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]

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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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [B]  time = 0.48, size = 168, normalized size = 3.43 \[ -\frac {2 \, \sqrt {2} {\left (2 \, \sqrt {2} \cosh \relax (x)^{2} + 4 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x) + 2 \, \sqrt {2} \sinh \relax (x)^{2} - \sqrt {2}\right )} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 3 \, {\left (\sqrt {2} \cosh \relax (x)^{3} + 3 \, \sqrt {2} \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{2} + \sqrt {2} \sinh \relax (x)^{3}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right )}{24 \, {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + \sinh \relax (x)^{3}\right )}} \]

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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giac [B]  time = 0.16, size = 74, normalized size = 1.51 \[ -\frac {1}{24} \, \sqrt {2} {\left (\frac {2 \, {\left (3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} - 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} + 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) - 4\right )} \]

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)

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maple [A]  time = 0.06, size = 35, normalized size = 0.71 \[ \frac {\arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{4}-\frac {1}{2 \sqrt {1+\tanh \relax (x )}}+\frac {1}{3 \left (1+\tanh \relax (x )\right )^{\frac {3}{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{12} \, \sqrt {2} {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}} + \int \frac {e^{\left (-x\right )}}{2 \, {\left (\frac {\sqrt {2} e^{\left (-x\right )}}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}}} + \frac {\sqrt {2} e^{\left (-3 \, x\right )}}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}}}\right )}}\,{d x} \]

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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mupad [B]  time = 1.09, size = 32, normalized size = 0.65 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )}{4}-\frac {\frac {\mathrm {tanh}\relax (x)}{2}+\frac {1}{6}}{{\left (\mathrm {tanh}\relax (x)+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(tanh(x) + 1)^(3/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/4 - (tanh(x)/2 + 1/6)/(tanh(x) + 1)^(3/2)

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sympy [A]  time = 11.45, size = 82, normalized size = 1.67 \[ - \frac {\begin {cases} - \frac {\sqrt {2} \operatorname {acoth}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 > 2 \\- \frac {\sqrt {2} \operatorname {atanh}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 < 2 \end {cases}}{2} - \frac {1}{2 \sqrt {\tanh {\relax (x )} + 1}} + \frac {1}{3 \left (\tanh {\relax (x )} + 1\right )^{\frac {3}{2}}} \]

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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