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

Optimal. Leaf size=27 \[ \frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{35} b x^{7/2} \]

[Out]

(-4*b*x^(7/2))/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5

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Rubi [A]  time = 0.0098923, 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} \[ \frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{35} b x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(-4*b*x^(7/2))/35 + (2*x^(5/2)*ArcTanh[Tanh[a + b*x]])/5

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 x^{3/2} \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))-\frac{1}{5} (2 b) \int x^{5/2} \, dx\\ &=-\frac{4}{35} b x^{7/2}+\frac{2}{5} x^{5/2} \tanh ^{-1}(\tanh (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0255402, size = 23, normalized size = 0.85 \[ \frac{2}{35} x^{5/2} \left (7 \tanh ^{-1}(\tanh (a+b x))-2 b x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(-2*b*x + 7*ArcTanh[Tanh[a + b*x]]))/35

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Maple [A]  time = 0.037, size = 20, normalized size = 0.7 \begin{align*} -{\frac{4\,b}{35}{x}^{{\frac{7}{2}}}}+{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{5}{x}^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/35*b*x^(7/2)+2/5*x^(5/2)*arctanh(tanh(b*x+a))

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Maxima [A]  time = 0.977613, size = 26, normalized size = 0.96 \begin{align*} -\frac{4}{35} \, b x^{\frac{7}{2}} + \frac{2}{5} \, x^{\frac{5}{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(x^(3/2)*arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-4/35*b*x^(7/2) + 2/5*x^(5/2)*arctanh(tanh(b*x + a))

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Fricas [A]  time = 1.99508, size = 46, normalized size = 1.7 \begin{align*} \frac{2}{35} \,{\left (5 \, b x^{3} + 7 \, a x^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b*x^3 + 7*a*x^2)*sqrt(x)

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Sympy [A]  time = 26.3986, size = 26, normalized size = 0.96 \begin{align*} - \frac{4 b x^{\frac{7}{2}}}{35} + \frac{2 x^{\frac{5}{2}} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*x**(7/2)/35 + 2*x**(5/2)*atanh(tanh(a + b*x))/5

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Giac [A]  time = 1.13407, size = 18, normalized size = 0.67 \begin{align*} \frac{2}{7} \, b x^{\frac{7}{2}} + \frac{2}{5} \, a x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b*x^(7/2) + 2/5*a*x^(5/2)