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

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

[Out]

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

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Rubi [A]  time = 0.0352135, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3527, 3480, 206} \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-2 \sqrt{\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]*Sqrt[1 + Tanh[x]],x]

[Out]

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

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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]

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 \tanh (x) \sqrt{1+\tanh (x)} \, dx &=-2 \sqrt{1+\tanh (x)}+\int \sqrt{1+\tanh (x)} \, dx\\ &=-2 \sqrt{1+\tanh (x)}+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 )-2 \sqrt{1+\tanh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0441599, size = 32, normalized size = 1. \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-2 \sqrt{\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]*Sqrt[1 + Tanh[x]],x]

[Out]

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

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Maple [A]  time = 0.026, size = 26, normalized size = 0.8 \begin{align*}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+\tanh \left ( x \right ) }} \right ) \sqrt{2}-2\,\sqrt{1+\tanh \left ( x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\sqrt{2}}{\sqrt{e^{\left (-2 \, x\right )} + 1}} + \int \frac{\sqrt{2} e^{\left (-x\right )}}{{\left (e^{\left (-x\right )} + e^{\left (-3 \, x\right )}\right )} \sqrt{e^{\left (-2 \, x\right )} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)/sqrt(e^(-2*x) + 1) + integrate(sqrt(2)*e^(-x)/((e^(-x) + e^(-3*x))*sqrt(e^(-2*x) + 1)), x)

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Fricas [B]  time = 2.29843, size = 456, normalized size = 14.25 \begin{align*} -\frac{4 \, \sqrt{2}{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} -{\left (\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2} + \sqrt{2}\right )} \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 )}{2 \,{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \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)*tanh(x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.13934, size = 70, normalized size = 2.19 \begin{align*} - 2 \sqrt{\tanh{\left (x \right )} + 1} - 2 \left (\begin{cases} - \frac{\sqrt{2} \operatorname{acoth}{\left (\frac{\sqrt{2} \sqrt{\tanh{\left (x \right )} + 1}}{2} \right )}}{2} & \text{for}\: \tanh{\left (x \right )} + 1 > 2 \\- \frac{\sqrt{2} \operatorname{atanh}{\left (\frac{\sqrt{2} \sqrt{\tanh{\left (x \right )} + 1}}{2} \right )}}{2} & \text{for}\: \tanh{\left (x \right )} + 1 < 2 \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(tanh(x) + 1) - 2*Piecewise((-sqrt(2)*acoth(sqrt(2)*sqrt(tanh(x) + 1)/2)/2, tanh(x) + 1 > 2), (-sqrt(2)
*atanh(sqrt(2)*sqrt(tanh(x) + 1)/2)/2, tanh(x) + 1 < 2))

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Giac [B]  time = 1.26433, size = 72, normalized size = 2.25 \begin{align*} \frac{1}{2} \, \sqrt{2}{\left (\frac{4}{\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1} - \log \left (-2 \, \sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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