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

Optimal. Leaf size=136 \[ -\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 \sqrt{b}}+\frac{5}{8} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{5}{12} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \]

[Out]

(-5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*Sqrt[b]) + (5
*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/8 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a +
b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/12 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2))/3

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Rubi [A]  time = 0.069098, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2169, 2165} \[ -\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 \sqrt{b}}+\frac{5}{8} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac{5}{12} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^3)/(8*Sqrt[b]) + (5
*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/8 - (5*Sqrt[x]*(b*x - ArcTanh[Tanh[a +
b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/12 + (Sqrt[x]*ArcTanh[Tanh[a + b*x]]^(5/2))/3

Rule 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rule 2165

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}} \, dx &=\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{1}{6} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{\sqrt{x}} \, dx\\ &=-\frac{5}{12} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2}+\frac{1}{8} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \, dx\\ &=\frac{5}{8} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5}{12} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2}-\frac{1}{16} \left (5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}{8 \sqrt{b}}+\frac{5}{8} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{5}{12} \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}+\frac{1}{3} \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.057602, size = 101, normalized size = 0.74 \[ \frac{1}{24} \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-40 b x \tanh ^{-1}(\tanh (a+b x))+33 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )+\frac{5 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{8 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 40*b*x*ArcTanh[Tanh[a + b*x]] + 33*ArcTanh[Tanh[a + b*x]]^
2))/24 + (5*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/(8*Sqrt
[b])

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Maple [B]  time = 0.046, size = 286, normalized size = 2.1 \begin{align*}{\frac{1}{3}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{5\,a}{12}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}}{8}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{5\,{a}^{3}}{8}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{\frac{15\,{a}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{8}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{\frac{5\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{4}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{15\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{8}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}}+{\frac{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx-5\,a}{12}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{8}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{5\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{3}}{8}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/12*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5/8*a^2*x^(1/2)*arctanh(tanh(
b*x+a))^(1/2)+5/8/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^3+15/8*a^2/b^(1/2)*ln(b^(1/2)*x^(1/
2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+5/4*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh
(tanh(b*x+a))^(1/2)+15/8*a/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)
^2+5/12*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5/8*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(
1/2)*arctanh(tanh(b*x+a))^(1/2)+5/8/b^(1/2)*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a
))-b*x-a)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}{\sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.16983, size = 365, normalized size = 2.68 \begin{align*} \left [\frac{15 \, a^{3} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{48 \, b}, -\frac{15 \, a^{3} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{24 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)
*sqrt(b*x + a)*sqrt(x))/b, -1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*b^3*x^2 + 26
*a*b^2*x + 33*a^2*b)*sqrt(b*x + a)*sqrt(x))/b]

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

[Out]

Timed out

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

[Out]

Timed out