\(\int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx\) [127]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{\sqrt {1+\tanh (x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3607, 3561, 212} \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{\sqrt {\tanh (x)+1}} \]

[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 212

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

Rule 3561

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 3607

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{\sqrt {1+\tanh (x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}+\frac {1}{\sqrt {1+\tanh \left (x \right )}}\)	\(25\)
default	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}+\frac {1}{\sqrt {1+\tanh \left (x \right )}}\)	\(25\)

[In]

int(tanh(x)/(1+tanh(x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.83 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {{\left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right )\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right ) + 4 \, \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}}}{4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]

[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)
)

Sympy [A] (verification not implemented)

Time = 1.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{4} + \frac {1}{\sqrt {\tanh {\left (x \right )} + 1}} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {\tanh \left (x\right ) + 1}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {1}{4} \, \sqrt {2} {\left (\frac {2}{\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}} - \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

[In]

integrate(tanh(x)/(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))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\tanh (x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{2}+\frac {1}{\sqrt {\mathrm {tanh}\left (x\right )+1}} \]

[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)