3.3.4 \(\int (1-\tanh ^2(x))^{3/2} \, dx\) [204]

Optimal. Leaf size=22 \[ \frac {1}{2} \text {ArcSin}(\tanh (x))+\frac {1}{2} \sqrt {\text {sech}^2(x)} \tanh (x) \]

[Out]

1/2*arcsin(tanh(x))+1/2*(sech(x)^2)^(1/2)*tanh(x)

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3738, 4207, 201, 222} \begin {gather*} \frac {1}{2} \text {ArcSin}(\tanh (x))+\frac {1}{2} \tanh (x) \sqrt {\text {sech}^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Tanh[x]^2)^(3/2),x]

[Out]

ArcSin[Tanh[x]]/2 + (Sqrt[Sech[x]^2]*Tanh[x])/2

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 3738

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[b*(ff/
f), Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 29, normalized size = 1.32 \begin {gather*} \frac {\text {sech}(x) \left (2 \text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right )+\text {sech}(x) \tanh (x)\right )}{2 \sqrt {\text {sech}^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Tanh[x]^2)^(3/2),x]

[Out]

(Sech[x]*(2*ArcTan[Tanh[x/2]] + Sech[x]*Tanh[x]))/(2*Sqrt[Sech[x]^2])

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Maple [A]
time = 0.38, size = 21, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\tanh \left (x \right ) \sqrt {1-\left (\tanh ^{2}\left (x \right )\right )}}{2}+\frac {\arcsin \left (\tanh \left (x \right )\right )}{2}\)	\(21\)
default	\(\frac {\tanh \left (x \right ) \sqrt {1-\left (\tanh ^{2}\left (x \right )\right )}}{2}+\frac {\arcsin \left (\tanh \left (x \right )\right )}{2}\)	\(21\)
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}+\frac {i \sqrt {\frac {{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}+i\right )}{2}-\frac {i \sqrt {\frac {{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}-i\right )}{2}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-tanh(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 28, normalized size = 1.27 \begin {gather*} \frac {e^{\left (3 \, x\right )} - e^{x}}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1} + \arctan \left (e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(3*x) - e^x)/(e^(4*x) + 2*e^(2*x) + 1) + arctan(e^x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (16) = 32\).
time = 0.34, size = 140, normalized size = 6.36 \begin {gather*} \frac {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} + {\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 )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + {\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right ) - \cosh \left (x\right )}{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (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)*arctan(cosh(x) + sinh(x)) + (3*cosh(x)^2 -
1)*sinh(x) - cosh(x))/(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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (1 - \tanh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((1 - tanh(x)**2)**(3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (16) = 32\).
time = 0.40, size = 45, normalized size = 2.05 \begin {gather*} \frac {1}{4} \, \pi - \frac {e^{\left (-x\right )} - e^{x}}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4} + \frac {1}{2} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*pi - (e^(-x) - e^x)/((e^(-x) - e^x)^2 + 4) + 1/2*arctan(1/2*(e^(2*x) - 1)*e^(-x))

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Mupad [B]
time = 0.10, size = 20, normalized size = 0.91 \begin {gather*} \frac {\mathrm {asin}\left (\mathrm {tanh}\left (x\right )\right )}{2}+\frac {\mathrm {tanh}\left (x\right )\,\sqrt {1-{\mathrm {tanh}\left (x\right )}^2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - tanh(x)^2)^(3/2),x)

[Out]

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

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