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

Optimal. Leaf size=63 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{3/2}}+\frac{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]

[Out]

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

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Rubi [A]  time = 0.0283892, antiderivative size = 63, 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} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{3/2}}+\frac{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0539003, size = 66, normalized size = 1.05 \[ \frac{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right ) \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.173, size = 80, normalized size = 1.3 \begin{align*}{\frac{1}{b}\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 ){b}^{-{\frac{3}{2}}}}-{({\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 ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b-1/b^(3/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a-1/b^(3/2)*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{x}}{\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(x^(1/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15033, size = 255, normalized size = 4.05 \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^{2}}, \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^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

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

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Giac [A]  time = 1.17339, size = 51, normalized size = 0.81 \begin{align*} \frac{a \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{3}{2}}} + \frac{\sqrt{b x + a} \sqrt{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(3/2) + sqrt(b*x + a)*sqrt(x)/b