3.3.19 \(\int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx\) [219]

Optimal. Leaf size=35 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

2/3*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]
time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 = {2198} \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(5/2),x]

[Out]

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

Rule 2198

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] &&
EqQ[m + n + 2, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 34, normalized size = 0.97 \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{x^{3/2} \left (3 b x-3 \tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(5/2),x]

[Out]

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

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Maple [A]
time = 0.12, size = 29, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}\)	\(29\)
default	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(3/2)

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Maxima [A]
time = 0.49, size = 15, normalized size = 0.43 \begin {gather*} -\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, a x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

-2/3*(b*x + a)^(3/2)/(a*x^(3/2))

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Fricas [A]
time = 0.36, size = 15, normalized size = 0.43 \begin {gather*} -\frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, a x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

-2/3*(b*x + a)^(3/2)/(a*x^(3/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{x^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**(5/2),x)

[Out]

Integral(sqrt(atanh(tanh(a + b*x)))/x**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).
time = 0.41, size = 59, normalized size = 1.69 \begin {gather*} \frac {4 \, {\left (3 \, b^{\frac {3}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.71, size = 210, normalized size = 6.00 \begin {gather*} \frac {2\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}-2\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{\sqrt {x}\,\left (3\,x\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-3\,x\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2)/x^(5/2),x)

[Out]

(2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) +
1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2) - 2*log(2/(exp(2*a)*exp(2*b*x) + 1))*(log((2*exp(2*a)*exp(2*
b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(x^(1/2)*(3*x*log(2/(exp(2*a)*
exp(2*b*x) + 1)) - 3*x*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 6*b*x^2))

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