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

Optimal. Leaf size=27 \[ \frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{15} b x^{5/2} \]

[Out]

(-4*b*x^(5/2))/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]])/3

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Rubi [A]  time = 0.0080778, antiderivative size = 27, 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} \[ \frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{15} b x^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*ArcTanh[Tanh[a + b*x]],x]

[Out]

(-4*b*x^(5/2))/15 + (2*x^(3/2)*ArcTanh[Tanh[a + b*x]])/3

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 \sqrt{x} \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))-\frac{1}{3} (2 b) \int x^{3/2} \, dx\\ &=-\frac{4}{15} b x^{5/2}+\frac{2}{3} x^{3/2} \tanh ^{-1}(\tanh (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0259446, size = 23, normalized size = 0.85 \[ \frac{2}{15} x^{3/2} \left (5 \tanh ^{-1}(\tanh (a+b x))-2 b x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(-2*b*x + 5*ArcTanh[Tanh[a + b*x]]))/15

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Maple [A]  time = 0.036, size = 20, normalized size = 0.7 \begin{align*} -{\frac{4\,b}{15}{x}^{{\frac{5}{2}}}}+{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*b*x^(5/2)+2/3*x^(3/2)*arctanh(tanh(b*x+a))

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Maxima [A]  time = 0.988791, size = 26, normalized size = 0.96 \begin{align*} -\frac{4}{15} \, b x^{\frac{5}{2}} + \frac{2}{3} \, x^{\frac{3}{2}} \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]

-4/15*b*x^(5/2) + 2/3*x^(3/2)*arctanh(tanh(b*x + a))

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Fricas [A]  time = 1.98817, size = 43, normalized size = 1.59 \begin{align*} \frac{2}{15} \,{\left (3 \, b x^{2} + 5 \, a x\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]

2/15*(3*b*x^2 + 5*a*x)*sqrt(x)

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

[Out]

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

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Giac [A]  time = 1.11668, size = 18, normalized size = 0.67 \begin{align*} \frac{2}{5} \, b x^{\frac{5}{2}} + \frac{2}{3} \, a x^{\frac{3}{2}} \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]

2/5*b*x^(5/2) + 2/3*a*x^(3/2)