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

Optimal. Leaf size=107 \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{5/2}}+\frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^2}+\frac{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0515712, antiderivative size = 107, 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 = {2169, 2165} \[ \frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{4 b^{5/2}}+\frac{3 \sqrt{x} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{4 b^2}+\frac{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.075713, size = 88, normalized size = 0.82 \[ \frac{\sqrt{b} \sqrt{x} \left (5 b x-3 \tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}+3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\sqrt{b} \sqrt{\tanh ^{-1}(\tanh (a+b x))}+b \sqrt{x}\right )}{4 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.167, size = 174, normalized size = 1.6 \begin{align*}{\frac{1}{2\,b}{x}^{{\frac{3}{2}}}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}-{\frac{3\,a}{4\,{b}^{2}}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{3\,{a}^{2}}{4}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){b}^{-{\frac{5}{2}}}}+{\frac{3\,a \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{2}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){b}^{-{\frac{5}{2}}}}-{\frac{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx-3\,a}{4\,{b}^{2}}\sqrt{x}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}+{\frac{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) ^{2}}{4}\ln \left ( \sqrt{b}\sqrt{x}+\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) } \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

1/2*x^(3/2)*arctanh(tanh(b*x+a))^(1/2)/b-3/4/b^2*a*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+3/4/b^(5/2)*ln(b^(1/2)*x
^(1/2)+arctanh(tanh(b*x+a))^(1/2))*a^2+3/2/b^(5/2)*a*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2))*(arctanh(t
anh(b*x+a))-b*x-a)-3/4/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+3/4/b^(5/2)*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{x^{\frac{3}{2}}}{\sqrt{\operatorname{artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.29574, size = 316, normalized size = 2.95 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b^{3}}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{4 \, 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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\sqrt{\operatorname{atanh}{\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))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.16519, size = 70, normalized size = 0.65 \begin{align*} \frac{1}{4} \, \sqrt{b x + a} \sqrt{x}{\left (\frac{2 \, x}{b} - \frac{3 \, a}{b^{2}}\right )} - \frac{3 \, a^{2} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{4 \, 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))^(1/2),x, algorithm="giac")

[Out]

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