3.189 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ 16 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}}-\frac{32}{3} b^3 x^{3/2} \]

[Out]

(-32*b^3*x^(3/2))/3 + 16*b^2*Sqrt[x]*ArcTanh[Tanh[a + b*x]] - (4*b*ArcTanh[Tanh[a + b*x]]^2)/Sqrt[x] - (2*ArcT
anh[Tanh[a + b*x]]^3)/(3*x^(3/2))

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Rubi [A]  time = 0.0382775, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 30} \[ 16 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}}-\frac{32}{3} b^3 x^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b^3*x^(3/2))/3 + 16*b^2*Sqrt[x]*ArcTanh[Tanh[a + b*x]] - (4*b*ArcTanh[Tanh[a + b*x]]^2)/Sqrt[x] - (2*ArcT
anh[Tanh[a + b*x]]^3)/(3*x^(3/2))

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))^3}{x^{5/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}+(2 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{x^{3/2}} \, dx\\ &=-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}+\left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))}{\sqrt{x}} \, dx\\ &=16 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}-\left (16 b^3\right ) \int \sqrt{x} \, dx\\ &=-\frac{32}{3} b^3 x^{3/2}+16 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{\sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0294446, size = 55, normalized size = 0.85 \[ -\frac{2 \left (-24 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+6 b x \tanh ^{-1}(\tanh (a+b x))^2+\tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(16*b^3*x^3 - 24*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 6*b*x*ArcTanh[Tanh[a + b*x]]^2 + ArcTanh[Tanh[a + b*x]]^
3))/(3*x^(3/2))

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Maple [A]  time = 0.043, size = 55, normalized size = 0.9 \begin{align*} -{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{3}{x}^{-{\frac{3}{2}}}}+4\,b \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{\sqrt{x}}}+4\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \sqrt{x}-2/3\,b{x}^{3/2} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.06879, size = 74, normalized size = 1.14 \begin{align*} -\frac{4 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{\sqrt{x}} - \frac{16}{3} \,{\left (2 \, b^{2} x^{\frac{3}{2}} - 3 \, b \sqrt{x} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )\right )} b - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="maxima")

[Out]

-4*b*arctanh(tanh(b*x + a))^2/sqrt(x) - 16/3*(2*b^2*x^(3/2) - 3*b*sqrt(x)*arctanh(tanh(b*x + a)))*b - 2/3*arct
anh(tanh(b*x + a))^3/x^(3/2)

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Fricas [A]  time = 1.97198, size = 74, normalized size = 1.14 \begin{align*} \frac{2 \,{\left (b^{3} x^{3} + 9 \, a b^{2} x^{2} - 9 \, a^{2} b x - a^{3}\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="fricas")

[Out]

2/3*(b^3*x^3 + 9*a*b^2*x^2 - 9*a^2*b*x - a^3)/x^(3/2)

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Sympy [A]  time = 26.9452, size = 66, normalized size = 1.02 \begin{align*} - \frac{32 b^{3} x^{\frac{3}{2}}}{3} + 16 b^{2} \sqrt{x} \operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )} - \frac{4 b \operatorname{atanh}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}{\sqrt{x}} - \frac{2 \operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}{3 x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3/x**(5/2),x)

[Out]

-32*b**3*x**(3/2)/3 + 16*b**2*sqrt(x)*atanh(tanh(a + b*x)) - 4*b*atanh(tanh(a + b*x))**2/sqrt(x) - 2*atanh(tan
h(a + b*x))**3/(3*x**(3/2))

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Giac [A]  time = 1.16433, size = 46, normalized size = 0.71 \begin{align*} \frac{2}{3} \, b^{3} x^{\frac{3}{2}} + 6 \, a b^{2} \sqrt{x} - \frac{2 \,{\left (9 \, a^{2} b x + a^{3}\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(5/2),x, algorithm="giac")

[Out]

2/3*b^3*x^(3/2) + 6*a*b^2*sqrt(x) - 2/3*(9*a^2*b*x + a^3)/x^(3/2)