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

Optimal. Leaf size=152 \[ \frac{1}{4 b^2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{1}{4 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt{b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac{3}{4 b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2} \]

[Out]

(-3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*Sqrt[b]*(b*x - ArcTanh[Tanh[a + b*x]])^(
5/2)) + 3/(4*b*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^2) + 1/(4*b^2*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])) -
1/(2*b*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2) + 1/(4*b^2*x^(3/2)*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.0974084, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{1}{4 b^2 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{1}{4 b^2 x^{3/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \sqrt{b} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{5/2}}+\frac{3}{4 b \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b \sqrt{x} \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

Mathematica [A]  time = 0.0785903, size = 118, normalized size = 0.78 \[ \frac{\sqrt{x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}+\frac{3 \sqrt{x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \sqrt{b} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

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

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Maple [A]  time = 0.132, size = 112, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( 2\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -2\,bx \right ) \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}+{\frac{3}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }\sqrt{x}}+{\frac{3}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{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(1/arctanh(tanh(b*x+a))^3/x^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.1019, size = 423, normalized size = 2.78 \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 (3 \, a b^{2} x + 5 \, a^{2} b\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\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 (3 \, a b^{2} x + 5 \, a^{2} b\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arctanh(tanh(b*x+a))^3/x^(1/2),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*(3*a*b^2*x +
5*a^2*b)*sqrt(x))/(a^3*b^3*x^2 + 2*a^4*b^2*x + a^5*b), -1/4*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a*b)*arctan(sqr
t(a*b)/(b*sqrt(x))) - (3*a*b^2*x + 5*a^2*b)*sqrt(x))/(a^3*b^3*x^2 + 2*a^4*b^2*x + a^5*b)]

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

[Out]

Integral(1/(sqrt(x)*atanh(tanh(a + b*x))**3), x)

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Giac [A]  time = 1.16266, size = 63, normalized size = 0.41 \begin{align*} \frac{3 \, \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{2}} + \frac{3 \, b x^{\frac{3}{2}} + 5 \, a \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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