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

Optimal. Leaf size=101 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 \sqrt{b}}-\frac{3}{4} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{2} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]

[Out]

(3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/(4*Sqrt[b]) - (3*
Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(3/2)
)/2

________________________________________________________________________________________

Rubi [A]  time = 0.0476427, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 \sqrt{b}}-\frac{3}{4} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{2} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/(4*Sqrt[b]) - (3*
Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(3/2)
)/2

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{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{\sqrt{x}} \, dx &=\frac{1}{2} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{1}{4} \left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \, dx\\ &=-\frac{3}{4} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{1}{2} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{8} \left (3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac{3 \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 )^2}{4 \sqrt{b}}-\frac{3}{4} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{1}{2} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}\\ \end{align*}

Mathematica [A]  time = 0.0576044, size = 83, normalized size = 0.82 \[ \frac{1}{4} \left (\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (5 \tanh ^{-1}(\tanh (a+b x))-3 b x\right )+\frac{3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{\sqrt{b}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-3*b*x + 5*ArcTanh[Tanh[a + b*x]]) + (3*(-(b*x) + ArcTanh[Tanh[a + b*x]
])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/Sqrt[b])/4

________________________________________________________________________________________

Maple [B]  time = 0.042, size = 165, normalized size = 1.6 \begin{align*}{\frac{1}{2}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{3\,a}{4}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{3\,{a}^{2}}{4}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{\frac{3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{2}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{\frac{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx-3\,a}{4}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{4}\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))^(3/2)/x^(1/2),x)

[Out]

1/2*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+3/4*a*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+3/4/b^(1/2)*ln(b^(1/2)*x^(1/2)
+arctanh(tanh(b*x+a))^(1/2))*a^2+3/2*a/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)+3/4*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+3/4/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)^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.27321, size = 311, normalized size = 3.08 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{2} x + 5 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{4 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*a^2*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x + 5*a*b)*sqrt(b*x + a)*sqrt(
x))/b, -1/4*(3*a^2*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x + 5*a*b)*sqrt(b*x + a)*sqrt(
x))/b]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{\frac{3}{2}}{\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))**(3/2)/x**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out