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

Optimal. Leaf size=148 \[ \frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{32 b^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{12 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

(32*b^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*Sqrt[ArcTanh[Tan
h[a + b*x]]])/(35*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (12*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*x^(5/2)*
(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(7*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])
)

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Rubi [A]  time = 0.0830278, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{32 b^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{12 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(9/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(32*b^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*Sqrt[ArcTanh[Tan
h[a + b*x]]])/(35*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (12*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*x^(5/2)*
(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(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{1}{x^{9/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(6 b) \int \frac{1}{x^{7/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{12 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (24 b^2\right ) \int \frac{1}{x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{12 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (16 b^3\right ) \int \frac{1}{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac{32 b^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{12 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0522516, size = 82, normalized size = 0.55 \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-35 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+21 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+35 b^3 x^3\right )}{35 x^{7/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(9/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(35*b^3*x^3 - 35*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 21*b*x*ArcTanh[Tanh[a + b*x]
]^2 - 5*ArcTanh[Tanh[a + b*x]]^3))/(35*x^(7/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.169, size = 151, normalized size = 1. \begin{align*} -{\frac{2}{7\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -7\,bx}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{7}{2}}}}-{\frac{12\,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}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{5}{2}}}}-{\frac{4\,b}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx} \left ( -{\frac{1}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{3}{2}}}}+{\frac{2\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{\frac{1}{\sqrt{x}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(1/2)-12/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(ar
ctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-4/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(arctanh(tan
h(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(1/2)+2/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*arctanh(tanh(b*x+
a))^(1/2)))

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Maxima [A]  time = 1.50141, size = 74, normalized size = 0.5 \begin{align*} \frac{2 \,{\left (16 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} + a^{3} b x - 5 \, a^{4}\right )}}{35 \, \sqrt{b x + a} a^{4} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(16*b^4*x^4 + 8*a*b^3*x^3 - 2*a^2*b^2*x^2 + a^3*b*x - 5*a^4)/(sqrt(b*x + a)*a^4*x^(7/2))

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Fricas [A]  time = 2.09876, size = 109, normalized size = 0.74 \begin{align*} \frac{2 \,{\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt{b x + a}}{35 \, a^{4} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x + a)/(a^4*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(1/x**(9/2)/atanh(tanh(b*x+a))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17013, size = 139, normalized size = 0.94 \begin{align*} \frac{64 \,{\left (35 \,{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{6} - 21 \, a{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} + 7 \, a^{2}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a^{3}\right )} b^{\frac{7}{2}}}{35 \,{\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(1/x^(9/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

64/35*(35*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 - 21*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 7*a^2*(sqrt(b)*sqrt
(x) - sqrt(b*x + a))^2 - a^3)*b^(7/2)/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7