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

Optimal. Leaf size=97 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt{b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{1}{b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]/(Sqrt[b]*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2)) -
1/(b*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(b*Sqrt[x]*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.0537793, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt{b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{1}{b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2),x]

[Out]

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

Mathematica [A]  time = 0.0603814, size = 80, normalized size = 0.82 \[ \frac{\sqrt{x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt{b} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2),x]

[Out]

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

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Maple [A]  time = 0.132, size = 82, normalized size = 0.9 \begin{align*}{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ){\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}+{\frac{1}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}\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(1/arctanh(tanh(b*x+a))^2/x^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.05718, size = 274, normalized size = 2.82 \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^{2} b^{2} x + a^{3} b\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^{2} b^{2} x + a^{3} b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arctanh(tanh(b*x+a))^2/x^(1/2),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^2*b^2*x + a^3*b
), (a*b*sqrt(x) - sqrt(a*b)*(b*x + a)*arctan(sqrt(a*b)/(b*sqrt(x))))/(a^2*b^2*x + a^3*b)]

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

[Out]

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

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Giac [A]  time = 1.18221, size = 47, normalized size = 0.48 \begin{align*} \frac{\arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a} + \frac{\sqrt{x}}{{\left (b x + a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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