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

Optimal. Leaf size=75 \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0419205, antiderivative size = 75, 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} \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0640778, size = 78, normalized size = 1.04 \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.118, size = 59, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,b}{x}^{{\frac{3}{2}}} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{\sqrt{x}}{{b}^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}+2\,{\frac{\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }{{b}^{5/2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.12632, size = 451, normalized size = 6.01 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{2 \,{\left (3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}\right )}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} 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))^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(4*b^2*x + 3*a*
b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -2/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-b)*arctan(s
qrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (4*b^2*x + 3*a*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^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(x**(3/2)/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.28144, size = 66, normalized size = 0.88 \begin{align*} -\frac{2 \, \sqrt{x}{\left (\frac{4 \, x}{b} + \frac{3 \, a}{b^{2}}\right )}}{3 \,{\left (b x + a\right )}^{\frac{3}{2}}} - \frac{2 \, \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(x)*(4*x/b + 3*a/b^2)/(b*x + a)^(3/2) - 2*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(5/2)