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

Optimal. Leaf size=167 \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}+\frac{5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{64 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^4 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4} \]

[Out]

(5*b^4*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(64*(b*x - ArcTanh[Tanh[a + b*
x]])^(3/2)) - (5*b^3)/(64*x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^4)/(64*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[Ar
cTanh[Tanh[a + b*x]]]) - (5*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(32*x^2) - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(2
4*x^3) - ArcTanh[Tanh[a + b*x]]^(5/2)/(4*x^4)

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Rubi [A]  time = 0.116921, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}+\frac{5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^3}{64 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^4 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^4*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(64*(b*x - ArcTanh[Tanh[a + b*
x]])^(3/2)) - (5*b^3)/(64*x*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*b^4)/(64*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[Ar
cTanh[Tanh[a + b*x]]]) - (5*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(32*x^2) - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(2
4*x^3) - ArcTanh[Tanh[a + b*x]]^(5/2)/(4*x^4)

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 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[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^5} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac{1}{8} (5 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^4} \, dx\\ &=-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac{1}{16} \left (5 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^3} \, dx\\ &=-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}+\frac{1}{64} \left (5 b^3\right ) \int \frac{1}{x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{5 b^3}{64 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac{1}{128} \left (5 b^4\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=-\frac{5 b^3}{64 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}-\frac{\left (5 b^4\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{128 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5 b^4 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{5 b^3}{64 x \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^4}{64 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{32 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{24 x^3}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.11106, size = 134, normalized size = 0.8 \[ \frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{64 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (10 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+8 b x \tanh ^{-1}(\tanh (a+b x))^2-48 \tanh ^{-1}(\tanh (a+b x))^3+15 b^3 x^3\right )}{192 x^4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^4*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(64*(-(b*x) + ArcTanh[Tanh
[a + b*x]])^(3/2)) - (Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^3*x^3 + 10*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 8*b*x*Arc
Tanh[Tanh[a + b*x]]^2 - 48*ArcTanh[Tanh[a + b*x]]^3))/(192*x^4*(b*x - ArcTanh[Tanh[a + b*x]]))

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Maple [A]  time = 0.118, size = 169, normalized size = 1. \begin{align*} 2\,{b}^{4} \left ({\frac{1}{{x}^{4}{b}^{4}} \left ( -{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{7/2}}{128\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -128\,bx}}-{\frac{73\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{384}}+ \left ({\frac{55\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{384}}-{\frac{55\,bx}{384}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( -{\frac{5\,{a}^{2}}{128}}-{\frac{5\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{64}}-{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{128}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }+{\frac{5}{128\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3/2}}{\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^5,x)

[Out]

2*b^4*((-5/128/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(7/2)-73/384*arctanh(tanh(b*x+a))^(5/2)+(55/384
*arctanh(tanh(b*x+a))-55/384*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-5/128*a^2-5/64*a*(arctanh(tanh(b*x+a))-b*x-a)-5
/128*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^4/x^4+5/128/(arctanh(tanh(b*x+a))-b*x)^(3/2
)*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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.16081, size = 413, normalized size = 2.47 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{4} x^{4} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt{b x + a}}{384 \, a^{2} x^{4}}, -\frac{15 \, \sqrt{-a} b^{4} x^{4} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt{b x + a}}{192 \, a^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 +
136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4)]

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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**5,x)

[Out]

Timed out

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Giac [A]  time = 1.16573, size = 146, normalized size = 0.87 \begin{align*} -\frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} b^{5} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{\sqrt{2}{\left (15 \,{\left (b x + a\right )}^{\frac{7}{2}} b^{5} + 73 \,{\left (b x + a\right )}^{\frac{5}{2}} a b^{5} - 55 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{2} b^{5} + 15 \, \sqrt{b x + a} a^{3} b^{5}\right )}}{a b^{4} x^{4}}\right )}}{384 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*sqrt(2)*(15*sqrt(2)*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + sqrt(2)*(15*(b*x + a)^(7/2)*b^5 +
73*(b*x + a)^(5/2)*a*b^5 - 55*(b*x + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b