3.137 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=113 \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0675483, antiderivative size = 113, 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 = {2168, 2161} \[ -\frac{5 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{8 x}+\frac{5 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{12 x^2}-\frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]

[Out]

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

Mathematica [A]  time = 0.0703738, size = 107, normalized size = 0.95 \[ \frac{1}{24} \left (-\frac{15 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{15 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^2}-\frac{8 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^4,x]

[Out]

((-15*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/x - (10*b*ArcTanh[Tanh[a + b*x]]^(3/2))/x^2 - (8*ArcTanh[Tanh[a + b*x]
]^(5/2))/x^3 - (15*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/Sqrt[-(b*x
) + ArcTanh[Tanh[a + b*x]]])/24

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Maple [A]  time = 0.118, size = 144, normalized size = 1.3 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ( -{\frac{11\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{16}}+ \left ( 5/6\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5/6\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( -{\frac{5\,{a}^{2}}{16}}-5/8\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) -{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{16}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{5}{16\,\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^4,x)

[Out]

2*b^3*((-11/16*arctanh(tanh(b*x+a))^(5/2)+(5/6*arctanh(tanh(b*x+a))-5/6*b*x)*arctanh(tanh(b*x+a))^(3/2)+(-5/16
*a^2-5/8*a*(arctanh(tanh(b*x+a))-b*x-a)-5/16*(arctanh(tanh(b*x+a))-b*x-a)^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x
^3-5/16/(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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^4, x)

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Fricas [A]  time = 2.12887, size = 350, normalized size = 3.1 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a x^{3}}, \frac{15 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(33*a*b^2*x^2 + 26*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a*x^3), 1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (33*a*b^2*x^2 + 26*a^2*b
*x + 8*a^3)*sqrt(b*x + a))/(a*x^3)]

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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**4,x)

[Out]

Timed out

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Giac [A]  time = 1.17132, size = 119, normalized size = 1.05 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{2}{\left (33 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} + 15 \, \sqrt{b x + a} a^{2} b^{4}\right )}}{b^{3} x^{3}}\right )}}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^4,x, algorithm="giac")

[Out]

1/48*sqrt(2)*(15*sqrt(2)*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - sqrt(2)*(33*(b*x + a)^(5/2)*b^4 - 40*(b
*x + a)^(3/2)*a*b^4 + 15*sqrt(b*x + a)*a^2*b^4)/(b^3*x^3))/b