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

Optimal. Leaf size=70 \[ 2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac{2 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0422076, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2168, 2165} \[ 2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )-\frac{2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac{2 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0419714, size = 74, normalized size = 1.06 \[ -\frac{2 \left (-3 b^{3/2} x^{3/2} \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )+3 b x \sqrt{\tanh ^{-1}(\tanh (a+b x))}+\tanh ^{-1}(\tanh (a+b x))^{3/2}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.135, size = 315, normalized size = 4.5 \begin{align*} -{\frac{2}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{3}{2}}}}-{\frac{4\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{x}}}}+{\frac{4\,{b}^{2}}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}\sqrt{x} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{3}{2}}}}+2\,{\frac{a{b}^{2}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}}+2\,{\frac{{b}^{3/2}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){a}^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}}+4\,{\frac{{b}^{3/2}a\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}}+2\,{\frac{{b}^{2} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}}+2\,{\frac{{b}^{3/2}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(5/2)-4/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*
arctanh(tanh(b*x+a))^(5/2)+4/3*b^2/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+2*b^2/(arct
anh(tanh(b*x+a))-b*x)^2*a*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+2*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^2+4*b^(3/2)/(arctanh(tanh(b*x+a))-b*x)^2*a*ln(b^(1/2)*x^(1/2)+arctanh(t
anh(b*x+a))^(1/2))*(arctanh(tanh(b*x+a))-b*x-a)+2*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))^(1/2)+2*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)^2

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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{3}{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))^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.31197, size = 302, normalized size = 4.31 \begin{align*} \left [\frac{3 \, b^{\frac{3}{2}} x^{2} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (4 \, b x + a\right )} \sqrt{b x + a} \sqrt{x}}{3 \, x^{2}}, -\frac{2 \,{\left (3 \, \sqrt{-b} b x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (4 \, b x + a\right )} \sqrt{b x + a} \sqrt{x}\right )}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out