\(\int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx\) [130]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-\frac {2}{3} (1+\tanh (x))^{3/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3624, 3561, 212} \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{3} (\tanh (x)+1)^{3/2} \]

[In]

Int[Tanh[x]^2*Sqrt[1 + Tanh[x]],x]

[Out]

Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (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 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 3624

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

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-\frac {2}{3} (1+\tanh (x))^{3/2} \]

[In]

Integrate[Tanh[x]^2*Sqrt[1 + Tanh[x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}{3}\)	\(26\)
default	\(\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}{3}\)	\(26\)

[In]

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

[Out]

arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(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. 237 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 237, normalized size of antiderivative = 6.97 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=-\frac {8 \, \sqrt {2} {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3}\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \, {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt {2} \cosh \left (x\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} + \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt {2}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{6 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \]

[In]

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

[Out]

-1/6*(8*sqrt(2)*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)^2*sinh(x) + 3*sqrt(2)*cosh(x)*sinh(x)^2 + sqrt(2)*sinh(
x)^3)*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*c
osh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(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 = 1.00 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{2} - \frac {2 \left (\tanh {\left (x \right )} + 1\right )^{\frac {3}{2}}}{3} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(tanh(x) + 1)*tanh(x)^2, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (25) = 50\).

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.82 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=\frac {1}{6} \, \sqrt {2} {\left (\frac {8 \, {\left (3 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 3 \, e^{\left (2 \, x\right )} + 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{3}} - 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

[In]

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

[Out]

1/6*sqrt(2)*(8*(3*(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) - 1)^3 - 3*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \tanh ^2(x) \sqrt {1+\tanh (x)} \, dx=\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )-\frac {2\,{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}}{3} \]

[In]

int(tanh(x)^2*(tanh(x) + 1)^(1/2),x)

[Out]

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