\(\int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx\) [128]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 49 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}+\frac {1}{3 (1+\tanh (x))^{3/2}}-\frac {1}{2 \sqrt {1+\tanh (x)}} \]

[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] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3607, 3560, 3561, 212} \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {1}{2 \sqrt {\tanh (x)+1}}+\frac {1}{3 (\tanh (x)+1)^{3/2}} \]

[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 212

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

Rule 3560

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 3561

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 3607

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*} \text {integral}& = \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} \text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\tanh (x)}\right ) \\ & = \frac {\text {arctanh}\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 [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {2-3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+\tanh (x))\right ) (1+\tanh (x))}{6 (1+\tanh (x))^{3/2}} \]

[In]

Integrate[Tanh[x]/(1 + Tanh[x])^(3/2),x]

[Out]

(2 - 3*Hypergeometric2F1[-1/2, 1, 1/2, (1 + Tanh[x])/2]*(1 + Tanh[x]))/(6*(1 + Tanh[x])^(3/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71

method result size
derivativedivides \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{4}-\frac {1}{2 \sqrt {1+\tanh \left (x \right )}}+\frac {1}{3 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}\) \(35\)
default \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{4}-\frac {1}{2 \sqrt {1+\tanh \left (x \right )}}+\frac {1}{3 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}\) \(35\)

[In]

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

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (34) = 68\).

Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 3.43 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=-\frac {2 \, \sqrt {2} {\left (2 \, \sqrt {2} \cosh \left (x\right )^{2} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sqrt {2} \sinh \left (x\right )^{2} - \sqrt {2}\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{24 \, {\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{8} - \frac {1}{2 \sqrt {\tanh {\left (x \right )} + 1}} + \frac {1}{3 \left (\tanh {\left (x \right )} + 1\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-sqrt(2)*(log(sqrt(tanh(x) + 1) - sqrt(2)) - log(sqrt(tanh(x) + 1) + sqrt(2)))/8 - 1/(2*sqrt(tanh(x) + 1)) + 1
/(3*(tanh(x) + 1)**(3/2))

Maxima [F]

\[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=\int { \frac {\tanh \left (x\right )}{{\left (\tanh \left (x\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (34) = 68\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.49 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=-\frac {1}{24} \, \sqrt {2} {\left (\frac {2 \, {\left (3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} - 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} + 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65 \[ \int \frac {\tanh (x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{4}-\frac {\frac {\mathrm {tanh}\left (x\right )}{2}+\frac {1}{6}}{{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}} \]

[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)