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

Optimal. Leaf size=86 \[ \frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{5/2}}-\frac{2 x^{3/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.0487386, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ \frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{b^2}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{5/2}}-\frac{2 x^{3/2}}{b \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0784536, size = 81, normalized size = 0.94 \[ \frac{\sqrt{x} \left (3 \tanh ^{-1}(\tanh (a+b x))-2 b x\right )}{b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.121, size = 130, normalized size = 1.5 \begin{align*}{\frac{1}{b}{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}+3\,{\frac{a\sqrt{x}}{{b}^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-3\,{\frac{a\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ) }{{b}^{5/2}}}+3\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \sqrt{x}}{{b}^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}-3\,{\frac{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) \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))^(3/2),x)

[Out]

x^(3/2)/b/arctanh(tanh(b*x+a))^(1/2)+3/b^2*a*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-3/b^(5/2)*a*ln(b^(1/2)*x^(1/2)
+arctanh(tanh(b*x+a))^(1/2))+3/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-3/b^(5/2)*(
arctanh(tanh(b*x+a))-b*x-a)*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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.2193, size = 363, normalized size = 4.22 \begin{align*} \left [\frac{3 \,{\left (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 (b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{2 \,{\left (b^{4} x + a b^{3}\right )}}, \frac{3 \,{\left (a b x + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{b^{4} x + a b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*(a*b*x + a^2)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(b^2*x + 3*a*b)*sqrt(b*x +
a)*sqrt(x))/(b^4*x + a*b^3), (3*(a*b*x + a^2)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (b^2*x + 3
*a*b)*sqrt(b*x + a)*sqrt(x))/(b^4*x + a*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}^{\frac{3}{2}}{\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))**(3/2),x)

[Out]

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

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Giac [A]  time = 1.20785, size = 65, normalized size = 0.76 \begin{align*} \frac{\sqrt{x}{\left (\frac{x}{b} + \frac{3 \, a}{b^{2}}\right )}}{\sqrt{b x + a}} + \frac{3 \, a \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))^(3/2),x, algorithm="giac")

[Out]

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