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

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

[Out]

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

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Rubi [A]  time = 0.0689825, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2159, 2161} \[ -\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x}+\frac{5}{3} b \tanh ^{-1}(\tanh (a+b x))^{3/2}-5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+5 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2161

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTan[Sqrt[v
]/Rt[(b*u - a*v)/a, 2]])/(a*Rt[(b*u - a*v)/a, 2]), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; Piecewis
eLinearQ[u, v, x]

Rubi steps

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

Mathematica [A]  time = 0.0530697, size = 106, normalized size = 0.96 \[ \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\frac{14}{3} b \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )-\frac{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{x}+\frac{2 b^2 x}{3}\right )-5 b \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5*b*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*
x]])^(3/2) + Sqrt[ArcTanh[Tanh[a + b*x]]]*((2*b^2*x)/3 + (14*b*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/3 - (-(b*x)
+ ArcTanh[Tanh[a + b*x]])^2/x)

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Maple [A]  time = 0.116, size = 193, normalized size = 1.8 \begin{align*} 2\,b \left ( 1/3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+2\,a\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+{\frac{ \left ( -1/2\,{a}^{2}-a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) -1/2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{bx}}-5/2\,{\frac{{a}^{2}+2\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) + \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}{\it Artanh} \left ({\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1134, size = 309, normalized size = 2.81 \begin{align*} \left [\frac{15 \, a^{\frac{3}{2}} b x \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt{b x + a}}{6 \, x}, \frac{15 \, \sqrt{-a} a b x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (2 \, b^{2} x^{2} + 14 \, a b x - 3 \, a^{2}\right )} \sqrt{b x + a}}{3 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a^(3/2)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*b^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x
+ a))/x, 1/3*(15*sqrt(-a)*a*b*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (2*b^2*x^2 + 14*a*b*x - 3*a^2)*sqrt(b*x + a
))/x]

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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))**(5/2)/x**2,x)

[Out]

Timed out

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Giac [A]  time = 1.14152, size = 120, normalized size = 1.09 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} a^{2} b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 2 \, \sqrt{2}{\left (b x + a\right )}^{\frac{3}{2}} b^{2} + 12 \, \sqrt{2} \sqrt{b x + a} a b^{2} - \frac{3 \, \sqrt{2} \sqrt{b x + a} a^{2} b}{x}\right )}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(2)*(15*sqrt(2)*a^2*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2*sqrt(2)*(b*x + a)^(3/2)*b^2 + 12*s
qrt(2)*sqrt(b*x + a)*a*b^2 - 3*sqrt(2)*sqrt(b*x + a)*a^2*b/x)/b