3.157 \(\int \frac{1}{x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=245 \[ -\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{35 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

(-35*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]])^(9/2)) - (5*b^2)/(8*x*ArcTanh[Tanh[a + b*x]]^(7/2)) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[
Tanh[a + b*x]]^(7/2)) + b/(4*x^2*ArcTanh[Tanh[a + b*x]]^(5/2)) - (7*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*A
rcTanh[Tanh[a + b*x]]^(5/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b
*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*Sqrt[ArcTanh[Tanh[a + b*x
]]])

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Rubi [A]  time = 0.182859, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ -\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{35 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-35*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]])^(9/2)) - (5*b^2)/(8*x*ArcTanh[Tanh[a + b*x]]^(7/2)) + (5*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[
Tanh[a + b*x]]^(7/2)) + b/(4*x^2*ArcTanh[Tanh[a + b*x]]^(5/2)) - (7*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^2*A
rcTanh[Tanh[a + b*x]]^(5/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b
*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*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^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{2} b \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\\ &=\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{1}{8} \left (5 b^2\right ) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{1}{16} \left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (35 b^3\right ) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{35 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 )^{9/2}}-\frac{5 b^2}{8 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{5 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{b}{4 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{7 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.113197, size = 133, normalized size = 0.54 \[ \frac{35 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{8 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}-\frac{87 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-38 b x \tanh ^{-1}(\tanh (a+b x))^2+8 \tanh ^{-1}(\tanh (a+b x))^3+48 b^3 x^3}{24 x^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(35*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(8*(-(b*x) + ArcTanh[Tanh
[a + b*x]])^(9/2)) - (48*b^3*x^3 + 87*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 38*b*x*ArcTanh[Tanh[a + b*x]]^2 + 8*Arc
Tanh[Tanh[a + b*x]]^3)/(24*x^3*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.128, size = 186, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ( -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ({\frac{19\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{16}}+ \left ( -{\frac{17\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{6}}+{\frac{17\,bx}{6}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ({\frac{29\,{a}^{2}}{16}}+{\frac{29\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{8}}+{\frac{29\, \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{35}{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 ) }-{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

2*b^3*(-1/(arctanh(tanh(b*x+a))-b*x)^4*((19/16*arctanh(tanh(b*x+a))^(5/2)+(-17/6*arctanh(tanh(b*x+a))+17/6*b*x
)*arctanh(tanh(b*x+a))^(3/2)+(29/16*a^2+29/8*a*(arctanh(tanh(b*x+a))-b*x-a)+29/16*(arctanh(tanh(b*x+a))-b*x-a)
^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-35/16/(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)))-1/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \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^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x^4*arctanh(tanh(b*x + a))^(3/2)), x)

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Fricas [A]  time = 2.12223, size = 478, normalized size = 1.95 \begin{align*} \left [\frac{105 \,{\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{48 \,{\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, -\frac{105 \,{\left (b^{4} x^{4} + a b^{3} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (105 \, a b^{3} x^{3} + 35 \, a^{2} b^{2} x^{2} - 14 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{b x + a}}{24 \,{\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(105*a*b^3*x^3 + 35*
a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3), -1/24*(105*(b^4*x^4 + a*b^3*x^3)*sqrt(
-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (105*a*b^3*x^3 + 35*a^2*b^2*x^2 - 14*a^3*b*x + 8*a^4)*sqrt(b*x + a))/(a
^5*b*x^4 + a^6*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \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**4/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(1/(x**4*atanh(tanh(a + b*x))**(3/2)), x)

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Giac [A]  time = 1.1462, size = 128, normalized size = 0.52 \begin{align*} -\frac{35 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{4}} - \frac{2 \, b^{3}}{\sqrt{b x + a} a^{4}} - \frac{57 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{3} - 136 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{3} + 87 \, \sqrt{b x + a} a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-35/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) - 2*b^3/(sqrt(b*x + a)*a^4) - 1/24*(57*(b*x + a)^(5/2)
*b^3 - 136*(b*x + a)^(3/2)*a*b^3 + 87*sqrt(b*x + a)*a^2*b^3)/(a^4*b^3*x^3)