\(\int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 13, antiderivative size = 49 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {1}{3 (1+\tanh (x))^{3/2}}+\frac {3}{2 \sqrt {1+\tanh (x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3621, 3607, 3561, 212} \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{2 \sqrt {2}}+\frac {3}{2 \sqrt {\tanh (x)+1}}-\frac {1}{3 (\tanh (x)+1)^{3/2}} \]

[In]

Int[Tanh[x]^2/(1 + Tanh[x])^(3/2),x]

[Out]

ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/(2*Sqrt[2]) - 1/(3*(1 + Tanh[x])^(3/2)) + 3/(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 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]

Rule 3621

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^m/(2*a^3*f*m)), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*
Simp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a
*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 (1+\tanh (x))^{3/2}}-\frac {1}{2} \int \frac {1-2 \tanh (x)}{\sqrt {1+\tanh (x)}} \, dx \\ & = -\frac {1}{3 (1+\tanh (x))^{3/2}}+\frac {3}{2 \sqrt {1+\tanh (x)}}+\frac {1}{4} \int \sqrt {1+\tanh (x)} \, dx \\ & = -\frac {1}{3 (1+\tanh (x))^{3/2}}+\frac {3}{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 {3}{2 \sqrt {1+\tanh (x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {14+18 \tanh (x)+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right ) (1+\tanh (x))^{3/2}}{12 (1+\tanh (x))^{3/2}} \]

[In]

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

[Out]

(14 + 18*Tanh[x] + 3*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]*(1 + Tanh[x])^(3/2))/(12*(1 + Tanh[x])^(3/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71

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

[In]

int(tanh(x)^2/(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)+3/2/(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 = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 3.43 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {2 \, \sqrt {2} {\left (8 \, \sqrt {2} \cosh \left (x\right )^{2} + 16 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \sqrt {2} \sinh \left (x\right )^{2} - \sqrt {2}\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} + 3 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{24 \, {\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3}\right )}} \]

[In]

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

[Out]

1/24*(2*sqrt(2)*(8*sqrt(2)*cosh(x)^2 + 16*sqrt(2)*cosh(x)*sinh(x) + 8*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 = 5.55 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{8} + \frac {3}{2 \sqrt {\tanh {\left (x \right )} + 1}} - \frac {1}{3 \left (\tanh {\left (x \right )} + 1\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.94 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {1}{24} \, \sqrt {2} {\left (\frac {2 \, {\left (6 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} + 3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} - 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} - 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

[In]

integrate(tanh(x)^2/(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))

Mupad [B] (verification not implemented)

Time = 1.69 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.63 \[ \int \frac {\tanh ^2(x)}{(1+\tanh (x))^{3/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{4}+\frac {\frac {3\,\mathrm {tanh}\left (x\right )}{2}+\frac {7}{6}}{{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}} \]

[In]

int(tanh(x)^2/(tanh(x) + 1)^(3/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/4 + ((3*tanh(x))/2 + 7/6)/(tanh(x) + 1)^(3/2)