3.119 \(\int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx\)

Optimal. Leaf size=179 \[ \frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3} \]

[Out]

(b^3*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(8*(b*x - ArcTanh[Tanh[a + b*x]]
)^(5/2)) + b^2/(24*x*ArcTanh[Tanh[a + b*x]]^(3/2)) - b^3/(24*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(3/2)) - b/(12*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[T
anh[a + b*x]]]) - Sqrt[ArcTanh[Tanh[a + b*x]]]/(3*x^3)

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Rubi [A]  time = 0.120568, antiderivative size = 179, 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} \[ \frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^4,x]

[Out]

(b^3*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(8*(b*x - ArcTanh[Tanh[a + b*x]]
)^(5/2)) + b^2/(24*x*ArcTanh[Tanh[a + b*x]]^(3/2)) - b^3/(24*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b
*x]]^(3/2)) - b/(12*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + b^3/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[T
anh[a + b*x]]]) - Sqrt[ArcTanh[Tanh[a + b*x]]]/(3*x^3)

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{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x^4} \, dx &=-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac{1}{6} b \int \frac{1}{x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac{1}{24} b^2 \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\\ &=\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}+\frac{1}{16} b^3 \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac{b^3 \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}-\frac{b^3 \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac{b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{b}{12 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0981389, size = 115, normalized size = 0.64 \[ \frac{1}{24} \left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (14 b x \tanh ^{-1}(\tanh (a+b x))-8 \tanh ^{-1}(\tanh (a+b x))^2-3 b^2 x^2\right )}{x^3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{3 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^4,x]

[Out]

((-3*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a
+ b*x]])^(5/2) + (Sqrt[ArcTanh[Tanh[a + b*x]]]*(-3*b^2*x^2 + 14*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a
+ b*x]]^2))/(x^3*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2))/24

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Maple [A]  time = 0.129, size = 185, normalized size = 1. \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ( 1/16\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{{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}}}-1/6\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}-1/16\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-1/16\,{\frac{1}{ \left ({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} \right ) \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))^(1/2)/x^4,x)

[Out]

2*b^3*((1/16/(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))^(5/2)-
1/6/(arctanh(tanh(b*x+a))-b*x)*arctanh(tanh(b*x+a))^(3/2)-1/16*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-1/16/(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(arcta
nh(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{\sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(arctanh(tanh(b*x + a)))/x^4, x)

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Fricas [A]  time = 1.85557, size = 347, normalized size = 1.94 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a^{3} x^{3}}, \frac{3 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sq
rt(b*x + a))/(a^3*x^3), 1/24*(3*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b^2*x^2 - 2*a^2*b*x -
8*a^3)*sqrt(b*x + a))/(a^3*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\operatorname{atanh}{\left (\tanh{\left (a + b x \right )} \right )}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**4,x)

[Out]

Integral(sqrt(atanh(tanh(a + b*x)))/x**4, x)

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Giac [A]  time = 1.16832, size = 113, normalized size = 0.63 \begin{align*} \frac{\frac{3 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 8 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} - 3 \, \sqrt{b x + a} a^{2} b^{4}}{a^{2} b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^4,x, algorithm="giac")

[Out]

1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*b^4 - 8*(b*x + a)^(3/2)*a*b^4 -
3*sqrt(b*x + a)*a^2*b^4)/(a^2*b^3*x^3))/b