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

Optimal. Leaf size=145 \[ \frac{5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{5 b}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.10639, antiderivative size = 145, 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, 2162} \[ \frac{5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}-\frac{5}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{b x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{b x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{5 b}{\sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.198996, size = 120, normalized size = 0.83 \[ \frac{b^2 \sqrt{x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}+\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}}+\frac{2 \left (\tanh ^{-1}(\tanh (a+b x))-7 b x\right )}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.18, size = 128, normalized size = 0.9 \begin{align*} -{\frac{2}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}{x}^{-{\frac{3}{2}}}}+4\,{\frac{b}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{x}}}+{\frac{{b}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}+5\,{\frac{{b}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\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^(5/2)/arctanh(tanh(b*x+a))^2,x)

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)+4/(arctanh(tanh(b*x+a))-b*x)^3*b/x^(1/2)+1/(arctanh(tanh(b*x+a))-b*x
)^3*b^2*x^(1/2)/arctanh(tanh(b*x+a))+5/(arctanh(tanh(b*x+a))-b*x)^3*b^2/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*a
rctan(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^(5/2)/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08098, size = 402, normalized size = 2.77 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{3} + a b x^{2}\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 (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt{x}}{6 \,{\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac{15 \,{\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (15 \, b^{2} x^{2} + 10 \, a b x - 2 \, a^{2}\right )} \sqrt{x}}{3 \,{\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(b^2*x^3 + a*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(15*b^2*x^2 + 10
*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2), -1/3*(15*(b^2*x^3 + a*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*s
qrt(x))) - (15*b^2*x^2 + 10*a*b*x - 2*a^2)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{5}{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**(5/2)/atanh(tanh(b*x+a))**2,x)

[Out]

Integral(1/(x**(5/2)*atanh(tanh(a + b*x))**2), x)

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Giac [A]  time = 1.12245, size = 78, normalized size = 0.54 \begin{align*} \frac{5 \, b^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{3}} + \frac{b^{2} \sqrt{x}}{{\left (b x + a\right )} a^{3}} + \frac{2 \,{\left (6 \, b x - a\right )}}{3 \, a^{3} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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