3.236 \(\int \frac{\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=93 \[ 2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}}-\frac{2 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0587725, antiderivative size = 93, 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 = {2168, 2165} \[ 2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}}-\frac{2 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0513485, size = 95, normalized size = 1.02 \[ -\frac{2 \left (15 b^2 x^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}-15 b^{5/2} x^{5/2} \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )+5 b x \tanh ^{-1}(\tanh (a+b x))^{3/2}+3 \tanh ^{-1}(\tanh (a+b x))^{5/2}\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]

[Out]

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

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Maple [B]  time = 0.124, size = 532, normalized size = 5.7 \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^(7/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(7/2)-4/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)
*arctanh(tanh(b*x+a))^(7/2)-16/15*b^2/(arctanh(tanh(b*x+a))-b*x)^3/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+16/15*b^
3/(arctanh(tanh(b*x+a))-b*x)^3*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+4/3*b^3/(arctanh(tanh(b*x+a))-b*x)^3*a*x^(1/
2)*arctanh(tanh(b*x+a))^(3/2)+2*b^3/(arctanh(tanh(b*x+a))-b*x)^3*a^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+2*b^(5
/2)/(arctanh(tanh(b*x+a))-b*x)^3*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^3+6*b^(5/2)/(arctanh(tanh(b*
x+a))-b*x)^3*a^2*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+4*b^3/(arctanh(ta
nh(b*x+a))-b*x)^3*a*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+6*b^(5/2)/(arctanh(tanh(b*
x+a))-b*x)^3*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)^2+4/3*b^3/(arctanh(
tanh(b*x+a))-b*x)^3*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+2*b^3/(arctanh(tanh(b*x+a)
)-b*x)^3*(arctanh(tanh(b*x+a))-b*x-a)^2*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+2*b^(5/2)/(arctanh(tanh(b*x+a))-b*x
)^3*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{7}{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^(7/2),x, algorithm="maxima")

[Out]

integrate(arctanh(tanh(b*x + a))^(5/2)/x^(7/2), x)

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Fricas [A]  time = 2.08364, size = 365, normalized size = 3.92 \begin{align*} \left [\frac{15 \, b^{\frac{5}{2}} x^{3} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{15 \, x^{3}}, -\frac{2 \,{\left (15 \, \sqrt{-b} b^{2} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}\right )}}{15 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*b^(5/2)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqr
t(b*x + a)*sqrt(x))/x^3, -2/15*(15*sqrt(-b)*b^2*x^3*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (23*b^2*x^2 +
11*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x))/x^3]

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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**(7/2),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^(7/2),x, algorithm="giac")

[Out]

Timed out