3.3.53 \(\int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\) [253]

Optimal. Leaf size=33 \[ -\frac {2 \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

-2*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))/arctanh(tanh(b*x+a))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 33, 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 \sqrt {x}}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 32, normalized size = 0.97 \begin {gather*} \frac {2 \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 29, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^(1/2)/(arctanh(tanh(b*x+a))-b*x)/arctanh(tanh(b*x+a))^(1/2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 22, normalized size = 0.67 \begin {gather*} \frac {2 \, \sqrt {b x + a} \sqrt {x}}{a b x + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + a)*sqrt(x)/(a*b*x + a^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.39, size = 15, normalized size = 0.45 \begin {gather*} \frac {2 \, \sqrt {x}}{\sqrt {b x + a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x)/(sqrt(b*x + a)*a)

________________________________________________________________________________________

Mupad [B]
time = 1.74, size = 163, normalized size = 4.94 \begin {gather*} \frac {4\,x\,\sqrt {\frac {\ln \left (\frac {{\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 {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{\left (\frac {\sqrt {x}\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2\,b}-\frac {\sqrt {x}\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2\,b}\right )\,\left (b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-b\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b^2\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________