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

Optimal. Leaf size=94 \[ \frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

[Out]

(b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/
2) - 1/(x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b/((b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.0528121, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(3/
2) - 1/(x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b/((b*x - ArcTanh[Tanh[a + b*x]])*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 2163

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

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

Mathematica [A]  time = 0.0559497, size = 78, normalized size = 0.83 \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.138, size = 95, normalized size = 1. \begin{align*} 2\,b \left ( 2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) bx}}-2\,{\frac{1}{ \left ( -4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) +4\,bx \right ) \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(1/x^2/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

2*b*(2*arctanh(tanh(b*x+a))^(1/2)/(-4*arctanh(tanh(b*x+a))+4*b*x)/b/x-2/(-4*arctanh(tanh(b*x+a))+4*b*x)/(arcta
nh(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{1}{x^{2} \sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15636, size = 232, normalized size = 2.47 \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^{2} x}, -\frac{\sqrt{-a} b x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + \sqrt{b x + a} a}{a^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/arctanh(tanh(b*x+a))^(1/2),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^2*x), -(sqrt(-a)*b*x*ar
ctan(sqrt(b*x + a)*sqrt(-a)/a) + sqrt(b*x + a)*a)/(a^2*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{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(1/x**2/atanh(tanh(b*x+a))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.14033, size = 63, normalized size = 0.67 \begin{align*} -\frac{\frac{b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{\sqrt{b x + a} b}{a x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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