3.182 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{x^{7/2}} \, dx\)

Optimal. Leaf size=48 \[ -\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac{16 b^2}{15 \sqrt{x}} \]

[Out]

(-16*b^2)/(15*Sqrt[x]) - (8*b*ArcTanh[Tanh[a + b*x]])/(15*x^(3/2)) - (2*ArcTanh[Tanh[a + b*x]]^2)/(5*x^(5/2))

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Rubi [A]  time = 0.023184, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}-\frac{16 b^2}{15 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^2/x^(7/2),x]

[Out]

(-16*b^2)/(15*Sqrt[x]) - (8*b*ArcTanh[Tanh[a + b*x]])/(15*x^(3/2)) - (2*ArcTanh[Tanh[a + b*x]]^2)/(5*x^(5/2))

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))^2}{x^{7/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac{1}{5} (4 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^{5/2}} \, dx\\ &=-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}+\frac{1}{15} \left (8 b^2\right ) \int \frac{1}{x^{3/2}} \, dx\\ &=-\frac{16 b^2}{15 \sqrt{x}}-\frac{8 b \tanh ^{-1}(\tanh (a+b x))}{15 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0428595, size = 40, normalized size = 0.83 \[ -\frac{2 \left (4 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^2/x^(7/2),x]

[Out]

(-2*(8*b^2*x^2 + 4*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2))

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Maple [A]  time = 0.043, size = 38, normalized size = 0.8 \begin{align*} -{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{5}{x}^{-{\frac{5}{2}}}}+{\frac{8\,b}{5} \left ( -{\frac{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{3}{x}^{-{\frac{3}{2}}}}-{\frac{2\,b}{3}{\frac{1}{\sqrt{x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^2/x^(7/2),x)

[Out]

-2/5*arctanh(tanh(b*x+a))^2/x^(5/2)+8/5*b*(-1/3*arctanh(tanh(b*x+a))/x^(3/2)-2/3*b/x^(1/2))

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Maxima [A]  time = 1.05763, size = 49, normalized size = 1.02 \begin{align*} -\frac{16 \, b^{2}}{15 \, \sqrt{x}} - \frac{8 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{15 \, x^{\frac{3}{2}}} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(7/2),x, algorithm="maxima")

[Out]

-16/15*b^2/sqrt(x) - 8/15*b*arctanh(tanh(b*x + a))/x^(3/2) - 2/5*arctanh(tanh(b*x + a))^2/x^(5/2)

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Fricas [A]  time = 2.08503, size = 63, normalized size = 1.31 \begin{align*} -\frac{2 \,{\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(7/2),x, algorithm="fricas")

[Out]

-2/15*(15*b^2*x^2 + 10*a*b*x + 3*a^2)/x^(5/2)

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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(atanh(tanh(b*x+a))**2/x**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14518, size = 32, normalized size = 0.67 \begin{align*} -\frac{2 \,{\left (15 \, b^{2} x^{2} + 10 \, a b x + 3 \, a^{2}\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^2/x^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*b^2*x^2 + 10*a*b*x + 3*a^2)/x^(5/2)