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

Optimal. Leaf size=146 \[ \frac{32 b^2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{16 b^2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{4 b}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0794707, antiderivative size = 146, 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} \[ \frac{32 b^2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{16 b^2 \sqrt{x}}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{2}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{4 b}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

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

Mathematica [A]  time = 0.058781, size = 79, normalized size = 0.54 \[ -\frac{2 \left (-9 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-9 b x \tanh ^{-1}(\tanh (a+b x))^2+\tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3\right )}{3 x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(-2*(b^3*x^3 - 9*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 9*b*x*ArcTanh[Tanh[a + b*x]]^2 + ArcTanh[Tanh[a + b*x]]^3))/
(3*x^(3/2)*ArcTanh[Tanh[a + b*x]]^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.118, size = 150, normalized size = 1. \begin{align*} -{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}{x}^{-{\frac{3}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{b}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx} \left ( -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}-4\,{\frac{b}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx} \left ( 1/3\,{\frac{\sqrt{x}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}+2/3\,{\frac{\sqrt{x}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)/arctanh(tanh(b*x+a))^(3/2)-4*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh
(tanh(b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(3/2)-4*b/(arctanh(tanh(b*x+a))-b*x)*(1/3*x^(1/2)/(arctanh(tan
h(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(3/2)+2/3/(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 = 1.51364, size = 76, normalized size = 0.52 \begin{align*} \frac{2 \,{\left (16 \, b^{4} x^{4} + 40 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x - a^{4}\right )}}{3 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{4} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/3*(16*b^4*x^4 + 40*a*b^3*x^3 + 30*a^2*b^2*x^2 + 5*a^3*b*x - a^4)/((b*x + a)^(5/2)*a^4*x^(3/2))

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Fricas [A]  time = 2.0808, size = 150, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + 6 \, a^{2} b x - a^{3}\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*(16*b^3*x^3 + 24*a*b^2*x^2 + 6*a^2*b*x - a^3)*sqrt(b*x + a)*sqrt(x)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^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**(5/2)/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.32203, size = 161, normalized size = 1.1 \begin{align*} \frac{2 \, \sqrt{x}{\left (\frac{8 \, b^{3} x}{a^{4}} + \frac{9 \, b^{2}}{a^{3}}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} - \frac{8 \,{\left (3 \, b^{\frac{3}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 9 \, a b^{\frac{3}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + 4 \, a^{2} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{3} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

2/3*sqrt(x)*(8*b^3*x/a^4 + 9*b^2/a^3)/(b*x + a)^(3/2) - 8/3*(3*b^(3/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 9
*a*b^(3/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + 4*a^2*b^(3/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^3*a
^3)