3.165 \(\int \frac{1}{x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=224 \[ \frac{35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{35 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \]

[Out]

(35*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]])^(9/2)) + (5*b)/(4*x*ArcTanh[Tanh[a + b*x]]^(7/2)) - (5*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tan
h[a + b*x]]^(7/2)) - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2)) + (7*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcT
anh[Tanh[a + b*x]]^(5/2)) - (35*b^2)/(12*(b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*
b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^4*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.15198, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}+\frac{35 b^2 \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(35*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]])^(9/2)) + (5*b)/(4*x*ArcTanh[Tanh[a + b*x]]^(7/2)) - (5*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tan
h[a + b*x]]^(7/2)) - 1/(2*x^2*ArcTanh[Tanh[a + b*x]]^(5/2)) + (7*b^2)/(4*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcT
anh[Tanh[a + b*x]]^(5/2)) - (35*b^2)/(12*(b*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (35*
b^2)/(4*(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^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{4} (5 b) \int \frac{1}{x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{1}{8} \left (35 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{9/2}} \, dx\\ &=\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{\left (35 b^2\right ) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx}{8 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{\left (35 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 )^2}\\ &=\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{\left (35 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 )^3}\\ &=\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (35 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 )^3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{35 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 )^{9/2}}+\frac{5 b}{4 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{5 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}-\frac{1}{2 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{7 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{35 b^2}{12 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{35 b^2}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.115992, size = 133, normalized size = 0.59 \[ \frac{80 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+39 b x \tanh ^{-1}(\tanh (a+b x))^2-6 \tanh ^{-1}(\tanh (a+b x))^3-8 b^3 x^3}{12 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac{35 b^2 \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(-35*b^2*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(4*(-(b*x) + ArcTanh[Tan
h[a + b*x]])^(9/2)) + (-8*b^3*x^3 + 80*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 39*b*x*ArcTanh[Tanh[a + b*x]]^2 - 6*Ar
cTanh[Tanh[a + b*x]]^3)/(12*x^2*ArcTanh[Tanh[a + b*x]]^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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Maple [A]  time = 0.128, size = 157, normalized size = 0.7 \begin{align*} 2\,{b}^{2} \left ( 3\,{\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 ) }}}+1/3\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}+{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{4}} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ({\frac{11\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}{8}}+ \left ( -{\frac{13\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{8}}+{\frac{13\,bx}{8}} \right ) \sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }-{\frac{35}{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))^(5/2),x)

[Out]

2*b^2*(3/(arctanh(tanh(b*x+a))-b*x)^4/arctanh(tanh(b*x+a))^(1/2)+1/3/(arctanh(tanh(b*x+a))-b*x)^3/arctanh(tanh
(b*x+a))^(3/2)+1/(arctanh(tanh(b*x+a))-b*x)^4*((11/8*arctanh(tanh(b*x+a))^(3/2)+(-13/8*arctanh(tanh(b*x+a))+13
/8*b*x)*arctanh(tanh(b*x+a))^(1/2))/b^2/x^2-35/8/(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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x^3*arctanh(tanh(b*x + a))^(5/2)), x)

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Fricas [A]  time = 2.1548, size = 566, normalized size = 2.53 \begin{align*} \left [\frac{105 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} 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 (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt{b x + a}}{24 \,{\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, \frac{105 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (105 \, a b^{3} x^{3} + 140 \, a^{2} b^{2} x^{2} + 21 \, a^{3} b x - 6 \, a^{4}\right )} \sqrt{b x + a}}{12 \,{\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} 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))^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(105
*a*b^3*x^3 + 140*a^2*b^2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2), 1/12*
(105*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (105*a*b^3*x^3 + 140*a^
2*b^2*x^2 + 21*a^3*b*x - 6*a^4)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)]

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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**3/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.15422, size = 126, normalized size = 0.56 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a} a^{4}} + \frac{2 \,{\left (9 \,{\left (b x + a\right )} b^{2} + a b^{2}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}} + \frac{11 \,{\left (b x + a\right )}^{\frac{3}{2}} b^{2} - 13 \, \sqrt{b x + a} a b^{2}}{4 \, a^{4} 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))^(5/2),x, algorithm="giac")

[Out]

35/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) + 2/3*(9*(b*x + a)*b^2 + a*b^2)/((b*x + a)^(3/2)*a^4) +
1/4*(11*(b*x + a)^(3/2)*b^2 - 13*sqrt(b*x + a)*a*b^2)/(a^4*b^2*x^2)