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

Optimal. Leaf size=155 \[ \frac{5 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.0998613, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2161} \[ \frac{5 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{5 b}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{5 b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*ArcTan[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(
7/2) - 1/(x*ArcTanh[Tanh[a + b*x]]^(5/2)) + b/((b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(5/2)) -
(5*b)/(3*(b*x - ArcTanh[Tanh[a + b*x]])^2*ArcTanh[Tanh[a + b*x]]^(3/2)) + (5*b)/((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^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{1}{2} (5 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{7/2}} \, dx\\ &=-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{(5 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{5 b}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac{(5 b) \int \frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{5 b}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{(5 b) \int \frac{1}{x \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5 b \tan ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{1}{x \tanh ^{-1}(\tanh (a+b x))^{5/2}}+\frac{b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{5/2}}-\frac{5 b}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.127023, size = 113, normalized size = 0.73 \[ \frac{14 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2-2 b^2 x^2}{3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*ArcTanh[Sqrt[ArcTanh[Tanh[a + b*x]]]/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b
*x]])^(7/2) + (-2*b^2*x^2 + 14*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2)/(3*x*(b*x - ArcTanh[Ta
nh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2))

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Maple [A]  time = 0.123, size = 130, normalized size = 0.8 \begin{align*} 2\,b \left ( -{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}} \left ( 1/2\,{\frac{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{bx}}-5/2\,{\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 ) }-1/3\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{3/2}}}-2\,{\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 ) }}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(-1/(arctanh(tanh(b*x+a))-b*x)^3*(1/2*arctanh(tanh(b*x+a))^(1/2)/b/x-5/2/(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)^2/arctanh
(tanh(b*x+a))^(3/2)-2/(arctanh(tanh(b*x+a))-b*x)^3/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^{2} \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^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15793, size = 494, normalized size = 3.19 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\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} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt{b x + a}}{6 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}, -\frac{15 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 3 \, a^{3}\right )} \sqrt{b x + a}}{3 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^2*
x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x), -1/3*(15*(b^3*x^3 + 2*a*b^2*x^2
+ a^2*b*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*a*b^2*x^2 + 20*a^2*b*x + 3*a^3)*sqrt(b*x + a))/(a^4
*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)]

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

[Out]

Timed out

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Giac [A]  time = 1.15389, size = 88, normalized size = 0.57 \begin{align*} -\frac{5 \, b \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{2 \,{\left (6 \,{\left (b x + a\right )} b + a b\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{3}} - \frac{\sqrt{b x + a}}{a^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2/3*(6*(b*x + a)*b + a*b)/((b*x + a)^(3/2)*a^3) - sqrt(b*
x + a)/(a^3*x)