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

Optimal. Leaf size=106 \[ 5 b^2 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt{x}} \]

[Out]

-5*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]) + 5*b^2*Sqrt
[x]*Sqrt[ArcTanh[Tanh[a + b*x]]] - (10*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]
^(5/2))/(3*x^(3/2))

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Rubi [A]  time = 0.0632354, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ 5 b^2 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-5 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5*b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]) + 5*b^2*Sqrt
[x]*Sqrt[ArcTanh[Tanh[a + b*x]]] - (10*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]
^(5/2))/(3*x^(3/2))

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 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}}{x^{5/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}+\frac{1}{3} (5 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{3/2}} \, dx\\ &=-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \, dx\\ &=5 b^2 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}-\frac{1}{2} \left (5 b^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=-5 b^{3/2} \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 )+5 b^2 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}-\frac{10 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 \sqrt{x}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0635914, size = 97, normalized size = 0.92 \[ \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-10 b x \tanh ^{-1}(\tanh (a+b x))-2 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )}{3 x^{3/2}}+5 b^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right ) \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 10*b*x*ArcTanh[Tanh[a + b*x]] - 2*ArcTanh[Tanh[a + b*x]]^2))/(3*x^
(3/2)) + 5*b^(3/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]]

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Maple [B]  time = 0.119, size = 501, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(7/2)-8/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*
arctanh(tanh(b*x+a))^(7/2)+8/3*b^2/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+10/3*b^2/(a
rctanh(tanh(b*x+a))-b*x)^2*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+5*b^2/(arctanh(tanh(b*x+a))-b*x)^2*a^2*x^(1/2)
*arctanh(tanh(b*x+a))^(1/2)+5*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/
2))*a^3+15*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(ta
nh(b*x+a))-b*x-a)+10*b^2/(arctanh(tanh(b*x+a))-b*x)^2*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+
a))^(1/2)+15*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(ta
nh(b*x+a))-b*x-a)^2+10/3*b^2/(arctanh(tanh(b*x+a))-b*x)^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*
x+a))^(3/2)+5*b^2/(arctanh(tanh(b*x+a))-b*x)^2*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/
2)+5*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^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}}}{x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15302, size = 362, normalized size = 3.42 \begin{align*} \left [\frac{15 \, a b^{\frac{3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{6 \, x^{2}}, -\frac{15 \, a \sqrt{-b} b x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqr
t(b*x + a)*sqrt(x))/x^2, -1/3*(15*a*sqrt(-b)*b*x^2*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 1
4*a*b*x - 2*a^2)*sqrt(b*x + a)*sqrt(x))/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(atanh(tanh(b*x+a))**(5/2)/x**(5/2),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^(5/2),x, algorithm="giac")

[Out]

Timed out