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

Optimal. Leaf size=176 \[ \frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

[Out]

(-15*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(7/2)) + 15/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + 5/(4*b*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) +
3/(4*b^2*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2) + 3/(4*b^2*x^(5/2)
*ArcTanh[Tanh[a + b*x]])

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Rubi [A]  time = 0.128015, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2168, 2163, 2162} \[ \frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-15*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(4*(b*x - ArcTanh[Tanh[a + b*x]])^
(7/2)) + 15/(4*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + 5/(4*b*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) +
3/(4*b^2*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(2*b*x^(3/2)*ArcTanh[Tanh[a + b*x]]^2) + 3/(4*b^2*x^(5/2)
*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}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}-\frac{3 \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx}{4 b}\\ &=-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{15 \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b^2}\\ &=\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}-\frac{15 \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 b \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{15 \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{8 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}+\frac{(15 b) \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 )^2}\\ &=-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac{15}{4 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{5}{4 b x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{3}{4 b^2 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac{1}{2 b x^{3/2} \tanh ^{-1}(\tanh (a+b x))^2}+\frac{3}{4 b^2 x^{5/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end{align*}

Mathematica [A]  time = 0.166529, size = 141, normalized size = 0.8 \[ -\frac{b \sqrt{x}}{2 \tanh ^{-1}(\tanh (a+b x))^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}-\frac{7 b \sqrt{x}}{4 \tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{2}{\sqrt{x} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*ArcTanh[Tanh[a + b*x]]^3),x]

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(4*(-(b*x) + ArcTanh[Tanh[a + b*
x]])^(7/2)) - 2/(Sqrt[x]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3) - (7*b*Sqrt[x])/(4*ArcTanh[Tanh[a + b*x]]*(-(b*x
) + ArcTanh[Tanh[a + b*x]])^3) - (b*Sqrt[x])/(2*ArcTanh[Tanh[a + b*x]]^2*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)

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Maple [A]  time = 0.142, size = 181, normalized size = 1. \begin{align*} -{\frac{7\,{b}^{2}}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{9\,ab}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{9\,b \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx-a \right ) }{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{2}}\sqrt{x}}-{\frac{15\,b}{4\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}}\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}}}}-2\,{\frac{1}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{3}\sqrt{x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/4/(arctanh(tanh(b*x+a))-b*x)^3*b^2/arctanh(tanh(b*x+a))^2*x^(3/2)-9/4/(arctanh(tanh(b*x+a))-b*x)^3*b/arctan
h(tanh(b*x+a))^2*a*x^(1/2)-9/4/(arctanh(tanh(b*x+a))-b*x)^3*b/arctanh(tanh(b*x+a))^2*x^(1/2)*(arctanh(tanh(b*x
+a))-b*x-a)-15/4/(arctanh(tanh(b*x+a))-b*x)^3*b/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctan
h(tanh(b*x+a))-b*x)*b)^(1/2))-2/(arctanh(tanh(b*x+a))-b*x)^3/x^(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/x^(3/2)/arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13912, size = 466, normalized size = 2.65 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) - 2 \,{\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac{15 \,{\left (b^{2} x^{3} + 2 \, a b x^{2} + a^{2} x\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (15 \, b^{2} x^{2} + 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^
2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 + 2*a*b*x^2 + a^2*x)*
sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 + 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 + 2*a^4*b*x^
2 + a^5*x)]

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

[Out]

Integral(1/(x**(3/2)*atanh(tanh(a + b*x))**3), x)

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Giac [A]  time = 1.10973, size = 80, normalized size = 0.45 \begin{align*} -\frac{15 \, b \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a^{3}} - \frac{2}{a^{3} \sqrt{x}} - \frac{7 \, b^{2} x^{\frac{3}{2}} + 9 \, a b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*b*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/(a^3*sqrt(x)) - 1/4*(7*b^2*x^(3/2) + 9*a*b*sqrt(x))/((
b*x + a)^2*a^3)