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

Optimal. Leaf size=35 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0140935, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {2167} \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{\tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{7/2}} \, dx &=\frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end{align*}

Mathematica [A]  time = 0.0363926, size = 34, normalized size = 0.97 \[ \frac{2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{5/2} \left (5 b x-5 \tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.146, size = 29, normalized size = 0.8 \begin{align*} -{\frac{2}{5\,{\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) -5\,bx} \left ({\it Artanh} \left ( \tanh \left ( bx+a \right ) \right ) \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(5/2)

________________________________________________________________________________________

Maxima [A]  time = 1.48283, size = 20, normalized size = 0.57 \begin{align*} -\frac{2 \,{\left (b x + a\right )}^{\frac{5}{2}}}{5 \, a x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(b*x + a)^(5/2)/(a*x^(5/2))

________________________________________________________________________________________

Fricas [A]  time = 2.30816, size = 78, normalized size = 2.23 \begin{align*} -\frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b x + a}}{5 \, a x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/(a*x^(5/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.33761, size = 45, normalized size = 1.29 \begin{align*} -\frac{2 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{6}}{5 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{5}{2}} a{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(b*x + a)^(5/2)*b^6/(((b*x + a)*b - a*b)^(5/2)*a*abs(b))