3.218 \(\int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{3/2}} \, dx\)

Optimal. Leaf size=49 \[ 2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0285149, antiderivative size = 49, 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 = {2168, 2165} \[ 2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]] - (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/Sqrt[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 2165

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

Rubi steps

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

Mathematica [A]  time = 0.0381369, size = 52, normalized size = 1.06 \[ 2 \sqrt{b} \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )-\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.144, size = 149, normalized size = 3. \begin{align*} -2\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \sqrt{x}}}+2\,{\frac{b\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}+2\,{\frac{\sqrt{b}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) a}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}+2\,{\frac{\sqrt{b}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06312, size = 243, normalized size = 4.96 \begin{align*} \left [\frac{\sqrt{b} x \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \, \sqrt{b x + a} \sqrt{x}}{x}, -\frac{2 \,{\left (\sqrt{-b} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) + \sqrt{b x + a} \sqrt{x}\right )}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*sqrt(b*x + a)*sqrt(x))/x, -2*(sqrt(-b)*x*arct
an(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + sqrt(b*x + a)*sqrt(x))/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^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(atanh(tanh(a + b*x)))/x**(3/2), x)

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Giac [A]  time = 1.16806, size = 77, normalized size = 1.57 \begin{align*} -\sqrt{b} \log \left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2}\right ) + \frac{4 \, a \sqrt{b}}{{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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