\(\int \tanh (x) (1+\tanh (x))^{3/2} \, dx\) [125]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (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 = 45 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)}-\frac {2}{3} (1+\tanh (x))^{3/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, 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 = {3608, 3559, 3561, 212} \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{3} (\tanh (x)+1)^{3/2}-2 \sqrt {\tanh (x)+1} \]

[In]

Int[Tanh[x]*(1 + Tanh[x])^(3/2),x]

[Out]

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

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 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[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] && G
tQ[n, 1]

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 3608

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

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-\frac {2}{3} \sqrt {1+\tanh (x)} (4+\tanh (x)) \]

[In]

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

[Out]

2*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (2*Sqrt[1 + Tanh[x]]*(4 + Tanh[x]))/3

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 5.60 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=-\frac {2 \, \sqrt {2} {\left (5 \, \sqrt {2} \cosh \left (x\right )^{3} + 15 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + 5 \, \sqrt {2} \sinh \left (x\right )^{3} + 3 \, {\left (5 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right ) + 3 \, \sqrt {2} \cosh \left (x\right )\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \, {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt {2} \cosh \left (x\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} + \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt {2}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{3 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \]

[In]

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

[Out]

-1/3*(2*sqrt(2)*(5*sqrt(2)*cosh(x)^3 + 15*sqrt(2)*cosh(x)*sinh(x)^2 + 5*sqrt(2)*sinh(x)^3 + 3*(5*sqrt(2)*cosh(
x)^2 + sqrt(2))*sinh(x) + 3*sqrt(2)*cosh(x))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - 3*(sqrt(2)*cosh(x)^4 + 4*sqrt
(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + 2*(3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 2*sqrt(2)*cosh(x)^2
+ 4*(sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*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)^4 + 4*cosh(x)*sinh(x)^3 + s
inh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)

Sympy [A] (verification not implemented)

Time = 3.90 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=- \sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right ) - \frac {2 \left (\tanh {\left (x \right )} + 1\right )^{\frac {3}{2}}}{3} - 2 \sqrt {\tanh {\left (x \right )} + 1} \]

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

Maxima [F]

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

[In]

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

[Out]

-2/3*sqrt(2)/(e^(-2*x) + 1)^(3/2) + integrate(2*sqrt(2)*e^(-x)/((e^(-x) + 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. 96 vs. \(2 (34) = 68\).

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.13 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=\frac {1}{3} \, \sqrt {2} {\left (\frac {2 \, {\left (9 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 12 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 12 \, e^{\left (2 \, x\right )} + 5\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{3}} - 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

[In]

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

[Out]

1/3*sqrt(2)*(2*(9*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 12*sqrt(e^(4*x) + e^(2*x)) + 12*e^(2*x) + 5)/(sqrt(e
^(4*x) + e^(2*x)) - e^(2*x) - 1)^3 - 3*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))

Mupad [B] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int \tanh (x) (1+\tanh (x))^{3/2} \, dx=2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )-2\,\sqrt {\mathrm {tanh}\left (x\right )+1}-\frac {2\,{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}}{3} \]

[In]

int(tanh(x)*(tanh(x) + 1)^(3/2),x)

[Out]

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