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

Optimal. Leaf size=73 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2} \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

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Rubi [A]  time = 0.0341577, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 2162} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{3/2} \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]/(b^(3/2)*Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]))
- Sqrt[x]/(b*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 2162

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

Rubi steps

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

Mathematica [A]  time = 0.0598, size = 70, normalized size = 0.96 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{\sqrt{x}}{b \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.135, size = 61, normalized size = 0.8 \begin{align*} -{\frac{1}{b{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}+{\frac{1}{b}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12579, size = 277, normalized size = 3.79 \begin{align*} \left [-\frac{2 \, a b \sqrt{x} + \sqrt{-a b}{\left (b x + a\right )} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right )}{2 \,{\left (a b^{3} x + a^{2} b^{2}\right )}}, -\frac{a b \sqrt{x} + \sqrt{a b}{\left (b x + a\right )} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right )}{a b^{3} x + a^{2} 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))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*a*b*sqrt(x) + sqrt(-a*b)*(b*x + a)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)))/(a*b^3*x + a^2*b^
2), -(a*b*sqrt(x) + sqrt(a*b)*(b*x + a)*arctan(sqrt(a*b)/(b*sqrt(x))))/(a*b^3*x + a^2*b^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\operatorname{atanh}^{2}{\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))**2,x)

[Out]

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

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Giac [A]  time = 1.13166, size = 49, normalized size = 0.67 \begin{align*} \frac{\arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b} - \frac{\sqrt{x}}{{\left (b x + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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