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

Optimal. Leaf size=120 \[ -\frac{1}{b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{3/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{3}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0799972, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ -\frac{1}{b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{3/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}-\frac{3}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(5/2)
- 3/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2) - 1/(b*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(b*x^(3/2)*
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 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[u, v, x]

Rubi steps

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

Mathematica [A]  time = 0.104345, size = 104, normalized size = 0.87 \[ -\frac{b \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{2}{\sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.124, size = 105, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{x}}}-{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}-3\,{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)-1/(arctanh(tanh(b*x+a))-b*x)^2*b*x^(1/2)/arctanh(tanh(b*x+a))-3/(arcta
nh(tanh(b*x+a))-b*x)^2*b/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^
(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10007, size = 323, normalized size = 2.69 \begin{align*} \left [\frac{3 \,{\left (b x^{2} + a x\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) - 2 \,{\left (3 \, b x + 2 \, a\right )} \sqrt{x}}{2 \,{\left (a^{2} b x^{2} + a^{3} x\right )}}, \frac{3 \,{\left (b x^{2} + a x\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (3 \, b x + 2 \, a\right )} \sqrt{x}}{a^{2} b x^{2} + a^{3} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.15267, size = 66, normalized size = 0.55 \begin{align*} -\frac{3 \, b \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2}} - \frac{3 \, b x + 2 \, a}{{\left (b x^{\frac{3}{2}} + a \sqrt{x}\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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