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

Optimal. Leaf size=110 \[ \frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

(16*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(15*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*Sqrt[ArcTanh[Tanh[a
+ b*x]]])/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(5*x^(5/2)*(b*x -
ArcTanh[Tanh[a + b*x]]))

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Rubi [A]  time = 0.0558492, antiderivative size = 110, 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 = {2171, 2167} \[ \frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(15*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*Sqrt[ArcTanh[Tanh[a
+ b*x]]])/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(5*x^(5/2)*(b*x -
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} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{(4 b) \int \frac{1}{x^{5/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac{8 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac{\left (8 b^2\right ) \int \frac{1}{x^{3/2} \sqrt{\tanh ^{-1}(\tanh (a+b x))}} \, dx}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac{16 b^2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt{x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac{8 b \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0483617, size = 66, normalized size = 0.6 \[ \frac{2 \sqrt{\tanh ^{-1}(\tanh (a+b x))} \left (-10 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 10*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15
*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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Maple [A]  time = 0.171, size = 105, normalized size = 1. \begin{align*} -{\frac{2}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{5}{2}}}}-{\frac{8\,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}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{x}^{-{\frac{3}{2}}}}+{\frac{2\,b}{3\, \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -bx \right ) ^{2}}\sqrt{{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) }{\frac{1}{\sqrt{x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-8/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(arc
tanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(1/2)+2/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*arctanh(t
anh(b*x+a))^(1/2))

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Maxima [A]  time = 1.48781, size = 61, normalized size = 0.55 \begin{align*} -\frac{2 \,{\left (8 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - a^{2} b x + 3 \, a^{3}\right )}}{15 \, \sqrt{b x + a} a^{3} 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))^(1/2),x, algorithm="maxima")

[Out]

-2/15*(8*b^3*x^3 + 4*a*b^2*x^2 - a^2*b*x + 3*a^3)/(sqrt(b*x + a)*a^3*x^(5/2))

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Fricas [A]  time = 2.00205, size = 88, normalized size = 0.8 \begin{align*} -\frac{2 \,{\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt{b x + a}}{15 \, a^{3} 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))^(1/2),x, algorithm="fricas")

[Out]

-2/15*(8*b^2*x^2 - 4*a*b*x + 3*a^2)*sqrt(b*x + a)/(a^3*x^(5/2))

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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))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17435, size = 104, normalized size = 0.95 \begin{align*} \frac{32 \,{\left (10 \,{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{4} - 5 \, a{\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} + a^{2}\right )} b^{\frac{5}{2}}}{15 \,{\left ({\left (\sqrt{b} \sqrt{x} - \sqrt{b x + a}\right )}^{2} - a\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*(10*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 5*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^2)*b^(5/2)/((sqrt(
b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5