\(\int \sqrt {\sinh (x) \tanh (x)} \, dx\) [560]

Optimal result

Integrand size = 9, antiderivative size = 13 \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=2 \coth (x) \sqrt {\sinh (x) \tanh (x)} \]

[Out]

2*coth(x)*(sinh(x)*tanh(x))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4483, 4485, 2669} \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=2 \coth (x) \sqrt {\sinh (x) \tanh (x)} \]

[In]

Int[Sqrt[Sinh[x]*Tanh[x]],x]

[Out]

2*Coth[x]*Sqrt[Sinh[x]*Tanh[x]]

Rule 2669

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*(a*Sin[e
+ f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rule 4483

Int[(u_.)*((a_)*(v_))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v]}, Dist[a^IntPart[p]*
((a*vv)^FracPart[p]/vv^FracPart[p]), Int[uu*vv^p, x], x]] /; FreeQ[{a, p}, x] &&  !IntegerQ[p] &&  !InertTrigF
reeQ[v]

Rule 4485

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])), Int[uu*vv^(m*p)*ww^(n*p)
, x], x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  !InertTrigFreeQ[w])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\sinh (x) \tanh (x)} \int \sqrt {-\sinh (x) \tanh (x)} \, dx}{\sqrt {-\sinh (x) \tanh (x)}} \\ & = \frac {\sqrt {\sinh (x) \tanh (x)} \int \sqrt {i \sinh (x)} \sqrt {i \tanh (x)} \, dx}{\sqrt {i \sinh (x)} \sqrt {i \tanh (x)}} \\ & = 2 \coth (x) \sqrt {\sinh (x) \tanh (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=2 \coth (x) \sqrt {\sinh (x) \tanh (x)} \]

[In]

Integrate[Sqrt[Sinh[x]*Tanh[x]],x]

[Out]

2*Coth[x]*Sqrt[Sinh[x]*Tanh[x]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(41\) vs. \(2(11)=22\).

Time = 0.66 (sec) , antiderivative size = 42, normalized size of antiderivative = 3.23

method	result	size
risch	\(\frac {\sqrt {2}\, \sqrt {\frac {\left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-x}}{1+{\mathrm e}^{2 x}}}\, \left (1+{\mathrm e}^{2 x}\right )}{{\mathrm e}^{2 x}-1}\)	\(42\)

[In]

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

[Out]

2^(1/2)*((exp(2*x)-1)^2*exp(-x)/(1+exp(2*x)))^(1/2)/(exp(2*x)-1)*(1+exp(2*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (11) = 22\).

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 4.08 \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=\frac {2 \, \sqrt {\frac {1}{2}} {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )}}{\sqrt {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} + {\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + \cosh \left (x\right )}} \]

[In]

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

[Out]

2*sqrt(1/2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)/sqrt(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 +
(3*cosh(x)^2 + 1)*sinh(x) + cosh(x))

Sympy [F]

\[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=\int \sqrt {\sinh {\left (x \right )} \tanh {\left (x \right )}}\, dx \]

[In]

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

[Out]

Integral(sqrt(sinh(x)*tanh(x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (11) = 22\).

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.69 \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, x\right )}}{\sqrt {e^{\left (-2 \, x\right )} + 1}} - \frac {\sqrt {2} e^{\left (-\frac {3}{2} \, x\right )}}{\sqrt {e^{\left (-2 \, x\right )} + 1}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=\int { \sqrt {\sinh \left (x\right ) \tanh \left (x\right )} \,d x } \]

[In]

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

[Out]

integrate(sqrt(sinh(x)*tanh(x)), x)

Mupad [B] (verification not implemented)

Time = 2.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.69 \[ \int \sqrt {\sinh (x) \tanh (x)} \, dx=2\,\mathrm {coth}\left (x\right )\,\sqrt {-\left (\frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^x}{2}\right )\,\left ({\mathrm {e}}^{2\,x}-1\right )}\,\sqrt {\frac {1}{{\mathrm {e}}^{2\,x}+1}} \]

[In]

int((sinh(x)*tanh(x))^(1/2),x)

[Out]

2*coth(x)*(-(exp(-x)/2 - exp(x)/2)*(exp(2*x) - 1))^(1/2)*(1/(exp(2*x) + 1))^(1/2)