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

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

[Out]

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

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Rubi [A]  time = 0.079619, antiderivative size = 49, 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 = {3540, 3526, 3480, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{3}{2 \sqrt{\tanh (x)+1}}-\frac{1}{3 (\tanh (x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(1 + Tanh[x])^(3/2),x]

[Out]

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

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && 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]

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

Mathematica [A]  time = 0.095627, size = 53, normalized size = 1.08 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{(\cosh (x)-\sinh (x)) (9 \sinh (x)+7 \cosh (x))}{6 \sqrt{\tanh (x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/(2*Sqrt[2]) + ((Cosh[x] - Sinh[x])*(7*Cosh[x] + 9*Sinh[x]))/(6*Sqrt[1 + Tan
h[x]])

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Maple [A]  time = 0.021, 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{3}{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)^2/(1+tanh(x))^(3/2),x)

[Out]

1/4*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)+3/2/(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*} \int \frac{\tanh \left (x\right )^{2}}{{\left (\tanh \left (x\right ) + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.26444, size = 579, normalized size = 11.82 \begin{align*} \frac{2 \, \sqrt{2}{\left (8 \, \sqrt{2} \cosh \left (x\right )^{2} + 16 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \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)^2/(1+tanh(x))^(3/2),x, algorithm="fricas")

[Out]

1/24*(2*sqrt(2)*(8*sqrt(2)*cosh(x)^2 + 16*sqrt(2)*cosh(x)*sinh(x) + 8*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 = 11.4247, 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{3}{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)**2/(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 + 3/(2*sqrt(tanh(x) + 1)) - 1/(3*(tanh(x) + 1)**(3/2))

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Giac [B]  time = 1.26838, size = 136, normalized size = 2.78 \begin{align*} -\frac{1}{8} \, \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{2}{3} \, \sqrt{2} + \frac{\sqrt{2}{\left (6 \,{\left (\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} + 3 \, \sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} - 1\right )}}{12 \,{\left (\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1) - 2/3*sqrt(2) + 1/12*sqrt(2)*(6*(sqrt(e^(4*x) + e
^(2*x)) - e^(2*x))^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