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

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)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {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[(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]

Rubi steps

\begin {align*} \int \frac {1}{(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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.15, size = 53, normalized size = 1.08 \[ \frac {1}{12} \left (3 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2 (\cosh (x)-\sinh (x)) (3 \sinh (x)+5 \cosh (x))}{\sqrt {\tanh (x)+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tanh[x])^(-3/2),x]

[Out]

(3*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (2*(Cosh[x] - Sinh[x])*(5*Cosh[x] + 3*Sinh[x]))/Sqrt[1 + Tanh[
x]])/12

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 166, normalized size = 3.39 \[ -\frac {2 \, \sqrt {2} {\left (4 \, \sqrt {2} \cosh \relax (x)^{2} + 8 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x) + 4 \, \sqrt {2} \sinh \relax (x)^{2} + \sqrt {2}\right )} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 3 \, {\left (\sqrt {2} \cosh \relax (x)^{3} + 3 \, \sqrt {2} \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{2} + \sqrt {2} \sinh \relax (x)^{3}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right )}{24 \, {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + \sinh \relax (x)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(2*sqrt(2)*(4*sqrt(2)*cosh(x)^2 + 8*sqrt(2)*cosh(x)*sinh(x) + 4*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)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 96, normalized size = 1.96 \[ -\frac {1}{24} \, \sqrt {2} {\left (\frac {2 \, {\left (6 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 3 \, e^{\left (2 \, x\right )} + 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} + 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right ) - 8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt(2)*(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 + 3*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1) - 8)

________________________________________________________________________________________

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(1/(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)

________________________________________________________________________________________

maxima [B]  time = 0.42, size = 69, normalized size = 1.41 \[ -\frac {1}{12} \, \sqrt {2} {\left (\frac {3}{e^{\left (-2 \, x\right )} + 1} + 1\right )} {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 32, normalized size = 0.65 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )}{4}-\frac {\frac {\mathrm {tanh}\relax (x)}{2}+\frac {5}{6}}{{\left (\mathrm {tanh}\relax (x)+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tanh(x) + 1)^(3/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/4 - (tanh(x)/2 + 5/6)/(tanh(x) + 1)^(3/2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\tanh {\relax (x )} + 1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((tanh(x) + 1)**(-3/2), x)

________________________________________________________________________________________