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

Optimal. Leaf size=98 \[ -\frac{3 \sqrt{x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{5/2} \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0528765, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2168, 2162} \[ -\frac{3 \sqrt{x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{5/2} \sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}-\frac{x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*b^(5/2)*Sqrt[b*x - ArcTanh[Tanh[a + b*x]
]]) - x^(3/2)/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (3*Sqrt[x])/(4*b^2*ArcTanh[Tanh[a + b*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 2162

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

Rubi steps

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

Mathematica [A]  time = 0.0727301, size = 96, normalized size = 0.98 \[ -\frac{3 \sqrt{x}}{4 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 b^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}-\frac{x^{3/2}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.145, size = 85, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}} \left ( -5/8\,{\frac{{x}^{3/2}}{b}}-3/8\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) \sqrt{x}}{{b}^{2}}} \right ) }+{\frac{3}{4\,{b}^{2}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.06684, size = 423, normalized size = 4.32 \begin{align*} \left [-\frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) + 2 \,{\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt{x}}{8 \,{\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, -\frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (5 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt{x}}{4 \,{\left (a b^{5} x^{2} + 2 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(5*a*b^2*x +
3*a^2*b)*sqrt(x))/(a*b^5*x^2 + 2*a^2*b^4*x + a^3*b^3), -1/4*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a*b)*arctan(sqr
t(a*b)/(b*sqrt(x))) + (5*a*b^2*x + 3*a^2*b)*sqrt(x))/(a*b^5*x^2 + 2*a^2*b^4*x + a^3*b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\operatorname{atanh}^{3}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/atanh(tanh(b*x+a))**3,x)

[Out]

Integral(x**(3/2)/atanh(tanh(a + b*x))**3, x)

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Giac [A]  time = 1.11162, size = 63, normalized size = 0.64 \begin{align*} \frac{3 \, \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{2}} - \frac{5 \, b x^{\frac{3}{2}} + 3 \, a \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) - 1/4*(5*b*x^(3/2) + 3*a*sqrt(x))/((b*x + a)^2*b^2)