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

Optimal. Leaf size=21 \[ \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3480, 206} \[ \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Tanh[x]],x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 21, normalized size = 1.00 \[ \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Tanh[x]],x]

[Out]

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

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fricas [B]  time = 0.57, size = 50, normalized size = 2.38 \[ \frac {1}{2} \, \sqrt {2} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 27, normalized size = 1.29 \[ -\frac {1}{2} \, \sqrt {2} \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 17, normalized size = 0.81 \[ \arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 43, normalized size = 2.05 \[ -\frac {1}{2} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 16, normalized size = 0.76 \[ \sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x) + 1)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tanh(x) + 1), x)

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