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

Optimal. Leaf size=38 \[ -\frac{4}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0140306, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ -\frac{4}{3 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{2 x}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0552531, size = 31, normalized size = 0.82 \[ -\frac{2 \left (2 \tanh ^{-1}(\tanh (a+b x))+b x\right )}{3 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 42, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{{b}^{2}} \left ( -{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-1/3\,{\frac{bx-{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{ \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/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.76718, size = 42, normalized size = 1.11 \begin{align*} -\frac{2 \,{\left (3 \, b^{2} x^{2} + 5 \, a b x + 2 \, a^{2}\right )}}{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*b^2*x^2 + 5*a*b*x + 2*a^2)/((b*x + a)^(5/2)*b^2)

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Fricas [A]  time = 2.05453, size = 89, normalized size = 2.34 \begin{align*} -\frac{2 \,{\left (3 \, b x + 2 \, a\right )} \sqrt{b x + a}}{3 \,{\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 85.0997, size = 51, normalized size = 1.34 \begin{align*} \begin{cases} - \frac{2 x}{3 b \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}} - \frac{4}{3 b^{2} \sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 \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/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Piecewise((-2*x/(3*b*atanh(tanh(a + b*x))**(3/2)) - 4/(3*b**2*sqrt(atanh(tanh(a + b*x)))), Ne(b, 0)), (x**2/(2
*atanh(tanh(a))**(5/2)), True))

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Giac [A]  time = 1.14186, size = 27, normalized size = 0.71 \begin{align*} -\frac{2 \,{\left (3 \, b x + 2 \, a\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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