3.136 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=110 \[ \frac{15}{4} b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{15}{4} b^2 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x} \]

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b
*x]]])/4 + (15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(4*x) - ArcTanh[Tanh[a
+ b*x]]^(5/2)/(2*x^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0667365, 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} \[ \frac{15}{4} b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{15}{4} b^2 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))} \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^3,x]

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh[a + b
*x]]])/4 + (15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 - (5*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(4*x) - ArcTanh[Tanh[a
+ b*x]]^(5/2)/(2*x^2)

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^3} \, dx &=-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}+\frac{1}{4} (5 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^2} \, dx\\ &=-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}+\frac{1}{8} \left (15 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x} \, dx\\ &=\frac{15}{4} b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}-\frac{1}{8} \left (15 b^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{15}{4} b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right ) \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}+\frac{15}{4} b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.043991, size = 108, normalized size = 0.98 \[ -\frac{-15 b^2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}+15 b^2 x^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}+2 \tanh ^{-1}(\tanh (a+b x))^{5/2}+5 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^3,x]

[Out]

-(-15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] + 5*b*x*ArcTanh[Tanh[a + b*x]]^(3/2) + 2*ArcTanh[Tanh[a + b*x]]^(5/
2) + 15*b^2*x^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]]*Sqrt[-(b*x) + ArcT
anh[Tanh[a + b*x]]])/(4*x^2)

________________________________________________________________________________________

Maple [A]  time = 0.115, size = 142, normalized size = 1.3 \begin{align*} 2\,{b}^{2} \left ( \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }+{\frac{1}{{b}^{2}{x}^{2}} \left ( \left ( -{\frac{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{8}}+{\frac{9\,bx}{8}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ({\frac{7\,{a}^{2}}{8}}+7/4\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) +{\frac{7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{8}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{15\,\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}{8}{\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^3,x)

[Out]

2*b^2*(arctanh(tanh(b*x+a))^(1/2)+((-9/8*arctanh(tanh(b*x+a))+9/8*b*x)*arctanh(tanh(b*x+a))^(3/2)+(7/8*a^2+7/4
*a*(arctanh(tanh(b*x+a))-b*x-a)+7/8*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-15/8*(
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)))

________________________________________________________________________________________

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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^3, x)

________________________________________________________________________________________

Fricas [A]  time = 2.11054, size = 320, normalized size = 2.91 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{8 \, x^{2}}, \frac{15 \, \sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a}}{4 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/8*(15*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2, 1/4*(15*sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2]

________________________________________________________________________________________

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**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.15224, size = 124, normalized size = 1.13 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} a b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 8 \, \sqrt{2} \sqrt{b x + a} b^{3} - \frac{\sqrt{2}{\left (9 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{3} - 7 \, \sqrt{b x + a} a^{2} b^{3}\right )}}{b^{2} x^{2}}\right )}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^3,x, algorithm="giac")

[Out]

1/8*sqrt(2)*(15*sqrt(2)*a*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(2)*sqrt(b*x + a)*b^3 - sqrt(2)*
(9*(b*x + a)^(3/2)*a*b^3 - 7*sqrt(b*x + a)*a^2*b^3)/(b^2*x^2))/b