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

Optimal. Leaf size=191 \[ -\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b
*x]])^(7/2)) + (3*b)/(4*x*ArcTanh[Tanh[a + b*x]]^(5/2)) - (3*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Ta
nh[a + b*x]]^(5/2)) - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) + (5*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^2*Arc
Tanh[Tanh[a + b*x]]^(3/2)) - (15*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.127917, antiderivative size = 191, 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} \[ -\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-15*b^2*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b
*x]])^(7/2)) + (3*b)/(4*x*ArcTanh[Tanh[a + b*x]]^(5/2)) - (3*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Ta
nh[a + b*x]]^(5/2)) - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)) + (5*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^2*Arc
Tanh[Tanh[a + b*x]]^(3/2)) - (15*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]])

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{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{4} (3 b) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{1}{8} \left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (15 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (15 b^2\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{3 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{3 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{15 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.116435, size = 115, normalized size = 0.6 \[ \frac{1}{4} \left (\frac{9 b x \tanh ^{-1}(\tanh (a+b x))-2 \tanh ^{-1}(\tanh (a+b x))^2+8 b^2 x^2}{x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

((-15*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(7/2) + (8*b^2*x^2 + 9*b*x*ArcTanh[Tanh[a + b*x]] - 2*ArcTanh[Tanh[a + b*x]]^2)/(x^2*Sqrt[ArcTanh[T
anh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3))/4

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Maple [A]  time = 0.127, size = 131, normalized size = 0.7 \begin{align*} 2\,{b}^{2} \left ({\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}+{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ({\frac{7\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{8}}+ \left ( -{\frac{9\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{8}}+{\frac{9\,bx}{8}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{15}{8\,\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 ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x^3*arctanh(tanh(b*x + a))^(3/2)), x)

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Fricas [A]  time = 2.13596, size = 420, normalized size = 2.2 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt{a} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{8 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac{15 \,{\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x + a}}{4 \,{\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*a*b^2*x^2 + 5*a^2*
b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2), 1/4*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(-a)*arctan(sqrt(b*x + a)
*sqrt(-a)/a) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \operatorname{atanh}^{\frac{3}{2}}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(1/(x**3*atanh(tanh(a + b*x))**(3/2)), x)

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Giac [A]  time = 1.1765, size = 108, normalized size = 0.57 \begin{align*} \frac{15 \, b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{3}} + \frac{2 \, b^{2}}{\sqrt{b x + a} a^{3}} + \frac{7 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2} - 9 \, \sqrt{b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*b^2/(sqrt(b*x + a)*a^3) + 1/4*(7*(b*x + a)^(3/2)*b^
2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^2)