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

Optimal. Leaf size=92 \[ \frac{3 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{2 x^2}-\frac{3 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.0490936, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 2161} \[ \frac{3 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{2 x^2}-\frac{3 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0594817, size = 88, normalized size = 0.96 \[ \frac{1}{4} \left (-\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^2}-\frac{3 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/x - (2*ArcTanh[Tanh[a + b*x]]^(3/2))/x^2 - (3*b^2*ArcTanh[Sqrt[ArcTanh[Ta
nh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]])/4

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Maple [A]  time = 0.115, size = 91, normalized size = 1. \begin{align*} 2\,{b}^{2} \left ({\frac{-5/8\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ( 3/8\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3/8\,bx \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{{b}^{2}{x}^{2}}}-3/8\,{\frac{1}{\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))^(3/2)/x^3,x)

[Out]

2*b^2*((-5/8*arctanh(tanh(b*x+a))^(3/2)+(3/8*arctanh(tanh(b*x+a))-3/8*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2
-3/8/(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)))

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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{3}{2}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14147, size = 296, normalized size = 3.22 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{8 \, a x^{2}}, \frac{3 \, \sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (5 \, a b x + 2 \, a^{2}\right )} \sqrt{b x + a}}{4 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(5*a*b*x + 2*a^2)*sqrt(b*x + a))/(a*x
^2), 1/4*(3*sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (5*a*b*x + 2*a^2)*sqrt(b*x + a))/(a*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(3/2)/x**3,x)

[Out]

Integral(atanh(tanh(a + b*x))**(3/2)/x**3, x)

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Giac [A]  time = 1.17312, size = 99, normalized size = 1.08 \begin{align*} \frac{\sqrt{2}{\left (\frac{3 \, \sqrt{2} b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{2}{\left (5 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{3} - 3 \, \sqrt{b x + a} a 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))^(3/2)/x^3,x, algorithm="giac")

[Out]

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