3.234 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{15}{4} b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{15}{4} \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

[Out]

(15*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/4 - (15*
b*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 + (5*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]
^(3/2))/2 - (2*ArcTanh[Tanh[a + b*x]]^(5/2))/Sqrt[x]

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Rubi [A]  time = 0.0660079, antiderivative size = 121, 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} \[ -\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{15}{4} b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{15}{4} \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(3/2),x]

[Out]

(15*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]])^2)/4 - (15*
b*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])/4 + (5*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]
^(3/2))/2 - (2*ArcTanh[Tanh[a + b*x]]^(5/2))/Sqrt[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 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^{3/2}} \, dx &=-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}+(5 b) \int \frac{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{\sqrt{x}} \, dx\\ &=\frac{5}{2} b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}-\frac{1}{4} \left (15 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \, dx\\ &=-\frac{15}{4} b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{5}{2} b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}+\frac{1}{8} \left (15 b \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac{1}{\sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx\\ &=\frac{15}{4} \sqrt{b} \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 )^2-\frac{15}{4} b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\frac{5}{2} b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{\sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0714554, size = 101, normalized size = 0.83 \[ \frac{15}{4} \sqrt{b} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )-\frac{\sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-25 b x \tanh ^{-1}(\tanh (a+b x))+8 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )}{4 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(3/2),x]

[Out]

-(Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 25*b*x*ArcTanh[Tanh[a + b*x]] + 8*ArcTanh[Tanh[a + b*x]]^2))/(4*S
qrt[x]) + (15*Sqrt[b]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]
])/4

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Maple [B]  time = 0.116, size = 460, normalized size = 3.8 \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^(3/2),x)

[Out]

-2/(arctanh(tanh(b*x+a))-b*x)/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+2*b/(arctanh(tanh(b*x+a))-b*x)*x^(1/2)*arctan
h(tanh(b*x+a))^(5/2)+5/2*b/(arctanh(tanh(b*x+a))-b*x)*a*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+15/4*b/(arctanh(tan
h(b*x+a))-b*x)*a^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+15/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*ln(b^(1/2)*x^(1/
2)+arctanh(tanh(b*x+a))^(1/2))*a^3+45/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh
(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+15/2*b/(arctanh(tanh(b*x+a))-b*x)*a*(arctanh(tanh(b*x+a))-b*x-a)*
x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+45/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b
*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2+5/2*b/(arctanh(tanh(b*x+a))-b*x)*(arctanh(tanh(b*x+a))-b*x-a)*x^(
1/2)*arctanh(tanh(b*x+a))^(3/2)+15/4*b/(arctanh(tanh(b*x+a))-b*x)*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arcta
nh(tanh(b*x+a))^(1/2)+15/4*b^(1/2)/(arctanh(tanh(b*x+a))-b*x)*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{3}{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^(3/2),x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^(3/2), x)

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Fricas [A]  time = 2.1141, size = 351, normalized size = 2.9 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{8 \, x}, -\frac{15 \, a^{2} \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{4 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt
(b*x + a)*sqrt(x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x^2 + 9*a*b*
x - 8*a^2)*sqrt(b*x + a)*sqrt(x))/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(atanh(tanh(b*x+a))**(5/2)/x**(3/2),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^(3/2),x, algorithm="giac")

[Out]

Timed out