3.54 \(\int \frac {1}{\sqrt {1+\tanh (x)}} \, dx\)

Optimal. Leaf size=32 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\tanh (x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3479, 3480, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Tanh[x]],x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Tanh[x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 32, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Tanh[x]],x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Tanh[x]]

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fricas [B]  time = 0.52, size = 85, normalized size = 2.66 \[ \frac {{\left (\sqrt {2} \cosh \relax (x) + \sqrt {2} \sinh \relax (x)\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right ) - 4 \, \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}}}{4 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x))^(1/2),x, algorithm="fricas")

[Out]

1/4*((sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x))
- 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*sinh(x)^2 - 1) - 4*sqrt(cosh(x)/(cosh(x) - sinh(x))))/(cosh(x) + sinh(x)
)

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giac [A]  time = 0.14, size = 50, normalized size = 1.56 \[ -\frac {1}{4} \, \sqrt {2} {\left (\frac {2}{\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}} + \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x))^(1/2),x, algorithm="giac")

[Out]

-1/4*sqrt(2)*(2/(sqrt(e^(4*x) + e^(2*x)) - e^(2*x)) + log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))

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maple [A]  time = 0.07, size = 27, normalized size = 0.84 \[ \frac {\arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}-\frac {1}{\sqrt {1+\tanh \relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+tanh(x))^(1/2),x)

[Out]

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

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maxima [B]  time = 0.41, size = 57, normalized size = 1.78 \[ -\frac {1}{4} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {e^{\left (-2 \, x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x))^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(2)*log(-(sqrt(2) - sqrt(2)/sqrt(e^(-2*x) + 1))/(sqrt(2) + sqrt(2)/sqrt(e^(-2*x) + 1))) - 1/2*sqrt(2)
*sqrt(e^(-2*x) + 1)

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mupad [B]  time = 0.13, size = 26, normalized size = 0.81 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )}{2}-\frac {1}{\sqrt {\mathrm {tanh}\relax (x)+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x) + 1)^(1/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/2 - 1/(tanh(x) + 1)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\tanh {\relax (x )} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(x))**(1/2),x)

[Out]

Integral(1/sqrt(tanh(x) + 1), x)

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