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

Optimal. Leaf size=72 \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0316608, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{4 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

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

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0405726, size = 46, normalized size = 0.64 \[ -\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-3 b x\right )}{3 x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

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

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Maple [A]  time = 0.168, size = 59, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{3}{2}}}}+{\frac{4\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.48287, size = 45, normalized size = 0.62 \begin{align*} \frac{2 \,{\left (2 \, b^{2} x^{2} + a b x - a^{2}\right )}}{3 \, \sqrt{b x + a} a^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.12291, size = 61, normalized size = 0.85 \begin{align*} \frac{2 \,{\left (2 \, b x - a\right )} \sqrt{b x + a}}{3 \, a^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/atanh(tanh(b*x+a))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.15603, size = 74, normalized size = 1.03 \begin{align*} \frac{8 \,{\left (3 \,{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )} b^{\frac{3}{2}}}{3 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

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