3.145 \(\int \frac {1}{x \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\)

Optimal. Leaf size=49 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2161} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2161

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(2*ArcTan[Sqrt[v
]/Rt[(b*u - a*v)/a, 2]])/(a*Rt[(b*u - a*v)/a, 2]), x] /; NeQ[b*u - a*v, 0] && PosQ[(b*u - a*v)/a]] /; Piecewis
eLinearQ[u, v, x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 47, normalized size = 0.96 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 56, normalized size = 1.14 \[ \left [\frac {\log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right )}{\sqrt {a}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right )}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x)/sqrt(a), 2*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a)/a]

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giac [A]  time = 0.17, size = 21, normalized size = 0.43 \[ \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a)

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maple [A]  time = 0.17, size = 42, normalized size = 0.86 \[ -\frac {2 \arctanh \left (\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )-b x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sqrt {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.19, size = 285, normalized size = 5.82 \[ \frac {\sqrt {2}\,\ln \left (\frac {\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,\left (\frac {\sqrt {2}\,b\,x}{2}-\frac {\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2}+\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{x}\right )\,1{}\mathrm {i}}{\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*log(((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)^(1/2)*((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)*(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)^(1/2)*1
i - (2^(1/2)*(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*b*
x))/2 + (2^(1/2)*b*x)/2)*1i)/x)*1i)/(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)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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