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

Optimal. Leaf size=34 \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-\frac{2}{3} (\tanh (x)+1)^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.0441479, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3543, 3480, 206} \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-\frac{2}{3} (\tanh (x)+1)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2*Sqrt[1 + Tanh[x]],x]

[Out]

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

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

Mathematica [A]  time = 0.067778, size = 34, normalized size = 1. \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-\frac{2}{3} (\tanh (x)+1)^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 26, normalized size = 0.8 \begin{align*}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+\tanh \left ( x \right ) }} \right ) \sqrt{2}-{\frac{2}{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((1+tanh(x))^(1/2)*tanh(x)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.20589, size = 818, normalized size = 24.06 \begin{align*} -\frac{8 \, \sqrt{2}{\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 )} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \,{\left (\sqrt{2} \cosh \left (x\right )^{4} + 4 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt{2} \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \sqrt{2} \cosh \left (x\right )^{2} + \sqrt{2}\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right )^{2} + 4 \,{\left (\sqrt{2} \cosh \left (x\right )^{3} + \sqrt{2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt{2}\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 )}{6 \,{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(8*sqrt(2)*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)^2*sinh(x) + 3*sqrt(2)*cosh(x)*sinh(x)^2 + sqrt(2)*sinh(
x)^3)*sqrt(cosh(x)/(cosh(x) - sinh(x))) - 3*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)
^4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 2*sqrt(2)*cosh(x)^2 + 4*(sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x
))*sinh(x) + sqrt(2))*log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) - 2*cosh(x)^2 - 4*c
osh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^
2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [A]  time = 2.99924, size = 71, normalized size = 2.09 \begin{align*} - \frac{2 \left (\tanh{\left (x \right )} + 1\right )^{\frac{3}{2}}}{3} - 2 \left (\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}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(tanh(x) + 1)**(3/2)/3 - 2*Piecewise((-sqrt(2)*acoth(sqrt(2)*sqrt(tanh(x) + 1)/2)/2, tanh(x) + 1 > 2), (-sq
rt(2)*atanh(sqrt(2)*sqrt(tanh(x) + 1)/2)/2, tanh(x) + 1 < 2))

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Giac [B]  time = 1.22259, size = 130, normalized size = 3.82 \begin{align*} \frac{1}{6} \, \sqrt{2}{\left (\frac{8 \,{\left (3 \,{\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 )}}{{\left (\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\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 )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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