x + 8*a^3)/((b*x + a)^(5/2)*b^3)

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Fricas [A]  time = 1.95248, size = 111, normalized size = 1.88 \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt{b x + a}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*b^2*x^2 + 12*a*b*x + 8*a^2)*sqrt(b*x + a)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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Sympy [A]  time = 92.1434, size = 71, normalized size = 1.2 \begin{align*} \begin{cases} - \frac{2 x^{2}}{3 b \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{8 x}{3 b^{2} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}} + \frac{16 \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}{3 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3}}{3 \operatorname{atanh}^{\frac{5}{2}}{\left (\tanh{\left (a \right )} \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Piecewise((-2*x**2/(3*b*atanh(tanh(a + b*x))**(3/2)) - 8*x/(3*b**2*sqrt(atanh(tanh(a + b*x)))) + 16*sqrt(atanh
(tanh(a + b*x)))/(3*b**3), Ne(b, 0)), (x**3/(3*atanh(tanh(a))**(5/2)), True))

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Giac [A]  time = 1.16516, size = 53, normalized size = 0.9 \begin{align*} \frac{2 \, \sqrt{b x + a}}{b^{3}} + \frac{2 \,{\left (6 \,{\left (b x + a\right )} a - a^{2}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

2*sqrt(b*x + a)/b^3 + 2/3*(6*(b*x + a)*a - a^2)/((b*x + a)^(3/2)*b^3)