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

Optimal. Leaf size=25 \[ 2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{3} b x^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.007859, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2168, 30} \[ 2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{3} b x^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]/Sqrt[x],x]

[Out]

(-4*b*x^(3/2))/3 + 2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0189722, size = 23, normalized size = 0.92 \[ \frac{2}{3} \sqrt{x} \left (3 \tanh ^{-1}(\tanh (a+b x))-2 b x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 20, normalized size = 0.8 \begin{align*} -{\frac{4\,b}{3}{x}^{{\frac{3}{2}}}}+2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989077, size = 26, normalized size = 1.04 \begin{align*} -\frac{4}{3} \, b x^{\frac{3}{2}} + 2 \, \sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99092, size = 34, normalized size = 1.36 \begin{align*} \frac{2}{3} \,{\left (b x + 3 \, a\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x + 3*a)*sqrt(x)

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

[Out]

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

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Giac [A]  time = 1.13399, size = 18, normalized size = 0.72 \begin{align*} \frac{2}{3} \, b x^{\frac{3}{2}} + 2 \, a \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b*x^(3/2) + 2*a*sqrt(x)