3.168 \(\int x^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\)

Optimal. Leaf size=27 \[ \frac{2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{63} b x^{9/2} \]

[Out]

(-4*b*x^(9/2))/63 + (2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/7

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Rubi [A]  time = 0.0090049, 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} \[ \frac{2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{4}{63} b x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(-4*b*x^(9/2))/63 + (2*x^(7/2)*ArcTanh[Tanh[a + b*x]])/7

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^{5/2} \tanh ^{-1}(\tanh (a+b x)) \, dx &=\frac{2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))-\frac{1}{7} (2 b) \int x^{7/2} \, dx\\ &=-\frac{4}{63} b x^{9/2}+\frac{2}{7} x^{7/2} \tanh ^{-1}(\tanh (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0260056, size = 23, normalized size = 0.85 \[ \frac{2}{63} x^{7/2} \left (9 \tanh ^{-1}(\tanh (a+b x))-2 b x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*ArcTanh[Tanh[a + b*x]],x]

[Out]

(2*x^(7/2)*(-2*b*x + 9*ArcTanh[Tanh[a + b*x]]))/63

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Maple [A]  time = 0.038, size = 20, normalized size = 0.7 \begin{align*} -{\frac{4\,b}{63}{x}^{{\frac{9}{2}}}}+{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{7}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*arctanh(tanh(b*x+a)),x)

[Out]

-4/63*b*x^(9/2)+2/7*x^(7/2)*arctanh(tanh(b*x+a))

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Maxima [A]  time = 0.982933, size = 26, normalized size = 0.96 \begin{align*} -\frac{4}{63} \, b x^{\frac{9}{2}} + \frac{2}{7} \, x^{\frac{7}{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^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="maxima")

[Out]

-4/63*b*x^(9/2) + 2/7*x^(7/2)*arctanh(tanh(b*x + a))

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Fricas [A]  time = 2.04117, size = 46, normalized size = 1.7 \begin{align*} \frac{2}{63} \,{\left (7 \, b x^{4} + 9 \, a x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="fricas")

[Out]

2/63*(7*b*x^4 + 9*a*x^3)*sqrt(x)

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Sympy [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(x**(5/2)*atanh(tanh(b*x+a)),x)

[Out]

Timed out

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Giac [A]  time = 1.12183, size = 18, normalized size = 0.67 \begin{align*} \frac{2}{9} \, b x^{\frac{9}{2}} + \frac{2}{7} \, a x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*arctanh(tanh(b*x+a)),x, algorithm="giac")

[Out]

2/9*b*x^(9/2) + 2/7*a*x^(7/2)