3.256 \(\int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=148 \[ -\frac{32 b^3 \sqrt{x}}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 b^2}{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

(-32*b^3*Sqrt[x])/(5*(b*x - ArcTanh[Tanh[a + b*x]])^4*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*b^2)/(5*Sqrt[x]*(b*x
- ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (4*b)/(5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2
*Sqrt[ArcTanh[Tanh[a + b*x]]]) + 2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])

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Rubi [A]  time = 0.0797879, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ -\frac{32 b^3 \sqrt{x}}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 b^2}{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-32*b^3*Sqrt[x])/(5*(b*x - ArcTanh[Tanh[a + b*x]])^4*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*b^2)/(5*Sqrt[x]*(b*x
- ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (4*b)/(5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2
*Sqrt[ArcTanh[Tanh[a + b*x]]]) + 2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[ArcTanh[Tanh[a + b*x]]])

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{(6 b) \int \frac{1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{\left (8 b^2\right ) \int \frac{1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2}{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}-\frac{\left (16 b^3\right ) \int \frac{1}{\sqrt{x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac{32 b^3 \sqrt{x}}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{16 b^2}{5 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}+\frac{2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt{\tanh ^{-1}(\tanh (a+b x))}}\\ \end{align*}

Mathematica [A]  time = 0.0622312, size = 80, normalized size = 0.54 \[ -\frac{2 \left (15 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-5 b x \tanh ^{-1}(\tanh (a+b x))^2+\tanh ^{-1}(\tanh (a+b x))^3+5 b^3 x^3\right )}{5 x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.119, size = 151, normalized size = 1. \begin{align*} -{\frac{2}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}-{\frac{12\,b}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx} \left ( -{\frac{1}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}-{\frac{4\,b}{3\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -3\,bx} \left ( -{\frac{1}{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}}}-2\,{\frac{b\sqrt{x}}{ \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)/arctanh(tanh(b*x+a))^(1/2)-12/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(ar
ctanh(tanh(b*x+a))-b*x)/x^(3/2)/arctanh(tanh(b*x+a))^(1/2)-4/3*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh(tanh(
b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-2*b/(arctanh(tanh(b*x+a))-b*x)^2*x^(1/2)/arctanh(tanh(b*x+a))^
(1/2)))

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Maxima [A]  time = 1.50686, size = 73, normalized size = 0.49 \begin{align*} -\frac{2 \,{\left (16 \, b^{4} x^{4} + 24 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - a^{3} b x + a^{4}\right )}}{5 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

-2/5*(16*b^4*x^4 + 24*a*b^3*x^3 + 6*a^2*b^2*x^2 - a^3*b*x + a^4)/((b*x + a)^(3/2)*a^4*x^(5/2))

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Fricas [A]  time = 1.99808, size = 128, normalized size = 0.86 \begin{align*} -\frac{2 \,{\left (16 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt{b x + a} \sqrt{x}}{5 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*b^3*x^3 + 8*a*b^2*x^2 - 2*a^2*b*x + a^3)*sqrt(b*x + a)*sqrt(x)/(a^4*b*x^4 + a^5*x^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.40087, size = 217, normalized size = 1.47 \begin{align*} -\frac{2 \, b^{3} \sqrt{x}}{\sqrt{b x + a} a^{4}} + \frac{4 \,{\left (5 \, b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{8} - 30 \, a b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{6} + 80 \, a^{2} b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 50 \, a^{3} b^{\frac{5}{2}}{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + 11 \, a^{4} b^{\frac{5}{2}}\right )}}{5 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{5} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-2*b^3*sqrt(x)/(sqrt(b*x + a)*a^4) + 4/5*(5*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 - 30*a*b^(5/2)*(sqrt(b
)*sqrt(x) - sqrt(b*x + a))^6 + 80*a^2*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 50*a^3*b^(5/2)*(sqrt(b)*sq
rt(x) - sqrt(b*x + a))^2 + 11*a^4*b^(5/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5*a^3)