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

Optimal. Leaf size=278 \[ \frac{105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{105 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

[Out]

(105*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]])^(11/2)) - (35*b^2)/(24*x*ArcTanh[Tanh[a + b*x]]^(9/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^(9/2)) + (5*b)/(12*x^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - (15*b^3)/(8*(b*x - ArcTanh[Tanh[a +
b*x]])^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(5/2)) + (21*b^3)/(8*(b*x - ArcTanh[
Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(5/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*ArcTanh[Tanh[a
+ b*x]]^(3/2)) + (105*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^5*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.221521, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{105 b^3 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{11/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(105*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]])^(11/2)) - (35*b^2)/(24*x*ArcTanh[Tanh[a + b*x]]^(9/2)) + (35*b^3)/(24*(b*x - ArcTanh[Tanh[a + b*x]])*Arc
Tanh[Tanh[a + b*x]]^(9/2)) + (5*b)/(12*x^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - (15*b^3)/(8*(b*x - ArcTanh[Tanh[a +
b*x]])^2*ArcTanh[Tanh[a + b*x]]^(7/2)) - 1/(3*x^3*ArcTanh[Tanh[a + b*x]]^(5/2)) + (21*b^3)/(8*(b*x - ArcTanh[
Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(5/2)) - (35*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^4*ArcTanh[Tanh[a
+ b*x]]^(3/2)) + (105*b^3)/(8*(b*x - ArcTanh[Tanh[a + b*x]])^5*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))^{5/2}} \, dx &=-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{6} (5 b) \int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{1}{24} \left (35 b^2\right ) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{16} \left (105 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{11/2}} \, dx\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{\left (105 b^3\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx}{16 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{\left (105 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 )^2}\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{\left (105 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 )^3}\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (105 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 )^4}\\ &=-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (105 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 )^4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{105 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 )^{11/2}}-\frac{35 b^2}{24 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{35 b^3}{24 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{9/2}}+\frac{5 b}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{15 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{3 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{21 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{105 b^3}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.195169, size = 150, normalized size = 0.54 \[ \frac{1}{24} \left (\frac{208 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+165 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-50 b x \tanh ^{-1}(\tanh (a+b x))^3+8 \tanh ^{-1}(\tanh (a+b x))^4-16 b^4 x^4}{x^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{315 b^3 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{11/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

((315*b^3*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[
a + b*x]])^(11/2) + (-16*b^4*x^4 + 208*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 165*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 -
50*b*x*ArcTanh[Tanh[a + b*x]]^3 + 8*ArcTanh[Tanh[a + b*x]]^4)/(x^3*(b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[T
anh[a + b*x]]^(3/2)))/24

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Maple [A]  time = 0.125, size = 211, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ( -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5}} \left ({\frac{1}{{x}^{3}{b}^{3}} \left ({\frac{41\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{5/2}}{16}}+ \left ( -{\frac{35\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{6}}+{\frac{35\,bx}{6}} \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}+ \left ({\frac{55\,{a}^{2}}{16}}+{\frac{55\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{8}}+{\frac{55\, \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{105}{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 ) }-1/3\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}-4\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{5}\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))^(5/2),x)

[Out]

2*b^3*(-1/(arctanh(tanh(b*x+a))-b*x)^5*((41/16*arctanh(tanh(b*x+a))^(5/2)+(-35/6*arctanh(tanh(b*x+a))+35/6*b*x
)*arctanh(tanh(b*x+a))^(3/2)+(55/16*a^2+55/8*a*(arctanh(tanh(b*x+a))-b*x-a)+55/16*(arctanh(tanh(b*x+a))-b*x-a)
^2)*arctanh(tanh(b*x+a))^(1/2))/b^3/x^3-105/16/(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)))-1/3/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(3/2)-4/(arctanh
(tanh(b*x+a))-b*x)^5/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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x^4*arctanh(tanh(b*x + a))^(5/2)), x)

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Fricas [A]  time = 2.17269, size = 613, normalized size = 2.21 \begin{align*} \left [\frac{315 \,{\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} 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 (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt{b x + a}}{48 \,{\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, -\frac{315 \,{\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (315 \, a b^{4} x^{4} + 420 \, a^{2} b^{3} x^{3} + 63 \, a^{3} b^{2} x^{2} - 18 \, a^{4} b x + 8 \, a^{5}\right )} \sqrt{b x + a}}{24 \,{\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} 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))^(5/2),x, algorithm="fricas")

[Out]

[1/48*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(315
*a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4
+ a^8*x^3), -1/24*(315*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (315*
a*b^4*x^4 + 420*a^2*b^3*x^3 + 63*a^3*b^2*x^2 - 18*a^4*b*x + 8*a^5)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 +
a^8*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(1/x**4/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.15118, size = 155, normalized size = 0.56 \begin{align*} -\frac{105 \, b^{3} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{5}} - \frac{315 \,{\left (b x + a\right )}^{4} b^{3} - 840 \,{\left (b x + a\right )}^{3} a b^{3} + 693 \,{\left (b x + a\right )}^{2} a^{2} b^{3} - 144 \,{\left (b x + a\right )} a^{3} b^{3} - 16 \, a^{4} b^{3}}{24 \,{\left ({\left (b x + a\right )}^{\frac{3}{2}} - \sqrt{b x + a} a\right )}^{3} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-105/8*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) - 1/24*(315*(b*x + a)^4*b^3 - 840*(b*x + a)^3*a*b^3 +
693*(b*x + a)^2*a^2*b^3 - 144*(b*x + a)*a^3*b^3 - 16*a^4*b^3)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)