3.118 \(\int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac{b}{4 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.0716427, antiderivative size = 125, 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, 2163, 2161} \[ \frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac{b}{4 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^3,x]

[Out]

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

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{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx &=-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}+\frac{1}{4} b \int \frac{1}{x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{b}{4 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac{1}{8} b^2 \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac{b}{4 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}-\frac{b^2 \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{b}{4 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0970263, size = 89, normalized size = 0.71 \[ \frac{1}{4} \left (\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}+\frac{\left (\frac{b x}{b x-\tanh ^{-1}(\tanh (a+b x))}-2\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^3,x]

[Out]

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

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Maple [A]  time = 0.127, size = 92, normalized size = 0.7 \begin{align*} 2\,{b}^{2} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}-1/8\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }+1/8\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3/2}}{\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^3,x)

[Out]

2*b^2*((-1/8/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(3/2)-1/8*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2+1/8
/(arctanh(tanh(b*x+a))-b*x)^(3/2)*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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(arctanh(tanh(b*x + a)))/x^3, x)

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Fricas [A]  time = 1.72048, size = 292, normalized size = 2.34 \begin{align*} \left [\frac{\sqrt{a} b^{2} x^{2} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{8 \, a^{2} x^{2}}, -\frac{\sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{4 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/8*(sqrt(a)*b^2*x^2*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2
), -1/4*(sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2)]

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**3,x)

[Out]

Integral(sqrt(atanh(tanh(a + b*x)))/x**3, x)

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Giac [A]  time = 1.16984, size = 89, normalized size = 0.71 \begin{align*} -\frac{\frac{b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (b x + a\right )}^{\frac{3}{2}} b^{3} + \sqrt{b x + a} a b^{3}}{a b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^3,x, algorithm="giac")

[Out]

-1/4*(b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + ((b*x + a)^(3/2)*b^3 + sqrt(b*x + a)*a*b^3)/(a*b^2*x^2
))/b