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

Optimal. Leaf size=121 \[ 2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{2}{3} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0702375, antiderivative size = 121, 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 = {2159, 2161} \[ 2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{2}{3} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{2}{5} \tanh ^{-1}(\tanh (a+b x))^{5/2}-2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x,x]

[Out]

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

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

Mathematica [A]  time = 0.0699247, size = 99, normalized size = 0.82 \[ \frac{2}{15} \left (15 b^2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}+23 \tanh ^{-1}(\tanh (a+b x))^{5/2}-35 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}-15 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] - 35*b*x*ArcTanh[Tanh[a + b*x]]^(3/2) + 23*ArcTanh[Tanh[a + b*x]]^
(5/2) - 15*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(5/2)))/15

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Maple [B]  time = 0.109, size = 222, normalized size = 1.8 \begin{align*}{\frac{2}{5} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,a}{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx-2\,a}{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+2\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{a}^{2}+4\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+2\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }-2\,{\frac{{a}^{3}+3\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}+ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctanh(tanh(b*x+a))^(5/2)+2/3*arctanh(tanh(b*x+a))^(3/2)*a+2/3*arctanh(tanh(b*x+a))^(3/2)*(arctanh(tanh(b
*x+a))-b*x-a)+2*arctanh(tanh(b*x+a))^(1/2)*a^2+4*a*(arctanh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)+2*(
arctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/2)-2*(a^3+3*a^2*(arctanh(tanh(b*x+a))-b*x-a)+3*a*(arctan
h(tanh(b*x+a))-b*x-a)^2+(arctanh(tanh(b*x+a))-b*x-a)^3)/(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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0917, size = 288, normalized size = 2.38 \begin{align*} \left [a^{\frac{5}{2}} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}, 2 \, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) + \frac{2}{15} \,{\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt{b x + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[a^(5/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/15*(3*b^2*x^2 + 11*a*b*x + 23*a^2)*sqrt(b*x + a), 2*
sqrt(-a)*a^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + 2/15*(3*b^2*x^2 + 11*a*b*x + 23*a^2)*sqrt(b*x + a)]

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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,x)

[Out]

Timed out

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Giac [A]  time = 1.17054, size = 99, normalized size = 0.82 \begin{align*} \frac{1}{15} \, \sqrt{2}{\left (\frac{15 \, \sqrt{2} a^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 3 \, \sqrt{2}{\left (b x + a\right )}^{\frac{5}{2}} + 5 \, \sqrt{2}{\left (b x + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{2} \sqrt{b x + a} a^{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*(15*sqrt(2)*a^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 3*sqrt(2)*(b*x + a)^(5/2) + 5*sqrt(2)*(
b*x + a)^(3/2)*a + 15*sqrt(2)*sqrt(b*x + a)*a^2)