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

Optimal. Leaf size=33 \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-2 \sqrt{\tanh (x)+1} \]

[Out]

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

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Rubi [A]  time = 0.0211645, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3478, 3480, 206} \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-2 \sqrt{\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tanh[x])^(3/2),x]

[Out]

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

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[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] && G
tQ[n, 1]

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

Mathematica [A]  time = 0.0665317, size = 33, normalized size = 1. \[ 2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\tanh (x)+1}}{\sqrt{2}}\right )-2 \sqrt{\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tanh[x])^(3/2),x]

[Out]

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

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Maple [A]  time = 0.013, size = 27, normalized size = 0.8 \begin{align*} 2\,{\it Artanh} \left ( 1/2\,\sqrt{1+\tanh \left ( x \right ) }\sqrt{2} \right ) \sqrt{2}-2\,\sqrt{1+\tanh \left ( x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.56577, size = 77, normalized size = 2.33 \begin{align*} -\sqrt{2} \log \left (-\frac{\sqrt{2} - \frac{\sqrt{2}}{\sqrt{e^{\left (-2 \, x\right )} + 1}}}{\sqrt{2} + \frac{\sqrt{2}}{\sqrt{e^{\left (-2 \, x\right )} + 1}}}\right ) - \frac{2 \, \sqrt{2}}{\sqrt{e^{\left (-2 \, x\right )} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.25509, size = 451, normalized size = 13.67 \begin{align*} -\frac{2 \, \sqrt{2}{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \sqrt{\frac{\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} -{\left (\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2} + \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 )}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*sqrt(2)*(sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - (sqrt(2)*cosh(x)^2 + 2*sqr
t(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 + sqrt(2))*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)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 +
1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tanh{\left (x \right )} + 1\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((tanh(x) + 1)**(3/2), x)

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Giac [A]  time = 1.17263, size = 70, normalized size = 2.12 \begin{align*} \sqrt{2}{\left (\frac{2}{\sqrt{e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1} - \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))^(3/2),x, algorithm="giac")

[Out]

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