3.128 \(\int \frac{\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx\)

Optimal. Leaf size=49 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{1}{2 \sqrt{\tanh (x)+1}}+\frac{1}{3 (\tanh (x)+1)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0488047, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3526, 3479, 3480, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}}-\frac{1}{2 \sqrt{\tanh (x)+1}}+\frac{1}{3 (\tanh (x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(1 + Tanh[x])^(3/2),x]

[Out]

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

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]

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 (x)}{(1+\tanh (x))^{3/2}} \, dx &=\frac{1}{3 (1+\tanh (x))^{3/2}}+\frac{1}{2} \int \frac{1}{\sqrt{1+\tanh (x)}} \, dx\\ &=\frac{1}{3 (1+\tanh (x))^{3/2}}-\frac{1}{2 \sqrt{1+\tanh (x)}}+\frac{1}{4} \int \sqrt{1+\tanh (x)} \, dx\\ &=\frac{1}{3 (1+\tanh (x))^{3/2}}-\frac{1}{2 \sqrt{1+\tanh (x)}}+\frac{1}{2} \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 )}{2 \sqrt{2}}+\frac{1}{3 (1+\tanh (x))^{3/2}}-\frac{1}{2 \sqrt{1+\tanh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.11948, size = 51, normalized size = 1.04 \[ \frac{1}{12} \left (3 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-\frac{2 (\cosh (x)-\sinh (x)) (3 \sinh (x)+\cosh (x))}{\sqrt{\tanh (x)+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(1 + Tanh[x])^(3/2),x]

[Out]

(3*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (2*(Cosh[x] - Sinh[x])*(Cosh[x] + 3*Sinh[x]))/Sqrt[1 + Tanh[x]
])/12

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Maple [A]  time = 0.016, size = 35, normalized size = 0.7 \begin{align*}{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+\tanh \left ( x \right ) }} \right ) }-{\frac{1}{2}{\frac{1}{\sqrt{1+\tanh \left ( x \right ) }}}}+{\frac{1}{3} \left ( 1+\tanh \left ( x \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(1+tanh(x))^(3/2),x)

[Out]

1/4*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-1/2/(1+tanh(x))^(1/2)+1/3/(1+tanh(x))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{12} \, \sqrt{2}{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}} + \int \frac{e^{\left (-x\right )}}{2 \,{\left (\frac{\sqrt{2} e^{\left (-x\right )}}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}} + \frac{\sqrt{2} e^{\left (-3 \, x\right )}}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac{3}{2}}}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(1+tanh(x))^(3/2),x, algorithm="maxima")

[Out]

1/12*sqrt(2)*(e^(-2*x) + 1)^(3/2) + integrate(1/2*e^(-x)/(sqrt(2)*e^(-x)/(e^(-2*x) + 1)^(3/2) + sqrt(2)*e^(-3*
x)/(e^(-2*x) + 1)^(3/2)), x)

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Fricas [B]  time = 2.24738, size = 579, normalized size = 11.82 \begin{align*} -\frac{2 \, \sqrt{2}{\left (2 \, \sqrt{2} \cosh \left (x\right )^{2} + 4 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sqrt{2} \sinh \left (x\right )^{2} - \sqrt{2}\right )} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \,{\left (\sqrt{2} \cosh \left (x\right )^{3} + 3 \, \sqrt{2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt{2} \sinh \left (x\right )^{3}\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 )}{24 \,{\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(1+tanh(x))^(3/2),x, algorithm="fricas")

[Out]

-1/24*(2*sqrt(2)*(2*sqrt(2)*cosh(x)^2 + 4*sqrt(2)*cosh(x)*sinh(x) + 2*sqrt(2)*sinh(x)^2 - sqrt(2))*sqrt(cosh(x
)/(cosh(x) - sinh(x))) - 3*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)^2*sinh(x) + 3*sqrt(2)*cosh(x)*sinh(x)^2 + sq
rt(2)*sinh(x)^3)*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)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 + sinh(x)^3)

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Sympy [A]  time = 10.0321, size = 82, normalized size = 1.67 \begin{align*} - \frac{\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}}{2} - \frac{1}{2 \sqrt{\tanh{\left (x \right )} + 1}} + \frac{1}{3 \left (\tanh{\left (x \right )} + 1\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [B]  time = 1.26935, size = 100, normalized size = 2.04 \begin{align*} -\frac{1}{24} \, \sqrt{2}{\left (\frac{2 \,{\left (3 \, \sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} - 1\right )}}{{\left (\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} + 3 \, \log \left (-2 \, \sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) - 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(1+tanh(x))^(3/2),x, algorithm="giac")

[Out]

-1/24*sqrt(2)*(2*(3*sqrt(e^(4*x) + e^(2*x)) - 3*e^(2*x) - 1)/(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^3 + 3*log(-2*
sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1) - 4)