3.131 \(\int \frac{\tanh ^2(x)}{\sqrt{1+\tanh (x)}} \, dx\)

Optimal. Leaf size=42 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{\sqrt{2}}-2 \sqrt{\tanh (x)+1}-\frac{1}{\sqrt{\tanh (x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.0558897, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3543, 3479, 3480, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{\sqrt{2}}-2 \sqrt{\tanh (x)+1}-\frac{1}{\sqrt{\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/Sqrt[1 + Tanh[x]],x]

[Out]

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

Rule 3543

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 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 \frac{\tanh ^2(x)}{\sqrt{1+\tanh (x)}} \, dx &=-2 \sqrt{1+\tanh (x)}+\int \frac{1}{\sqrt{1+\tanh (x)}} \, dx\\ &=-\frac{1}{\sqrt{1+\tanh (x)}}-2 \sqrt{1+\tanh (x)}+\frac{1}{2} \int \sqrt{1+\tanh (x)} \, dx\\ &=-\frac{1}{\sqrt{1+\tanh (x)}}-2 \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)}}-2 \sqrt{1+\tanh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0877899, size = 37, normalized size = 0.88 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{\sqrt{2}}+\frac{-2 \tanh (x)-3}{\sqrt{\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/Sqrt[1 + Tanh[x]],x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] + (-3 - 2*Tanh[x])/Sqrt[1 + Tanh[x]]

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Maple [A]  time = 0.039, size = 35, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}}{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+\tanh \left ( x \right ) }} \right ) }-{\frac{1}{\sqrt{1+\tanh \left ( x \right ) }}}-2\,\sqrt{1+\tanh \left ( x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tanh(x)^2/sqrt(tanh(x) + 1), x)

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Fricas [B]  time = 2.34995, size = 639, normalized size = 15.21 \begin{align*} -\frac{2 \, \sqrt{2}{\left (5 \, \sqrt{2} \cosh \left (x\right )^{2} + 10 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + 5 \, \sqrt{2} \sinh \left (x\right )^{2} + \sqrt{2}\right )} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} -{\left (\sqrt{2} \cosh \left (x\right )^{3} + 3 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt{2} \sinh \left (x\right )^{3} +{\left (3 \, \sqrt{2} \cosh \left (x\right )^{2} + \sqrt{2}\right )} \sinh \left (x\right ) + \sqrt{2} \cosh \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 \,{\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} +{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*(5*sqrt(2)*cosh(x)^2 + 10*sqrt(2)*cosh(x)*sinh(x) + 5*sqrt(2)*sinh(x)^2 + sqrt(2))*sqrt(cosh(x
)/(cosh(x) - sinh(x))) - (sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)*sinh(x)^2 + sqrt(2)*sinh(x)^3 + (3*sqrt(2)*cos
h(x)^2 + sqrt(2))*sinh(x) + sqrt(2)*cosh(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))/(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (3*co
sh(x)^2 + 1)*sinh(x) + cosh(x))

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Sympy [A]  time = 3.40278, size = 78, normalized size = 1.86 \begin{align*} - 2 \sqrt{\tanh{\left (x \right )} + 1} - \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} - \frac{1}{\sqrt{\tanh{\left (x \right )} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25964, size = 73, normalized size = 1.74 \begin{align*} -\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{5 \, \sqrt{2} e^{\left (2 \, x\right )} + \sqrt{2}}{2 \, \sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(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*(5*sqrt(2)*e^(2*x) + sqrt(2))/sqrt(e^(4*x)
+ e^(2*x))