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

Optimal. Leaf size=72 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

2/3*arctanh(tanh(b*x+a))^(1/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))+4/3*b*arctanh(tanh(b*x+a))^(1/2)/(b*x-arctan
h(tanh(b*x+a)))^2/x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 46, normalized size = 0.64 \[ -\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-3 b x\right )}{3 x^{3/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

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

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fricas [A]  time = 0.56, size = 23, normalized size = 0.32 \[ \frac {2 \, {\left (2 \, b x - a\right )} \sqrt {b x + a}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 55, normalized size = 0.76 \[ \frac {8 \, {\left (3 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )} b^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 59, normalized size = 0.82 \[ -\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {4 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(arctanh(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)^2/x^(1/2)*
arctanh(tanh(b*x+a))^(1/2)

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maxima [A]  time = 0.43, size = 33, normalized size = 0.46 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} + a b x - a^{2}\right )}}{3 \, \sqrt {b x + a} a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.57, size = 218, normalized size = 3.03 \[ \frac {\sqrt {2}\,\left (\frac {\frac {4\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\frac {4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}+\frac {8\,b\,x}{3}}{{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {16\,b\,x}{3\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )\,\sqrt {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}}{2\,x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*(((4*log(2/(exp(2*a)*exp(2*b*x) + 1)))/3 - (4*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))
/3 + (8*b*x)/3)/(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
*b*x)^2 + (16*b*x)/(3*(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*b*x)^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) + 1)))^(1
/2))/(2*x^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {5}{2}} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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