\(\int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx\) [602]

Optimal result

Integrand size = 13, antiderivative size = 9 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {e^{2 x}}{2} \]

[Out]

1/2*exp(2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2320, 30} \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {e^{2 x}}{2} \]

[In]

Int[E^x/(Cosh[x] - Sinh[x]),x]

[Out]

E^(2*x)/2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int x \, dx,x,e^x\right ) \\ & = \frac {e^{2 x}}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {e^{2 x}}{2} \]

[In]

Integrate[E^x/(Cosh[x] - Sinh[x]),x]

[Out]

E^(2*x)/2

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {{\mathrm e}^{2 x}}{2}\)	\(7\)
gosper	\(\frac {{\mathrm e}^{x}}{2 \cosh \left (x \right )-2 \sinh \left (x \right )}\)	\(14\)
default	\(\frac {2}{\tanh \left (\frac {x}{2}\right )-1}+\frac {2}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}\)	\(22\)

[In]

int(exp(x)/(cosh(x)-sinh(x)),x,method=_RETURNVERBOSE)

[Out]

1/2*exp(2*x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (6) = 12\).

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.78 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {\cosh \left (x\right ) + \sinh \left (x\right )}{2 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}} \]

[In]

integrate(exp(x)/(cosh(x)-sinh(x)),x, algorithm="fricas")

[Out]

1/2*(cosh(x) + sinh(x))/(cosh(x) - sinh(x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12 vs. \(2 (5) = 10\).

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.33 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {e^{x}}{- 2 \sinh {\left (x \right )} + 2 \cosh {\left (x \right )}} \]

[In]

integrate(exp(x)/(cosh(x)-sinh(x)),x)

[Out]

exp(x)/(-2*sinh(x) + 2*cosh(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {1}{2} \, e^{\left (2 \, x\right )} \]

[In]

integrate(exp(x)/(cosh(x)-sinh(x)),x, algorithm="maxima")

[Out]

1/2*e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {1}{2} \, e^{\left (2 \, x\right )} \]

[In]

integrate(exp(x)/(cosh(x)-sinh(x)),x, algorithm="giac")

[Out]

1/2*e^(2*x)

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {e^x}{\cosh (x)-\sinh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{2} \]

[In]

int(exp(x)/(cosh(x) - sinh(x)),x)

[Out]

exp(2*x)/2