3.190 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{32 b^3 \sqrt{x}}{5} \]

[Out]

(32*b^3*Sqrt[x])/5 - (16*b^2*ArcTanh[Tanh[a + b*x]])/(5*Sqrt[x]) - (4*b*ArcTanh[Tanh[a + b*x]]^2)/(5*x^(3/2))
- (2*ArcTanh[Tanh[a + b*x]]^3)/(5*x^(5/2))

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Rubi [A]  time = 0.0376807, antiderivative size = 69, 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} \[ -\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{32 b^3 \sqrt{x}}{5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]

[Out]

(32*b^3*Sqrt[x])/5 - (16*b^2*ArcTanh[Tanh[a + b*x]])/(5*Sqrt[x]) - (4*b*ArcTanh[Tanh[a + b*x]]^2)/(5*x^(3/2))
- (2*ArcTanh[Tanh[a + b*x]]^3)/(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))^3}{x^{7/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} (6 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^2}{x^{5/2}} \, dx\\ &=-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} \left (8 b^2\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\\ &=-\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac{1}{5} \left (16 b^3\right ) \int \frac{1}{\sqrt{x}} \, dx\\ &=\frac{32 b^3 \sqrt{x}}{5}-\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt{x}}-\frac{4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0308873, size = 57, normalized size = 0.83 \[ \frac{2 \left (-8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-2 b x \tanh ^{-1}(\tanh (a+b x))^2-\tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]

[Out]

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

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Maple [A]  time = 0.044, size = 56, normalized size = 0.8 \begin{align*} -{\frac{2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3}}{5}{x}^{-{\frac{5}{2}}}}+{\frac{12\,b}{5} \left ( -{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{3}{x}^{-{\frac{3}{2}}}}+{\frac{4\,b}{3} \left ( -{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ){\frac{1}{\sqrt{x}}}}+2\,b\sqrt{x} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^(7/2),x)

[Out]

-2/5*arctanh(tanh(b*x+a))^3/x^(5/2)+12/5*b*(-1/3*arctanh(tanh(b*x+a))^2/x^(3/2)+4/3*b*(-arctanh(tanh(b*x+a))/x
^(1/2)+2*b*x^(1/2)))

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Maxima [A]  time = 1.08763, size = 74, normalized size = 1.07 \begin{align*} \frac{16}{5} \,{\left (2 \, b^{2} \sqrt{x} - \frac{b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt{x}}\right )} b - \frac{4 \, b \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac{3}{2}}} - \frac{2 \, \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="maxima")

[Out]

16/5*(2*b^2*sqrt(x) - b*arctanh(tanh(b*x + a))/sqrt(x))*b - 4/5*b*arctanh(tanh(b*x + a))^2/x^(3/2) - 2/5*arcta
nh(tanh(b*x + a))^3/x^(5/2)

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Fricas [A]  time = 1.96389, size = 78, normalized size = 1.13 \begin{align*} \frac{2 \,{\left (5 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 5 \, a^{2} b x - a^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="fricas")

[Out]

2/5*(5*b^3*x^3 - 15*a*b^2*x^2 - 5*a^2*b*x - a^3)/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))**3/x**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.12904, size = 46, normalized size = 0.67 \begin{align*} 2 \, b^{3} \sqrt{x} - \frac{2 \,{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x + a^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="giac")

[Out]

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