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

Optimal. Leaf size=30 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{\sqrt {\tanh (x)+1}} \]

[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 = 0.04, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {3526, 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[Tanh[x]/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 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 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\tanh (x)}{\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 = 0.06, size = 30, 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[Tanh[x]/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 = 0.46, size = 85, normalized size = 2.83 \[ \frac {{\left (\sqrt {2} \cosh \relax (x) + \sqrt {2} \sinh \relax (x)\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right ) + 4 \, \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}}}{4 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(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 [B]  time = 0.18, size = 58, normalized size = 1.93 \[ -\frac {1}{4} \, \sqrt {2} \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \sqrt {2} + \frac {\sqrt {2}}{2 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 25, normalized size = 0.83 \[ \frac {\arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}+\frac {1}{\sqrt {1+\tanh \relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, \sqrt {2} \sqrt {e^{\left (-2 \, x\right )} + 1} + \int \frac {e^{\left (-x\right )}}{\frac {\sqrt {2} e^{\left (-x\right )}}{\sqrt {e^{\left (-2 \, x\right )} + 1}} + \frac {\sqrt {2} e^{\left (-3 \, x\right )}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 24, normalized size = 0.80 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )}{2}+\frac {1}{\sqrt {\mathrm {tanh}\relax (x)+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(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 [A]  time = 2.65, size = 66, normalized size = 2.20 \[ - \begin {cases} - \frac {\sqrt {2} \operatorname {acoth}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 > 2 \\- \frac {\sqrt {2} \operatorname {atanh}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 < 2 \end {cases} + \frac {1}{\sqrt {\tanh {\relax (x )} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)) + 1/sqrt(tanh(x) + 1)

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