3.221 \(\int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{9/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

(16*b^2*ArcTanh[Tanh[a + b*x]]^(3/2))/(105*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*ArcTanh[Tanh[a + b
*x]]^(3/2))/(35*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(7*x^(7/2)*(b*x -
ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0543659, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(9/2),x]

[Out]

(16*b^2*ArcTanh[Tanh[a + b*x]]^(3/2))/(105*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*ArcTanh[Tanh[a + b
*x]]^(3/2))/(35*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(7*x^(7/2)*(b*x -
ArcTanh[Tanh[a + b*x]]))

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{9/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(4 b) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (8 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.056038, size = 66, normalized size = 0.6 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-42 b x \tanh ^{-1}(\tanh (a+b x))+15 \tanh ^{-1}(\tanh (a+b x))^2+35 b^2 x^2\right )}{105 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(9/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(35*b^2*x^2 - 42*b*x*ArcTanh[Tanh[a + b*x]] + 15*ArcTanh[Tanh[a + b*x]]^2))/(1
05*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.143, size = 105, normalized size = 1. \begin{align*} -{\frac{2}{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -7\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{7}{2}}}}-{\frac{8\,b}{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -7\,bx} \left ( -{\frac{1}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{5}{2}}}}+{\frac{2\,b}{15\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x)

[Out]

-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(3/2)-8/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(arc
tanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+2/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)*arctanh(
tanh(b*x+a))^(3/2))

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Maxima [A]  time = 1.4809, size = 61, normalized size = 0.55 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt{b x + a}}{105 \, a^{3} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="maxima")

[Out]

-2/105*(8*b^3*x^3 - 4*a*b^2*x^2 + 3*a^2*b*x + 15*a^3)*sqrt(b*x + a)/(a^3*x^(7/2))

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Fricas [A]  time = 2.25493, size = 112, normalized size = 1.02 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt{b x + a}}{105 \, a^{3} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(8*b^3*x^3 - 4*a*b^2*x^2 + 3*a^2*b*x + 15*a^3)*sqrt(b*x + a)/(a^3*x^(7/2))

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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))**(1/2)/x**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18521, size = 186, normalized size = 1.69 \begin{align*} \frac{32 \,{\left (70 \, b^{\frac{7}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{8} + 35 \, a b^{\frac{7}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{6} + 21 \, a^{2} b^{\frac{7}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 7 \, a^{3} b^{\frac{7}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + a^{4} b^{\frac{7}{2}}\right )}}{105 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(9/2),x, algorithm="giac")

[Out]

32/105*(70*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 + 35*a*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 + 21
*a^2*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 7*a^3*b^(7/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^4*b^(
7/2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7