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]

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

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Rubi [A]  time = 0.0114613, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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])

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*}

Mathematica [A]  time = 0.0450765, size = 21, normalized size = 1. \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right ) \]

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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Maple [A]  time = 0.032, size = 17, normalized size = 0.8 \begin{align*}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+\tanh \left ( x \right ) }} \right ) \sqrt{2} \end{align*}

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 = 1.60106, size = 58, normalized size = 2.76 \begin{align*} -\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 ) \end{align*}

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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Fricas [B]  time = 2.21971, size = 182, normalized size = 8.67 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-2 \, \sqrt{2} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tanh{\left (x \right )} + 1}\, dx \end{align*}

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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Giac [A]  time = 1.20234, size = 36, normalized size = 1.71 \begin{align*} -\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 ) \end{align*}

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)