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

Optimal. Leaf size=66 \[ \frac{b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x} \]

[Out]

(b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]
- Sqrt[ArcTanh[Tanh[a + b*x]]]/x

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Rubi [A]  time = 0.0315191, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 2161} \[ \frac{b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]
- Sqrt[ArcTanh[Tanh[a + b*x]]]/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 2161

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTan[Sqrt[v
]/Rt[(b*u - a*v)/a, 2]])/(a*Rt[(b*u - a*v)/a, 2]), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; Piecewis
eLinearQ[u, v, x]

Rubi steps

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

Mathematica [A]  time = 0.0371004, size = 65, normalized size = 0.98 \[ -\frac{b \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.126, size = 63, normalized size = 1. \begin{align*} 2\,b \left ( -1/2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{bx}}-1/2\,{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}{\it Artanh} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(arctanh(tanh(b*x + a)))/x^2, x)

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Fricas [A]  time = 1.77652, size = 225, normalized size = 3.41 \begin{align*} \left [\frac{\sqrt{a} b x \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \, \sqrt{b x + a} a}{2 \, a x}, \frac{\sqrt{-a} b x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - \sqrt{b x + a} a}{a x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*sqrt(b*x + a)*a)/(a*x), (sqrt(-a)*b*x*arcta
n(sqrt(b*x + a)*sqrt(-a)/a) - sqrt(b*x + a)*a)/(a*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18025, size = 55, normalized size = 0.83 \begin{align*} \frac{\frac{b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{b x + a} b}{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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