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

Optimal. Leaf size=61 \[ \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{\sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0293383, antiderivative size = 61, 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 = {2169, 2165} \[ \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

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

Mathematica [A]  time = 0.0393929, size = 62, normalized size = 1.02 \[ \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right ) \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.069, size = 75, normalized size = 1.2 \begin{align*} \sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+{a\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a)\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a+1/b^(1/2)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15892, size = 251, normalized size = 4.11 \begin{align*} \left [\frac{a \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \, \sqrt{b x + a} b \sqrt{x}}{2 \, b}, -\frac{a \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) - \sqrt{b x + a} b \sqrt{x}}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}}{\sqrt{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14933, size = 49, normalized size = 0.8 \begin{align*} -\frac{a \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{\sqrt{b}} + \sqrt{b x + a} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/sqrt(b) + sqrt(b*x + a)*sqrt(x)