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

Optimal. Leaf size=125 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{1}{4 b^2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{4 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))}-\frac{\sqrt{x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]/(4*b^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))
- 1/(4*b^2*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])) - Sqrt[x]/(2*b*ArcTanh[Tanh[a + b*x]]^2) - 1/(4*b^2*Sqrt[x
]*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.0706845, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 b^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{3/2}}-\frac{1}{4 b^2 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{4 b^2 \sqrt{x} \tanh ^{-1}(\tanh (a+b x))}-\frac{\sqrt{x}}{2 b \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]/(4*b^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^(3/2))
- 1/(4*b^2*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])) - Sqrt[x]/(2*b*ArcTanh[Tanh[a + b*x]]^2) - 1/(4*b^2*Sqrt[x
]*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 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

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

Mathematica [A]  time = 0.131997, size = 107, normalized size = 0.86 \[ \frac{1}{4} \left (\frac{\sqrt{x}}{b \tanh ^{-1}(\tanh (a+b x))^2-b^2 x \tanh ^{-1}(\tanh (a+b x))}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{b^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{3/2}}-\frac{2 \sqrt{x}}{b \tanh ^{-1}(\tanh (a+b x))^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.135, size = 98, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}} \left ( 1/8\,{\frac{{x}^{3/2}}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}}-1/8\,{\frac{\sqrt{x}}{b}} \right ) }+{\frac{1}{ \left ( 4\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -4\,bx \right ) b}\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^(1/2)/arctanh(tanh(b*x+a))^3,x)

[Out]

2*(1/8/(arctanh(tanh(b*x+a))-b*x)*x^(3/2)-1/8*x^(1/2)/b)/arctanh(tanh(b*x+a))^2+1/4/(arctanh(tanh(b*x+a))-b*x)
/b/((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^(1/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03657, size = 412, normalized size = 3.3 \begin{align*} \left [-\frac{{\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 (a b^{2} x - a^{2} b\right )} \sqrt{x}}{8 \,{\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac{{\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 (a b^{2} x - a^{2} b\right )} \sqrt{x}}{4 \,{\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.13072, size = 70, normalized size = 0.56 \begin{align*} \frac{\arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a b} + \frac{b x^{\frac{3}{2}} - a \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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